1

Chapter 5

Flow Analysis Using Control Volume

(Finite Control Volume Analysis )

2

• Many practical problems in fluid mechanics require analysis of the behavior of the contents of a finite region in space (a control volume).

for example, to determine the amount of time to allow for complete

filling of a large storage tank. to estimate of how much power it would take to move

water from one location to another at a higher elevation and several miles away may be sought.

3

• The bases of this analysis method are some fundamental

principle of physics , namely ,

Conservation of mass

Newton’s second law of motion , and

the first and second laws of thermodynamics.

• The finite control volume formulas are easy to interpret physically and are not difficult to use.

4

§ 5.1 Conservation of Mass-The continuity Equation

• §5.1.1 Derivation of the continuity Equation

syssys

syssys

dV

dt

dmor

dt

dM

m where

00

is system afor principle mass ofon Conservati

5

-(5.3)---)(or

)(

1b m B

)(or

)()(

t theorem transporReynolds theFrom

. .

. .

..

vc scsys

vc sc

sys

scvc

sys

cv outcs incs

sys

dAnVdVt

dVdt

d

dAnVdtdt

dmm

mm

B

dAnVbdAbtdt

dB

dAnVbdAnVbdVbtdt

dB

time rate of change of the mass of the coincident system

time rate of change of the mass of the contents of the coincident c.v.

net rate of flow of mass through the control surface

= +

6

t = t-δt t = t t = t+δt

-(5.3)---)(or

)(

. .

. .

vc scsys

vc sc

sys

dAnVdt

ddt

d

dAnVdtdt

dm

time rate of change of the mass of the coincident system

time rate of change of the mass of the contents of the coincident c.v.

net rate of flow of mass through the control surface

= +

7

(5.5)-----------------0)(

mass) ofion (conservat 0since

cscv

sys

dAnVdVt

dt

dm

velocityaverageVA

dAnVV

dAnVAVm

VAVAdVt

mass

mmdVt

dAnVdAnVdVt

A

A

i iiniouticv

cvinout

cv incsoutcs

)(

)( since

0)()(

VAQ

c.s. ofsection a through rate flow m where

(5.4)-----------------0

0)()(

continuity ofEquation

.

..

