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Module-5
Flow Through Pipes
98795 10743
Fluid Mechanics & Hydraulics (2140611)
10-02-2020
Prof. Mehul Pujara
2Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Fluid:
β’ The term fluid refers to both aliquid and a gas; it is generallydefined as a state of matter inwhich the material flows freelyunder the action of a shearstress.
3Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
CONTINUITY EQUATION:
It follows βPrinciple of conservationof massβ
βMass neither be created nor bedestroyed, total mass of systemremains constant.β
Mass offluidenteringunit time
Mass offluidmovingout perunit time
Rate ofincreaseof massper unittime
- =
4Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
CONTINUITY EQUATION:
Considering flow though the stream tube.
Mass of fluid entering into stream tube perunit time is
= Density * Discharge
= Density * Area * velocity
= π π΄ π£ (1)
Mass of fluid coming out from stream tubeper unit time is
= π π΄ π£ +π (π π΄ π£)
ππ ds (2)
Net mass of fluid reaming in stream tubeper unit time is
= -π (π π΄ π£)
ππ ds (3)
π π΄ π£
π π΄ π£ +π (π π΄ π£)
ππ ds
5Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
The mass of fluid in stream tube is
= density * volume of fluid in tube
= π π΄ ds (4)
Rate of increase of mass of fluid with timein stream tube is
=π (π π΄ππ )
ππ‘
=π (π π΄ )
ππ‘ππ (5)
Net mass remained in the stream tube per unittime is equal to rate of increase of mass withthe time.
Equation (3) = Equation (5)
-π (π π΄ π£)
ππ ππ =
π (π π΄ )
ππ‘ππ
CONTINUITY EQUATION:
-π (π π΄ π£)
ππ = π (π π΄ )
ππ‘
π (π π΄ )
ππ‘+ π (π π΄ π£)
ππ = 0 (6)
Equation (6) is known as continuityequation for one dimensional flow, it isapplicable to all cases of flow.
6Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
For Steady flowπ (π π΄ )
ππ‘= 0
Thereforeπ (π π΄ π£)
ππ = 0
π π΄ π£ = Constant
At section 1 and 2 we can write continuityequation
π1 π΄1 π£1 = π2 π΄2 π£2
In case of incompressible flow
π1 = π2
π΄1 π£1 = π΄2 π£2
CONTINUITY EQUATION:
1
2
7Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Hydrodynamics
Aerodynamics
Electromagnetism
Quantum mechanics
Flow of fluid through pipe, ducts or tubes
Rivers
Process plants
Power plants
Dairies
Logistics in general
CONTINUITY EQUATION APPLICATION:
1
2
8Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Consider Steady flow
Incompressible flow
Momentum at section 1-1
= π1 π π£π₯1
Momentum at section 2-2
= π2 π π£π₯2
Change of momentum
= (π2 π π£π₯2) β (π1 π π£π₯1)
For incompressible flow
π1 =π2
Change of momentum = π π (π£π₯2- π£π₯1)
MOMENTUM EQUATION:
β’ By impulse momentum principleβ¦F = π π (π£π₯2- π£π₯1)
9Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Now, components of velocity
π£π₯1 = π£1 cosΞΈ1 , π£π₯2 = π£2 cosΞΈ2
π£π¦1 = π£1 sinΞΈ1 , π£π¦2 = π£2 sinΞΈ2
Component of force along x-axisand y-axis
Fx= Ο Q (π£2 cosΞΈ2 - π£1 cosΞΈ1 )
Fy= Ο Q (π£2 sinΞΈ2 - π£1 sinΞΈ1 )
Above equations represent the forceexerted by the pipe bend on fluidmass.
Same but opposite force applied byfluid on pipe bend so equation is
Fx= Ο Q (π£1 cosΞΈ1 - π£2 cosΞΈ2 )
Fy= Ο Q (π£1 sinΞΈ1 - π£2 sinΞΈ2 )
MOMENTUM EQUATION:
The resultant force exerted by the fluid is
F = πΉπ₯2 + πΉπ¦2
The direction of resultant force is
Ξ = tan-1 (FyFx
)
10Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
LOSS OF ENERGY IN PIPE:
11Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Consider uniform horizontal pipe with steady flow.
Let 1-1 and 2-2 are two section of pipe.
