CHAPTER 1VISCOUS FLOW IN PIPES

Why this chapter is so importantWhy study this topic

Fluid problem ndash flow in pipes

Viscous fluid

Shear stress and friction

uniform velocity profile to parabolic velocity profile

other flow properties also changed

CHAPTER 1 VISCOUS FLOW IN PIPES

Introduction

1048707 Piping systems are encountered in almost every engineering area

1048707 Problems are related to flow in ducts or pipes with various velocities fluids duct and pipe shapes and sizes

1048707 When lsquoreal worldrsquo (viscous effect) effects are important it is difficult to use theoretical method to obtain the desired result

1048707 A combination or experimental data with theoretical considerations and dimensional analysis provide the desired results

Pipe Flow Characteristics

1048707 Not all conduits used to transport fluid are round in cross section

1048707 Heating and air conditioning ducts are often of rectangular cross section Why

1048707 For heating and air conditioning pressure difference between inside and outside is relatively small and basic principle involved are independent of the cross sectional shape

1048707 Assume involved in this chapter- The pipe is round in cross section- The pipe is completely filled with fluid- Viscous fluid

- Incompressible fluid

1048707 Pipe flow vs Open channel flow - Pipe flow ndash pressure gradient in the driving

force (gravity may be important)- Open channel flow - gravity is the driving force

1048707 Steady and unsteady flow- Steady flows occur when flow parameters such

as pressure velocity temperature etc do not vary with time

- If flow parameters vary with time it is called unsteady

1048707 Laminar and turbulent flow- Flow is said to be laminar when adjacent fluid

layers move at same velocity and paths of individual particles of fluid do not cross each other Occur at low velocities and high viscosity (Re le 2100)

- Flow is turbulent when streamlines cross each other and mixing of fluid flow occur Occur at high velocities and low viscosity (Re ge 4000)

- 2100 Re 4000

Reynoldrsquos Experiment Dye Streaks

Fluid velocity at a point

1048707 Compressible and incompressible - Fluid is incompressible when its density does

not depend on pressure (Volume does not change when pressure is applied)

- When density changes when pressure is applied it is called compressible

1048707 Example Water at temperature 16oC flows through a pipe of diameter 0018 m a Determine the minimum time taken to fill a

354 x 10-4 m3 glass with water if the flow in pipe is to be laminar

b Determine the maximum time taken to fill a 354 x 10-4 m3 glass with water if the flow in pipe is to be turbulent

Solution

Given D = 0018 mV = 354 x 10-4 m3

a Minimum time occur when Reynolds number

is maximum allowed for laminar flow (maximum velocity)

v = Re (micro)D= 2100 (112 x 10-3)[1000(0018)]= 0131 ms

Q = vA= 0131[π(0018)24]= 333 x 10-5 m3s

t = VQ= 354 x 10-4333 x 10-5

= 1063 s

b Maximum time occur when Reynolds number is minimum allowed for turbulent flow (minimum velocity)

v = Re (micro)D= 4000 (112 x 10-3)[1000(0018)]= 0249 ms

Q = vA= 0249 [π(0018)24]= 633 x 10-5 m3s

t = VQ= 354 x 10-4633 x 10-5

= 559 s

Entrance Region and Fully Developed Flow

1048707 Entrance region - the region of flow near where the fluid enters the pipe

1048707 The fluid enters the pipe with nearly uniform velocity

profile [section (1)]

Entrance region developing flow and fully developed flow

1048707 As the fluid move through the pipe viscous effects cause it to stick to the pipe wall Thus boundary layer is produced along the pipe such that the initial velocity profile changes with distance along the pipe until the fluid reaches the end of the entrance length [section (2)]

1048707 From section (2) to section (3) the velocity profile does not vary with pipe length and the boundary layer is fully developed (fully developed flow)

1048707 The shape of velocity profile and the dimensionless entrance length leD depends on whether the flow is laminar or turbulence

- leD = 006 Re for laminar flow(11)

- leD = 44 (Re)16 for turbulent flow(12)

1048707 Because of the character of the pipe changes from section (3) to section (4) the flow gradually begin its return to its fully developed character (section (5))

1048707 Example Water flows through a 15m pipe with 13 cm diameter at 20 lmin Determine the length of entrance region le

Solution

Given L = 15mD = 13 cm = 0013 mQ = 20 lmin = 20(1000 x 60)m3s = 333 x

10-4 m3s

v = QA = 333 x 10-4[π(0013)24] = 333 x 10-4133 x 10-4

= 250 ms

Re = vDmicro= 1000(25)(0013)1 x 10-3

= 32500 (gt 4000 turbulent flow)

therefore

leD = 44 (Re)16

le = 44 (Re)16Dle = 44(32500)16(0013)

= 032 m

Pressure and Shear Stress

1048707 Pressure different between 2 sectionpoint forces the fluid through the horizontal pipe and viscous effects provide the restraining force that exactly balances the pressure force

1048707 In the entrance region fluid accelerate or decelerate as it flow thus there is a balance between pressure viscous and inertia (acceleration)

1048707 The magnitude of the pressure gradient px is larger in the entrance region than in the fully develop flow where it is a constantpx = -pl 0

Pressure distribution along the pipe

Equation for Fully Developed Laminar Flow in Pipes

1048707 Fully developed laminar flow - velocity profile is the same at any cross section of the pipe

1048707 From the velocity profile we can get other information regarding the flow such as pressure drop flow rate shear stress etc

1048707 3 method can be used to derive equations pertaining to fully developed laminar flow in pipes

- Applying F = ma to a fluid element- Dimensional analysis- Navier-Stokes equation of motion

Applying F = ma to a Fluid Element

1048707 Consider fluid element at time t ndash circular cylinder of fluid of length l and radius r

1048707 Even though the fluid is moving it is not accelerating

so ax = 0

Free body diagram of fluid element

1048707 Apply F = ma

(p1)πr2 ndash (p1 - p)πr2 ndash ()2πrl = 0

pl = 2r (13)

This equation represents the basic balance in force needed to drive each fluid particle along the pipe with constant velocity

1048707 is independent to r

= Cr

where C is a constant

at r = 0 there is no shear stress ( = 0)

at r = D2 the shear stress is maximum ( = w)

C = 2wD

therefore

= 2wrD(14)

and

p = 4lwD (15)

These equation (13 14 15) valid for both laminar and turbulent flow

1048707 How shear stress related to velocity

Two governing laws for fully developed laminar flow- pl = 2r- = -microur

Combine this two equation

ur = - (p2microl)r

velocity profile

u = - (p2microl)r r

u = - (p4microl)r 2 + C1

at r = D2 u = 0 and C1 = (p16microl)D 2

ur = (pD216microl)1 ndash (2rD) 2

at r = 0 centerline velocity Vc

Vc = (pD216microl)

therefore

ur = Vc 1 ndash (2rD) 2(16)

combine equation 15 and 16 and D2 = R

ur = w D4micro 1 ndash (rR) 2

1048707Flowrate

Q = u dA

Q = u 2πr dr

Q = 2π Vc 1 ndash (rR) 2r dr

Q = π R2 Vc2

knowing that average velocity V = QA = QπR2

V = (π R2 Vc2) πR2 = Vc2 = pD232microl(17)

and

Q = πD4p128microl (18)

