Control Volume Part 1

7/29/2019 Control Volume Part 1

1/42

FINITE CONTROL VOLUME ANALYSIS

Fluid mechanics require analysis of the behaviour of the contents of a

finite region in space

anchoring force required to hold a jet engine in place during test

amount of time to allow for complete filling of a large storage tank

power required to move water from location to another at a higherelevation

Control volume relations are derived from equations representing

basic laws applied to a system REYNOLDS TRANSPORT THEOREM


2/42

CONSERVATION OF MASS OR CONTINUITY EQUATION

TIME RATE OF CHANGE OF SYSTEM MASS = ZERO

0tD

DMsys

B = mb; B = mass; b =1

Time rate of change

of the mass of the

coincident system

Time rate of change

of the mass of the

contents of the

coincident controlvolume

Net rate of flow of

mass through the

control surface

cscv

sysdAnbVVbd

ttD

DB

0dAnV

Vdt

cscv


3/42

Steady Flow 0Vdt

cv

cs

dAnV Product of the component of the velocity V,perpendicular to small portion of the control surface,

differential area Mass flow rate through dA

n

V Positive for the flow out of the control volume

nV Negative for the flow into the control volume

Considered positive when it is pointing out of the control volumen

incs

out mmdAnV

Integral is positive net flow is out of the control volume

Integral is negative net flow is into the control volume


4/42

A

dAnVm

VAm

Incompressible flow - is distributed uniformly over area A

V

Average value of the component of velocity normal to the section area

involved

A

dAnV

V A

If the velocity is uniformly distributed (one dimensional flow) over the

section A, then

VA

dAnV

V A


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FIXED AND NON-DEFORMING CONTROL VOLUME

Seawater flows steadily through a simple conical-shaped nozzle at the

end of a fire hose as illustrated in Fig. If the nozzle exit velocity must

be at least 20 m/s. determine the minimum pumping capacity required

in m3/s.


6/42

0dAnVVdt cscv

cv

Vdt

Zero flow is steady

0mmdAnV 1

cs

2

12 mm

Q1 = Q2 = V2A2 Density is constant

s/m0251.010404

20 323


7/42

Moist air (a mixture of dry air and water vapor) enters a dehumidifier at the

rate of 324 kg/hr. Liquid water drains out of the dehumidifier at a rate of 7.3

kg/hr Determine the mass flowrate of the dry air and the water vapor

leaving the dehumidifier. A simplified sketch of the process is provided in Fig.

hr/kg324m1

hr/kg3.7m3


8/42

0dAnVVdt cscv

cv

Vdt

Zero flow is steady

0mmmdAnV 321

cs

hr/kg7.3163.7324mmm 312

The answer is same regardless of which control volume is chosen. If

we select the control volume as before except that we include the

cooling coils to be within the control volume

5454312mmandmmmmm


9/42

Incompressible laminar water flow develops in a straight pipe having radius R as in

Fig. At section (1), the velocity profile is uniform; the velocity is equal to constant

value U and is parallel to the pipe axis everywhere. At section (2), the velocity

profile is axisymmetric and parabolic, with zero velocity at the pipe wall and a

maximum value of umax at the center line. How are U and umax related ? How are the

average velocity atsection (2) and maximum velocity related.

0dAnVVdt

cscv


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0

cs

dAnV

02211

cs

dAnVUA

R

cs

rdrudAnV0

22222

2

42

20

2

2

22242212 R

RRudrR

rrudAn

V max

R

max

cs

242

2

2

2

222

RuRudAnV maxmax

cs

02

0

2

2

112211

RuUAdAnVUA max

cs

0

2

2

22Ru

UR max

2

maxuU

2

2

2

22

RudAnV max

cs


11/42

2

2

2

22

RudAnV max

cs

A

dAnV

V A

2

22

2

max

max

A u

R

Ru

A

dAnV

V

2

maxuV


12/42

MOVING CONTROL VOLUMES

V1 velocity of the fluid which strikes the vane

Vo velocity of the moving vane

Determine the force that puts on the vane

steam, gas and hydraulic turbines

To find Reynolds Transport theorem for moving control volumes


13/42

Moving control volume and system

Shape, size and orientation of control volume do not change with time

Control volume translates with constant velocity Vcv

FIXED CONTROL VOLUMEAbsolute velocity V that carries

fluid across the fixed controlvolume

MOVING CONTROL VOLUMERelative velocity W that carries

fluid across the moving controlvolume


14/42

Relative velocity is the fluids velocity relative to the moving control volume the

