Fluid MechanicsCHEM 218

Zin-Eddine Dadach

2006-2007

Goals of the course

• 1) Fluid statics and its application

• 2) Fluid Flow concepts and related experiments

• 3) Flow of incompressible fluids in pipes and related experiments

• 4) Flow of compressible fluids

• 5) Pumping and metering systems of fluids

• 6) Fluid mixing and fluidization—principles and applications

Fluid Mechanics?

• 1) Fluid Mechanics is the study of the

behavior of fluids at rest ( Fluid statics) or in motion ( Fluid dynamics)

• 2) In fluid statics, specific weight ( unit weight) is the important property

• 3) In fluid dynamics, density and viscosity are the predominant properties

Fluid?

• 1) Capable of flowing and conform to the shape of the container

• 2) Fluid can be liquid or gas

• 3) Liquids are incompressible whereas gases are compresible

Effects of temperature and pressure on the density of fluids?

• 1) Density of all fluids depends on the temperature and pressure

• 2) Incompressible fluids: slight change on the density –> example 1.4

• 3) Compressible fluids:

• T ↑→ V↑→ d↓

• P↑ →V↓ → d↑

What is compressibility?

• 1) Change in volume V of a fluid with change of pressure

• 2) The bulk modulus E is used:

E= (-ΔP) / (ΔV/V)

Examples for values of E at 20°C:

Ethyl Alcohol : 130.000

Water : 316.000

Mercury : 3.590.000

What is density?

• 1) Amount of mass per unit volume

• ρ = m/ V

• ASTM method of measure : pycnometers

• Units :

SI system: ( Kilograms per cubic meter)

US system: (slugs per cubic foot)

What is specific weight?

• Amount of weight per unit volume

• γ= ω/ V

• V is the volume of a substance having the weight ω

• SI system: newtons per cubic meter

• US system: pounds per cubic foot

What is specific gravity?

• Definitions:

1) Ratio of the density of a substance to the density of water at 4°C

2) Ratio of the specific weight of a substance to the specific weight of water at 4°C

3) Mathematically:sg= ( γs / γw at 4°C) = (ρs/ρw at 4°C)

4) γw at 4°C= 9.81 kN/ m3 or 62.4 lb/ft3

5) ρw at 4°C = 1000kg/m3 or 1.94 slugs / ft3

Density and specific gravity relationship ?

• γ=ρg with g= acceleration of gravity

• Do examples 1.5 to 1.9

Class work

• Do problems : 1.17 -1.20 page 22

• Do problems : 1.48-1.54 page 23

• Do problems : 1.58-1.67 page 23

• Do problems : 1.70-1.76 page 24

What is pressure?

Amount of force exerted on a unit area of a substance

p=F/A Units: SI Pascal or US lb/ft2

Absolute and gage pressure:

Pabs=Pgage + Patm

Do examples 3.1 to 3.4

Pressure and elevation relationship?

For homogeneous liquid at rest:

Δp= γ.h where

Δp = Change in pressure

γ = specific weight of liquid

h = change in elevation

Do examples 3.5 to 3.7

Example 3.7 on page 56

From figure 3.3 on page 57 -

Calculate the pressure at:

Point A: Point B:

Point C: Point D:

Point E: Point F:

What is a manometer?

• Pressure measurement device

• Uses the relationship : Δp=γ.h where γ is the specific weight and h is the height

• Simplest kind of manometer is the U-tube (Figures 3.9 and 3.10 on pages 55 and 56)

• The tube contains a liquid called the gage fluid which should not mix with the fluid which pressure is to be measured

• Fluid can be water, mercury and colored light oils

Calculation of pressure usingthe U-tube

• In calculations , it is convenient to start with the open end

• Class work:

• Given the γmercury= 132.8 kN/m3 and

γwater = 9.81 kN/m3

What is the gage pressure at point A of Figure 3.10 page 63?

CLASS WORK

• Work examples : 3-8 to 3-13

• Work Problems : 3.52, 3.54, 3.56, 3.57, 3.58 and 3.59

What is a gravity decanter?