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

CHAPTER 1 VISCOUS FLOW IN PIPES

Introduction

1048707 Piping systems are encountered in almost every engineering area

1048707 Problems are related to flow in ducts or pipes with various velocities fluids duct and pipe shapes and sizes

1048707 When lsquoreal worldrsquo (viscous effect) effects are important it is difficult to use theoretical method to obtain the desired result

1048707 A combination or experimental data with theoretical considerations and dimensional analysis provide the desired results

Pipe Flow Characteristics

1048707 Not all conduits used to transport fluid are round in cross section

1048707 Heating and air conditioning ducts are often of rectangular cross section Why

1048707 For heating and air conditioning pressure difference between inside and outside is relatively small and basic principle involved are independent of the cross sectional shape

1048707 Assume involved in this chapter- The pipe is round in cross section- The pipe is completely filled with fluid- Viscous fluid

- Incompressible fluid

1048707 Pipe flow vs Open channel flow - Pipe flow ndash pressure gradient in the driving

force (gravity may be important)- Open channel flow - gravity is the driving force

1048707 Steady and unsteady flow- Steady flows occur when flow parameters such

as pressure velocity temperature etc do not vary with time

- If flow parameters vary with time it is called unsteady

1048707 Laminar and turbulent flow- Flow is said to be laminar when adjacent fluid

layers move at same velocity and paths of individual particles of fluid do not cross each other Occur at low velocities and high viscosity (Re le 2100)

- Flow is turbulent when streamlines cross each other and mixing of fluid flow occur Occur at high velocities and low viscosity (Re ge 4000)

- 2100 Re 4000

Reynoldrsquos Experiment Dye Streaks

Fluid velocity at a point

1048707 Compressible and incompressible - Fluid is incompressible when its density does

not depend on pressure (Volume does not change when pressure is applied)

- When density changes when pressure is applied it is called compressible

1048707 Example Water at temperature 16oC flows through a pipe of diameter 0018 m a Determine the minimum time taken to fill a

354 x 10-4 m3 glass with water if the flow in pipe is to be laminar

b Determine the maximum time taken to fill a 354 x 10-4 m3 glass with water if the flow in pipe is to be turbulent

Solution

Given D = 0018 mV = 354 x 10-4 m3

a Minimum time occur when Reynolds number

is maximum allowed for laminar flow (maximum velocity)

v = Re (micro)D= 2100 (112 x 10-3)[1000(0018)]= 0131 ms

Q = vA= 0131[π(0018)24]= 333 x 10-5 m3s

t = VQ= 354 x 10-4333 x 10-5

= 1063 s

b Maximum time occur when Reynolds number is minimum allowed for turbulent flow (minimum velocity)

v = Re (micro)D= 4000 (112 x 10-3)[1000(0018)]= 0249 ms

Q = vA= 0249 [π(0018)24]= 633 x 10-5 m3s

t = VQ= 354 x 10-4633 x 10-5

= 559 s

Entrance Region and Fully Developed Flow

1048707 Entrance region - the region of flow near where the fluid enters the pipe

1048707 The fluid enters the pipe with nearly uniform velocity

profile [section (1)]

Entrance region developing flow and fully developed flow

1048707 As the fluid move through the pipe viscous effects cause it to stick to the pipe wall Thus boundary layer is produced along the pipe such that the initial velocity profile changes with distance along the pipe until the fluid reaches the end of the entrance length [section (2)]

1048707 From section (2) to section (3) the velocity profile does not vary with pipe length and the boundary layer is fully developed (fully developed flow)

1048707 The shape of velocity profile and the dimensionless entrance length leD depends on whether the flow is laminar or turbulence

- leD = 006 Re for laminar flow(11)

- leD = 44 (Re)16 for turbulent flow(12)

1048707 Because of the character of the pipe changes from section (3) to section (4) the flow gradually begin its return to its fully developed character (section (5))

1048707 Example Water flows through a 15m pipe with 13 cm diameter at 20 lmin Determine the length of entrance region le

Solution

Given L = 15mD = 13 cm = 0013 mQ = 20 lmin = 20(1000 x 60)m3s = 333 x

10-4 m3s

v = QA = 333 x 10-4[π(0013)24] = 333 x 10-4133 x 10-4

= 250 ms

Re = vDmicro= 1000(25)(0013)1 x 10-3

= 32500 (gt 4000 turbulent flow)

therefore

leD = 44 (Re)16

le = 44 (Re)16Dle = 44(32500)16(0013)

= 032 m

Pressure and Shear Stress

1048707 Pressure different between 2 sectionpoint forces the fluid through the horizontal pipe and viscous effects provide the restraining force that exactly balances the pressure force

1048707 In the entrance region fluid accelerate or decelerate as it flow thus there is a balance between pressure viscous and inertia (acceleration)

1048707 The magnitude of the pressure gradient px is larger in the entrance region than in the fully develop flow where it is a constantpx = -pl 0

Pressure distribution along the pipe

Equation for Fully Developed Laminar Flow in Pipes

1048707 Fully developed laminar flow - velocity profile is the same at any cross section of the pipe

1048707 From the velocity profile we can get other information regarding the flow such as pressure drop flow rate shear stress etc

1048707 3 method can be used to derive equations pertaining to fully developed laminar flow in pipes

- Applying F = ma to a fluid element- Dimensional analysis- Navier-Stokes equation of motion

Applying F = ma to a Fluid Element

1048707 Consider fluid element at time t ndash circular cylinder of fluid of length l and radius r

1048707 Even though the fluid is moving it is not accelerating

so ax = 0

Free body diagram of fluid element

1048707 Apply F = ma

(p1)πr2 ndash (p1 - p)πr2 ndash ()2πrl = 0

pl = 2r (13)

This equation represents the basic balance in force needed to drive each fluid particle along the pipe with constant velocity

1048707 is independent to r

= Cr

where C is a constant

at r = 0 there is no shear stress ( = 0)

at r = D2 the shear stress is maximum ( = w)

C = 2wD

therefore

= 2wrD(14)

and

p = 4lwD (15)

These equation (13 14 15) valid for both laminar and turbulent flow

1048707 How shear stress related to velocity

Two governing laws for fully developed laminar flow- pl = 2r- = -microur

Combine this two equation

ur = - (p2microl)r

velocity profile

u = - (p2microl)r r

u = - (p4microl)r 2 + C1

at r = D2 u = 0 and C1 = (p16microl)D 2

ur = (pD216microl)1 ndash (2rD) 2

at r = 0 centerline velocity Vc

Vc = (pD216microl)

therefore

ur = Vc 1 ndash (2rD) 2(16)

combine equation 15 and 16 and D2 = R

ur = w D4micro 1 ndash (rR) 2

1048707Flowrate

Q = u dA

Q = u 2πr dr

Q = 2π Vc 1 ndash (rR) 2r dr

Q = π R2 Vc2

knowing that average velocity V = QA = QπR2

V = (π R2 Vc2) πR2 = Vc2 = pD232microl(17)

and

Q = πD4p128microl (18)

