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Pharos UniversityPharos University

ME 259 Fluid Mechanics for ME 259 Fluid Mechanics for Electrical StudentsElectrical Students

Application of Bernoulli Application of Bernoulli EquationEquation

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• Euler’s equation of motion Euler’s equation of motion

• Bernoulli equationBernoulli equation

3

INTRODUCTIONINTRODUCTION The three equations commonly used in fluid mechanics are:

the mass, Bernoulli, and energy equations.

The mass equation is an expression of the conservation of mass principle.

The Bernoulli equation: Conservation of kinetic, potential, and flow energies ( viscous forces are negligible.)

4

Fluid MotionFluid Motion

• Two ways to describe fluid motion– Lagrangian

• Follow particles around

– Eularian• Watch fluid pass by a

point or an entire region

– Flow pattern• Streamlines – velocity

is tangent to them

kjiVdt

dz

dt

dy

dt

dx

kjiV wvu

5

Steady flow: the flow in which conditions at any point do not change with time is called steady flow.

Then, …etc.

• Unsteady flow: the flow in which conditions at any point change with time, is called unsteady flow.

Then, …etc.

,0,0,0

tt

V

t

P

,0,0,0

tt

V

t

P

STEADY AND UNSTEADY FLOWSTEADY AND UNSTEADY FLOW

6

UNIFORM AND NON-UNIFORMUNIFORM AND NON-UNIFORM

• The flow in which the conditions at all points are the same at the same instant is uniform flow.uniform flow.

• The flow in which the conditions vary from point to point at the

same instant is non-uniform flow.non-uniform flow.

,0,0,0

ss

V

s

P

,0,0,0

ss

V

s

P

7

ACCELERATIONACCELERATION

• Acceleration = rate of change of velocity

• Components: – Normal – changing direction

– Tangential – changing speed

dt

dV

dt

dV

dt

d

tsV

tt

t

ee

Va

eV

),(

nt

nt

r

V

t

V

s

VV

r

V

dt

dt

V

s

VV

dt

dV

eea

ee

2)(

Fluid Mechanics I 8

ACCELERATIONACCELERATION

• Cartesian coordinates

• In steady flow ∂u/∂t = 0 , local acceleration is zero.• In unsteady flow ∂u/∂t ≠ 0 ; local acceleration Occurs.• Other terms u ∂u/∂x, v ∂u/∂y,.. are called convective

accelerations. Convective acceleration Occurs when the velocity varies with position.

• Uniform flow: convective acceleration = 0• Non-uniform flow: convective acceleration ≠0

kjiV

wvu

Convective Local

t

ww

z

wv

y

wu

x

w

t

w

dt

dz

z

w

dt

dy

y

w

dt

dx

x

w

dt

dwa

t

vw

z

vv

y

vu

x

v

t

v

dt

dz

z

v

dt

dy

y

v

dt

dx

x

v

dt

dva

t

uw

z

uv

y

uu

x

u

t

u

dt

dz

z

u

dt

dy

y

u

dt

dx

x

u

dt

dua

z

y

x

.

.

.

kjia

zyx aaa

Fluid Mechanics I 9

Example

• Valve at C is opened slowly

• The flow at B is non uniform

• The flow at A is uniform

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Laminar vs Turbulent Flow

• Laminar • Turbulent

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Flow Rate

• Volume rate of flow– Constant velocity over

cross-section

– Variable velocity

• Mass flow rate

VAQ

AVdAQ

QVdAVdAmAA

12

Flow RateFlow Rate

• Only x-direction component of velocity (u) contributes to flow through cross-section

AV

V

Q

or

dAQ

or

dAVudAVdAQ

A

A AAcos

13

CV Inflow & Outflow

Area vector always points outward from CV

AV

Q

CS

inout AVAVQQ

AV

AVAV

1122

1122

14

Examples

• Discharge in a 25-cm pipe is 0.03 m3/s. What is the average velocity?

smd

Q

A

QV

VAQ

/611.0)25.0(

4

03.0

422

• A pipe whose diameter is 8 cm transports air with a temp. of 20oC and pressure of 200 kPa abs. At 20 m/s. What is the mass flow rate?

skg

VAm

mkgRT

p

/239.0)08.0(4

*20*378.2

/378.2293*287

200000

2

3

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Example:Example: The velocity distribution in a circular duct is , where r is the radial location in the duct, R is the duct radius, and Vo is the velocity on the axis. Find the ratio of the mean velocity to the velocity on the axis.