8

§ 5.1.2 Fixed, non-deforming Control Volume

?)V(or

?Q ：Determine

40D

20V

flowsteady figure, :Given

5.1 Example

11

3

1

2

2

As

m

mms

m

9

smm

smAVQ

QQAVA

AVAV

flowsteadydVbt

AVAVdVbtdt

dm

bmB

AVAVdVbtdt

dB

cv

iiniiicv

ioutiii

sys

iiniiicv

ioutiii

sys

323

221

121122

21

111222

0251.02

)(104020

V

0

0

0)()(

1,

)()(

t theorem transporReynolds

:Solution

10

?:min

)(5.0

)por water va&air dry ,airmoist (

22

figure :Given

5.3 Example

2

3

1

meDeter

waterhrslugsm

hrslugsm

hrslug

hrslug

hrslugmmm

mmm

mmAVAV

flowsteadydVbt

AVAVdVtdt

dm

bmB

AVAVdVbtdt

dB

i

iin

i

iout

iiniii

ioutiii

cv

iiniiicv

ioutiii

sys

iiniiicv

ioutiii

sys

5.21)(5.0)(22

0

00)()(

0

0)()(

1

)()(

t theorem transporReynolds The :Solution

312

132

11

min/?:Determine

min9

figureRight :Given

5.5 Example

incht

h

GalQin

iiniiicv

ioutiii

sys

iiniiicv

ioutiii

sys

AVAVdVtdt

dm

bmB

AVbAVbdVbtdt

dB

)()(

1

)()(

t theorem transporReynolds The

:Solution

cv outairairair

sysair

cv inwateroutairwaterwatercv airair

syswatersysair

inwateroutaircv waterair

syswatersysair

VAdVtdt

dm

VAVAdVt

dVtdt

dm

dt

dm

VAVAdVdVtdt

dm

dt

dm

waterair

0)(

0)()(

)()(])()[(

12

min44.1min1203.010

1614.5

421

min9

10sin1010

010

0])5.1()52([

0)(

0)(

2

3

2

inftft

bblft

galbblGal

ftAceQ

A

Q

t

h

Qt

hA

t

h

QAhht

VAdVtdt

dm

VAdVtdt

dm

jinwater

j

inwater

inwaterwaterjwaterwater

inwaterwaterjwaterwater

cv inwaterwaterwaterwater

cv outairairair

sysair

13

§ 5.1.3 Moving, Non-deforming Control Volume • Example of moving, non-deforming control volume ---Control volumes containing a gas turbine engine on an

aircraft in flight, and gasoline tank of an automobile passing.

0)(

)(

C.V. with movingobserver an by seen velocityfluid the

system coordinate fixed a fromseen as C.V. theof velocitythe

system coordinate fixed an observatio stationary aby seen velocityfluid

. .

..

vc sc

sys

scvc

sys

cv

cv

cv

dAnWdtdt

dmor

dAnVdVtdt

dm

W

V

absoluteV

VVW

VWV

14

?:

1050

515.0;736.0

558.0;8.0

;971

figureRight :Given

5.6 Example

2

3231

22

21

infuel

airair

cv

mFind

hrkmV

mkg

mkg

mAmA

hrkmV

hrkg

AWAWm

mmor

mAWAW

dVt

stateSteady

AWAWdtdt

dm

infuel

i jinjouti

infuel

CV

i iiniiioutiiicv

sys

/9100

8.01000971736.0558.010002021515.0

0

0

0

0)()(

:Solution

111222

111222

15

§ 5.1.4. Deforming Control Volume

• A deforming Control volume

~Changing volume size & control surface movement.

cv cs

sys

cscs

cscv

sys

dAnWdVtdt

dm

bmBFor

VWVVVW

twhere

dAnWbdVbtdt

dB

0)(

1

0

)(

16

? :Determine

figureRight :Given

5.8 Example

pV

0)()(

/1051.0

/105)10(

)1(

60

min1

min300

?

105)1(

)10(500

:Solution

372111

3632

33

222

1

242

232

1

jinjjj

ioutiiicv

sys

leak

p

p

AVAVdVtdt

dm

smQAVQQ

smcm

m

s

cmAVQ

VV

mmm

mmmAA

17

min/66.0min1

60

1

10101.1

/101.1/105105105

11

1

0)(

0.

0][

32

23672421

211

21122111

22111

mms

m

mm

s

m

smsmm

QQAt

lV

QQAt

l

QQt

lAAVAVlA

t

needleandconstSince

AVAVlAt

p

needle

0)()(

j

injjji

outiiicv

sys AVAVdVtdt

dm

18

§ 5.2 Newton’s Second Law

--The linear momentum and moment-of-momentum equations.

• Newton’s second law of motion for a system is

vm

BbvmB

dVVdt

d

dt

vmdF

dt

vmdFor

dt

vdmForamF

sys

syssys

;

)19.5()(

=>Any reference or coordinate system for which this statement is true is called inertial. A fixed coordinate system is inertial.

A coordinate system that moves in a straight line with constant velocity and is thus without acceleration is also inertial.

19

• When a control volume is coincident with a system at an instant of time, the forces acting on the system and the forces acting on the contents of the coincident control volume are instantaneously identical, that is,

)20.5(.... umeControlVolcoincidenttheofcontentssys FF

20

• Furthermore, for a system and the contents of a coincident control volume that is fixed and non-deforming, the Reynolds transport theorem allows us to conclude that

surface control the through momentumlinear of flow of ratenet )(

volumecontrol

theof contents theof momentumlinear theof change of rate time

system theof momentumlinear of change of rate time

)21.5()()(

cs

cv

sys

cscvsys

sys

dAnvv

dVvt

dVvdt

d

dAnvvdVvt

dVvdt

d

dt

mvd

21

• For a control volume that is fixed (inertial) and non-deforming, Eq.(5.19),(5.20), and (5.21) suggest that an appropriate mathematical statement of Newton’s Second law of motion is

• Eq(5.22) is the linear momentum equation for a fixed, non-deforming control volume.