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

- Incompressible fluid

1048707 Pipe flow vs Open channel flow - Pipe flow ndash pressure gradient in the driving

force (gravity may be important)- Open channel flow - gravity is the driving force

1048707 Steady and unsteady flow- Steady flows occur when flow parameters such

as pressure velocity temperature etc do not vary with time

- If flow parameters vary with time it is called unsteady

1048707 Laminar and turbulent flow- Flow is said to be laminar when adjacent fluid

layers move at same velocity and paths of individual particles of fluid do not cross each other Occur at low velocities and high viscosity (Re le 2100)

- Flow is turbulent when streamlines cross each other and mixing of fluid flow occur Occur at high velocities and low viscosity (Re ge 4000)

- 2100 Re 4000

Reynoldrsquos Experiment Dye Streaks

Fluid velocity at a point

1048707 Compressible and incompressible - Fluid is incompressible when its density does

not depend on pressure (Volume does not change when pressure is applied)

- When density changes when pressure is applied it is called compressible

1048707 Example Water at temperature 16oC flows through a pipe of diameter 0018 m a Determine the minimum time taken to fill a

354 x 10-4 m3 glass with water if the flow in pipe is to be laminar

b Determine the maximum time taken to fill a 354 x 10-4 m3 glass with water if the flow in pipe is to be turbulent

Solution

Given D = 0018 mV = 354 x 10-4 m3

a Minimum time occur when Reynolds number

is maximum allowed for laminar flow (maximum velocity)

v = Re (micro)D= 2100 (112 x 10-3)[1000(0018)]= 0131 ms

Q = vA= 0131[π(0018)24]= 333 x 10-5 m3s

t = VQ= 354 x 10-4333 x 10-5

= 1063 s

b Maximum time occur when Reynolds number is minimum allowed for turbulent flow (minimum velocity)

v = Re (micro)D= 4000 (112 x 10-3)[1000(0018)]= 0249 ms

Q = vA= 0249 [π(0018)24]= 633 x 10-5 m3s

t = VQ= 354 x 10-4633 x 10-5

= 559 s

Entrance Region and Fully Developed Flow

1048707 Entrance region - the region of flow near where the fluid enters the pipe

1048707 The fluid enters the pipe with nearly uniform velocity

profile [section (1)]

Entrance region developing flow and fully developed flow

1048707 As the fluid move through the pipe viscous effects cause it to stick to the pipe wall Thus boundary layer is produced along the pipe such that the initial velocity profile changes with distance along the pipe until the fluid reaches the end of the entrance length [section (2)]

1048707 From section (2) to section (3) the velocity profile does not vary with pipe length and the boundary layer is fully developed (fully developed flow)

1048707 The shape of velocity profile and the dimensionless entrance length leD depends on whether the flow is laminar or turbulence

- leD = 006 Re for laminar flow(11)

- leD = 44 (Re)16 for turbulent flow(12)

1048707 Because of the character of the pipe changes from section (3) to section (4) the flow gradually begin its return to its fully developed character (section (5))

1048707 Example Water flows through a 15m pipe with 13 cm diameter at 20 lmin Determine the length of entrance region le

Solution

Given L = 15mD = 13 cm = 0013 mQ = 20 lmin = 20(1000 x 60)m3s = 333 x

10-4 m3s

v = QA = 333 x 10-4[π(0013)24] = 333 x 10-4133 x 10-4

= 250 ms

Re = vDmicro= 1000(25)(0013)1 x 10-3

= 32500 (gt 4000 turbulent flow)

therefore

leD = 44 (Re)16

le = 44 (Re)16Dle = 44(32500)16(0013)

= 032 m

Pressure and Shear Stress

1048707 Pressure different between 2 sectionpoint forces the fluid through the horizontal pipe and viscous effects provide the restraining force that exactly balances the pressure force

1048707 In the entrance region fluid accelerate or decelerate as it flow thus there is a balance between pressure viscous and inertia (acceleration)

1048707 The magnitude of the pressure gradient px is larger in the entrance region than in the fully develop flow where it is a constantpx = -pl 0

Pressure distribution along the pipe

Equation for Fully Developed Laminar Flow in Pipes

1048707 Fully developed laminar flow - velocity profile is the same at any cross section of the pipe

1048707 From the velocity profile we can get other information regarding the flow such as pressure drop flow rate shear stress etc

1048707 3 method can be used to derive equations pertaining to fully developed laminar flow in pipes

- Applying F = ma to a fluid element- Dimensional analysis- Navier-Stokes equation of motion

Applying F = ma to a Fluid Element

1048707 Consider fluid element at time t ndash circular cylinder of fluid of length l and radius r

1048707 Even though the fluid is moving it is not accelerating

so ax = 0

Free body diagram of fluid element

1048707 Apply F = ma

(p1)πr2 ndash (p1 - p)πr2 ndash ()2πrl = 0

pl = 2r (13)

This equation represents the basic balance in force needed to drive each fluid particle along the pipe with constant velocity

1048707 is independent to r

= Cr

where C is a constant

at r = 0 there is no shear stress ( = 0)

at r = D2 the shear stress is maximum ( = w)

C = 2wD

therefore

= 2wrD(14)

and

p = 4lwD (15)

These equation (13 14 15) valid for both laminar and turbulent flow

1048707 How shear stress related to velocity

Two governing laws for fully developed laminar flow- pl = 2r- = -microur

Combine this two equation

ur = - (p2microl)r

velocity profile

u = - (p2microl)r r

u = - (p4microl)r 2 + C1

at r = D2 u = 0 and C1 = (p16microl)D 2

ur = (pD216microl)1 ndash (2rD) 2

at r = 0 centerline velocity Vc

Vc = (pD216microl)

therefore

ur = Vc 1 ndash (2rD) 2(16)

combine equation 15 and 16 and D2 = R

ur = w D4micro 1 ndash (rR) 2

1048707Flowrate

Q = u dA

Q = u 2πr dr

Q = 2π Vc 1 ndash (rR) 2r dr

Q = π R2 Vc2

knowing that average velocity V = QA = QπR2

V = (π R2 Vc2) πR2 = Vc2 = pD232microl(17)

and

Q = πD4p128microl (18)

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

Fluid velocity at a point

1048707 Compressible and incompressible - Fluid is incompressible when its density does

not depend on pressure (Volume does not change when pressure is applied)

- When density changes when pressure is applied it is called compressible

1048707 Example Water at temperature 16oC flows through a pipe of diameter 0018 m a Determine the minimum time taken to fill a

354 x 10-4 m3 glass with water if the flow in pipe is to be laminar

b Determine the maximum time taken to fill a 354 x 10-4 m3 glass with water if the flow in pipe is to be turbulent

Solution

Given D = 0018 mV = 354 x 10-4 m3

a Minimum time occur when Reynolds number

is maximum allowed for laminar flow (maximum velocity)

v = Re (micro)D= 2100 (112 x 10-3)[1000(0018)]= 0131 ms

Q = vA= 0131[π(0018)24]= 333 x 10-5 m3s

t = VQ= 354 x 10-4333 x 10-5

= 1063 s

b Maximum time occur when Reynolds number is minimum allowed for turbulent flow (minimum velocity)

v = Re (micro)D= 4000 (112 x 10-3)[1000(0018)]= 0249 ms

Q = vA= 0249 [π(0018)24]= 633 x 10-5 m3s

t = VQ= 354 x 10-4633 x 10-5

= 559 s

Entrance Region and Fully Developed Flow

1048707 Entrance region - the region of flow near where the fluid enters the pipe

1048707 The fluid enters the pipe with nearly uniform velocity

profile [section (1)]