• Find:

)1()(R

rVrv o oV

V

3

131

3

1

)32

(2)32

(2

2)/1(

2

2

2

22

0

32

0

o

o

oo

o

o

R

o

R

o

A

VR

RV

AV

Q

V

V

RV

RRV

R

rrV

rdrRrVVdAQ

)1()(R

rVrv o

16

Example: Air (ρ =1.2 kg/m^3) enters the duct shown:

• Find:

• V/10=y/.5 V=20y

• dA=1*dy

mVQ ,,

skgQm

smA

QV

smy

ydyVdAQ

/65*2.1

/51

5

/52

40

2022

35.0

0

2

5.0

0

5.0

0

Fluid Mechanics I 17

Example:Example: In this flow passage the discharge is varying with time according to the following expression: . At time t=0.5 s, it is known that at section A-A the velocity gradient in the S direction is +2m/s per meter. Given that Qo, Q1, and to are constants with values of 0.985 m^3/s, 0.5 m^3/s, and 1 s, respectively, and assuming that one-dimensional flow, answer the following questions for time t=0.5 s.a. What is the velocity a A-A?b. What is the local acceleration at A-A?c. What is the convective acceleration at A-A?

2

2

22

1

22

1

1

/49.72*743.3

/55.2)1()5.0(

4

5.0

4

)/(

/4743.3)5.0(

4

)5.0(5.0985.0

4

/2

5.0985.0

sms

VVa

smtd

Q

t

AQ

t

Va

smd

t

tQQ

A

QV

sms

V

tt

tQQQ

C

o

L

oo

oo

oo t

tQQQ 1

18

ExampleExample

a

kjiV

on,Accelerati:Find

3:Given 2tyxzt

kjikjia

)2()3(3 22 yxyzttxytzaaa zyx

2;;3 tywxzvtu

222

22

2

2)(0)(2)3(0

30)()(0)3(

33)(0)(0)3(0

yxyztytyxztytt

ww

z

wv

y

wu

x

wa

txyzttyxxztzt

vw

z

vv

y

vu

x

va

tyxztt

uw

z

uv

y

uu

x

ua

z

y

x

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Systems, Control Volume, and Control SurfaceSystems, Control Volume, and Control Surface

System (sys)A fluid system: contains the same fluid particles. Mass does not cross the system boundaries. Thus the mass of the system is constant.

Control Volume (C.V)

A control volume is a selected volumetric region in space. It’s shape and position may change with time.(Open System)

Control Surface (C.S)

The surface enclosing the control volume is called the control surface.

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Systems, Control Volume, and Control Surface (continued )Systems, Control Volume, and Control Surface (continued )

Consider the tank shown, assume:

• the control volume is defined by the tank walls and the top of the liquid.

• The control surface that encloses the control volume is designated by the dashed line.

• The liquid in the tank at time t is elected as the system and is indicated by the solid line.

At this instant in time, the system completely occupies the control volume and is contained by the control surface. Thus, at this time:

After a time some liquid has flowed out of the control volume to the right. The amount that flowed out is:

During the same period some liquid has entered the control volume from the left, the amount being:

)1.........()()( tMtM cvsys t

tmM outout . tmM inin

.

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Systems, Control Volume, and Control Surface (continued )Systems, Control Volume, and Control Surface (continued )Now the system has been deformed as shown in Fig. (b). Part of the system is the liquid that has flowed out across the control surface. The system remaining in the control volume has been deformed by the mass that has flowed in across the control surface. Also, the height of the control volume has changed to accommodate the net flow into the tank. The mass of the system at time t + Δt can be determined by taking the mass in the control volume, subtracting the mass that entered, and adding the mass that left.

)2......()()( inoutcvsys MMttMttM

Subtracting (1) from (2), we have:

inoutcvcvsyssys MMtMttMtMttM )()()()(

tmmtMttMtMttM inoutcvcvsyssys )()()()()(..

Dividing by Δt and taking the limit as Δt 0 yields

)3.......(..

inoutcvsys mm

dt

dM

dt

dM

The equation relates the rate of change of the mass of the system to the rate of change of mass in the control volume plus the net outflow (efflux) across the control surface

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Systems, Control Volume, and Control Surface (continued )Systems, Control Volume, and Control Surface (continued )

By definition, the mass of the system is constant so

Lagrangian statement

And Eq. (3) becomes

The corresponding Eulerian statement and can be written as:

0dt

dM sys

0..

inoutcv mm

dt

dM

outincv mm

dt

dM ..

This equation states that there is an increasing mass in the control volume if there is a net mass influx through the control surface and decreasing mass if there is a net mass efflux. This is identified as the continuity equation.