Body force -- gravity only

Surface force -- exerted on the contents of the control

volume by material just outside the control volume

in contact with material just inside the control volume

)22.5()(....

cscvVolumeControltheofcontents dAnvvdvt

F

VolumeControltheofcontentsF ....

22

faucet.sink

laboratory of end the toattached nozzle conical a

placein hold torequired )(F force Anchoring

:Determine

464;30

5;16

)(1.0

/6.0

figureRight :Given

5.10 Example

A

1

21

kpapmmh

mmDmmD

nozzleofmasskgw

sliterQ

n

0:

)()()(

,

)()(

t theorem transporReynold The

:Solution

cv

cv

j iiinjjj

ioutiicv

jjjjjj

iiiiiicv

sys

dVvt

statesteadyNote

FAnvvAnvvdVvtdt

vdm

vbvmB

AnvbAnvbdVbtdt

dBout

23

mmm

ratemassmmNote

APAPWWFmwmwor

rateFlowQQNote

APAPWWFQwQwor

APAPWWFAwwAww

kwv

WW

WW

AFNote

APAPWWF

AnvvAnvv

dVvt

statesteadyNote

FAnvvAnvvdVvtdt

vdm

wnA

wnA

wnA

w

n

A

wnA

cv

cv

j iiinjjj

ioutiicv

21

21

22111122

21

2211111222

221111112222

2211

1111122222

mass ofon Conservati

: ,:

)()(

: ,:

)()(

))1()(())((

waterofeight

nozzle ofeight

force nchoring:

)()(

0:

)()()(

24

)(8.770)10

1(

4

)16(

1

1000464

0278.098.0)6.3098.2(6.0

0278.081.9})10

1)](5)(16(

)5()16[(10

130

12{/1000

)}(12

{

98.0)/8.9)(1.0(

/6.30

/98.2]

101

16[4

101

6.0

4

/6.010

1)/6.()/1000()(

0;0;:)(

2

3

2

22

3

22

3

3

21

2

2

2

1

2

2

2

2

3

3

3

2

11

1

3

33

2211

12221121

wpwardNmm

mmm

kpa

pakpa

NNsm

sm

skgF

Ns

mmm

mmmmm

mmmmmm

mmmmkg

gDDDDh

gW

NewtonsmkggmW

smA

Qw

sm

mmm

mm

literm

sliter

D

Q

A

Qw

skgliter

mslitersomkgAworAwQm

WWwwnoteAPAPWWwwmF

A

ww

nn

wnwnA

25

Several important notes:

(1) 1-D flow problem when the flow is uniformly distributed over a section of the C.S.(2) Linear momentum is directional –three orthogonal coordinate directions.(3) Flux term is linear momentum─

(4) for Steady flow (In the textbook ,it is aussmed: Steady flow for the momentam problem)

(5) If control surface direction of flow⊥ Surface force exerted at these locations by fluid outsid

e the C.V. on fluid inside will be due to pressure.

dAnVV )(

signshavenVV )(&

0CV

dVVt

26

(6) Uniform pressure on control volume

(7) Positive external force if the force is in the assigned positive coordinate direction. Negative otherwise.

(8) Only external forces acting on the contents of the control volume are considered in the linear momentum equation.(Eq.5.22)

CS CS gageaCS

CS a

dsnpdsnppdsnp

dsnp

)())(()(

0)(

27

placein bend thehold torequired force

anchoring theof components y)(x, Horizontal

:Calculate

psia 24P psia 30P

ft2 0.1A ft/s 50 V

bend pipe180

flowfor water figureright As :Give

5.11 Example

21

0

)()(

,

)()(

t theorem transporReynold The

:Solution

11112222

dVvt

statesteady

FFFAVVAVVdVvt

VAvVAvdvt

Fdt

vdm

VbVmB

VABVAbdbtdt

dB

surfacebody

i jjiniout

CVi j

jrimiatsys

28

lb

ft

in

sftftslugsft

APPAPPmVF

APPAPPFmV

vvv

APPAPPFAvVAvV

atmatmay

atmatmay

atmatmay

1324

1

144]1.0)7.1424(

1.0)7.1430[()1.0/50/94.1(/502

)()(2

)()(2

on conservati mass From

)()()(

direction y In

2

2

3

2211

2211

21

221111112222

29

(2) & (1) sectionsbetween flowair on the

wallpipe by the exerted force frictional The

:Determine

flowair Steady 4.

