thermal science ch 05
TRANSCRIPT
72
FIVE
The Second Law of Thermodynamics
5.1FE D
5.2FE D
5.3FE A max283
1 1 0.05667.300
L
H
TT
293
COP 7.325.293 253
H
H L
TT T
5.4FE C293 10
1 1 0.3805 . 16.3 kW473 10
LL
H L L
T WQ
T Q W Q
5.5FE C
5.6FE B
5.7FE A20
COP 2.10
LQW
293From Eq. 5.11, COP 2. 439.5 K
293
L
HH L H
TT
T T T
5.8FE C Use Eq. 5.10:293
COP 7.325.293 253
H
H L
TT T
2000COP 7.325 . 273 kJ/hr HQ
WW W
5.9FE A293
1 1 0.214373
L
H
TT
5.10FE C
5.11FE C 2
1
573ln 1 0.717 ln 0.481 kJ/K
293 v
TS mc
T
5.12FE D 2 1( ) 10 0.717 (356.6 773) 2986 kJvW mc T T
where1/ 0.2857
22 1
1
400773 356.6 K
6000
k kp
T Tp
5.13FE D The heat gained by the ice is lost by the water: melt ,i w p w wh m m c T . (We
assume not all the ice melts.) The entropy change is:
2,
5.02 333 273ln 20 4.18 ln 0.213 kJ/K
293 273 293
i w w p wi
TQS S m c
T
73
5.14FE A The final temperature is 0C. The heat lost by the iron is:
Fe 1 2
2Fe
1 ice
( ) 10 0.448 300 1344 kJ
273 1344ln 10 0.448 ln 0.881 kJ/K
673 273
p
p
Q m c T T
T QS m c
T T
5.15FE C
5.16FE D First, find T2:1/ 0.2857
22 1
1
2000673 1066 K
400
k kp
T Tp
.
With Q = 0: 2 1( ) 2 0.717 (1066 673) 563 kJ vW mc T T
5.17FE D State 1 is at the inlet and state 2 at the outlet. Use enthalpies found in Tables C.3 and C.2:
1 2 3658 2637 1021 kJ/kg Tw h h
5.18FE C State 2is at the same entropy as state 1:
2 1 2' 2'7.1677 1.0259 6.6441. 0.9244s s x x
2' ,max317.6 0.9244 2319 2462 kJ/kg. 3658 2462 1197 kJ/kg Th w
5.19FE B It’s the temperature of vaporization found in Table C.2.
5.20FE: A ,actual
,isentropic
10200.85
1200 T
T
w
w
5.130000 / 3600output
COP . 4 . 2.083 kW or 2.79 hpinput
WW
5.275000 / 3600output
COP 5.21input 4
5.3500 10 2545
1 1 0.256. 99,400 Btu/hr672 0.256
HH
L
T WQ
T
5.4output 500
0.3input 100 6000 / 3600
5.5 100 1000 / 3600 27.78 m/sV 2
1 22
3
11.23 27.78 2 0.28 27.78Drag power 2 0.519
Inputpower 1 100 1000 10740 9000
13 3600
DV AC V
W
74
5.6 A Kelvin-Planck violation is sketched as the first block in the figure below. Thedashed box encloses a refrigerator which is a violation of the Clausius statement:
1 2.H H LQ W Q W Q 1HQ
2 1Let H H HQ Q Q Then L HQ Q with no work required.This is a violation of the Clausiusstatement of the 2nd law.
5.7 No. It is not operating on a cycle.
5.8
net netAssume . Then ( ) ( ) .
Impossible.R C L HW W Q Q
net netAssume . Then ( ) ( ) .
Impossible. It violates the Clausius statement.R C L HW W Q Q
5.9 The maximum temperature drop for the seawater is 17
20 4.18 17 1421 kWH pQ mc T The efficiency of the proposed engine is
1000.0704 or 7.04%
1421WQ
The efficiency of the Carnot engine would be283
1 1 0.0567 or 5.67%300
L
H
TT
The inventor’s claim is impossible since it exceeds the Carnot efficiency.
5.10 The maximum possible efficiency is the Carnot efficiency:293
1 1 0.2038368
L
H
TT
assuming the water is rejected at atmospheric temperature. Then
0.2 4.18 (95 20) 62.7 kWH pQ mc T Then
max 0.2038 62.7 12.8 kWHW Q
W
QH1 QH2
QL
W
QH1 QH2
QL
QH1 QH2
QL2QL1
WR
WC
75
5.11 a)473
1 1 0.6791473
L
H
TT
b)473
COP 0.4731000
L L
H L H L
Q TQ Q T T
c)1473
COP 1.4731000
H H
H L H L
Q TQ Q T T
5.12 First, we find the power produced:80 1000
40 17.78 kW3600H LW Q Q
17.78 2830.4444 1 . 509.4 or 236.4 C
40 HHH
WT
TQ
5.13 Let T be the unknown intermediate temperature. Then
1 2560
1 and 11060
TT
It is given that 1 2 20.2 . Substitute and find
25601 1.2 1 . 212 712,300 0. 745 R or 285 F
1060T
T T TT
5.1420
1 1 0.6. 0.450
L L
H H
q Tq T
For the adiabatic process (see Fig. 5.8)1
1/0.42 2
3 3
. 0.4 0.1012k
L
H
T v vT v v
Then 32 30.1012 0.1012 10 1.012 m /kg. The high temperature is thenv v
2 2 200 1.012 705 K or 432.2 C0.287H
p vT
R
The low temperature is
0.4 0.4 705.2 282.1 K or 9.1 CL HT T
5.15 The maximum possible efficiency is
actual293 output 43 0.746
1 1 0.75. 0.77. Impossible1173 input 2500 / 60
L
H
TT
5.16 a)510
1 1 0.0642 or 6.42%545
L
H
TT
b)545
COP 15.57545 510
H H
H L H L
Q TQ Q T T
76
5.17 1 2 11 2 3 1 2 1 2 1 2
2 3 3
1 , 1 , 1 1 (1 )(1 )T T TT T T
As an example let 1 2 3100 C, 200 C, and 300 C. ThenT T T
1 2 3373 473 373
1 0.2114, 1 0.1745, 1 0.349473 573 573
or 3 0.2114 0.1745 0.2114 0.1745 0.349
5.18 engine273
1 0.75. 1092 K1 0.75
LH
H
TT
T
273