velocity seen by an observer ridingalong on the control volume

Absolute velocity is the fluid velocity as seen by a stationary observer in a fixed

coordinatesystem

V = W + Vcv

In general, absolute velocity V and control volume velocity Vcv will

have different directions, hence we should use VECTOR ADDITION


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cscv

sysdAnbWVbd

ttDDB

REYNOLDS TRANSPORT THEOREM FOR MOVING CONTROL VOLUMES

cscv

sys dAnbVVbdttD

DB

FOR FIXED CONTROL VOLUME

FOR MOVING CONTROL VOLUME

Replace V by W because Relative velocity W that carries fluid across

the moving control volume


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Moving and non-deforming control volume

When a moving control volume is used, fluid velocity relative to the

moving control volume (relative velocity) is important

V = W + Vcv

W relative velocity is the fluid velocity seen by the observer moving

with the control volumeVcv velocity of the control volume as seen from a fixed coordinate

system

V Absolute velocity i.e, the fluid velocity seen by a stationary

observer in a fixed co-ordinate system

0dAnWVdt cscv

i l f d d f 9 k h h i i h f l i k


17/42

An airplane moves forward at a speed of 971 kmph as shown in Fig. The frontal intake

area of the jet engine is 0.8m2 and the entering air density is 0.736 kg/m3. A stationary

observer determines that relative to the earth, the jet engine exhaust gases move away

from the engine with a speed of 1050 kmph. The engine exhaust area is 0.558 m2 and

the exhaust gas density is 0.515kg/m3. Estimate the mass flowrate of fuel into the

engine.


18/42

0dAnWVdt cscv

cv

Vdt

Zero flow relative to the moving control volume is

considered steady

0WAWAm 222111infuel

111222infuel WAWAm

V2 = W2 + Vplane; W2 = V2 Vplane = 1050 + 971 = 2021 kmph

)s/m3600/10002021)(m558.0)(m/kg515.0(m 33infuel

)s/m3600/1000971)(m8.0)(m/kg736.0( 33

s/kg5278.2minfuel

N S d L Th Li M E i


19/42

Newtons Second Law The Linear Momentum Equation

Time rate of change of the

linear momentum of the system

Newtons second law of motion for a system is

Sum of external forces

acting on the system

=

sys

sys

FVdVtD

D

Any reference or coordinate system for which this statement is true is called

INERTIAL

Fixed co-ordinate system

Coordinate system which moves in a straight line with constantvelocity

cvsys FF

D


20/42

cscvsys

dAnVVVVdt

VVdtD

D

Time rate of change oflinear momentum of the

system

Time rate of change of

linear momentum of the

contents of the control

volume

= +

Net rate of flow of

linear momentum

through the control

surface

volumecontrol

ofcontents

cscv

FdAnVVVVdt

For a control volume that is fixed (and thus inertial) and non-deforming


21/42

Surface forces

Body forces - weight

Net Pressure Force on a Closed Control Surface dAF


22/42

Net Pressure Force on a Closed Control Surface

UNIFORM PRESSURE

0

cs

aUP dAnpF

cs

press dAnpF

cs

a

cs

aUPSSUREUNIFORMPRE dAnpdAnpFF

cs

aUP dAnpF

This result is independent of the shape of the surface as long as the surface is closed

and all our control volumesare closed.

Thus a seemingly complicated pressure-force problem can be simplified by

subtracting any convenient uniform pressure Patm and working only with the piecesof gage pressure which remain

NON UNIFORMPRESSURE


23/42

NON-UNIFORM PRESSURE

cs

a dAnppF

csgage

dAnpF

This trick can mean quite a saving in computation

A control volume of a nozzle section has surface pressures of 2 8 atm (absolute) at


24/42

A control volume of a nozzle section has surface pressures of 2.8 atm (absolute) at

section 1 and atmospheric pressure of 1 atm (absolute) at section 2 and on the

external rounded part of the nozzle as in Fig. Compute the net pressure force if D1 =

75 mm and D2 = 25 mm.