ondistributi temp. Uniform3.

(2)&(1)section at

on distributiVelocity Uniform2.

I.D. in. 4=D figureRight 1.

:Given

5.12 Example

0

)()()(

assuch ,t theorem transporReynold theFrom

:Solution

CV

inj

jjjouti

iiii

CV

sys

dVvt

statesteady

FAjvVAvVdVvtdt

Vmd

30

fff

22

2

2f

2

2

2f

222

2

2

222

2222111

222111

221111112222

lb 793 lb 232 - lb 1025

)2

1

12

4()

144in

in

lb 18.2-

144in

in

lb(100

)sslugs(0.297)s

ft(921-sslugs0.297)s

ft(1000

/297.0)2

1

12

4()/1000(

)(4531716

1

1444.18

mass ofon conservati From

)()(

x

x

o

f

x

R

ftftR

sslugsftsftR

Rslug

lbftft

in

in

lb

AVRT

PmAVAV

AvAvm

APAPRAvVAvV

directionxIn

31

? :Determine

]1[2 , uniform is (3)

flowater laminar w ibleIncompress (2)

figureRight (1)

:Given

5.13 Example

12

2

2

121

ppR

rWWW

0

)()()(

assuch ,t theorem transporReynold theFrom

:Solution

CV

inj

jjjouti

iiii

CV

sys

dVvt

statesteady

FAjvVAvVdVvtdt

Vmd

3211

2121

221

2212211

221122

1

221

221

01

322

1

0

1

22

21

2

22

2

0 2

221

0

22

221

22

2

122222

2211111222

3

1

)(3

)(4

34

34]

34

)2

(8

2

21)1(8

2sin)2()1(4

]}1[2{

direction -zIn

2 2

2

A

R

A

Wwpp

RWRWR

wApAp

RWApApRwR

w

Rw

xRw

dxxR

w

dxr

Rdrdr

R

rdx

R

rxdr

R

rrw

rdrdAcerdrR

rw

dAR

rwdAwdAww

RwAPAPAwwdAww

z

z

z

R

R

cs A

cs z

at

at

33

open? isit or when closed is gate the

en larger wh placein gate tohold

torequired force anchoring theIs

:Question

figuresRight

:Given

5.16 Example

bgHbHH

gApR

RApF

dAnvvdVvt

flowno

FdAnvvdVvtdt

vmd

CGx

xCG

cscv

cscv

sys

2

2

1)(

2

00

0)(&0

)()(

close is gate e when th(a)

:Solution

34gateopenxgateclosex

fx

fx

xf

xf

cs xfCG

cv

cscv

RR

Ans

hbuFbghbgHR

Hbuuce

HbuhbuFbghbgHR

RFbghbgHHbuhbu

RFbghbgHAuuAuu

RFAPAPdAnvv

dVvt

statesteady

FdAnvvdVvtdt

vmd

..

22

22

211

21

22

22

2221

22

22111222

221

||

.2

1

2

1

00sin

2

1

2

12

1

2

12

1

2

1

)(

0

)()(

t theorem transporReynold The

open is gate e when th(b)

35

• For a system and

an inertial , moving , non-deforming control volume that are both coincident at an instant of time , the Reynolds transport theorem leads to

This is the linear momentum equation for an inertial, moving, non-deforming control volume that involves steady flow.

cs VCofcontent

VCofcontentcvcs

cs VCofcontentcv

cv

cs VCofcontentscvcv cv

cv

VCofcontentscscv

icscvsys

sys

FdAnww

dAnw

FdAnwVdAnww

FdAnwvw

V

FdAnwvwdVvwt

vvwwhere

FdAnwvdVvt

FdAnwvdVvt

dVvdt

d

dt

vmd

....