COP 0.3331092 273
L
H L
TT T
5.19 221 2 2 2
2
313. 1 1 . 773 313. 492 K or 219 C
773T
T TT
5.20 2293COP . . 138600 or 372 K =99 C
293 473L
H L
T TT T
T T T T
5.21 4 4 200 0.03 20.920.9 K. 1 0.956 or 95.6%0.287 473L
p vT
R
0.956 30 28.7 kJ/kgHw q
5.22 Refer to Fig. 5.7: 3 3L
p VT
15 144 250 /1728586.3 R
0.01 53.3mR
400 144 25 /17281173 R
0.01 53.3HT
5861 0.500 or 50%
1173
31
2 4
ln ln
300 150.01 53.3 1173ln 586ln 177 ft-lbf
170.2 26.44
H L H L
ppW Q Q mR T T
p p
where
/ 1 3.5 3.52
2 3 43
1173 58615 170.2 psia, 300 26.44 psia
586 1173
k kT
p p pT
77
5.230.4 /1.41 273.4
600 273.4 K. 1 0.54415 600LT
3.53
3 40.287 273.4 273.40.7847 m /kg. 4698 300 kPa
100 600v p
3.5
22 1 2
1
600100 1566 kPa. 1566 3 4698 kPa273.4
vp p pv
31net
2 4
4698 100ln ln 0.287 600ln 273.4ln 103 kJ/kg
1566 300H L
ppw RT RT
p p
5.24 pump530
COP 26.5530 510
H
H L
TT T
75,000. 72,170 Btu/hr
75,000H
LH L L
Q Q Q
water water water. 72,170 4.18 12. 6014 lbm/hrL pQ m c T m m
75,000COP . 26.5 . 2830 Btu/hr or 1.11 hpHQ
WW W
5.25293 1800 / 60
COP 6.511 . 4.61 kW or 6.18 hp293 248
H H
H L
T QW
T T W W
5.261800 2
4.61 kW. (296 )60 20 ( 25) 3L L
TW Q T
2ac
2(296 )
3COP . . 599 87 620 0296 4.61
LL L L
L LH L L
TT Q TT T
T T TW
345 K or 72 CLT
5.27 a) cool500 / 60
COP 2.365 550 / 778
LQW
b) heat500 / 60 5 550 / 778
COP 3.365 550 / 778
HQW
5.285 6
ref 6
10 2 10COP . . 1465 kJ
293 2 10L L
H L
Q TW
W T T W
78
5.29 max255
COP 6.71293 255
L
H L
TT T
actual3000 / 60
COP 3.73. Yes, it's possible10 0.746
LQW
5.30 11
1
( )278COP 13.9
298 278LQ
W
22 2 1
2
( )253COP 6.325 . But
298 253LQ
W WW
2 12 1
( ) ( ) 12000. ( ) 0.455( ) 0.455 5 1.055 8 kJ/s
6.325 13.9 360L L
L LQ Q
Q Q
5.31373 100
1 1 0.707. 141.4 kW.1273 0.707
LH
H
T WQ
T
141.4 100 41.4 kWL HQ Q W
141.4 20 60133.3 kJ/K
1273H
HH
Q tS
T
41.4 20 60133.2 kJ/K
373L
LL
Q tS
T
net 133.2 133.3 0.1 kJ/KS which is zero except for round-off error.
5.32100 100
a) 0.0932 kJ/K. b) 0.341 kJ/K1073 293
SRR S
R S
QQS S
T T
5.33200
COP . 10 . 20 kJ. 220 kJLH
QW Q
W W
200a) 0.76 kJ/K
263LS
and b)220
0.76 kJ/K289.3HS
where263
COP 10 and 289.3 K.263 H
H
TT
5.3450,000 / 3600 530
COP 4.91 . 422.1 R4 550 / 778 530
HL
L
QT
TW
50,000 1/ 6 39,820 1/ 615.72 Btu/ R and 15.72 Btu/ R
530 422.1S LS S
where 50,000 4 (550 / 778) 3600 39,820 Btu/hr.L HQ Q W
5.35 From Problem 5.28 1465 kJ.W 1465 0.00001 1465 kJ.HQ 5
6
1465 105 kJ/K and 5 kJ/K
293 2 10H SS S
79
5.36/
2 2 2 2 2 2
1 1 1 1 1 1
0 ln ln or ln ln or ln lnvR c
vv
T v T R v T vc RT v T c v T v
( 1) 1
2 2 1
1 1 2
But 1 (see Eq. 4.32).k k
v
R T v vkc T v v
.
To show Eq. 5.29, use Eqs. 5.28:( 1)/1 ( 1)
12 21 1 1
2 1 1 2 2
.
kk kk k kkp pv v v
v p p v v
5.37 11
p VT 1 200 0.8
278.7 K2 0.287mR
2 2
1 1
ln lnp
T pS m c R
T p
7732 1.0 ln 2.04 kJ/K
278.7
5.38 V 11
1
andmRT
Vp
2 V1/
11
2
kpp
a) V 31
0.2 0.287 313=0.1198 m ,
150V
1/1.43
2150
0.1198 0.0445 m600
b) V 31
0.2 0.1889 313=0.0788 m ,
150V
1/1.2893
2150
0.0788 0.0269 m600
c) V 31
0.2 0.2968 313=0.1239 m ,
150V
1/1.43
2150
0.1239 0.0460 m600
d) V 31
0.2 4.124 313=1.769 m ,
150V
1/1.4093
2150
1.769 0.246 m600
5.391/ 0.4 /1.4
22 1
1
2000520 2117 R or 1657 F
14.7
k kp
T Tp
5.40 2 2
1 1
ln lnvT vS m c RT v
22 1
1
, sincepT T Vp
const
a) 21500 600
2 0.717 ln 2.31 kJ/K since 300 1500 K300 120
S T
b) 21500 600
2 0.653ln 2.10 kJ/K since 300 1500 K300 120
S T
c) 21500 600
2 0.745ln 2.40 kJ/K since 300 1500 K300 120
S T
d) 21500 600
2 10.08ln 32.4 kJ/K since 300 1500 K300 120
S T
80
5.41 1 1
2
lnp p V
Q W mRTp
1
RT 11
RT 61
2
6000ln 6000 500 10 ln 10.2 kJ
200pp
2
1
lnp
TS mc
T 1p V
6
1 2
1 1
6000 500 10 200ln ln 9.51 kJ/K
1073 6000p
T p
5.42 1p Vm 1 2 2
1 1 1
, ln lnvT vS m c R
RT T v 2
1
lnvTmcT
a)27.2 144 10 100
1.592 lbm, 1.592 0.171ln 0.349 Btu/ R53.3 470 27.7
m S
b)27.2 144 10 100
2.42 lbm, 2.42 0.156ln 0.485 Btu/ R35.1 470 27.7
m S
c)27.2 144 10 100
1.54 lbm, 1.54 0.178ln 0.352 Btu/ R55.15 470 27.7
m S
d)27.2 144 10 100
0.111 lbm, 0.111 2.40 ln 0.342 Btu/ R766.4 470 27.7
m S
5.43 2 2400 0.2
. ( 200) 0.717( 313). 626.2 K0.287 313vW mc T T T
2 2
1 1
ln lnvT vS m c RT v
2
1
400 .2 626.2ln .717 ln 0.443 kJ/K.287 313 313v
TmcT
5.44 1p Vm 1 2
1 1
, ln lnv
T VS m c R
RT T 2
V 1
a)80 4 400
3.72 kg, 5.2 3.72 .717 ln .287 ln.287 300 300
Vm
2 .