Entrance region developing flow and fully developed flow

1048707 As the fluid move through the pipe viscous effects cause it to stick to the pipe wall Thus boundary layer is produced along the pipe such that the initial velocity profile changes with distance along the pipe until the fluid reaches the end of the entrance length [section (2)]

1048707 From section (2) to section (3) the velocity profile does not vary with pipe length and the boundary layer is fully developed (fully developed flow)

1048707 The shape of velocity profile and the dimensionless entrance length leD depends on whether the flow is laminar or turbulence

- leD = 006 Re for laminar flow(11)

- leD = 44 (Re)16 for turbulent flow(12)

1048707 Because of the character of the pipe changes from section (3) to section (4) the flow gradually begin its return to its fully developed character (section (5))

1048707 Example Water flows through a 15m pipe with 13 cm diameter at 20 lmin Determine the length of entrance region le

Solution

Given L = 15mD = 13 cm = 0013 mQ = 20 lmin = 20(1000 x 60)m3s = 333 x

10-4 m3s

v = QA = 333 x 10-4[π(0013)24] = 333 x 10-4133 x 10-4

= 250 ms

Re = vDmicro= 1000(25)(0013)1 x 10-3

= 32500 (gt 4000 turbulent flow)

therefore

leD = 44 (Re)16

le = 44 (Re)16Dle = 44(32500)16(0013)

= 032 m

Pressure and Shear Stress

1048707 Pressure different between 2 sectionpoint forces the fluid through the horizontal pipe and viscous effects provide the restraining force that exactly balances the pressure force

1048707 In the entrance region fluid accelerate or decelerate as it flow thus there is a balance between pressure viscous and inertia (acceleration)

1048707 The magnitude of the pressure gradient px is larger in the entrance region than in the fully develop flow where it is a constantpx = -pl 0

Pressure distribution along the pipe

Equation for Fully Developed Laminar Flow in Pipes

1048707 Fully developed laminar flow - velocity profile is the same at any cross section of the pipe

1048707 From the velocity profile we can get other information regarding the flow such as pressure drop flow rate shear stress etc

1048707 3 method can be used to derive equations pertaining to fully developed laminar flow in pipes

- Applying F = ma to a fluid element- Dimensional analysis- Navier-Stokes equation of motion

Applying F = ma to a Fluid Element

1048707 Consider fluid element at time t ndash circular cylinder of fluid of length l and radius r

1048707 Even though the fluid is moving it is not accelerating

so ax = 0

Free body diagram of fluid element

1048707 Apply F = ma

(p1)πr2 ndash (p1 - p)πr2 ndash ()2πrl = 0

pl = 2r (13)

This equation represents the basic balance in force needed to drive each fluid particle along the pipe with constant velocity

1048707 is independent to r

= Cr

where C is a constant

at r = 0 there is no shear stress ( = 0)

at r = D2 the shear stress is maximum ( = w)

C = 2wD

therefore

= 2wrD(14)

and

p = 4lwD (15)

These equation (13 14 15) valid for both laminar and turbulent flow

1048707 How shear stress related to velocity

Two governing laws for fully developed laminar flow- pl = 2r- = -microur

Combine this two equation

ur = - (p2microl)r

velocity profile

u = - (p2microl)r r

u = - (p4microl)r 2 + C1

at r = D2 u = 0 and C1 = (p16microl)D 2

ur = (pD216microl)1 ndash (2rD) 2

at r = 0 centerline velocity Vc

Vc = (pD216microl)

therefore

ur = Vc 1 ndash (2rD) 2(16)

combine equation 15 and 16 and D2 = R

ur = w D4micro 1 ndash (rR) 2

1048707Flowrate

Q = u dA

Q = u 2πr dr

Q = 2π Vc 1 ndash (rR) 2r dr

Q = π R2 Vc2

knowing that average velocity V = QA = QπR2

V = (π R2 Vc2) πR2 = Vc2 = pD232microl(17)

and

Q = πD4p128microl (18)

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

v = Re (micro)D= 2100 (112 x 10-3)[1000(0018)]= 0131 ms

Q = vA= 0131[π(0018)24]= 333 x 10-5 m3s

t = VQ= 354 x 10-4333 x 10-5

= 1063 s

b Maximum time occur when Reynolds number is minimum allowed for turbulent flow (minimum velocity)

v = Re (micro)D= 4000 (112 x 10-3)[1000(0018)]= 0249 ms

Q = vA= 0249 [π(0018)24]= 633 x 10-5 m3s

t = VQ= 354 x 10-4633 x 10-5

= 559 s

Entrance Region and Fully Developed Flow

1048707 Entrance region - the region of flow near where the fluid enters the pipe

1048707 The fluid enters the pipe with nearly uniform velocity

profile [section (1)]

Entrance region developing flow and fully developed flow

1048707 As the fluid move through the pipe viscous effects cause it to stick to the pipe wall Thus boundary layer is produced along the pipe such that the initial velocity profile changes with distance along the pipe until the fluid reaches the end of the entrance length [section (2)]

1048707 From section (2) to section (3) the velocity profile does not vary with pipe length and the boundary layer is fully developed (fully developed flow)

1048707 The shape of velocity profile and the dimensionless entrance length leD depends on whether the flow is laminar or turbulence

- leD = 006 Re for laminar flow(11)

- leD = 44 (Re)16 for turbulent flow(12)

1048707 Because of the character of the pipe changes from section (3) to section (4) the flow gradually begin its return to its fully developed character (section (5))

1048707 Example Water flows through a 15m pipe with 13 cm diameter at 20 lmin Determine the length of entrance region le

Solution

Given L = 15mD = 13 cm = 0013 mQ = 20 lmin = 20(1000 x 60)m3s = 333 x

10-4 m3s

v = QA = 333 x 10-4[π(0013)24] = 333 x 10-4133 x 10-4

= 250 ms

Re = vDmicro= 1000(25)(0013)1 x 10-3

= 32500 (gt 4000 turbulent flow)

therefore

leD = 44 (Re)16

le = 44 (Re)16Dle = 44(32500)16(0013)

= 032 m

Pressure and Shear Stress

1048707 Pressure different between 2 sectionpoint forces the fluid through the horizontal pipe and viscous effects provide the restraining force that exactly balances the pressure force

1048707 In the entrance region fluid accelerate or decelerate as it flow thus there is a balance between pressure viscous and inertia (acceleration)

1048707 The magnitude of the pressure gradient px is larger in the entrance region than in the fully develop flow where it is a constantpx = -pl 0

Pressure distribution along the pipe

Equation for Fully Developed Laminar Flow in Pipes

1048707 Fully developed laminar flow - velocity profile is the same at any cross section of the pipe

1048707 From the velocity profile we can get other information regarding the flow such as pressure drop flow rate shear stress etc

1048707 3 method can be used to derive equations pertaining to fully developed laminar flow in pipes