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Systems

• Laws of Mechanics– Written for systems

– System = arbitrary quantity of mass of fixed identity

– Fixed quantity of mass, m

0dt

dm

dt

md )( VF

dt

dW

dt

dQ

dt

dE

• Conservation of Mass– Mass is conserved and

does not change

• Momentum– If surroundings

exert force on system, mass will accelerate

• Energy– If heat is added to

system or work is done by system, energy will change

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Control Volumes

• Solid Mechanics– Follow the system, determine

what happens to it• Fluid Mechanics

– Consider the behavior in a specific region or Control Volume

• Convert System approach to CV approach– Look at specific regions, rather

than specific masses• Reynolds Transport Theorem

– Relates time derivative of system properties to rate of change of property in CV

)(extensiveenergymomentum,mass,

CVCV

dbbdmB

)(intensivemassunitperofamount Bdm

dBb

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CV Inflow & Outflow

CSCS

inoutinoutnet mbbmbmbBBB

AV

Bmb

inoutttCVttsys

inoutttCVttsys

BBBB

MMMM

,,

,,

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Continuity EquationContinuity Equation

• In the case of the continuity equation, the extensive property in the control volume equation is the mass of the system, Msys, and the corresponding intensive variable, b, is the mass per unit mass, or

• Substituting b equal to unity in the control volume equation yields the general form of the continuity equation.

• This is sometimes called the integral form of the continuity equation.

cscv

sys dddt

d

dt

dMAV

1sys

sys

M

Mb

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Continuity Equation• The term on the left is the rate of change of the mass of the system. However, by

definition, the mass of a system is constant. Therefore the left-hand side of the equation is zero, and the equation can be written as

• This is the general form of the continuity equation. It states that the net rate of the outflow of mass from the control volume is equal to the rate of decrease of mass within the control volume.

• The continuity equation involving flow streams having a uniform velocity across the flow section is given as

cvcs

ddt

dd AV

cv

cs

ddt

d AV

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Example: at a certain time rate of Example: at a certain time rate of rising is 0.1 cm/s:rising is 0.1 cm/s:• Continuity equation

smV

gVx

AVAVdt

dhA

AVAVhAdt

d

ddt

d

in

in

outoutinintank

outoutinintank

CSCV

/47.4

)0025.0(1*2)0025.0(101.0*1.0

)(

0

2

AV

0..

outincv mm

dt

dM

Fluid Mechanics I 29

Example:Example: Both pistons are moving to the left, but piston A has a speed twice as great as that of piston B. Then the water level in the tank is: a) rising, b) not moving up or down, c) falling.

• Select a CV that moves up and down with the water surface

• Continuity EquationContinuity Equation

CS

VA=2VBVB

risingissurface0

)(20

;6;3

20

0

21

41

412

42

4

BB

BBBB

BABA

BBABCV

CSCV

AVdt

dhA

AVAVdt

dhA

AAAA

AVAVddt

d

ddt

d

AV

h

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Euler EquationEuler Equation• Fluid element accelerating in l

direction & acted on by pressure and weight forces only (no friction)

• Newton’s 2nd Law

g

az

p

dl

d

gadl

dzg

dl

dp

lalppp

AlW

AalWAppAp

MaF

l

l

l

l

ll

)(

sin)(

sin)(

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Example 1:Example 1:

2.0

)5.03.0(

30sin)3.0(

)(

dl

dp

gg

dl

dza

gdl

dp

g

az

p

dl

d

o

l

l

Flow

30o

l

• Given: Steady flow. Liquid is decelerating at a rate of 0.3g.

• Find: Pressure gradient in flow direction in terms of specific weight.

32

Example 2:Example 2:• Given: = 10 kN/m3, pB-pA=12 kPa.

• Find: Direction of fluid acceleration.A

B

1 m

vertical

up)ision(accelerat0)12.1(

)1000,10

000,12(

)1(

)1

(

)(

ga

ga

ppga

dz

dz

dz

dpga

g

az

p

dz

d

z

z

BAz

z

z

33

Example 3:Example 3:• Given: Steady flow. Velocity varies

linearly with distance through the nozzle.

• Find: Pressure gradient ½-way through the nozzle

ftftlbf

ftsftsftftslugs

dxdV

Vadxdp

g

az

pdxd

x

x

//355,5

)//50(*)/55(*)/94.1(

)(

)(

2

3

V1/2=(80+30)/2 ft/s = 55 ft/s

dV/dx = (80-30) ft/s /1 ft = 50 ft/s/ft

34

Bernoulli EquationBernoulli Equation

• Consider steady flow along streamline

• s is along streamline, and t is tangent to streamline

Constant2

02

2

1

1)(

2

2

2

gV

zp

gV

zp

dsd

gV

dsd

dsdV

Vg

ag

zp

dsd

t

headdynamicVelocityg

V

headcPiezometrizp

)(2

2

gV

zp

gV

zp

22

22

22

21

11

35

The Bernoulli equation states that the sum of the kinetic, potential, and flow energies of a fluid particle is constant along a streamline during steady flow.

An alternative form of the Bernoulli equation is expressed in terms of heads as: The sum of the pressure, velocity, and elevation heads is constant along a streamline.