4V
3
2 254 m
b)80 4 400
5.64 kg, 5.2 5.64 .653ln .189 ln.189 300 300
Vm
2 .
4V
3
2 195 m
c)80 4 400
3.59 kg, 5.2 3.59 .745ln .297 ln.297 300 300
Vm
2 .
4V
3
2 255 m
d)80 4 400
.259 kg, 5.2 .259 10.08ln 4.12 ln4.12 300 300
Vm
2 .
4V
3
2 259 m
5.45 1p Vm 1
1
, VRT
2 2 31 0.1 0.2 0.00628 m , pr h Q mc T
V 2 V 21
1
, (T
W m p VT
2 V 21
1
), lnp
TS mc
T
81
a) 2 2200 0.00628
0.0137 kg, 9 0.0137 1.0( 320). 978 K0.287 320
m T T
V 32
978.00628 .0192 m , .0137 200(.0192 .00628) 0.0354 kJ
320W
9780.0137 1.0ln 0.154 kJ/K
320S
b) 2 2200 0.00628
0.0208 kg, 9 0.0208 0.824( 320). 843 K0.189 320
m T T
V 32
834.00628 .0164 m , .0208 200(.0164 .00628) 0.0420 kJ
320W
8340.0208 0.842ln 0.168 kJ/K
320S
c) 2 2200 0.00628
0.0132 kg, 9 0.0132 1.04( 320). 976 K0.297 320
m T T
V 32
976.00628 .0191 m , .0132 200(.0191 .00628) 0.0340 kJ
320W
9760.0132 1.04 ln 0.153 kJ/K
320S
d) 2 2200 0.00628
0.000953 kg, 9 0.000953 14.21( 320). 984 K4.124 320
m T T
V 32
984.00628 .0193 m , .000953 200(.0193 .00628) 0.00258 kJ
320W
9840.000953 14.21ln 0.0152 kJ/K
320S
5.46 2 2
1 1
ln , lnp p
q w RT s Rp p
a)4000
0.287 300ln 377 kJ/kg50
q w
40000.287 ln 1.26 kJ/kg K
50s
b)4000
0.189 300ln 248 kJ/kg50
q w
40000.189ln 0.828 kJ/kg K
50s
c)4000
0.297 300ln 390 kJ/kg50
q w
40000.297 ln 1.30 kJ/kg K
50s
82
d)4000
4.124 300ln 5420 kJ/kg50
q w
40004.124 ln 18.1 kJ/kg K
50s
5.47 2
1
, lnp p
Tq c T s c
T
a)1360
0.24 (900 60) 202 Btu/lbm, 0.24 ln 0.231 Btu/lbm- R520
q s
b)1360
0.202 (900 60) 170 Btu/lbm, 0.202 ln 0.194 Btu/lbm- R520
q s
c)1360
0.248 (900 60) 208 Btu/lbm, 0.248 ln 0.238 Btu/lbm- R520
q s
d)1360
3.42 (900 60) 2870 Btu/lbm, 3.42 ln 3.29 Btu/lbm- R520
q s
5.48 1p Vm 1
1
100 2 502.38 kg, 2.38 0.287 ln 0.473 kJ/K
0.287 293 100S
RT
where 2 1V
p p 1
V 2
2100 50 kPa
4 .
5.49 a) 1p Vm 1
1
100 22.11 kg,
0.287 330Q
RT
vW mc T
2 2594
( 400) 2.11 .717( 330). 594 K. 2.11 .717 ln 0.889 kJ/K330
T T S
b)100 2
3.2 kg,0.189 330
m Q
2 23.2 0.653( 300). 511 KvW mc T T T
5113.2 .653ln 0.914 kJ/K
330S
c)100 2
2.04 kg,.297 330
m Q
2 22.04 .745( 300). 593 KvW mc T T T
5932.04 .745ln 0.891 kJ/K
330S
d)100 2
.147 kg,4.12 330
m Q
2 2.147 10.1( 300). 600 KvW mc T T T
6000.147 10.1ln 0.886 kJ/K
330S
83
5.50 22 1 2
1
20001.2147 165. Interpolate from Table F.1E: 1974 R.
14.7r r
pp p T
p
Compare with the constant specific heat prediction:1/ 0.2857
22 1
1
2000520 2117 R, an error of 7.24%.
14.7
k kp
T Tp
5.51 1p Vm 1
1
200 24.36 kg.
0.287 320Q W m u
RT
.
a) 2 240 40
500 10 60 4.36 0.717( 320). 787 K1000
T T
2
1
787ln 4.36 0.717 ln 2.81 kJ/K
320v
TS mc
T
b) From Table F.1 1 1228.4 kJ/kg and 1.767 kJ/kg K.u
2 240 40
500 10 60 4.36( 228.4). 563 kJ/kg1000
u u
.
22 2 2 1
1
764764 K and 2.667. Then 200 478 kPa
320T
T p pT
4784.36 2.667 1.767 0.287 ln 2.83 kJ/K
200S
5.52 a) Q W 2 2200,000
. 2 0.171( 520). 1272 R or 812 F778vmc T T T
/ 1 3.52
2 11
127216 366 psia
520
k kT
p pT
b) 2 2 2200,000
2 ( 88.62). 217.2 Btu/lbm. 1245 R or 785 F778
u u T
1 2 227.5
1.215, 27.5. 16 362 psia1.215rp p p
5.536600 200 10
0.001427 kg0.287 293
m
. 2 1V
T T 2
V 1
1000293 1465 K
200
a) 0.001427 1.0(1465 293) 1.67 kJpQ mc T
b) Find the enthalpies in Table F.1:
2 1( ) 0.001427 (1593.7 293.2) 1.86 kJQ m h h
84
5.54 a)1/ 0.2857
22 1
1
2000300 670.2 K
120
k kp
T Tp
0.2 0.717(670.2 300) 53.1 kJW
b) 2 2 22000
( 1.702) 0.287 ln . 2.51 kJ/kg K. 481 kJ/kg120
u
0.2 (481 214) 53.4 kJW
5.55 a)1/ 0.2857
22 1
1
500500 199 K. 4 .717(199 500) 863 kJ
20
k kp
T T Wp
b) 2 2 220
2.220 0.287 ln . 1.296. 200 K500
T
4 (143 359) 864 kJW
5.56 a) 1 12.047 0.85(4.617) 5.971, 721 0.85(2048) 2462s h
22 1 2 2 2
2
800. 2000 2462. 4462. 8.877
4462p
q h h h h sh
28.877 5.971 2.91 kJ/kg K and 829 Cs T
b) TK solution:Rule Sheet
; This is a closed system, so the first law is q = u2 - u1 + p2*v2 - p1 * v1 orq = h2 - h1dels = s2 - s1
; Steam tables based on NBS/NRC Steam Tables by Lester Haar,; John S. Gallagher, and George S. Kell, Hemisphere Publishing Corp., 1984.; STEAM STEAM8SI.TKW
Variable Sheet
Status Input Name Output Unit Comment` Thermal Sciences, Potter & Scott
P5-56.tkw Problem 5-56*STEAM8SI.TKW Steam, 1-8 States, SI Units
T1 170 C Temperature800 p1 kPa Pressure
h1 2460 kJ/kg Enthalpys1 5.97 kJ/(kg*K) Entropyv1 0.204 m^3/kg Specific Volume
0.85 x1 Qualityphase1 'SAT PhaseT2 927 C Temperature (Enter a guess value.)