- Applying F = ma to a fluid element- Dimensional analysis- Navier-Stokes equation of motion

Applying F = ma to a Fluid Element

1048707 Consider fluid element at time t ndash circular cylinder of fluid of length l and radius r

1048707 Even though the fluid is moving it is not accelerating

so ax = 0

Free body diagram of fluid element

1048707 Apply F = ma

(p1)πr2 ndash (p1 - p)πr2 ndash ()2πrl = 0

pl = 2r (13)

This equation represents the basic balance in force needed to drive each fluid particle along the pipe with constant velocity

1048707 is independent to r

= Cr

where C is a constant

at r = 0 there is no shear stress ( = 0)

at r = D2 the shear stress is maximum ( = w)

C = 2wD

therefore

= 2wrD(14)

and

p = 4lwD (15)

These equation (13 14 15) valid for both laminar and turbulent flow

1048707 How shear stress related to velocity

Two governing laws for fully developed laminar flow- pl = 2r- = -microur

Combine this two equation

ur = - (p2microl)r

velocity profile

u = - (p2microl)r r

u = - (p4microl)r 2 + C1

at r = D2 u = 0 and C1 = (p16microl)D 2

ur = (pD216microl)1 ndash (2rD) 2

at r = 0 centerline velocity Vc

Vc = (pD216microl)

therefore

ur = Vc 1 ndash (2rD) 2(16)

combine equation 15 and 16 and D2 = R

ur = w D4micro 1 ndash (rR) 2

1048707Flowrate

Q = u dA

Q = u 2πr dr

Q = 2π Vc 1 ndash (rR) 2r dr

Q = π R2 Vc2

knowing that average velocity V = QA = QπR2

V = (π R2 Vc2) πR2 = Vc2 = pD232microl(17)

and

Q = πD4p128microl (18)

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

Entrance region developing flow and fully developed flow

1048707 As the fluid move through the pipe viscous effects cause it to stick to the pipe wall Thus boundary layer is produced along the pipe such that the initial velocity profile changes with distance along the pipe until the fluid reaches the end of the entrance length [section (2)]

1048707 From section (2) to section (3) the velocity profile does not vary with pipe length and the boundary layer is fully developed (fully developed flow)

1048707 The shape of velocity profile and the dimensionless entrance length leD depends on whether the flow is laminar or turbulence

- leD = 006 Re for laminar flow(11)

- leD = 44 (Re)16 for turbulent flow(12)

1048707 Because of the character of the pipe changes from section (3) to section (4) the flow gradually begin its return to its fully developed character (section (5))

1048707 Example Water flows through a 15m pipe with 13 cm diameter at 20 lmin Determine the length of entrance region le

Solution

Given L = 15mD = 13 cm = 0013 mQ = 20 lmin = 20(1000 x 60)m3s = 333 x

10-4 m3s

v = QA = 333 x 10-4[π(0013)24] = 333 x 10-4133 x 10-4

= 250 ms

Re = vDmicro= 1000(25)(0013)1 x 10-3

= 32500 (gt 4000 turbulent flow)

therefore

leD = 44 (Re)16

le = 44 (Re)16Dle = 44(32500)16(0013)

= 032 m

Pressure and Shear Stress

1048707 Pressure different between 2 sectionpoint forces the fluid through the horizontal pipe and viscous effects provide the restraining force that exactly balances the pressure force

1048707 In the entrance region fluid accelerate or decelerate as it flow thus there is a balance between pressure viscous and inertia (acceleration)

1048707 The magnitude of the pressure gradient px is larger in the entrance region than in the fully develop flow where it is a constantpx = -pl 0

Pressure distribution along the pipe

Equation for Fully Developed Laminar Flow in Pipes

1048707 Fully developed laminar flow - velocity profile is the same at any cross section of the pipe

1048707 From the velocity profile we can get other information regarding the flow such as pressure drop flow rate shear stress etc

1048707 3 method can be used to derive equations pertaining to fully developed laminar flow in pipes

- Applying F = ma to a fluid element- Dimensional analysis- Navier-Stokes equation of motion

Applying F = ma to a Fluid Element

1048707 Consider fluid element at time t ndash circular cylinder of fluid of length l and radius r

1048707 Even though the fluid is moving it is not accelerating

so ax = 0

Free body diagram of fluid element

1048707 Apply F = ma

(p1)πr2 ndash (p1 - p)πr2 ndash ()2πrl = 0

pl = 2r (13)

This equation represents the basic balance in force needed to drive each fluid particle along the pipe with constant velocity

1048707 is independent to r

= Cr

where C is a constant

at r = 0 there is no shear stress ( = 0)

at r = D2 the shear stress is maximum ( = w)

C = 2wD

therefore

= 2wrD(14)

and

p = 4lwD (15)

These equation (13 14 15) valid for both laminar and turbulent flow

1048707 How shear stress related to velocity

Two governing laws for fully developed laminar flow- pl = 2r- = -microur

Combine this two equation

ur = - (p2microl)r

velocity profile

u = - (p2microl)r r

u = - (p4microl)r 2 + C1

at r = D2 u = 0 and C1 = (p16microl)D 2

ur = (pD216microl)1 ndash (2rD) 2

at r = 0 centerline velocity Vc

Vc = (pD216microl)

therefore

ur = Vc 1 ndash (2rD) 2(16)

combine equation 15 and 16 and D2 = R

ur = w D4micro 1 ndash (rR) 2

1048707Flowrate

Q = u dA

Q = u 2πr dr

Q = 2π Vc 1 ndash (rR) 2r dr

Q = π R2 Vc2

knowing that average velocity V = QA = QπR2

V = (π R2 Vc2) πR2 = Vc2 = pD232microl(17)

and

Q = πD4p128microl (18)

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

1048707 Example Water flows through a 15m pipe with 13 cm diameter at 20 lmin Determine the length of entrance region le

Solution

Given L = 15mD = 13 cm = 0013 mQ = 20 lmin = 20(1000 x 60)m3s = 333 x

10-4 m3s

v = QA = 333 x 10-4[π(0013)24] = 333 x 10-4133 x 10-4

= 250 ms

Re = vDmicro= 1000(25)(0013)1 x 10-3

= 32500 (gt 4000 turbulent flow)

therefore

leD = 44 (Re)16

le = 44 (Re)16Dle = 44(32500)16(0013)

= 032 m

Pressure and Shear Stress

1048707 Pressure different between 2 sectionpoint forces the fluid through the horizontal pipe and viscous effects provide the restraining force that exactly balances the pressure force

1048707 In the entrance region fluid accelerate or decelerate as it flow thus there is a balance between pressure viscous and inertia (acceleration)

1048707 The magnitude of the pressure gradient px is larger in the entrance region than in the fully develop flow where it is a constantpx = -pl 0

Pressure distribution along the pipe

Equation for Fully Developed Laminar Flow in Pipes

1048707 Fully developed laminar flow - velocity profile is the same at any cross section of the pipe

1048707 From the velocity profile we can get other information regarding the flow such as pressure drop flow rate shear stress etc

1048707 3 method can be used to derive equations pertaining to fully developed laminar flow in pipes