36

Example 4:Example 4:

• Given: Velocity in outlet pipe from reservoir is 6 m/s and h = 15 m.

• Find: Pressure at A.• Solution: Bernoulli equation

kPap

gV

hp

gVp

gh

gV

zp

gV

zp

A

AA

AA

AA

A

2.129

)81.9

1815(9810)

2(

20

200

22

2

2

221

11

Point 1

Point A

37

Example 5:Example 5:

• Given: D=30 in, d=1 in, h=4 ft• Find: VA

• Solution: Bernoulli equation

sft

ghV

gV

gh

gV

zp

gV

zp

A

A

AA

A

/16

2

20

0200

222

221

11

Point A

Point 1

38

Static, Stagnation, Dynamic, and Total Pressure: Bernoulli Equation

Static Pressure

Dynamic Pressure

Hydrostatic Pressure

Static Pressure: moves along the fluid “static” to the motion.

Hydrostatic Pressure: potential energy due to elevation changes.

Dynamic Pressure: due to the mean flow going to forced stagnation.

Follow a Streamline from point 1 to 2

hp 1

12

1122

22 2

1

2

1zVpzVp

Following a streamline:

0 0, no elevation 0, no elevation

2112 2

1Vpp Hp 2

H > hNote:

hHV 1In this way we obtain a measurement of the centerline flow with piezometer tube.

“Total Pressure = Dynamic Pressure + Static Pressure”

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• Bernoulli equation: The sum of flow, kinetic, and potential energies of a fluid particle along a streamline is constant.

• Each term in this equation has pressure units, and thus each term represents Energy per unit volune

• P is the static pressure.• ϱV2/2 is the dynamic pressure.• ϱ gz is the hydrostatic pressure• The sum of the static, dynamic, and hydrostatic pressures

is called the total pressure. Bernoulli equation states that the total pressure along a streamline is constant.

• The sum of the static and dynamic pressures is called the stagnation pressure, and it is expressed as

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• The static, dynamic, and stagnation pressures are shown.

• When static and stagnation pressures are measured at a specified location, the fluid velocity at that location can be calculated from

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Stagnation Tube

glV

ddl

ppV

p

g

Vp

g

Vz

p

g

Vz

p

2

))((2

)(2

2

22

1

122

1

22

11

22

22

21

11

42

Stagnation Tube in a Pipe

p

g

V

2

2

z

Flow

Pipe

0z

g

Vz

pH

2

2

1

2

43

Pitot Tube

)(2

)()[(2

22

22

21

11

11

2

222

211

22

22

21

11

hhgV

zp

zp

gV

gVp

gVp

gV

zp

gV

zp

Fluid Mechanics I 44

Example – Venturi Tube

• Given: Water 20oC, V1=2 m/s, p1=50 kPa, D=6 cm, d=3 cm

• Find: p2 and p3

• Solution: Continuity Eq.

• Bernoulli Eq.

2

12

112

2211

dD

VAA

VV

AVAV

D Dd

1

2

3

kPap

Pa

VdDp

VVpp

g

Vz

p

g

Vz

p

120

2]3/61[2

1000000,150

]/1[2

)(2

22

2

24

21

41

22

2112

22

22

21

11

Similarly for 2 3, or 1 3

Pressure drop is fully recovered, since we assumed no frictional losses

kPap 1503

Nozzle: velocity increases, pressure decreases

Diffuser: velocity decreases, pressure increases

]/1[

)(24

212

Dd

ppV

Knowing the pressure drop 1 2 and d/D, we can calculate the velocity and flow rate

45

Ex • Given: Velocity in circular duct = 30

m/s, air density = 1.2 kg/m3.• Find: Pressure change between circular

and square section.• Solution: Continuity equation

• Bernoulli equation

)(2

22

22

22

cssc

ss

scc

c

VVpp

gV

zp

gV

zp

smV

DVD

AVAV

s

s

sscc

/

)4

(30 22

2/ mNpp sc

Air conditioning (~ 60 oF)

46

Ex

• Given: = 1000 kg/m3 V1= 30 m/s, and A2/A1=0.5, m=20000 N/m3

• Find: h• Solution: Continuity equation

• Bernoulli equation

smA

AVV

AVAV

/2

112

2211

)(2

22

21

2221

22

22

21

11

VVpp

gV

zp

gV

zp

Heating (~ 170 oF)

2

21

/ mN

pp

• Manometer equation

mh

hpp airm

)(21

47

Pitot Tube Application

1

2

V

y

l

z1-z2

smgyV

yhh

yzz

pp

zzypp

pylylzzp

kHg

kHg

k

kHg

k

kkHg

kHgkk

/)1/(2

)1/(

)(

)()(

)()(

21

2121

2121

2211