800 p2 kPa Pressureh2 4460 kJ/kg Enthalpys2 8.87 kJ/(kg*K) Entropy
85
v2 0.691 m^3/kg Specific Volumex2 'mngless Qualityphase2 'SH Phase
2000 q kJ/kg Heat transferdels 2.9 kJ/(kg*K) Entropy change
5.57 1 1 2 1600 psia, 486 F, 0.672 0.4(0.774) 0.982p T T s
31 1506.6 0.4(608.4) 750 kJ/kg, 0.0201 0.4(0.7501) 0.320 m / kgu v
22 2 2
2
300 psia1.56, 1153, 1.73
486 F
2 (1.56 0.982) 1.158 Btu/lbm- R
600 144(0.77 0.32) 450 144(1.733 0.77)2 (1153 750) 2 1066 Btu778
ps u v
T
S m sQ m u W
Note: The work was estimated using graphical integration (a straight line was assumedbetween the saturated vapor point and state 2).
5.58 a) 1 2 1at 20 C. ( ) 400(0.7726 0.001) 309 kJ/kgfv v w p v v
2 1
2 1
2964.4 83.9 309 3189 kJ/kg
or3273 84 3189 kJ/kg
q u u w
q h h
2 1 7.899 0.2965 7.604 kJ/kg Ks s s b) TK solution:
Rule Sheet
;Assume that the water is in a closed system. The first law is q = u2 - u1 + p2*v2 - p1 * v1 orq = h2 - h1dels = s2 - s1; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and GeorgeS. Kell, Hemisphere Publishing Corp., 1984. ;STEAM STEAM8SI.TKW
Variable Sheet
Input Name Output Unit Comment` Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI UnitsP5-58.tkw Problem 5-58
20 T1 C Temperature400 p1 kPa Pressure
h1 84.2 kJ/kg Enthalpys1 0.293 kJ/(kg*K) Entropyv1 0.001 m^3/kg Specific Volumex1 'mngless Qualityphase1 'CL Phase
86
400 T2 C Temperature400 p2 kPa Pressure
h2 3270 kJ/kg Enthalpys2 7.9 kJ/(kg*K) Entropyv2 0.773 m^3/kg Specific Volumex2 'mngless Qualityphase2 'SH Phase
*THUNITS.TKW Units for thermoq 3190 kJ/kg Heat transfer per unit massdels 7.6 kJ/(kg*K) Entropy change
5.59 1Vv
3
1 16 10 0.001017 (7.671 0.001). 0.0002585
2x x
m
1 251.1 0.0002585(2456.6 251.1) 251.7 kJ/kgu
2 1( )Q m u u W 2 11000
. 251.7 751.7 kJ/kg2
Qu u
m
32 1 20.003 m / kg and 752 kJ/kg. Locate state 2 by trial-and-error:v v u
2 2 2
2 2
Guess 170 C : 0.003 0.0011 (0.2428 0.001). 0.00786751.7 718.3 (2576.5 718.3). 0.0178
T x xx x
2 2 2
2 2
Guess 177 C: 0.003 0.0011 (0.2087 0.0011). 0.00915751.7 750.0 (2581.5 750.0). 0.00093
T x xx x
A temperature of 176 C is chosen. We interpolate to find
2 2
2 2
0.003 0.0011 (0.2136 0.0011). 0.00894.
( ) 2 (2.101 0.00894 4.518) 4.28 kJ/Kf fg
x x
S m s x s
5.60 a) . . 0.1(3674 1087) 0.1(3279 1082)Q W m u m h W m u W
39 kJ and 0.1(7.370 2.797) 0.457 kJ/KW S b) TK solution:
Rule Sheet
;This is a closed system, so the work of a frictionless process is W = INT pdV, and for a; constant pressure process this becomesW = m* p1 * (v2 - v1)delS = m* (s2 - s1); Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and GeorgeS. Kell, Hemisphere Publishing Corp., 1984. ; STEAM8SI.TKW
Variable Sheet
Input Name Output Unit Comment` Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI UnitsP5-60.tkw Problem 5-60
T1 250 C Temperature
87
4000 p1 kPa Pressureh1 1090 kJ/kg Enthalpys1 2.8 kJ/(kg*K) Entropyv1 0.00125 m^3/kg Specific Volume
0 x1 Qualityphase1 'SAT Phase
600 T2 C Temperature4000 p2 kPa Pressure
h2 3670 kJ/kg Enthalpys2 7.37 kJ/(kg*K) Entropyv2 0.0988 m^3/kg Specific Volumex2 'mngless Qualityphase2 'SH Phase
*THUNITS.TKW Units for thermoW 39 kJ Work
0.1 m kg Mass of steamdelS 0.457 kJ/K Entropy change of system
5.61 1 1 2 2 22507 kJ/kg, 7.356 0.832 (7.077). 0.922.u s s x x
2 251 0.922(2205) 2284 kJ/kgu
1 2( ) 2(2507 2284) 447 kJW m u u
5.62 a) 1 1 1 1 12 0.4 0.001 3.992 . 0.1. Then 1.69 and 5335
v x x s u
22 1 2
2 1
5 MPa570 kJ/kg. ( ) 5(570 533) 185 kJ
1.69p
u W m u us s
b) TK solution:
Rule Sheet;This is a closed system, so for an adiabatic process, W = U1 - U2 orW = m * (u1 - u2) ; First lawv1 = V1/m ; Definition of specific volumes2 = s1 ; for an isentropic process; Steam tables based on NBS/NRC Steam Tables by Lester Haar, ; John S. Gallagher, and GeorgeS. Kell, Hemisphere Publishing Corp., 1984. ; STEAM8SI.TKW
Variable Sheet
Status Input Name Output Unit Comment` Thermal Sciences, Potter & Scott
*STEAM8SI.TKW Steam, 1-8 States, SI UnitsP5-62.tkw Problem 5-62
T1 75.9 C Temperature40 p1 kPa Pressure
h1 550 kJ/kg Enthalpys1 1.69 kJ/(kg*K) Entropy
88
0.4 v1 m^3/kg Specific Volume (transfer value to input)x1 0.1 Qualityphase1 'SAT PhaseT2 136 C Temperature (starting guess needed)
5000 p2 kPa Pressureh2 577 kJ/kg Enthalpys2 1.69 kJ/(kg*K) Entropyv2 0.00107 m^3/kg Specific Volumex2 'mngless Qualityphase2 'CL Phase
*THUNITS.TKW Units for thermoW -190 kJ Work
5 m kg Massu1 534 kJ/kg Internal energy, state 1u2 572 kJ/kg Internal Energy, state 2
2 V1 m^3 Specific volume, state 1
5.63 10(1150 8.02) 11, 420 Btu. 10(1.757 0.016) 17.4 Btu/ RQ m h S
5.64 The heat that enters the ice leaves the water:a) Assume that all the ice does not melt:
220 1.9 5 330 10 4.18(20). 1.96 kg and 0 Cm m T
2 2
1 1
( ) ln ( ) ln
273 1.96 2735 1.9 ln 330 10 4.18 ln 0.064 kJ/K
253 273 293
i i wi p i w p w
i i w
T Q TS m c m c
T T T
b) Assume that all the ice melts:
2 2 220 1.9 5 330 5 5 4.18( 0) 40 4.18(20 ). 8.0 CT T T
2 2 2
1 1 1
( ) ln ( ) ln ( ) ln
273 5 330 281 2815 1.9ln 5 4.18ln 40 4.18ln 0.378 kJ/K
253 273 273 293
i i iw wi p i iw p w w p w
i i iw w
T Q T TS m c m c m c
T T T T
5.655 1.2 /1728
0.199 lbm, 1 lbm0.01745i wm m
The heat that enters the ice leaves the water. Assume that not all of the ice melts:
melt melt0.199 0.49 (32 0) 143 1 1.0 (60 32). 0.178 lbmm m 492 0.178 143 492
0.199 0.49 ln 1 1.0 ln 0.166 Btu/lbm460 460 520
S
89
5.66293
1 1 0.489 or 48.9%. 300 kJ/kg573
LH L H L
H
Tw q q T s T s
T
1 1300
1.071 5.705 . 4.634 kJ/kg K573 293
s s s
1 1 14.634 3.254 (2.451). 0.563s x x
5.67 a) For the cycle, the work output equals the heat input:
2 2. 500 (275.6 45.8)( 3.027). 5.203 kJ/kg KW T s s s
2 2At 6 MPa 5.203 3.027 (2.863). 0.760x x b) TK solution:
Rule Sheet;The net work of a Carnot cycle is the enclosed area on a Ts diagram. The adiabatic ompression(process 1-2) is entirely within the saturated vapor region, so the specified pressures determine theupper and lower temperatures.w = (T2 - T1)*(s3 - s2) ; work = Ts diagram areas1 = s2T3 = T2; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and; George S. Kell, Hemisphere Publishing Corp., 1984. Stm8si.tkw
Variable SheetInput Name Output Unit Comment
` Thermal Sciences, Potter & Scott*Stm8si.tkw Steam, 1-8 States, SI UnitsP5-67.tkw Problem 5.67
T1 45.83 C Temperature10 p1 kPa Pressure3.027 s1 kJ/(kg*K) Entropy (transfer value to input)
v1 4.637 m^3/kg Specific Volumex1 0.317 Qualityphase1 'SAT PhaseT2 275.6 C Temperature
6000 p2 kPa Pressures2 3.027 kJ/(kg*K) Entropyv2 0.001319 m^3/kg Specific Volume
0 x2 Qualityphase2 'SAT Phase
275.6 T3 C Temperature (transfer value to input)p3 6000 kPa Pressure
5.203 s3 kJ/(kg*K) Entropy (transfer value to input)v3 0.02498 m^3/kg Specific Volumex3 0.7603 Qualityphase3 'SAT Phase
*THUNITS.tkw Units for thermo500 w kJ/kg
90
5.68 1 2.046 0.15(7.077) 2.738 kJ/kg K. 6.664 2.738 3.93 kJ/kg Ks s
net 3.93(170.4 60.1) 433 kJ/kgnetw q s T
60.1 2731 0.249 or 24.9%
170.4 273
5.69 a) 4 20.704 0.2(7.372) 2.178 kJ/kg K. 5.704 kJ/kg Ks s
573 (5.704 2.178) 2020 kJ/kgH Hq T s b) TK solution:
Rule Sheets3 = s2s1 = s4q12 = T1 * (s2 - s1); Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and; George S. Kell, Hemisphere Publishing Corp., 1984. Stm8si.tkw
Variable SheetInput Name Output Unit Comment
` Thermal Sciences, Potter & Scott*Stm8si.tkw Steam, 1-8 States, SI UnitsP5-69.tkw Problem 5.69
300 T1 C Temperaturep1 kPa Pressures1 2.18 kJ/(kg*K) Entropyx1 Qualityphase1 Phase
300 T2 C Temperaturep2 8580 kPa Pressures2 5.7 kJ/(kg*K) Entropy
1 x2 Qualityphase2 'SAT Phase
50 T3 C Temperaturep3 kPa Pressures3 5.7 kJ/(kg*K) Entropyx3 Qualityphase3 Phase
50 T4 C Temperaturep4 12.3 kPa Pressures4 2.18 kJ/(kg*K) Entropy
0.2 x4 Qualityphase4 'SAT Phase
*THUNITS.tkw Units for thermoq12 2020 kJ/kg
91
5.70253
COP 3.614 where323 253
L L L
H L H L
Q T qw s T
Q Q T T w
1 4 3 20.704 0.15(7.372) 1.81, 5.704 kJ/kg Ks s s s
a) out net 0.02 (5.704 1.81)(300 50) 19.5 kWW Q m s T
b) 3 35.704 0.704 (7.372). 0.678x x
5.71 a) A refrigeration cycle is a reversed power cycle. Heat is added to the R134a from4 to 1 and rejected from 2 to 3:
netnet
253COP 3.614 where
323 253L L L
H L H L
Q T qw s T
Q Q T T w
net [50 ( 20)] [0.901 0.434] 32.7 kJ/kgw
netCOP 3.614 32.7 118 kJ/kgLq w
4 3 4 40.434 0.0996 (0.9332 0.0996). 0.401s s x x c) TK solution:
Rule Sheetq23 = h3 - h2 ; for the constant pressure processq41/ q23 = T4/T3 ; for a Carnot refrigerators4 = s3; R134a tables based on 'Thermodynamic Properties of HFC-134a'; DuPont Technical Information, which is based upon the Modified; Benedict-Webb-Rubin equation of state. R134a8SI.tkw
Variable SheetInput Name Output Unit Comment
` Thermal Sciences, Potter & Scott*R1348si.tkw, R134a, 1-8 States, SI unitsP5-71.tkw Problem 5.71
-20 T1 C Temperaturep1 kPa Pressures1 kJ/(kg*K) Entropyx1 Qualityphase1 Phase
50 T2 C Temperaturep2 1320 kPa Pressureh2 276 kJ/kg Enthalpys2 0.913 kJ/(kg*K) Entropy
1 x2 Qualityphase2 'SAT Phase
50 T3 C Temperaturep3 1320 kPa Pressureh3 124 kJ/kg Enthalpys3 0.442 kJ/(kg*K) Entropy
92
0 x3 Qualityphase3 'SAT Phase
-20 T4 C Temperaturep4 133 kPa Pressure
0.442 s4 kJ/(kg*K) Entropy (transfer value to input)x4 0.402 Quality at beginning of heat addition processphase4 'SAT Phase
*THUNITS.tkw Units for thermoq23 -152 kJ/kg Amount of heat rejectedq41 -119 kJ/kg Amount of heat gained from refrigerated space
5.72 Refer to the numbers in Problem 5.12:8000040 40 60
0 which verifies the inequality of Clasius.509.4 283
H L
H L
Q QT T
5.73 net . 350 (250.4 75.9). 2.006 kJ/kg Kw s T s s
(250.4 273)(2.006) 1050 kJ/kgH Hq T s
(75.9 273)(2.006) 700 kJ/kgL Lq T s 1050 700
0.0002 which is essentially zero.523.4 348.9
H L
H L
q qT T
5.74 Using values from Problem 5.21:30 1.326
0473 20.9
H L
H L
q qT T
.
