1 chapter 5 flow analysis using control volume (finite control volume analysis )
Post on 20-Dec-2015
225 views
TRANSCRIPT
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
ininout
outout
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