• Used for continuous separation of two immiscible liquids (Figure 2.5 from note):

A) Feed enters at one end

B) slow flow through the decanter

C) Separation into two layers

D) Discharge trough overflow lines on the other hand

Fluid Statics analysis of gravity decanter

• Hypothesis:

A) Large overflow lines to neglect flow resistance

B) the discharge is at the same pressure as the one in the gas space above the liquid in the vessel:

Calculation of a gravity decanter

At point 1: P1 =0At point 4: P4 =P1+ ρA. ZA2= ρA. ZA2

At point 5: P5=P4= ρA. ZA2

At point 2: P2= P5 - ρA. ZA1= ρA. (ZA2- ZA1)At point 3: P3= 0= P2 - ρB. ZB

P3= ρA. (ZA2- ZA1) - ρB. ZB=0

ZT=ZA1 + ZB =} ρA. (ZA2- ZA1) - ρB.( ZT-ZA1)=0

And finally : ZA1= [ZA2 – ZT (ρB/ ρA)]/ [1- (ρB/ ρA)]

Separation time of the two fluids in the decanter

ZA1= [ZA2 – ZT (ρB/ ρA)]/ [1- (ρB/ ρA)]

This equation shows that the position of liquid-liquid interface depends on:

A) Ratio of densities

B) Elevation of overflow lines

By definition the time of separation is calculated by: t= (100.μ) / (ρA- ρB)

If ρA≈ ρB → t approaches ∞

What is a centrifugal decanter?

• We need it when the difference between the density of the two liquids is too small

• Centrifugal decanter is based on centrifugal force created by a vertical rotating cylinder

• Bowl at rest ( Figure 2.6a)

• Bowl rotating ( Figure 2.6b)

Goal 2: Fluid Flow

• Fluid flow concept:

One dimensional flow

Laminar Flow

Turbulent Flow

One dimensional flow

A characteristic of this flow is that the velocity becomes invariant in the flow direction as shown in Figure

It is readily seen that velocity at any location depends just on the radial distance from the centerline and is independent of distance, x or of the angular position. This represents a typical one-dimensional flow.

Laminar Flow

• Fluid flows uniformly with low velocity and with little or no mixing from one layer to another layer ( Figures 8.1 and 8.2)

• Velocity profile is parabolic

Velocity Profile in Laminar Flow

• To get the kinetic energy of laminar flow in a tube, an average of the square of the velocity must be taken to account for the velocity profile.

•

•

• V = Vm ( 1- (r/R)2) where Vm is the velocity at r=0

•

• V= 2Va (1- (r/R)2) where Va is the average velocity•• vm = 2va

•

Turbulent Flow

• Occurs at higher velocities

• Mixing is chaotic within the stream

• Flat velocity Profile ( Figures 8.3 and 8.5)

Flows and velocity gradients

• Velocity gradient is a measure of velocity changes or shear rate and is defined as:Δv/ Δy where y is the direction of the flow

• Example: a fluid is placed between two parallel plates that are 1.0 cm apart, the upper plate moving at a velocity of 1.0 cm/sec and the lower plate fixed.

• The velocity gradient is the rate of change of velocity with distance from the plates.

• This simple case shows the uniform velocity gradient with shear rate (v1 - v2)/h = shear rate = (cm/sec)/(cm/1) = 1/sec.

• Hence, shear rate units are reciprocal seconds or sec-1

What is shear Stress? IS the force per unit area required to sustain a constant rate of fluid

movement. Mathematically, shear stress can be defined as:

•

• If a fluid is placed between two parallel plates spaced 1.0 cm apart, and a force of 1.0 dyne is applied to each square centimeter of the surface of the upper plate to keep it in motion, the shear stress in the fluid is 1 dyne/cm2 at any point between the two plates

For Common Fluids

• Water, oil, gasoline, air ……

the shearing stress and rate of shearing strain ( velocity gradient) can be related by :

τ= μ ( dv /dy)

Where μ is the dynamic viscosity with unit

as [μ] = N.s/m2 or Pa.s and [τ] = N/m2

What is viscosity?

• The viscosity of a fluid is a very important property in the analysis of liquid behavior and fluid motion near a solid boundary.

• The knowledge of viscosity is often necessary for proper design of required temperatures for storage, pumping or injection of fluids.

Dynamic viscosity

The dynamic viscosity of a fluid is its resistance to shear or flow and is a measure of the fluids adhesive/cohesive or frictional properties.

τ= μ ( dv /dy)

Where μ is the absolute or dynamic viscosity with

unit as [μ] = N.s/m2 or Pa.s and [τ] = N/m2

Kinematic viscosity

• A coefficient which describes the diffusion of momentum.

• Let the dynamic viscosity be μ, then

ν= μ/ρ• the unit of kinematic viscosity is the Stoke,

equal to 1 cm2 s-1.