2.8 atm abs 1.0 atm abs

1.0 atm abs

1.0 bar

abs

1.8 atm gauge

0.0 atm gauge

0.0 atm gauge

0 atm gauge

2211 ApApF

2

23

010754

10132581 A.F

N.F 8805


25/42

Determine the anchoring force required to

hold in place a conical nozzle attached to

the end of a laboratory sink faucet shown

in Fig. when the water flowrate is 0.6

liters. The nozzle

mass is 0.1 kg. The nozzle inlet and exit

diameters are 16 mm and 5 mm,

respectively.

The nozzle axis is vertical and the axial

distance between sections (1) and (2) is 30

mm. The pressure at section (1) is 464 kPa.

The anchoring force sought is the reactionforce between the faucet and nozzle

threads. To evaluate this force, control

volume selected includes the nozzle and

the water contained in the nozzle


26/42

Wn

FA

p1A1

w1

Ww

p2A2

w2

z

FA anchoring force that holds the nozzle in

place

Wn weight of the nozzle

Ww weight of the water in the nozzle

P1 gage pressure at section (1)

A1 cross section area at section (1)

P2 gage pressure at section (2)A2 cross section area at section (2)

w1 z direction velocity at the control volume

entrance

w2 z direction velocity at the control volumeexit

The action of atmospheric pressure cancels out in

every direction and is not shown


27/42

volumecontrolofcontents

cscv

FdAnVVVVdtFlow is

steady

22w11nA

cs

ApWApWFdAnVw

n

V Positive for the flow out of the control volume

nV Negative for the flow into the control volume

2211wmwmdAnVw

cs

22112211ApWApWFwmwm wnA

22112211ApWApWwmwmF wnA

Conservationof mass;


28/42

221121ApWApWwwmF wnA

Conservation of mass;mmm 21

s/kg..QwAm 6010601000 311

2423211 m10011.21016

4D

4A

2523222 m10964.1105

4D

4A

s/m..

.

A

Qw 982

100112

1060

4

3

1

1

s/m..

.

A

Qw 630

109641

1060

5

3

2

2


29/42

N981.081.91.0gmW nn

gDDDD12

hgVW 21

22

21ww

N0278.081.9004.0016.0004.0016.012

03.01000W

22w

0p2 Atmospheric pressure

221121ApWApWwwmF wnA

00278.010011.2464000981.06.3098.26.0F 4A

00278.03104.93981.0572.16FA

N75.77FA

A horizontal jet of water exits a nozzle with a uniform speed of 3 m/s strikes a vane and is


30/42

A horizontal jet of water exits a nozzle with a uniform speed of 3 m/s strikes a vane and is

turned through a vane, and is turned through an . Determine the anchoring force needed to

hold the vane stationary. Neglect gravity and viscous effects. A1 = 0.06 m2

2211VAVA

21AA

21VV


31/42

volumecontrolofcontents

cscv

FdAnVVVVdt

Flow is steady

x

cs

FdAnVu zcs

FdAnVw

Axx

cs

FAVcosVAVVFdAnVu211111

Azz

cs

FAVsinVAVFdAnVw21111

0

cosVAFAVcosVAVV Ax 12

11211111

sinAVFAVsinVAV Az 12

1111110

cosVAFAx

12

11sinAVFAz 1

2

1

Water flows through a horizontal, pipe bend as illustrated in Fig. The flow cross-


32/42

Water flows through a horizontal, pipe bend as illustrated in Fig. The flow cross

sectional area is constant at a value of 0.09 m2 through the bend. The flow velocity

everywhere in the bend is axial and 15.24 m/s. The absolute pressures at the

entrance and exit of the bend are 2.07 bar and 1.66 bar, respectively. Calculate the

horizontal (x) and (y) components of the anchoring force required to hold the bend

in place

V =15.24 m/s

V

0.09m2

FA = 0 No net inflow or


33/42

FAx 0 No net inflow or

outflow in the x-direction

22112211ApApFvmvm Ay

x

cs

FdAnVu

y

cs

FdAnVv

z

cs

FdAnVw

mmm 21

221121ApApFvvm Ay

212211vvmApApFAy

2415241524150901000

0901001325110661090100132511025555

....