- Applying F = ma to a fluid element- Dimensional analysis- Navier-Stokes equation of motion

Applying F = ma to a Fluid Element

1048707 Consider fluid element at time t ndash circular cylinder of fluid of length l and radius r

1048707 Even though the fluid is moving it is not accelerating

so ax = 0

Free body diagram of fluid element

1048707 Apply F = ma

(p1)πr2 ndash (p1 - p)πr2 ndash ()2πrl = 0

pl = 2r (13)

This equation represents the basic balance in force needed to drive each fluid particle along the pipe with constant velocity

1048707 is independent to r

= Cr

where C is a constant

at r = 0 there is no shear stress ( = 0)

at r = D2 the shear stress is maximum ( = w)

C = 2wD

therefore

= 2wrD(14)

and

p = 4lwD (15)

These equation (13 14 15) valid for both laminar and turbulent flow

1048707 How shear stress related to velocity

Two governing laws for fully developed laminar flow- pl = 2r- = -microur

Combine this two equation

ur = - (p2microl)r

velocity profile

u = - (p2microl)r r

u = - (p4microl)r 2 + C1

at r = D2 u = 0 and C1 = (p16microl)D 2

ur = (pD216microl)1 ndash (2rD) 2

at r = 0 centerline velocity Vc

Vc = (pD216microl)

therefore

ur = Vc 1 ndash (2rD) 2(16)

combine equation 15 and 16 and D2 = R

ur = w D4micro 1 ndash (rR) 2

1048707Flowrate

Q = u dA

Q = u 2πr dr

Q = 2π Vc 1 ndash (rR) 2r dr

Q = π R2 Vc2

knowing that average velocity V = QA = QπR2

V = (π R2 Vc2) πR2 = Vc2 = pD232microl(17)

and

Q = πD4p128microl (18)

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

1048707 In the entrance region fluid accelerate or decelerate as it flow thus there is a balance between pressure viscous and inertia (acceleration)

1048707 The magnitude of the pressure gradient px is larger in the entrance region than in the fully develop flow where it is a constantpx = -pl 0

Pressure distribution along the pipe

Equation for Fully Developed Laminar Flow in Pipes

1048707 Fully developed laminar flow - velocity profile is the same at any cross section of the pipe

1048707 From the velocity profile we can get other information regarding the flow such as pressure drop flow rate shear stress etc

1048707 3 method can be used to derive equations pertaining to fully developed laminar flow in pipes

- Applying F = ma to a fluid element- Dimensional analysis- Navier-Stokes equation of motion

Applying F = ma to a Fluid Element

1048707 Consider fluid element at time t ndash circular cylinder of fluid of length l and radius r

1048707 Even though the fluid is moving it is not accelerating

so ax = 0

Free body diagram of fluid element

1048707 Apply F = ma

(p1)πr2 ndash (p1 - p)πr2 ndash ()2πrl = 0

pl = 2r (13)

This equation represents the basic balance in force needed to drive each fluid particle along the pipe with constant velocity

1048707 is independent to r

= Cr

where C is a constant

at r = 0 there is no shear stress ( = 0)

at r = D2 the shear stress is maximum ( = w)

C = 2wD

therefore

= 2wrD(14)

and

p = 4lwD (15)

These equation (13 14 15) valid for both laminar and turbulent flow

1048707 How shear stress related to velocity

Two governing laws for fully developed laminar flow- pl = 2r- = -microur

Combine this two equation

ur = - (p2microl)r

velocity profile

u = - (p2microl)r r

u = - (p4microl)r 2 + C1

at r = D2 u = 0 and C1 = (p16microl)D 2

ur = (pD216microl)1 ndash (2rD) 2

at r = 0 centerline velocity Vc

Vc = (pD216microl)

therefore

ur = Vc 1 ndash (2rD) 2(16)

combine equation 15 and 16 and D2 = R

ur = w D4micro 1 ndash (rR) 2

1048707Flowrate

Q = u dA

Q = u 2πr dr

Q = 2π Vc 1 ndash (rR) 2r dr

Q = π R2 Vc2

knowing that average velocity V = QA = QπR2

V = (π R2 Vc2) πR2 = Vc2 = pD232microl(17)

and

Q = πD4p128microl (18)

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

Applying F = ma to a Fluid Element

1048707 Consider fluid element at time t ndash circular cylinder of fluid of length l and radius r

1048707 Even though the fluid is moving it is not accelerating

so ax = 0

Free body diagram of fluid element

1048707 Apply F = ma

(p1)πr2 ndash (p1 - p)πr2 ndash ()2πrl = 0

pl = 2r (13)

This equation represents the basic balance in force needed to drive each fluid particle along the pipe with constant velocity

1048707 is independent to r

= Cr

where C is a constant

at r = 0 there is no shear stress ( = 0)

at r = D2 the shear stress is maximum ( = w)

C = 2wD

therefore

= 2wrD(14)

and

p = 4lwD (15)

These equation (13 14 15) valid for both laminar and turbulent flow

1048707 How shear stress related to velocity

Two governing laws for fully developed laminar flow- pl = 2r- = -microur

Combine this two equation

ur = - (p2microl)r

velocity profile

u = - (p2microl)r r

u = - (p4microl)r 2 + C1

at r = D2 u = 0 and C1 = (p16microl)D 2

ur = (pD216microl)1 ndash (2rD) 2

at r = 0 centerline velocity Vc

Vc = (pD216microl)

therefore

ur = Vc 1 ndash (2rD) 2(16)

combine equation 15 and 16 and D2 = R

ur = w D4micro 1 ndash (rR) 2

1048707Flowrate

Q = u dA

Q = u 2πr dr

Q = 2π Vc 1 ndash (rR) 2r dr

Q = π R2 Vc2

knowing that average velocity V = QA = QπR2

V = (π R2 Vc2) πR2 = Vc2 = pD232microl(17)

and

Q = πD4p128microl (18)

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

where C is a constant

at r = 0 there is no shear stress ( = 0)

at r = D2 the shear stress is maximum ( = w)

C = 2wD

therefore

= 2wrD(14)

and

p = 4lwD (15)

These equation (13 14 15) valid for both laminar and turbulent flow

1048707 How shear stress related to velocity

Two governing laws for fully developed laminar flow- pl = 2r- = -microur

Combine this two equation

ur = - (p2microl)r

velocity profile

u = - (p2microl)r r

u = - (p4microl)r 2 + C1

at r = D2 u = 0 and C1 = (p16microl)D 2

ur = (pD216microl)1 ndash (2rD) 2

at r = 0 centerline velocity Vc

Vc = (pD216microl)

therefore

ur = Vc 1 ndash (2rD) 2(16)

combine equation 15 and 16 and D2 = R

ur = w D4micro 1 ndash (rR) 2

1048707Flowrate

Q = u dA

Q = u 2πr dr

Q = 2π Vc 1 ndash (rR) 2r dr

Q = π R2 Vc2

knowing that average velocity V = QA = QπR2

V = (π R2 Vc2) πR2 = Vc2 = pD232microl(17)

and

Q = πD4p128microl (18)