•

• A force of 3 Newtons is used to move a plate of 1cm2 section. The variation of velocity is 3 cm s-1. The two plates are 1 cm apart.

• A) calculate the dynamic viscosity

• B) If the density is 1000 kg/m3, calculate the kinematic viscosity

For turbulent flow

• The dynamic viscosity is replaced by the Eddy viscosity ( Ev) and then

τ= Ev ( dv /dy)

Newtonian Fluid ?

A fluid that has a constant viscosity at all shearrates at a constant temperature and pressure, and can be described by a one-parameter rheological model.

•

Water, sugar solutions, glycerin, silicone oils, light-hydrocarbon oils, air and other gases are Newtonian fluids.

Non Newtonian flow?

Reynolds Number The Reynolds Number is important in analyzing any type of flow when there is

substantial velocity gradient - shear. The Reynolds Number indicates the relative significance of the viscous effect

compared to the inertia effect. The Reynolds number is proportional to inertial force divided by viscous force.• Reynolds Number can be express

• The viscosity above is dynamic viscosity also called absolute viscosity. For a pipe or duct the characteristic length is the pipe or duct diameter.

Values of NRe

• For laminar Flow :

NRe < 2000

• For turbulent Flow:

NRe > 4000

Do examples 8.1 to 8.3 of the book

Do problems 8.1 to 8.4 page 247

Continuity Equation

• When a fluid is in motion, it must move in such a way that mass is conserved.

• To see how mass conservation places restrictions on the velocity field, consider the steady flow of fluid through a duct (that is, the inlet and outlet flows do not vary with time).

• The inflow and outflow are one-dimensional, so that the velocity V and density are constant over the area A (figure 14).

FIGURE

•

M1=M2 This is the continuity equation

Volume Flow rate

volume of fluid passing trough a

section A per unit time

Q = A.V [Q] = m3/s

Example: If a section is 6.356.10-4m2 and the velocity equal 3m/s . Calculate Q

Response : Q= 1.907.10-3 m3/s

Specific weight flow rate

• W = γ. Q

Example: If the specific weight of water is 9.81 kN / m3 and its volume flow rate is 0.01 m3/s

W= 9.81x 0.01 = 0.0981 kN/s

Mass Flow rate

• M =ρ.Q

Example : If the density of water is 1000 kg/m3 and the volume flow rate is 0.01 m3/s

Mass Flow rate 1000 x 0.01 = 10 kg/s

Class work #1

• From figure given , The inside diameters of the pipe in sections 1 and 2 are 100mm and 50 mm respectively. Water at 200C flows with an average velocity of 2m/s at section 1, Calculate the following:

* Velocity at section 2 ( 8m/s)

* Volume flow rate ( 0.0157m3/s)

* Weight flow rate ( 0.154 kN/s)

* Mass Flow rate ( 15.7 kg/s)

Class work #2

• A) Determine the maximum velocity of a gasoline and water flowing at 200C in laminar flow manner in a 20mm pipe?

• Kinematic viscosity of gasoline and water are respectively 6.48 10-7m2/s and

• 1.02 10-6 m2/s• B) Determine the type of flow occuring in a 12”

pipe when water at 600F flows at velocity of 3.50 ft/s ? Same question for gasoline

• ( νwater= 1.217 10-5 ft2/s and νgasoline =221 10-5

ft2/s)

Class work #3• A) Water flows into a pipe of diameter 0.1 m

with a velocity of 4 m/s . Determine the diameter of the other hand of the pipe if the velocity becomes 2m/s

• B) If the velocity of liquid is 1.65ft/s in a 12” diameter pipe ..what would be the velocity at the other end where the diameter is 3”

• C) 2000L/min of water flows through 300 mm diameter pipe that reduces to 150 mm diameter .calculate the velocity at the two ends.

Review #2

1) Use Figure 3.57 page 80 and calculate the pressure PA.

2) Page 247

3) Page 187

Review

I) Analyze the figure given and calculate the differential pressure ( PA-PB)

II) Determine the minimum velocity of a gasoline and water flowing at 250C in turbulent flow manner in a 50mm pipe?Kinematic viscosity of gasoline and water are respectively 6.48 10-7m2/s and 1.02 10-6

m2/sIII) 1m3/min of water flows through 200 mm

diameter pipe that reduces to 50 mm diameter .calculate the velocity at the two ends

Bernoulli equation

• Bernoulli's Equation

• The Bernoulli equation states that,

• Restrictions:

• the fluid has constant density,

• the flow is steady, and

• there is no friction.