Let,
P1 = pressure intensity at section 1-1
v1 = velocity of flow at section 1-1
L = length of pipe between sections
d = diameter of the pipe
fβ = frictional resistance per unit wetted are per unit velocity
hf = loss of head due to friction
Darcy-Weisback equation for head loss due to friction:
L1
1
2
2
P1 P2F1
F1
d
P2 = pressure intensity at section 2-2v2 = velocity of flow at section 2-2
12Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Applying Bernoulliβs equation at section 1-1 and 2-2
p1/Οg + v12/2g + Z1 = p2/Οg + v2
2/2g + Z2 + hf
But pipe is horizontal and diameter same at both the section
so Z1 = Z2 and v1 = v2
p1/Οg = p2/Οg + hf
Therefore,
p1/Οg - p2/Οg = hf or (p1 - p2) = Ο g hf
Darcy-Weisback equation for head loss due to friction:
1
1
2
2
P1 P2F1
F1
d
13Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Now,
Frictional resistance = frictional resistance per unit wetted area per unit velocity * wetted area * (velocity)2
F1 = fβ * ΟdL * v2
= fβ * P * L * v2
Darcy-Weisback equation for head loss due to friction:
1
1
2
2
P1 P2F1
F1
d
The force acting on the fluid between the sections1. Pressure force at section 1-1 = p1 * A2. Pressure force at section 2-2 = p2 * A3. Frictional force as shown in the figure = F1
Resolving all the forces in horizontal direction, we have
p1 * A - p2 * A - F1 = 0, or ( p1 - p2 ) * A = F1
Putting F1 = fβ * P * L * v2 and (p1 - p2) = Ο g hf
14Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Rewriting the equation
Ο g hf A = fβ * P * L * v2
hf = fβ * P * L * v2 / A * Ο g
Now, P/A = wetted perimeter / area = 4/d
Darcy-Weisback equation for head loss due to friction:
1
1
2
2
P1 P2F1
F1
d
So,
hf = fβ * 4 * L * v2 / d * Ο g
hf = 4 βfβ β L β v2
Ο g d
Putting fβ/ Ο = f/2 , where f is co-efficient of friction.
hf = 4 βf β L β v2
2 g d
15Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Co-efficient of friction is given by
f = 16/Re for laminar flow
f = 0.0791/(Re)1/4 for turbulent flow (Reβ₯4000 but β€105)
f = 0.0008 + 0.5525/0.257 * Re (Re β₯ 105 but β€ 107)
Some times equation can be written as
f is called friction factor (f = 4f), which is dimensionless quantity.
Darcy-Weisback equation for head loss due to friction:
hf = f β L β v2
2 g d
16Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
In previous derivation we have seen that
hf = fβ * P * L * v2 / A * Ο g
P/A = wetted perimeter / area = 1/m
V2 = Ο g
fβ* m *
hf
L
V = Ο gfβ
* m βhf
L
V = C π β π
Chezyβs formula for head loss due to friction:
C = Chezyβs Constant, hf
L= i, i = loss of head per unit length
17Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Example:
18Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Losses due to
1. Sudden contraction
2. Sudden enlargement
3. Bend in pipe
4. At entrance of pipe
5. An obstruction
6. Various pipe fitting
7. At exit of pipe
MINOR LOSSES:
19Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Flow separation is takes place due to suddenchange in diameter of pipe, it results inturbulent eddies formation as shown in figure.
The loss of head takes place due to eddiesformation.
Let,
P1 = pressure intensity at section 1-1
v1 = velocity of flow at section 1-1
A1 = area of pipe at section 1-1
P2 , v2 , A2 corresponding value at section 2-2
Pβ = pressure intensity of liquid eddies on the area(A2 β A1 )
he = loss of head due to sudden enlargement
1. Loss of head due to sudden enlargement:
P1 A1 P2 A2
20Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Applying Bernoulliβs equation to section 1-1 and2-2
p1/Οg + v12/2g + Z1 = p2/Οg + v2
2/2g + Z2 + he
Pipe is horizontal so Z1 = Z2 rewriting the equ.