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

ur = (pD216microl)1 ndash (2rD) 2

at r = 0 centerline velocity Vc

Vc = (pD216microl)

therefore

ur = Vc 1 ndash (2rD) 2(16)

combine equation 15 and 16 and D2 = R

ur = w D4micro 1 ndash (rR) 2

1048707Flowrate

Q = u dA

Q = u 2πr dr

Q = 2π Vc 1 ndash (rR) 2r dr

Q = π R2 Vc2

knowing that average velocity V = QA = QπR2

V = (π R2 Vc2) πR2 = Vc2 = pD232microl(17)

and

Q = πD4p128microl (18)

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

1048707 1048707umm r1048707a 1048707 low prop rt s orF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

leD = 006 Re leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

pl = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = pD232microl

Flowrate Q = πD4p128microl

1048707 Adjustment to account for non horizontal pipe ndash gravity effect

Free body diagram of fluid element for non horizontal pipe

θ - angle between pipe centerline axis and horizontal axis

Apply F = ma

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

(p + p)πr2 ndash (p)πr2 ndash mgsinθ ndash ()2πrl = 0

(p + p)πr2 ndash (p)πr2 ndash (πr2)lgsinθ ndash ()2πrl = 0(p ndash γlsin θ)l = 2r(19)

effects of non horizontal pipe

p (p ndash γlsin θ)

therefore

V = (p ndash γlsin θ)D232microl(110)

and

Q = πD4(p ndash γlsin θ)128microl(111)

1048707 1048707umm r1048707a 1048707 lowprop rt s ornonF e ie f or ont lp ph iz a i e

Flow Properties

Equation Remarks

Entrance Length leD

- leD = 006 Re - leD = 44 (Re)16

Laminar flowTurbulent flow

Pressure drop per unit length

(p ndash γlsin θ)l = 2r Valid for both laminar and turbulent flow

Shear stress = 2wrD Valid for both laminar and turbulent flow

Pressure drop

p ndash γlsin θ = 4lwD Valid for both laminar and turbulent flow

Velocity profile

ur = Vc 1 ndash (2rD) 2

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

Average velocity

V = (π R2 Vc2) πR2

V= Vc2 V = (pndashγlsin θ)D232microl

Flowrate Q = πD4(p ndash γlsin θ)128microl

1048707 Example An oil with a viscosity of micro = 040 Nsm2

and density = 900 kgm3 flows in pipe of diameter D = 0020 ma What pressure drop is needed to produce a

flowrate of Q = 20 x 10-5 m3s if the pipe is horizontal and x1 = 0 m and x2 = 10 m

b How steep a hill θ must the pipe be on if the oil is to flow at the same rate as in part (a) but with p1 = p2

c For a condition of part (b) if p1 = 200 kPa what is the pressure at x3 = 5 m

Solution

a Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 m

from equation Q = πD4p128microl

p = Q(128microl)πD4

= [128(20 x 10-5)(040)10][314(002)4]= 20400 Nm2

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

b Given micro = 040 Nsm2

= 900 kgm3

D = 0020 mQ = 20 x 10-5 m3sx1 = 0 mx2 = 10 mp= 0

from equation Q = πD4(p ndash γlsin θ)128microl

p ndash γlsin θ = Q(128microl)πD4

sin θ = - 128QmicroπD4γ= - 128(20 x 10-5)(040)[314(002)4(900)

(981)]

θ = sin-1[[- 128(20 x 10-

5)(040)][314(002)4(900)(981)]]= -13340

c Condition as part (b) pressure different along the pipe p = 0 and p1 = p2 = p3

therefore at x3 = 5 m p3 = 200kPa

Pressure Drop and Head Loss

1048707 Important of pressure drop ndash it is related to power required by pump or fan to maintain fluid flow

1048707 Power W = gQhL

1048707 From energy equation

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL

for horizontal pipe v1 = v2 and z1 = z2 = kinetic energy coefficient and for uniform flow 1 = 2 = 1

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

hL = pg(112)

from previous researcher for laminar flow

hL = f(lD)(v22g)(113)

combine equation (112) and (113) and f = 64Re

pg = (64Re)(lD)(v22g)

p = 32microlvD2

(114)

combine equation (13)and (112)

hLg = 2lr

hL = 2lgr

hL = 4wlgD (115)

1048707 Example Water with a viscosity of micro = 1545 x 10-3

kgsm and density = 998 kgm3 is flowing through 0003 m diameter 9 m long horizontal pipe steadily at an average velocity of 09 ms Determine

a the head lossb the pressure dropc the pumping power requirement to

overcome this pressure drop

Solution

a head lossRe = vDmicro

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

= 998(09)(0003) 1545 x 10-3 = 1744 ( 2100 laminar flow)

for laminar flow

f = 64Re= 641744= 00367

and

hL = f(lD)(v22g)= (00367)(90003)[0922(981)]= 4545 m

b pressure drop

for laminar flow

pg = (64Re)(lD)(v22g)p = (64Re)(lD)( v22)

= 00367(90003)[998(092)2]= 445 kPa

c power required

P = gQhL

and

pg = hL

therefore

P = Qp= 09π(000324)(44500)= 0283 W

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

Concept of Turbulent Flow in Pipes

1048707Characteristics of Turbulent Flow in Pipes - Re gt 4000 - Random movements of eddies which mixes up

the layers of fluid- Particle path is irregular - Most common type of flow - Difficult mathematical analysis to describe the

flow 1048707 Important of turbulent flow

- mixing process- heat and mass transfer process

transition from laminar to turbulent flow in pipe

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

axial velocity measured at a given location

Turbulent Shear Stress

1048707 Random 3-dimensional fluid motions (eddies) produce shear force for the turbulent flow

1048707 3-dimensional eddies conveys mass with average velocity ū Therefore flow momentum exists The result of this momentum transfer is shear force

1048707 Shear stress in pipe is given as the summation of laminar shear stress and turbulent shear stress

τ = τlam + τturb

structure of turbulent flow in a pipe

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

1048707 Laminar shear stress is dominant near the pipe wall and the turbulent shear stress dominates the flow at center of pipe

1048707 The region where laminar shear force dominates is called the viscous sublayer or the viscous wall layer

1048707 The region where turbulent shear force dominates is called the outer turbulent layer or simply the outer layer

1048707 There is also a region where both laminar and shear are important This region is called the overlap region

1048707 The character of the each layers such as their velocity are different so we need different equations to describe them

Turbulent Velocity Profile

1048707 Velocity profile for viscous sublayer

ūu = yu (116)

where ū = average velocity y = distance measured from wall = R ndash r u = friction velocity = (τw )12

= kinematic viscosity

This equation is called the Law of Wall which is valid only near a smooth wall for 0 le yu le 5

1048707 Velocity profile for overlap region

ūu = 25 ln (yu) + 50(117)

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

where 25 and 50 are constants determined by experiments

1048707 For the outer layer the Power Law is used from the following expression

ūVc = [1 ndash (rR)]1n

(118)

The value of n which indicates the ldquopowerrdquo of the equation is a function of Re and determined experimentally

exponent n for power laws velocity profiles

1048707 The relationship between average velocity V volume flowrate Q and centerline velocity VC can be obtained by integrating the power law velocity profile