Bernoulli equation

Bernoulli equation

• Consider the steady, flow of a constant density fluid in a converging duct, without losses due to friction The flow therefore satisfies all the

restrictions governing the use of Bernoulli's equation.

• Upstream and downstream of the contraction we make the one-dimensional assumption that the velocity is constant over the inlet and outlet areas.

Bernoulli equation

Example

• Water at 100C is flowing from section 1 to section 2. At section 1, the gage pressure is 345 kPa, the diameter is 25 mm and the velocity of water is 3 m/s. Section 2 is above section 1 by 2 m and has a diameter of 50 mm.

• Calculate P2?

Bernoulli equation

• P1 + ½ ρ v12 + ρgh1= P2 + ½ ρ v2

2 + ρgh2

• P2= P1 + ½ ρ[ v12 - v2

2] + ρg[h1-h2]

• Calculate v2?

• Continuity equation A1V1= A2V2

• V2 = [A1/ A2] V1

• A1= ¼ ΠD12= ¼ x3.14x (25)2= 491 mm2

• A2 = ¼ ΠD22= ¼ x3.14x (50)2= 1963 mm2

• V2 = [ 491/1963] x 3 = 0.75 m/s

• ρ = 1000 kg/m3

Bernoulli equation

• P2 = P1 + ½ x1000 x [ 32- 0.752 ] +

1000X 9.81 x (-2)

• P2=P1 – 15.4 kPa

• P2 = 345 – 15.4 = 329.6 kPa

General Energy Equation

• Objective: We will apply Energy equations to real systems including pumps, fluid motors, turbines and energy losses from friction, valves and fittings

Analyze the energy in fluids flow

systems by adding terms to Bernoulli equation

What is a turbine?

• Turbines like fluid motors act in opposite way than pumps

• It takes energy from the fluid and deliver it in form of work to cause rotating of a shaft or a linear movement of a piston

What is fluid friction?

• A Fluid in motion will lose some energy due to frictional resistance to flow

• The magnitude of this energy loss depends on:

* properties of fluid

* flow velocity

* Pipe size

* Smoothness of pipe walls

Some of this energy is lost as heat through the pipe walls

What is valve and Fittings?

• Mechanical devices that control the direction of the flow

• they cause some energy loss usually small called minor losses

Energy Nomenclature

• We will call:

hA= Energy added to the fluid with a mechanical device such as a pump often called total head of the pump.

hR = Energy removed from the fluid with a mechanical device such as a turbine or fluid motor

hL = Energy losses from the system due to friction in pipes or minor losses due to valves or fittings

General Energy Equation

• E’1+ hA-hR-hL= E’

2• The terms E’1 and E’2 are the energy

possessed by the fluid per unit weight as seen in Bernoulli equation

•

g

vz

PE

2'

2

Be Carefull!!!!!

• This general energy equation can be applied only in the direction of the flow

An element of fluid at section 1 and

having an energy per unit weight E’1 may

have energy added ( hA) from a pump , an energy removed (-hR) from a turbine or energy loss ( -hL) before it reaches section 2 ( Example see Figure 7.6 page 197)

Class work

• Example 7.1 page 198

• Example 7.2 page 199

POWER REQUIRED BY PUMPS

• PA= hA. Wwhere W is called the weight flow rate and is

expressed as N/s• Since W=γ.Q PA= hA. γ.Q

• Units: SI 1Watt = 1N.m/s = 1 joule/sUS 1 hp= 550 lb.ft/s

1hp= 745.7 W1lb.ft/s = 1.356 W

Mechanical efficiency of pump

• is the ration between the power delivered to the fluid PA and the power received by the pump PI

• eM= PA/PI

• Example 7.3 page 203

Fluid motors and turbines

• Power delivered PR= hR.γ.Q

• Mechanical efficiency is equal to the ratio of the power output from motor or turbine Po to the power delivered by the fluid PR

• eM= Po/PR

• Example 7.4 page 205

Energy Losses Due to Friction

• Darcy’s equation:

• hL is the energy loss from the system

• One component of the energy loss is due to friction in the flowing fluid

g

vz

phhh

g

vz

pLRA

22

2

22

2

2

11

1

Friction in the flowing fluid

• Friction is proportional to:

* velocity head of the flow v2/2g

* Ratio of the length to the diameter of the flow stream L/D

• Mathematically, we can write:

g

v

D

LfhL

2..

