thermal science Ch 05

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

QQ

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


q s

c)1360


q s

d)1360


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


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


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


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


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


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



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


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


` 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


50 T3 C Temperaturep3 1320 kPa Pressureh3 124 kJ/kg Enthalpys3 0.442 kJ/(kg*K) Entropy

92


-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


;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



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


;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


` 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


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


*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


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



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


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


` 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


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


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


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




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


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


` 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


;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


` 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


` 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



*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

thermal science Ch 05

Technology