....

....

....

....

)(

mass ofon conservati 0)(

)()(

)()(

flowsteady and , velocityC.V.constant afor

)25.5()()()(

....

)24.5()(

)23.5()()(

36

. surface

vaneon the water of stream by the exerted

, F force, theof direction and magnitude The

:Determine

&...45 (3)

006.0

/20/100 (2)

figureRight (1)

:Given

5.17 Example

21

21

01

VVbetweean

ftA

sftVsftV

iinout

cv

cscs

cv

cv cs isys

FAnwwAnww

vvwnote

dAnwwdAnwv

statesteadydVvt

ce

FdAnwvdVvtdt

vmd

])([])([

:

)()(

0sin

)()(

t theorem transporReynolds The

:Solution

37

cos

)( and )( where

cos

direction In

])([])([

1122

011022

11112222

x

x

iinout

Rmwmw

vvwvvw

RAwwAww

x

FAnwwAnww

f

x

lbs

slug

s

ft

mwmwmwR

s

ftwwAA

s

slug

fts

ft

ft

slug

AwAwAvvmm

8.219312.0)45cos1(80

)cos1(cos

80 , , Also

9312.0

006.0)20100(94.1

)(

mass ofon Conservati

12211

212121

23

111222101121

38

3.6880.21

05.53tantan

35.5705.538.21

05.53

374.0/9312.045sin/80

374.0

1006.02.3294.1

sin

0sin

)()(

)(

direction-zIn

11

2222

2231

22

22

x

z

zx

f

z

z

z

zinzoutz

iics

R

R

lbfRRR

lb

sslugsftR

lbf

fts

ft

ft

sluglgAW

where

WmwR

RWmw

Fmwmw

FdAnww

39

§ 5.3 First Law of Thermodynamics-The Energy equation

§5.3.1 Derivation of the Energy Equation• The first law of thermodynamics for a system is, in word

system theinto

transfer by work addition

energy of rate net time

system theinto

sfer heat tranby addition

energy of rate net time

system theof

energy stored totalthe

of increase of rate time

system theinto going if0

0

senergy/mas -potential- gz

senergy/mas -kinetic-2

senergy/mas -internal-

system in the particleeach

for massunit per energy stored totalthee

)55.5()(

)()(

2

netin

netin

sysinnet

innetsys

sysoutinsysoutinsys

W

Q

v

u

where

WQdedt

dor

WWQQdedt

d

----(5.56)

40

• Eq.(5.55) is valid for inertial and non-inertial reference system

For the control volume that is coincident with the system at an instant of time

)55.5()(

)()(

sysinnet

innetsys

sysoutinsysoutinsys

WQdedt

dor

WWQQdedt

d

)57.5()()( volumecontralcoincitent

innet

innetsys

innet

innet WQWQ

41

surface control he through t volumecontrol

ofout energy stored totalof flow of ratenet the)(

volumecontrol theof contents

theofenergy stored totalof increase of rate timethe

system stored total theof increase of rate timethe

)58.5()(

EB e,b with ,t theorem transporReynolds theFrom

cs

cv

sys

sys cscv

sys

dAnve

dVet

dVedt

d

dAnvedVet

dVedt

d

dt

dE

42

positive considered is volumecontrol theintosfer Heat tran --

0

0 process Adiabatic--

possible are convectionor / and ,conduction radiation, Thus,

.difference ure temperata of because

gsurroundin & contents C.V. hebetween t exchanged isenergy --

rate,sfer heat tran The(1)

：Note

)]()[(

)59.5()()(

(5.58)& (5.57) (5.55), Eq From

outinnetin

cvoutinoutin

cvnetinnetincv cs

QQQ

Q

Q

WWQQ

WQdAnvedet

43

velocityangularTorque

W

W

shaft

shaftshaft

:;:

--

propellers and fans, , turbinesinclude devicesRotary shaft.

moving aby surface control theacross red transferis work --

kShaft wor (a)

worksof Types--

gsurroundin by the

C.V. theof contents on the done isrk when wopositive isIt --

power called also , rate,ansfer work trThe(2)

44

volume.control theleaves and enters fluid eonly wher0

0 because pipe theof surface inside wettedon the0:

)62.5(

)(

-pstress) normal (

distance aover acts

stress normal fluid with associated force an nsfer whe Work tra(b)

stressnormal

stressnormal

cscsstressnormal

stressnormal

W

nvWnote

dAnvPdAnvW

AnVpVAnpVAn

forcestressnormalFwhereVFWstressnormal

stressnormal

45equation.energy theis This

)64.5()2

(

][

)59.5(

(5.59) Eq. From

0

c.s. thecrosses fluid where(2)

0

pipe theof surface inside wettedon the everywhere 0 (1)：Note

materialshaft in the stress ialby tangent d transfereis shaft work Rotating

forces stress al tangentiof because c.s.at occur can nsfer Work tra(c)

....

2

tantan

tan

tantan

vcnetinshaft

vincnetcscv

netinshaft

cvinnetcscv

cvinnet

cvinnetcscv

stressgential

stressgential

stressgential

stressgential

stressgential

WQdAnvgzvp

udet

dAnvpWQdAnvedet

WQdAnvedet

WvF

W

v

vFW

46

§ 5.3.2 Application of the Energy Equation

0)(0)2

(

0)(0)2

(

(2)

) Cyclical (mean in thesteady is flow themean when in the 2.

steady is flow1.when

0(1):

)64.5()2

(

equationenergy The

2

2

....

2

nvifdAnvgzvp

u

nvifdAnvgzvp

u

dVet

note

WQdAnvgzvp

udVet

cs

cs

cv

vcnetinshaft

vincnetcscv

47)(][

NEm/sor lbf/s-]ft[

)69.5()](2

ˆˆ[

ˆˆ since

flow.mean -the-in-steadyfor equation energy ldimensiona-one theis This

)67.5()](2

)()(ˆˆ[

)66.5()22

ˆ()22

ˆ()2

(

C.V. theleaving and entering stream oneonly is thereIf

)65.5()22

ˆ()22

ˆ()2

(

involved area sectional-cross flow

over the eddistrilnit uniformly be toassumed are gz and,2

,,ˆ If

22

22

222

222

2

kg

Joules

kg

mNor

shug

lbfft

m

Wor

m

Q

WQzzgvv

hhm

enthalpyp

uh

WQzzgvvpp

uum

mgzvp

umgzvp

udAnvgzvp

u

mgzvp

umgzvp

udAnvgzvp

u

vpu

shaftnetinnetin

outnetshaft

innetinout

inoutinout

outnetshaft

innetinout

inoutinoutinout

ininoutoutcs

inflow

outflow

cs

48

pump by the requined (hp)power the:Determine

process Adiabatic

] f(T) waterofenergy internal [

/3000ˆˆ

0=z-z ; psi(g) 60=p ; psi(g) 18=p

inch 1=D ;inch 3.5=D ;gal/min 300=Q

flowwater

figure,Right ：Given

5.19 Example

12

1221

21

shuglbfftun

sslugsGal

ft

Min

Gal

ft

slugQm

vvppuumW

flowadiabaticQchangeelevationnoz

WQzzgvvpp

uum

WQdAnvgzvp

udVetdt

dE

inoutinoutinout

netinshaft

innetout

outnetshaft

innetinout

inoutinoutinout

netinshaft

innetcscv

sys

/30.160

min1

48.7

130094.1

]2

)()(ˆˆ[

00

)](2

)()(ˆˆ[

)2

(

3

3

22

22

2

49

hp 32.2

lb/s)]-(ft 500lbf/s[1hp/-17721.6ft

7514.5)1336.1-4453.61.30(3000

]2

)10()123(

94.1

144*18

/94.1

1

14460

/3000[30.1

123

/0.10)