2

• hL is the energy loss due to friction ( N.m/N or lb.ft/lb or m or ft)

• L is the length of the flow stream ( m or ft)

• D is the pipe diameter ( m or ft)

• v is the average velocity of flow ( m/s or ft/s)

• f friction factor ( dimensionless)

When can we use Darcy’s equation?

• The Darcy’s equation can be used to calculate energy loss due to friction in long straight sections of round pipes for both laminar and turbulent flow

• The difference between laminar and turbulent flow is in the evaluation of the friction factor f

Hagen- Poiseuille equation• The parameters involved in energy loss in laminar

flow are:

* Fluid properties : viscosity and specific gravity

* Dynamics of the flow

* geometrical features of length and diameter

• The Hagen- Poiseuille equation:

2

32

D

LvhL

Friction Factor in Laminar Flow

• The Reynolds Number is defined as :

• Therefore f=64/NRe

..Re

DvN

g

Example

• Problem 9.1 page 242

Friction Loss in Turbulent Flow

• There is no formula like in laminar flow

• Experimental data have shown that the friction factor depends on Reynolds number and on the roughness of the pipe

• Some values of the roughness are given in table 9.1 page 243

MOODY DIAGRAMFIGURE 9.2 page 244

• the diagram shows that the friction factor f plotted versus the Reynolds number with a series of parametric curves related to the relative roughness D/ε

Description of the Moody Diagram

• Both f and NRe are plotted on logaritmic scales

• At the left side of the graph, we have the laminar flow equation f=64/NRe

• In the critical region 2000<NRe< 4000 , there is no plot, the behavior can not be predicted

• Beyond 4000, the family of curves for different values of D/ε are plotted

Moody Diagram: Turbulent section

• For a given NRe : if D/ε increases f decreases

• For a given D/ε : if NRe increases f decreases

• In the zone of complete turbulence, the Reynolds number has no effects on the friction factor

• As D/ε increases, the value of NRe where the complete turbulence starts increases also

Examples

• Problems 9.2, 9.3, 9.4 in pages 246-248

Minor Losses

• What is valve and Fittings?

• Mechanical devices that control the direction of the flow

• they cause some energy loss usually small called minor losses

• Darcy’s equation:

• hL is the energy loss from the system

• One component of the energy loss is minor losses due to valves and fittings.

g

vz

phhh

g

vz

pLRA

22

2

22

2

2

11

1

Resistance Coefficient K

• Minor energy losses include elbows, enlargement of pipe, contraction of pipe and a valve

• These energy losses are proportional to the velocity head of the fluid:

)2/( 2 gvKhL

Sudden enlargement

• As shown in Figure 10.1 ( page 272), the minor loss energy is calculated by the equation:

• If the velocity v1 is close to 1.2 m/s , the value of K can be estimated by the relation:

)2/( 2

1 gvKhL

22

21 /1 DDK

• Use Figure 10.2 ( page 273) for a more precise value depending on the velocity

• Work example 10.1 page 273

EXIT LOSS

• This is the case where a fluid flows from a pipe to a tank ( Figure 10.3 page 275)

• The energy loss hL can be calculated as:

Work example 10.3 page 275

gvhL 2/2

1

GRADUAL ENLARGEMENT

• In this Figure ( Figure 10.4 page 276), the transition from small pipe to large pipe is smoother and the energy loss is therefore smaller.

• The resistance coefficient K depends on the ratio D1/D2 and on the value of the cone’s angle θ ( Table 10.2 page 277)

• Work example 10.4 page 277

Sudden Contraction

• The energy loss due to sudden contraction can be calculated by the relation ( Figure 10.6 page 278):

• The value of K are given in Table 10.7 page 279

• Work example 10.5 page 280

)2/( 2

2 gvKhL

Gradual Contraction

• Figures 10.10 and 10.11 ( pages 281-282) show that K depends on the ratio D1/D2 and on the angle θ

Entrance loss

• The values of K depends on the shape and geometry of the entrance.

• Figure 10.13 gives the values of K for different situations

• Work example 10.6 page 284

Valves and fittings

• K = fT(Le/D)

• The value of Le/D are given in table 10.4 page 287 and the values of fT are given in table 10.5 page 288

• Work example 10.7 page 288

Class Work

• Work problems 10.1M , 10.4M, 10.5E, 10.6 M, 10.15M, 10.17 E

Homework

• Work problems 10.2, 10.3, 10.8 and 10.16