p1/Οg + v12/2g = p2/Οg + v2
2/2g + he
he = (p1/Οg - p2/Οg) + (v12/2g - v2
2/2g )
1. Loss of head due to sudden enlargement:
P1 A1 P2 A2
Consider the control volume of liquid between sections 1-1 and 2-2. Force acting on the liquid in the direction of flow is given by,Fx = p1A1 +pβ(A2 β A1) β p2A2
Experimentally it is found that pβ = p1
So equation becomesβ¦Fx = p1A1 +p1(A2 β A1) β p2A2
21Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
1. Loss of head due to sudden enlargement:
P1 A1 P2 A2
Equation is simplified in the form of β¦.Fx = p1A2 β p2A2
= (p1β p2) A2
Momentum of liquid at section 1-1 = mass * velocity
= π1 π΄1 π£1 * π£1= π1 π΄1 π£1
2
Similarly momentum at section 2-2 = π2 π΄2 π£22
β΄ Change of momentum = (π2 π΄2 π£22 - π1 π΄1 π£1
2 )
From continuity equation π¨π ππ = π¨π ππSo π¨π = π¨π ππ / ππ
Change of momentum = (π2 π΄2 π£22 - π1 π΄1 π£1
2 )
= π2 π΄2 π£22 - π1 π΄2 π£1π£2
22Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
1. Loss of head due to sudden enlargement:
P1 A1 P2 A2
Change of momentum = π2 π΄2 π£22 - π1 π΄2 π£1π£2
= π2 π΄2 (π£22 - π£1π£2 )
Now net force acting on the fluid is equal to change of momentum,
Fx = π2 π΄2 (π£22 - π£1π£2 )
(p1β p2) A2 = π2 π΄2 (π£22 - π£1π£2 )
β΄ (p1β p2)/ π2 = (π£22 - π£1π£2 )
Fluid is incompressible so (π2 = π1 = π )β΄ (p1β p2)/ π = (π£2
2 - π£1π£2 )
Dividing both the sides with g,
(p1β p2)/ π g = (π£22 - π£1π£2 )/g
Rememberβ¦β¦he = (p1/Οg - p2/Οg) + (v1
2/2g - v22/2g )
he = (π£22 - π£1π£2 )/g + (v1
2/2g - v22/2g )
23Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
1. Loss of head due to sudden enlargement:
P1 A1 P2 A2
he = (π£22 - π£1π£2 )/g + (v1
2/2g - v22/2g )
=π£1
2 β2π£1π£2 + π£22
2g
β΄ he =(π£1βπ£ 2)
2
2g
The above equation is used for calculation ofhead loss due to sudden enlargement.
24Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
2. Loss of head due to sudden contraction:
C
C
P1 A1 P2 A2
1 2
Figure shows flow of liquid through section1-1 and 2-2.
Flow of liquid is from large pipe 1-1 tosmall pipe 2-2.
Flow at section c-c is minimum so sectionc-c is called vena-contract.
Let,
Ac = area of flow at section c-c
vc = velocity of flow at section c-c
A2 = area of flow at section 2-2
v2 = velocity of flow at section 2-2
hc = loss of head due to sudden enlargementfrom section c-c to 2-2
25Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
2. Loss of head due to sudden contraction:
C
C
P1 A1 P2 A2
1 2
Similar to loss of head due to suddenenlargement,
hc =(π£cβπ£
2)2
2g
=π£22
2g(π£c
π£2
β 1)2
hc =π£22
2g(π
πͺπ
β 1)2
=πΎ π£
22
2gwhere K = (
π
πͺπβ 1)2
From continuity equation π¨π ππ = π¨π ππSo ππ / ππ = π¨π / π¨π = 1/ Cc (Cc = π¨π / π¨π)
If Cc is assumed to be 0.62, K = 0.375
β΄ hc = 0.375 π£22
2gGenerally,
hc = 0.5π£22
2g
26Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
The loss of head when pipe is connected to a large tank or reservoir.
The loss is similar to loss of head due to sudden contraction.
In general,
hi or hent = πΎ π£
22
2g
The value of k is as per given table.
3. Loss of head at the entrance of pipe:
27Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
The loss of head at the exit of pipe as aresult of form of a free jet or it may beconnected to reservoir.
General equation
ho =π£2
2g
v = velocity at the outlet of the pipe.
4. Loss of head at the exit of pipe:
28Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
As shown in the figure if there is anyobstruction in pipe, it cause loss ofenergy.
. There is sudden enlargement of the areaof flow beyond the obstruction due towhich loss of head take place.
Loss of head due to obstruction is
=π£2
2g(
π΄
πΆπ(π΄βπ)
β 1)2
Where,
Cc = coefficient of contraction
A = are of pipe
a = maximum area of obstruction
v = velocity of liquid in pipe
5. Loss of head due to an obstruction in pipe:
29Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Due to bend in pipe, change in velocity as well formation of eddies take place.
It result in loss of energy due to bend in pipe.
It can be express as per below:
hf = πΎ π£
22
2g
The value of K is different for deferent bend angle and r/d ratio.
6. Loss of head due to bend in pipe:
30Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Example:
31Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
To analyze the pipe problem the concept of HEL or TEL is very useful.
Both are graphical representation of the longitudinal variation in the total head at salient points of pipe line.
Total head as per Bernoulliβs equation given by
p/πΎ + v2/2g + Z = Constant
p/πΎ = Pressure Head
v2/2g = Velocity Head
Z = Potential Head
HYDRAULIC GRADIENT AND TOTAL ENERGY LINE:
32Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Line representing the sum of pressurehead, datum head and velocity head withrespect to some reference line is calledtotal energy line (TEL).
It is represented by line connecting thevalues of the total head at successivepoints along a piping system.
For ideal fluid as there are no losses, totalenergy line would remain parallel to thedatum.
Total energy line (TEL):
33Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Hydraulic grade line (HGL):
Line representing sum of pressure headand datum head with respect to somereference line (datum) is called as hydraulicgradient line (HGL).