5.75 a) Using values from Problem 5.66, we have613.5 313.5 300
0, using 613.5 kJ/kg.573 293 0.489
H LH
H L
q qq
T T
b) TK solution:Rule Sheet
;Problem 5.66eta = (T1 - T4) /T1q12 = wnet/ etaq12 = T1 * (s2 - s1) ; the sought value of s1 is shown on the Variable SheetProblem 5.75: The cyclic integral of deltaq/T = q12/T1 + 0 + q34/T3 + 0 , but q12/T1 = - q34T3;therefore , the cyclic integral of delta q/T = 0, as called for by the Clausius inequality for areversible cycle.
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and; George S. Kell, Hemisphere Publishing Corp., 1984. Stm8si.tkw
Variable SheetInput Name Output Unit Comment
` Thermal Sciences, Potter & Scott*Stm8si.tkw Steam, 1-8 States, SI UnitsP5-75.tkw Problem 5.75
300 T1 C Temperature
93
p1 8580 kPa Pressure4.63 s1 kJ/(kg*K) Entropy (transfer value to input)
x1 0.563 Quality300 T2 C Temperature
p2 8580 kPa Pressures2 5.7 kJ/(kg*K) Entropy
1 x2 Quality20 T3 C Temperature
p3 kPa Pressures3 kJ/(kg*K) Entropyx3 Quality
20 T4 C Temperaturep4 kPa Pressures4 kJ/(kg*K) Entropy
*THUNITS.tkw Units for thermoeta 0.489 Thermal efficiency
614 q12 kJ/kg Heat input to cycle300 wnet kJ/kg Net work of cycle
5.76 a) 2 2 2( ) ( ) . 5 0.093(200 ) 10 1.0( 50). 56.7 Fc p c c w p w wm c T m c T T T T
2
1
516.7( ) ln 5 0.093ln 0.1138 Btu/ R
( ) 660c c p cc
TS m c
T
2
1
universe
516.7( ) ln 10 1.0 ln 0.1305 Btu/ R
( ) 510
S 0.1138 0.1305 0.0167 Btu/ R
w w p ww
c w
TS m c
T
S S
b) TK solution:Rule Sheet
;mc * cc * (Tc - T2) = mw * cw * (T2 - T1) ; This is the first law applied to system of copper and; water, under the assumption that there is no heat transfer.delSuniv = delSc + delSwdelSc = mc* cc * ln(T2/Tc) ; from delSc = INT (delQ/T) = INT(m * cc*dT/T)delSw = mw * (s2 -s1)
; Steam tables based on NBS/NRC Steam Tables by Haar, Gallagher,; and Kell, Hemisphere Publishing Corp., 1984. Stm8e.tkw
Variable SheetInput Name Output Unit Comment
` Thermal Sciences, Potter & Scott*Stm8e.tkw Steam, 1-8 States, English unitsP5-76.tkw Problem 5.76
50 T1 F Initial temperature of waterh1 18.1 B/lbm Enthalpys1 0.0361 B/(lbm*R) Entropy
0 x1 Qualityphase1 'SAT Phase
94
T2 56.6 F Final temp of water and copper (starting guess needed)h2 24.7 B/lbm Enthalpys2 0.0489 B/(lbm*R) Entropy
0 x2 Qualityphase2 'SAT Phase
*THUNITS.tkw Units for thermo5 mc lbm Mass of copper0.092 cc B/(lbm*R) Specific heat of copper200 Tc F Initial temperature of copper10 mw lbm Mass of water
delSuniv 0.0159 B/R Entropy change of the universedelSc -0.113 B/R Entropy change of the copperdelSw 0.129 B/R Entropy change of the water
1 cw B/(lbm*R) Specific heat of water
5.77 a) universe 0.264 0.156 0.104 Btu/ RS
31 2
0.2 0.1 m /kg and from Tables C.1 and C.2 we observe that 2 MPa2
v p
and 1 1 1212.4 C, 6.342 kJ/kg K, and 2600 kJ/kgT s u . Since for a rigid volume
2 1v v trial-and-error provides
2 2
2 2
At 0.4 MPa: 0.0011 0.2(0.4625 0.0011) 0.0934At 0.3 MPa: 0.0011 0.2(0.6058) 0.122
p vp v
Obviously, at 2 0.1v state 2 lies between 0.3 and 0.4 MPa. Interpolate:
20.122 0.1
0.1 0.3 0.377 MPa0.122 0.0934
p
Interpolate to find 2 2and :s u
2 21.753 0.2 5.166 2.786 kJ/kg K, 594 0.2 (2551 594) 986 kJ/kgs u Then
Q W 2 1( ) 2 (986 2600) 3230 kJm u u (heat flows to surr.)