Q = ū A

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

Q = Vc [1 ndash (rR)]1n A

Q = Vc [1 ndash (rR)]1n 2πr r

Q = 2πR2 Vc n2[(n + 1)(2n + 1)](119)

since Q = πR2V

πR2V πR2 Vc = 2n2[(n + 1)(2n + 1)]

VVc = 2n2[(n + 1)(2n + 1)]

V = 2n2Vc [(n + 1)(2n + 1)](120)

Head Loss

1048707 Energy equation for steady incompressible flow in horizontal pipes

p1g + 1v122g + z1 = p2g + 2v2

22g + z2 + hL+ w - q

where hL= head lossw = turbine headq = pump head

1048707 Head loss- major loss- minor loss

Major Losses

1048707 Major losses is caused by friction at walls and due to the resistance of fluid particles as they roll rub and slide each other

1048707 Losses cause by doing work against friction

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

hL = flv22gD

where f = friction factorl = lengthv = velocityd = gravity acceleration

1048707 For laminar flow

f = 64Re

1048707 For turbulent flow

1f05 = - 18 log [(69Re) + (37D)11]

1048707 Surface Roughness

PipeEquivalent Roughness (ft) (mm)

Riveted Steel 0003 ndash 003 09 ndash 90Concrete 0001 ndash 001 03 ndash 30Wood Stave 00006 ndash 0003 019 ndash 09Cast Iron 000085 026Galvanized Iron 00005 015Commercial Steel

000015 0045

Drawn Tubing 0000005 00015Plastic glass 00 (smooth) 00 (smooth)

1048707 Normally for pipe analysis we obtain the dependence of friction factor on Re amp εD through the Moody Chart

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

1048707 To construct this chart the equivalent roughness ε is usually obtained for lsquocleanrsquo and new pipes because after considerable use most pipes may have increased roughness

1048707 For high Re flows the viscous sub-layer is so thin that the surface roughness completely dominates the character of flow near the walls

1048707 For smooth pipes (ε = 0) we notice that friction factor (f) is not zero because there is still head loss These pipes are called ldquohydraulically smoothrdquo

1048707 The moody chart offers the relationships between εD f and Re for a very wide range of pipe flows including that for laminar flows as long as the flow is steady fully developed and incompressible

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

1048707 The Moody chart is useful because in real applications a large variety of D V ρ and μ exists but only for small ranges win the Moody chart

1048707 The Moody chart is valid for all steady fully developed incompressible pipe flows

Non Circular Conduits

1048707 Air Conditioner Ducting System

1048707 Hydraulic Radius

RH = Area Circumference

Circle cross section area

RH = Area Circumference= [πD24]πD= D4

D = 4 RH

therefore

hf = flv22gD= flv28gRH

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

D = 4 RH

vDmicro =4v RHmicro

Minor Losses

1048707 Minor losses are caused by the geometry of pipes such as the presence of valves and fittings such as elbows tees bends etc

1048707In equation form

hL = KLv22g

where KL is loss coefficient

KL = hL2gv2 = 2pv2

1048707 KL vary depending on the shapes involved

Entrance loss

1048707 Head loss when liquid enters pipe from a large tankreservoir

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

Exit Loss

1048707 Head loss produced when liquid leaves pipe and enters a large tankreservoir

1048707 The entire kinetic energy of exiting fluid v1 is dissipated through viscous effects and eventually becomes v2 = 0

1048707 Exit loss from (1) to (2) is equivalent to one velocity head

1048707 KL = 10

Sudden Expansion and Contraction

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

1048707 Losses that occur where there is a sudden increase in pipe diameter (expansion) or where there is a sudden decrease in pipe diameter (contraction)

1048707 Loss coefficient is a function of area ratio A2A1

1048707 For sudden expansionKL = [1 ndash (D1D2)2]2

1048707 A2A1 = 0 KL = 05 - extreme sharp edge entrance

1048707 A2A1 = 1 KL = 0 - no area change

Vena Contracta

1048707 Fluid entering a sharp corner

1048707 Fluid cannot through sharp corners At a sharp corner the flow will separate and reattaches at the pipe wall

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

1048707 This separation and reattachment forms a bubble (separation bubble) making the area of fluid flow smaller than the actual pipe area

1048707 This causes the velocity of fluid passing through this small area to increase

1048707 Maximum velocity exists at section with minimum area called the vena contracta

1048707 Because high speed flows cannot slow down efficiently the kinetic energy could not be fully converted into pressure

Pump

1048707 Pumps are used to increase energy of the fluid (liquid)

1048707 40 - 50 of industrial energy is used to drive pumps and compressors

1048707 Proper design construction and selection of pumps are economically significant

1048707 One of the most common pump is the centrifugal pump

1048707 It consist of rotating elements called impeller which is contained within the pump housing

1048707 The shaft transfers mechanical energy to the impeller

1048707 A system of bearings and seals are required to complete the design

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

1048707 Flow enters the machine nearly axial at some radius through the eye of the impeller and leaves radially outward

1048707 Energy is added to the fluid by rotating blades and both pressure and absolute velocity are increased as fluid flows from eye to the periphery of the blades

1048707 Fluid discharges into the housing which is designed to reduce velocity

1048707 Types of pumps - Reciprocating pistons or plunger- Gear pump

- Double screw pump - Sliding vane - Lobe pump - Differential piston - Flexible squeegee

1048707 In reality pumps cause losses

1048707 Normally pumps are driven by electric motors IC engines etc

1048707 In short we can say that pump draws kinetic energy

and delivers it to the fluid

1048707 If we include energy of pump in the energy equation we get

p1g + v122g + z1 + hP= p2g + v2

22g + z2 + hL

where hP is the energy added to fluid and hL is the head loss discussed earlier

Power Required by Pump

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

1048707 Power is the rate of work or the rate of which energy is being transferred and is given by

1048707 Power added to fluid

Power = pghPQ

Unit Watt Nms or Js

1048707 Example A tank of water empties by gravity through a horizontal pipe into another tank There is a sudden enlargement in the pipe At a certain time the difference in level is 3 m Each pipe is 2 m long and has a friction coefficient f = 0005 The inlet loss coefficient is 03 Calculate the flowrate at this point

Solution

energy equation

p1g + v122g + z1 = p2g + v2

22g + z2 + hL

3 m

60 mm

20 mm

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

p1 = p2 = 0 v1 = v2 = 0

hL = z1 - z2

= 3 m

major loss

hf 1 = f1lv22gD= 0005(2) [Q2π2(00224)2][2(981)(002)]= 258471Q2

hf 2 = f2lv22gD= 0005(2) [Q2π2(00624)2][2(981)(006)]= 1064Q2

minor losses

for entrance

hL1 = KL1v22g= 03[Q2π2(00224)2][2(981)]= 155083Q2

for sudden enlargement

KL2 = [1 ndash (D1D2)2]2

= [1ndash (002006)2]2

= 079therefore

hL2 = KL2v22g= 079[Q2π2(00224)2][2(981)]= 408384Q2

for exit

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s

hL3 = KL3v22g= 1[Q2π2(00624)2][2(981)]= 6382Q2

total head loss

hL = 258471Q2 + 1064Q2 +155083Q2 + 408384Q2 + 6382Q2

Q2 = 3596754Q = 2242 x 10-3 m3s