It the line obtained by connecting thevalues of the piezometric head (p/πΎ + Z) atsuccessive points along the piping system.
HGL is always vertically below TEL by andequal amount to the velocity head (v2/2g).
34Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Two or more pipes of different diameter are
connected end to end to form a single pipe line.
Pipes can be same or different length.
Discharge through all pipes is same.
Q = Q1= Q2 =Q3
Total loss of head is sum of losses in all individual pipes and fittings.
H = hf1 + hf2 + hf3+β¦β¦β¦.+ minor losses
For given figure
H = 0.5 π£
12
2g+ 4π
1πΏ1π£12
2ππ1
+0.5 π£
22
2g+ 4π
2πΏ2π£22
2ππ2
+(π£2βπ£
3)2
2g+ 4π
3πΏ3π£3
2ππ3
+ π£32
2g
If minor losses are neglected,
H = 4π
1πΏ1π£12
2ππ1
+ + 4π
2πΏ2π£22
2ππ2
+ 4π
3πΏ3π£3
2ππ3
(4f = f, f is friction factor)
PIPE IN SERIES OR COMPUND PIPE:
35Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Compound pipe is consisting of several pipes
of varying diameter and length.
Compound pipe is replaced by pipe of uniform diameter having
EQUIVALENT PIPE:
loss of head and discharge
loss of head and discharge of compound pipe
=
L1d1v1
L2d2v2
L3d3v3
Then, L= L1 + L2 + L3
Total head loss in the compound pipe neglecting minor losses,
H = ππ
ππ³ππππ
πππ π
+ ππ
ππ³ππππ
πππ π
+ ππ
ππ³ππππ
πππ π
(Assuming f = f1 = f2 = f3)
Discharge Q = A1v1 = A2 v2 = A3 v3
Q = π
πd1
2 v1 = π
πd2
2 v2 =π
πd3
2 v3
v1 = πQπ d1
2 , v2 = πQπ d2
2 , v3 = πQπ d3
2
36Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
EQUIVALENT PIPE:
L1d1v1
L2d2v2
L3d3v3
Putting the value of velocity in head loss equation
H = ππ
ππ³π(πQπ d1
2)π
πππ π
+ + ππ
ππ³π(πQπ d2
2)π
πππ π
+ ππ
ππ³π
πQπ d3
2
π
πππ π
H = πβπππQπ
π πβππ
π³π
π ππ +
π³π
π ππ +
π³π
π ππ
Head loss in equivalent pipe
H = πππ³ππ
πππ = πβπππQπ
π πβππ
π³
π π
Equating both head loss equation πβπππQπ
π πβππ
π³π
π ππ +
π³π
π ππ +
π³π
π ππ =
πβπππQπ
π πβππ
π³
π π
π³
π π = π³π
π ππ +
π³π
π ππ +
π³π
π ππ
The above equation is known as Dupuitβs equ.
37Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
The purpose of the parallel pipe is toincrease the discharge of the fluid.
For parallel pipes
Q = Q1 + Q2 + Q3+β¦β¦.
Head loss through each branch is same
hf1 = hf2 = hf3 =β¦β¦..
As per figure
Q = Q1 + Q2 &
hf1 = hf2
ππππ³ππππ
πππ π
= ππ
ππ³ππππ
πππ π
For f1 = f2
π³ππππ
π π
=π³ππππ
π π
PIPES IN PARALLELS:
38Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
The hydraulic power transmitted by a pipe depends on:
1. Discharge
2. Total head
Head available at the outlet of pipe isH -hf
The power available at the outlet of the pipe is
P = π π π (H βhf) , Watts
The maximum power transmission isobtained by
ππ
ππ£= 0
hf = H/3
POWER TRANSMISSION THROUGH THE PIPES:
39Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Ξ· = πππ€ππ πππππ£ππππ ππ‘ ππ’π‘πππ‘ ππ ππππ
πππ€ππ ππ£ππππππ ππ‘ πππππ‘ ππ ππππ
= π π π (H βhf)
π π π H
= (H βhf)
H
Under condition of maximum power the efficiency is given by
Ξ· = (H βH/3)
H= 2/3
= 66.67%
EFFICIENCY OF POWER TRANSMISSION:
40Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
Example:
41Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot
References:
1. Fluid Mechanics and Fluid Power Engineering by D.S. Kumar, S.K.Kataria & Sons
2. Fluid Mechanics and Hydraulic Machines by R.K. Bansal, LaxmiPublications
3. Fluid Mechanics and Hydraulic Machines by R.K. Rajput, S.Chand & Co
4. Fluid Mechanics; Fundamentals and Applications by John. M. CimbalaYunus A. Cengel, McGraw-Hill Publication