universe system3230
2 (2.786 6.342) 3.55 kJ/K30 273surrS m s S
b) TK solution:Rule Sheet
v1 = V1 / m ; Definition of specific volumev2 = v1 ; Volume of steam is constantQ12 = m * (u2 - u1) ;heat added to steamQ12 = - QsurrdelS = m* (s2 - s1) ; Entropy change of steamdelSsurr = Qsurr/Tsurr ; entropy change of surroundings; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and; George S. Kell, Hemisphere Publishing Corp., 1984. Stm8si.tkw
95
Variable SheetInput Name Output Unit Comment
` Thermal Sciences, Potter & Scott*Stm8si.tkw Steam, 1-8 States, SI UnitsP5-77.tkw Problem 5.77
T1 212 C Initial temperature of steamp1 1990 kPa Pressures1 6.34 kJ/(kg*K) Entropy
0.1 v1 m^3/kg Specific Volume (transfer value to input)1 x1 Quality
phase1 'SAT PhaseT2 141 C Temperaturep2 371 kPa Pressures2 2.78 kJ/(kg*K) Entropy
0.1 v2 m^3/kg Specific Volume (transfer value to input)0.2 x2 Quality
phase2 'SAT Phaseu1 2600 kJ/kg Initial internal energyu2 985 kJ/kg Final internal energy
*THUNITS.tkw Units for thermo0.2 V1 m^3 Volume of steam2 m kg Mass of steam
Q12 -3230 kJ Heat added to steamdelS -7.11 kJ/K Entropy change of steamdelSsurr 10.7 kJ/K Entropy change of the surroundingsQsurr 3230 kJ Heat added to surroundings
30 Tsurr C Temperature of the surroundingsdelSuniv 3.54 kJ/K Entropy change of the universe
5.78 1 230 144 6
486.3 R, 1459 R1 53.3
T T
, 1 0.24(1459 486.3) 233 BtuQ
a) air1459
1 0.24ln 0.264 Btu/ R486.3
S
b) surr233
0.156 Btu/ R1460
S
c) universe 0.264 0.156 0.104 Btu/ RS
5.79 1 22 0.287 573 120
164.5 kPa, 573 418 K2.0 164.5
p T
a) air418
2 0.717 ln 0.452 kJ/K573
S
b) universe222
2 0.717(573 418) 222 kJ. S 0.452 0.289 kJ/K300
Q
96
5.80 a) steam3(2793 852) 5821 kJ. S 3(2.331 6.433) 12.31 kJ/KQ m h
universe5821
12.31 7.56 kJ/K293
S
b) TK solution:Rule Sheet
Q12 = m * (h2 - h1) ; from the first law, Q12 = U2 - U1 + W = m* (u2 - u1 + p1 * (v2 - v1); for a quasiequilibrium constant pressure process of a closed system
delSuniv = delSsys+ delSsurrQ12 = -Q12surrdelSsurr = Q12surr/TsurrdelSsys = m * (s2 - s1)
; Steam tables based on NBS/NRC Steam Tables by Lester Haar, John S. Gallagher, and; George S. Kell, Hemisphere Publishing Corp., 1984. Stm8si.tkw
Variable SheetInput Name Output Unit Comment
` Thermal Sciences, Potter & ScottP5-80.tkw Problem 5.80
200 T1 C Temperaturep1 1554 kPa Pressureh1 2793 kJ/kg Enthalpys1 6.431 kJ/(kg*K) Entropyv1 0.1273 m^3/kg Specific Volume
1 x1 Qualityphase1 'SAT Phase
200 T2 C Temperaturep2 1554 kPa Pressureh2 852.4 kJ/kg Enthalpys2 2.331 kJ/(kg*K) Entropyv2 0.001156 m^3/kg Specific Volume
0 x2 Qualityphase2 'SAT Phase
3 m kg Mass of steam in systemdelSuniv 7.553 kJ/K Entropy change of the universedelSsys -12.3 kJ/K Entropy change of the system
delSsurr 19.85 kJ/K Entropy change of thesurroundings
Q12 -5820 kJ Heat added to the systemQ12surr 5820 kJ Heat added to the surroundings
20 Tsurr C Temperature of the surroundings
97
5.81 1 11.53 0.8(5.598) 6.008. 504.5 0.8(2025) 2124 kJ/kgs u
22 2 2
2 1
0.8 MPa650 C, u 3389, 8.391
.0011 .8(.8846) .709p
T sv v
6400 10(3389 2124) 0.714 kJ
0.709Q
64
universe400 10 0.714
(8.391 6.008) 6.11 10 kJ/K0.709 973
S
5.82 The enthalpy leaving the heater equals the enthalpy entering the heater:
262.2(8 ) 1283 8 48.1. 1.678 lbm/secs s sm m m out in
(8 1.678) 0.4273 8 0.0933 1.678 1.768 0.432 Btu/sec- R
S S S
5.83 a) 2 1( ) 2 (2609.7 3658.4) 2000 97.4 kJ/st TQ m h h W
surr 97.4 kJ/sTQ Q
surrprod c.v. 2 1
surr
97.4( ) 0 2(7.909 7.168) 1.80 kW/K
303Q
S S m s sT
b) TK solution:Rule Sheet
Wdot = mdot * (h1 - h2) +Qdot ; first law for steady flow turbine with zero change in ke and peQdot = - QdotsurrdelSuniv = delSsys + delSsurr = 0 + delSsurrdelSsurr = mdot* (s2 - s1) + Qdotsurr/Tsurr
Variable SheetInput Name Output Unit Comment
` Thermal Sciences, Potter & Scott*Stm8si.tkw Steam, 1-8 States, SI UnitsP5-83.tkw Problem 5.83
600 T1 C Temperature6000 p1 kPa Pressure
h1 3660 kJ/kg Enthalpys1 7.17 kJ/(kg*K) Entropyv1 0.0652 m^3/kg Specific Volumex1 'mngless Qualityphase1 'SH PhaseT2 60.1 C Temperature
20 p2 kPa Pressureh2 2610 kJ/kg Enthalpys2 7.91 kJ/(kg*K) Entropy
98
v2 7.65 m^3/kg Specific Volume1 x2 Quality
phase2 'SAT Phase*THUNITS.tkw Units for thermo
2000 Wdot kW Poer output2 mdot kg/s Mass rate of flow
Qdot -98.3 kJ/s Rate of heat transfer to the systemdelSuniv 0.001 kJ/(K*s) Rate of entropy increase of the universe
0 delSsys kJ/(K*s) Rate of entropy increase of the system
delSsurr 1.8 kJ/(K*s) Rate of entropy increase of thesurroundings
Qdotsurr 98.3 kJ/s Rate of heat transfer to the surroundings30 Tsurr C Temperature of the surroundings
5.840.4 /1.4
22 1 2
1
323 100100 118.3 kPa. 323 308 K
273 118.3T
p p TT
3 2 32 ( )pV c T T 2 1000 (323 308) 173.8 m/s
Note: The factor of 1000 converts kJ to J, so the units will work out.
5.850.4 /1.4 2
22 2
100300 272.5 K. 1.0(300 272.5). 234.5 m/s
140 2 1000V
T V 2100
( 0.0125 ) 234.5 0.147 kg/s0.287 272.5
m
Note: We assumed 2 100 kPap since the exiting pressure was not given.
5.860.4 /1.4 2 2
22 2
4085423 374.6 K. 1.0(423 374.6). 309 m/s
130 2 1000V
T V
5.87 a)0.4 /1.4
290
1173 628.4 K. 1.0(628.4 1173) 545 kJ/kg800 TT w
b) 2 2 290
3.152 0.287 ln . 2.525 kJ/kg K. 682 kJ/kg800
h
2 1( ) (682 1246) 564 kJ/kgTw h h
5.88 a) 21 2 1 2
2
800 psia1.636, 1180, 1449 kJ/kg
1.636p
s s h hs
2 1( ) 6(1449 1180) 778 / 550 2280 hpCW m h h Note: The factor 778 converts Btu to ft-lbf, and 550 converts ft-lbf/sec to hp.