2

1

12

5.3(

60

min1

48.2

1min/300

sin

22

3

2

2

2

22

22

3

11

ftslug

ft

in

in

lb

sluglbfts

slugW

sft

AQV

sftft

sGal

ftGal

AQV

AQVAVQce

f

fnetinshaft

50

? W:Determine

/2250ˆ;/60

/3348ˆ ; 30m/s=

figureight R：Given

5.20 Example

outshaft

21

22

11

zzprocessAdiabatic

KgKJhsmV

KgKJhV

51

out)(net kg-796,650J/ 1350103-798

)/s302)(m-1/2(602103(J/kg)3348-103(J/kg)2550

2ˆˆ

]2

ˆˆ[

)(0)(sin0

)](2

ˆˆ[

)](2

)()(ˆˆ[

flowmean -the-in-statesteady for eq.energy D-1

:Solution

22

22

22

21

22

22

inoutinout

netinshaft

netinshaft

inoutinout

netinshaft

innetinout

outnetshaft

innetinout

inoutinout

innetshaft

innetinout

inoutinoutinout

vvhh

m

WW

vvhhmW

flowadiabaticQzzcezz

WQzzgvv

hhmor

WQzzgvvpp

uum

J/kg=(N m)/kg‧ [=](kg m/s m)/kg‧ ‧ [=]m2/s2

52

?change eTemperatur :Determine

flowsteady 500ft =z figureRight ：Given

5.21 Example

R

lbsftlb

Btu

lbft

Rlb

Btu

ftsft

c

zzgTT

zzgcTT

zzgcTTm

workshaftnoWflowadiabaticQ

VVpp

waterofheatspecifictheRlbBtucwherecTTuuce

WQzzgvvpp

uum

f

mf

m

netinshaftnetin

m

innetshaft

innetinout

inoutinoutinout

642.0

1/2.32

1

7781

500)/(2.32ˆ

)(

0)(ˆ)(

0)](ˆ)[(

00

0;)()(

)/(1ˆˆ)(ˆˆsin

)](2

)()(ˆˆ[

flowmean -the-in-statesteady foequation energy ldimeusiona-one

:Solution

2

212

12

1212

1212

1212

1212

22

53

§ 5.3.3 Comparison of the Energy Equation with the Bornoulli Equation

massenergy

quugzVp

gzVp

m

QqLet

QzzgVVpp

uum

constW

WQzzgVVpp

uum

innetinoutin

ininout

outout

innet

innet

innetinoutinout

inoutinout

innetshaft

innetshaft

innetinoutinoutinoutinout

/][

)73.5(ˆˆ22

massunit per sfer heat tranor rate massper ratesfer heat tran

)72.5()](2

1ˆˆ[

.0

flow ibleincompress and shaftwork nowith

)](2

1)()(ˆˆ[

with equation energy flow-steady D-1 From

22

22

22

54)79.5(

22

ˆˆ

2

0ˆˆ

0ˆˆ)75.5.()73.5.(

)75.5(22

)74.5(22

assuch

) streamline along invicid, flow, ibleincompress state,(Steady equation Bernoulli From