.....FAy

N.FAy 86827104 WFAz

Air flows steadily between two cross sections in a long, straight portion of 100 mm


34/42

Air flows steadily between two cross sections in a long, straight portion of 100 mm

inside diameter pipe as indicated in Fig., where the uniformly distributed

temperature and pressure at each cross section are given. If the average velocity at

section (2) is 305 m/s. Assuming the uniform velocity distributions at sections (1) and

(2), determine the frictional force exerted by the pipe wall on the air flow between

sections (1) and (2).

P1 = 6.9 bar (abs)

T1 = 27 deg C

P2 = 1.27 bar (abs)

T2 = -21 deg C

x

cs

FdAnVu

22112211ApApRumum x

mmm 21

221121 ApApRuum x


35/42

P1 = 6.9 bar (abs)T1 = 27 deg C

P2 = 1.27 bar (abs)T2 = -21 deg C

3

5

1

1

10148

27327287

1096m/kg.

.

RT

P 35

2

2

27561

27321287

10271m/kg.

.

RT

P

22211121uAuAmmm

212211AAuu

s/m..

.uu 83660148

3057561

1

221 s/m.u 83661

s/m....uAm 206448366104

01482

111 s/m.m 20644

AAR


36/42

221121ApApRuum x

2510

41001325127101325196305836620644 .....R.. x

794421841001 .R. x

NRx 3420

A static thrust stand as sketched in Fig. is to be designed for testing a jet engine. The


37/42

g g g j g

following conditions are known for a typical test:

Intake air velocity = 200 m/s,

Exhaust gas velocity = 500 m/s,

Intake cross-sectional area= 1 m2;

Intake static pressure = -22.5 kPa (gauge) = 78.5 kPa (abs);Intake static temperature = 268 K;

Exhauststatic pressure= 101 kPa (abs).

Estimate the nominal thrust for which to design.

u1 = 200 m/s,


38/42

x

cs

FdAnVu

22112211ApApFumum th

mmm 21

1 / ,

u2 = 500 m/s,

A1 = 1 m2;

P1 = -22.5 kPa (gauge)= = 78.5 kPa

(abs);

T1 = 268 K;P2 = 101 kPa (abs).

3

1

1

10211

268287

1000578 m/kg..RTP

s/kg..uAm 220420010211111

201100052250022042002204 A.F.. th

NFth83760

LINEAR MOMENTUM EQUATION FOR MOVING NON-DEFORMING


39/42

CONTROL VOLUMES (MOVING AT A CONSTANT VELOCITY)

cscv

sysdAnbWVbd

ttD

DB

b = V

cscv

dAnWVVVdt volumecontrol

ofcontentsF

V = W + Vcv

cs Volumecontrolofcontentscv

cvcv FdAnWVWVdVW

t

Zero for steady flow

cs cs

cv

cs

cv dAnWVdAnWWdAnWVW

cs cscv

cscv dAnWVdAnWWdAnWVW

Zero for steady flow

cs VolumecontrolofcontentsFdAnWW

A vane on wheels moves with constant velocity Vo when a stream of water having a


40/42

y o g

nozzle exit velocity of V1 is turned 45o by the vane as indicated in Fig. Determine the

magnitude and direction of the force, F, exerted by the stream of water on the

vane surface. The speed of the water jet leaving the nozzle is 33 m/s, and the vane is

moving to the right with a constantspeed of 6 m/s.

A1 = 5.6 10-4 m2

0.3 m

W

d


41/42

cs

xRdAnVu

cs

zz WRdAnWW

x2211 R45cosWmWm

Conservation of mass; mmm 21

;WAm;WAm 22221111

x-direction

z-direction

WR45sinWm z22

Water flow is frictionless and that the change in water elevation across the vane is

negligible. Therefore, W is constant

The relative veloicty of the stream of water entering the control volume W1 = V1

Vo = 33- 6 = 27 m/s = W2

W1 = W2;

s/kg12.1527106.51000mm 421


42/42

s/kg12.1527106.51000mm 21

x2211 R45cosWmWm

xR45cos2712.152712.15N6.119Rx

WR45sinWm z22

s/kg65.13.0106.581.91000lAgW 41

65.1R45sin2712.15 z

N3.290Rz

N3143.2906.119RRR 222

z2x

43.23.290

TanR

Tan1z1 o6.67

Control Volume Part 1

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Transcript of Control Volume Part 1