99
b) TK solution:
Rule SheetWdotin = mdot * (h2 - h1) ; First law, assuming negligible change in ke and pe and no heattransfers2 = s1 ; for a reversible adiabatic process
; Steam tables based on NBS/NRC Steam Tables by Haar, Gallagher, ; and Kell, HemispherePublishing Corp., 198
Variable SheetInput Name Output Unit Comment
` Thermal Sciences, Potter & Scott*Stm8e.tkw Steam, 1-8 States, English unitsP5-88.tkw Problem 5.88
300 T1 F Temperaturep1 67 psi Pressureh1 1180 B/lbm Enthalpys1 1.64 B/(lbm*R) Entropy
1 x1 Qualityphase1 'SAT PhaseT2 888 F Temperature
800 p2 psi Pressureh2 1450 B/lbm Enthalpy
1.64 s2 B/(lbm*R) Entropy (transfer value to input)x2 'mngless Qualityphase2 'SH Phase
*THUNITS.tkw Units for thermoWdotin 2280 hp Power input to compressor
6 mdot lbm/s Mass flow rate
5.89 2 1 2 27.168 0.649 (7.502). 0.869s s x x
2 192 0.869(2393) 2271 kJ/kg. 2(3658 2271) 2774 kWTh W m h
5.90 a) 2 1 2 21.681 0.175 (1.745). 0.863s s x x
2 94 0.863(1022) 976 Btu/lbm. 1512 976 536 Btu/lbmTh w h
actual3000 550 / 778
382 Btu/lbm20,000 / 3600
TWw
m
actual 3820.713 or 71.3%
536Ts
ww
100
b) TK solution:Rule Sheet
;State 1 = turbine throttle state. State 2 = Ideal (isentropic turbine exhaust state. State 3 = actualturbine exhaust statewi = h1 - h2 ; first law for ideal turbine with negligible changes in ke and pes2 = s1 ; for isentropic turbine. Note that the rule below for a state identified by p2and ;s2 in model Stm8e.tkw is modified so that it is unnecessary to transfer value of s2 fromoutput to input manually..if and (given('p2),known('s2)) then call pands(p2,s2;T2,h2,v2,x2,phase2)wa = Wdot/ mdot ; actual turbine worketat = wa / wi ; definition of turbine efficiency
; Steam tables based on NBS/NRC Steam Tables by Haar, Gallagher,; and Kell, Hemisphere Publishing Corp., 1984. Stm8e.tkw
Variable SheetInput Name Output Unit Comment
` Thermal Sciences, Potter & Scott*Stm8e.tkw Steam, 1-8 States, EnglishP5-90.tkw Problem 5.90
1000 T1 F Temperature800 p1 psi Pressure
h1 1510 B/lbm Enthalpys1 1.68 B/(lbm*R) Entropyx1 'mngless Qualityphase1 'SH PhaseT2 126 F Temperature
2 p2 psi Pressureh2 976 B/lbm Enthalpys2 1.68 B/(lbm*R) Entropyx2 0.863 Qualityphase2 'SAT PhaseT3 F Temperature
2 p3 psi Pressureh3 B/lbm Enthalpys3 B/(lbm*R) Entropyx3 Qualityphase3 Phase
*THUNITS.tkw Units for thermowi 536 B/lbm Work of ideal turbinewa 382 B/lbm Work of actual turbine
3000 Wdot hp Power output of actual turbine20000 mdot lbm/h Mass rate of flow
etat 0.712 Turbine efficiency
101
5.91 2 1 2 2 17.839 0.832 (7.077). 0.990. 3694s s x x h kJ/kg
2 251 0.990(2358) 2585 kJ/kg. 3.5(3694 2585) 3880 kWTh W
5.92 a) 12 2 1
2 1
600 C2666, 1.0, 3677,
7.435T
h x hs s
1 2( ). 200 (3678 2666). 0.198 kg/sTW m h h m m b) TK solution:
Rule SheetWdot = mdot * (h1 - h2) ; First law, assuming negligible change in ke and pes1 = s2 ; for isentropic turbine. Note that the rule below for a state identified by T1 and s1 in;model Stm8e.tkw is altered to make it unnecessary to transfer value of s1 from output to input.if and (given('T1),known('s1)) then call Tands(T1,s1;p1,h1,v1,x1,phase1)
; Steam tables based on NBS/NRC Steam Tables by Haar, Gallagher,; and Kell, Hemisphere Publishing Corp., 1984. Stm8si.tkw
Variable SheetInput Name Output Unit Comment
` Thermal Sciences, Potter & Scott*Stm8si.tkw Steam, 1-8 States, SI uniitsP5-92.tkw Problem 5.92
600 T1 C Temperaturep1 3530 kPa Pressureh1 3680 kJ/kg Enthalpys1 7.43 kJ/(kg*K) Entropyx1 'mngless Qualityphase1 'SH PhaseT2 93.5 C Temperature
80 p2 kPa Pressureh2 2670 kJ/kg Enthalpys2 7.43 kJ/(kg*K) Entropy
1 x2 Qualityphase2 'SAT Phase
*thunits.tkw Units for thermo200 Wdot kW Power output
mdot 0.197 kg/s Mass rate of flow
5.93 1 2 1 2 2 21984 kJ/kg, 6.710 0.261 (8.464). 0.762. 2534ah s s x x h
Then 22923-2534
73.5 .762(2460) 1948 kJ/kg. .399 or 39.9%2923 1948sh
5.94 a) 1 1 2 2 21474, 1.759 0.2198 (1.6426), 0.937h s s x x
2 120.9 0.937(1006) 1064. 0.85(1474 1064) 348 Btu/lbms Th w
b) . 3000 550 / 778 348. 6.09 lbm/secT TW mw m m
102
5.95 The efficiency is the actual kinetic energy increase divided by the maximum possibleincrease:
2 22 12 2
2 10.2857 0.2857
22 1
1
2 2 22 1 2 1 2
2 2
2 2
( )
( )
100293 281.5 K
115
( ) 10 2 1000 (281.5 293). 152 m/s
150 100.97 or 97%
152 10
a a
s s
s
p
KE V VKE V V
pT T
p
V V c T T V
5.96 The 1st law: 2 22 1 1 2 22( ) 2 (2874 ) 1000V V h h h
To find h2 we use s2 = s1 = 7.759 kJ/kg.K and p2 = 100 kPa. Therefore h2 = 2841 kJ/kg.2 2 2 2
2 1 222 2
2 1
( ) 20. 0.85 . 238 m/s
2 (2874 2841) 1000( )a
s
V V VV
V V