sections. wobetween t streamline a along flowor section two

between fluid of stream single a of flow D-1 toapplicable is (5.73)equation This

)73.5(ˆˆ22

22

2

22

22

22

lossgzVp

gzVp

Lossquu

energyavailableorusefulgzVp

quufrictionwithflowaFrom

quuEqEqIf

gzVp

gzVp

gzV

pgzV

p

quugzVp

gzVp

ininin

outoutout

innetinout

innetinout

innetinout

ininin

outoutout

inin

inoutout

out

innetinoutin

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55

(b) & (a) case

:Compare

/1.23

figureright in (b) & (a) case

:Given

5.22 Example

2

3

inQ

mkg

)2

(sin)(

2

)(0

)0(02

)(0

)2(&)1(sec

22

(5.79)equition energy From

:Solution

22

212121

2

1

1

21

2

21

211

211

2

222

Vklossceloss

ppV

changeelevationnogz

VV

changeelevationnogz

tionbetweenlossthelosswhere

lossgzVp

gzVp

L

56

smQbcaseFor

smQacaseFor

k

ppDVAQ

k

ppV

Vk

ppV

Vklossceloss

ppV

L

LL

L

/445.0)(

/372.0)(

2

)1(4

2

)1(2

)(2

)2

(sin)(

2

32

32

2122

222

212

2221

2

22

212121

2

57

pLsp

TLsT

ff

L

innetshaft

innetshaft

innetshaft

s

Lsininin

outoutout

ff

innetshaftin

ininout

outout

shaftininin

outoutout

hhhpumpafor

hhhturbineaforNote

mNmNorftlblbftweightenergyhead

g

lossh

gQ

W

gm

W

g

Whwhere

hhzg

V

g

pz

g

V

g

p

mNmmNorftlbftlbftvolumeenergy

lossWgzV

pgzV

p

lossWgzVp

gzVp

:

/][/][/][""

)85.5(

)84.5(22

////][/][

)83.5()(22

)82.5(22

22

2323

22

22

5822

)7.5(

22

2

attention. special require willintegral following theuniform,not is

surface control thecrosses flow heresection wany at profile velocity theIf

)steady(0:

)64.5()2

(

(5.64)equation energy From

22

22

2

....

2

VmdAnV

V

A

dAnV

velocityaveragetheV

tcoefficienenergykineticthewhere

VVm

dAnVV

statedVet

note

WQdAnvgzvp

udVet

A

A

ininoutout

cs

cv

vcnetinshaft

vincnetcscv

59

m

WWwhere

hg

Wz

g

V

g

pz

g

V

g

p

lossWgzV

pgzV

p

lossWgzVp

gzVp

flowuniformfor

profilevelocityanyforwhere

Vm

dAnVV

VmdAnV

V

innetshaft

innetshaft

Lshaft

in

in

inin

out

out

outout

shaftin

in

ininout

out

outout

shaftin

inininout

outoutout

A

A

)89.5(22

) heador eight energy / w (equation Energy

)88.5()(22

) olumeenergy / v (equation Energy

)87.5(22

esteadystat ible,incompress ) massenergy / (equation Energy

1

1

)86.5(

2

2

22

22

22

22

2

2

22

60.)(

.min)(

/23.114.0

101.0

)2(sec

.)(08.1

30

)1(sec

.)(0.2

60

min/1.0

:Given

5.25 Example

3

312

2

2

1

1

distvelocityactualgConsiderinb

distvelocityuniformgassua

lossofvaluetheCompare

mkgWattsW

papp

ondistributivelocityUniform

flowTurbulenttion

coefenergykinetic

mmD

ondistributivelocityParabolic

flowLamianrtion

coefenergykinetic

mmD

kgm

airinnet

shaft

61sm

mmkg

skg

A

mV

smmmkg

skg

A

mV

smNWatt

smNWattW

s

kg

s

kgm

VVpp

m

Wloss

m

WW

negligibleisgzinchangegzgzwhere

WlossgzVp

gzVp

innetshaft

innetshaft

innetshaft

innetshaft

innetshaft

/92.1)10

2

30()/23.1(

/10667.1

/479.0)10

2

60()/23.1(

/10667.1

/14.01

/114.0

10667.160

min1

min1.0

22

)(0&0

22

flow statesteady & ibleincompressfor equation Energy

:Solution

2233

3

22

2233

3

11

3

222

21112

12

1

2111

2

2222

62

kgmN

VVpp

m

Wloss

kgmN

smsm

mkg

pa

skg

smN

VVpp

m

Wloss

innetshaft

innetshaft

/975.0

22

)0.1 ( profiles velocityuniform assumed for the (b)

/92.0

99.123.030.8198.83

2

/92.108.1

2

/479.02

/23.1

101.0

/10667.1

/14.0

22

)08.1,2 ( profiles velocityactual for the (a)

222

21112

21

22

3

3

3

222

21112

21