Problem as Term o

7/24/2019 Problem as Term o

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INSTITUTO TECNOLGICO DE PACHUCA

MAESTRA EN INGENIERA MECANICA

FUNDAMENTOS DE INGENIERA MECNICA(TERMOFLUDOS)

UNIDAD II CICLOS TERMODINMICOS

ASESOR:

DR. ABDIEL GMEZ MERCADO

YUNUN LPEZ GRIJALBA

30 NOVIEMBRE 2009


2/28

917E An air-standard cycle with variable specific heats is executed in a closed system

and is composed of the following four processes:1-2 v = constant heat addition from 14.7 psia and 80F in the amount of 300 Btu/lbm2-3 P = constant heat addition to 3200 R3-4 Isentropic expansion to 14.7 psia4-1 P= constant heat rejection to initial state(a) Show the cycle on P-v and T-s diagrams.(b) Calculate the total heat input per unit mass.(c) Determine the thermal efficiency.

Consideraciones:

Se aplica la condicin de are-estndar

Las energas cintica y potencial son despreciables, considerando un procesoreversible.

Se considera el are como gas ideal con calor especfico variable.

Diagramas P-V y T-s

V

P

1

2 3

4

Qin

Qin

Qout

14.7psi

57.61psi


3/28

T

1

23

4Qin

Qin

Qout340R

3400R

Partimos del Edo. 1 al Edo. 2 para obtener los datos para ste ltimo:

T1 = 80F = 540R

u1 = 92.04 BTU/lbm

h1 = 129.06 BTU/lbm

Edo.2 de la Tabla A-17E

T2 = 2116.04 R

u2 = 392.04 BTU/lbm

h2 = 537.103 BTU/lbm

Considerando el are como un gas ideal tenemos que:

y ;

Para el Edo. 3

T3 = 3200 R

u3 = 630.12 BTU/lbm

h3 = 849.48 BTU/lbm

Pr3 = 1242

u2 Q12 u1 300 BTU

lbm92.04

BTU

lbm392.04

BTU

lbm

P2V2

T2

P1V1

T1 V2 V1

P2

T2

P1

T1P2

P1 T2

T1

14.7 psi 2116.27 R

540 R57.6095 psi


4/28

Para el Edo.4

h4 = 593.176 BTU/lbm

Pr4 = 316.9165

Del balance de energa, tenemos que,

La eficiencia trmica del sistema es:

923 An air-standard Carnot cycle is executed in a closed system between thetemperature limits of 350 and 1200K. The pressures before and after the isothermal

compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and(c) the mass of air. Assume variable specific heats for air.Consideraciones:


Las energas cintica y potencial son despreciables, considerando un proceso

reversible.


Diagramas T-s

Pr4

P4

Pr3

P3Pr4

Pr3 P4

P3

1242 14.7 psi

57.6095 psi316.9165; P1 P4 y P3 P2

Q23in h3 h2 849.28 BTU

lbm537.103

BTU

lbm312.377

BTU

lbm

Qintotal Q12 Q23 300BTU

lbm

312.377BTU

lbm

612.377BTU

lbm

Qout h4 h1 593.176 BTU

lbm129.06

BTU

lbm464.116

BTU

lbm

TH 1 Qout

Qin1

464.116 BTUlbm

612.377 BTUlbm

0.2421 x 100 % 24.21 %


5/28

s

T

350R

1200R

1

2 3

4

Qin

Qout

De las tablas A-17 obtenemos los siguientes valores para obtener la presin

mxima del sistema, tenemos que;

T1 = 1200 K; Pr1 = 238 T4 = 350K; Pr4 = 2.379

La transferencia de calor del are es:

Para la masa del are:

964E An ideal Ericsson engine using helium as the working fluid operates betweentemperature limits of 550 and 3000 R and pressure limits of 25 and 200 psia. Assuming amass flow rate of 14lbm/s, determine (a) the thermal efficiency of the cycle, (b) theheat transfer rate in the regenerator, and (c) the power delivered.Consideraciones:


P1Pr1 P4

Pr4

238 300 kPa

2.37430, 012.61 kPa 30.012 MPa

Qinwnet

THTH 1

TL

TH1

350 K

1200 K0.7083 x 100 % 70.83 % ; Qin

0.5kJ

0.70830.7059 kJ

m Wnet

wnet; Wnet s2 s1 TH TL ;

considerandoque: s2 s1 s4 s3 y que; s4 s3 s4

s3

R lnP4

P3

s4 s3 0 0.287kJ

kg Kln

300 kPa

150 kPa0.1989

kJ

kg K

Wnet s4 s3 TH TL 0.1989kJ

kg K

1200 K 350 K 169.093kJ

kg

m Wnet

wnet

0.5 kJ

169.093 kJkg

0.002956kg


6/28


reversible.

Diagramas T-s

s

T

550R

3000R

1

23

4

Qin

Qout

.

.

Para el Helio tenemos que R =0.4961 BTU/lbmR; Cp = 1.25 BTU/lbmR

Para la eficiencia trmica:

Calor transferido en la regeneracin es el obtenido en el proceso del estado 4 al1;

Para la potencia liberada o el trabajo neto obtenido en el proceso;

984 A gas-turbine power plant operates on the simple Brayton cycle between thepressure limits of 100 and 1200 kPa. The working fluid is air, which enters thecompressor at 30C at a rate of 150 m3/min and leaves the turbine at 500C. Usingvariable specific heats for air and assuming a compressor isentropic efficiency of 82percent and a turbine isentropic efficiency of 88 percent, determine (a) the net poweroutput, (b) the back work ratio, and (c) the thermal efficiency.

TH 1 TL

TH1

550 R

3000 R0.8166 x 100 % 81.66 %

Qreg Q41in m h1 h4 Cp T1 T4 14 lbm

s 1.25 BTU

lbm R 3000 R 550 R 42 875 BTU

s

wout TH Qin Qin m TH s2 s1 ;

wout TH Qin Qin m TH s2 s1 ; s2 s1 Cp ln T2

T1R ln

P2

P1sabiendoque ln

T2

T10;

s2 s1 0.4961BTU

lbm Rln

25 psi

200 psi1.0316

BTU

lbm R

Qin 14 lbms

3000 R 1.0316 BTUlbm R

43 327.6598 BTUs

wout 0.8166 43327.6598 BTU

s35381.367

BTU

s


7/28

Consideraciones:



reversible.


s

T

550C

30C

2

1

3

4

Qin

Qout

P=12

00kP

a

P=100kPa

V

P

1200kPa

100kPa

3

4

2

1

Qin

Qout

s=cte

s=cte


8/28

1-2Compresin isoentrpica (en el compresor)

2-3Adicin de calor a presin constante

3-4Expansin isonetrpica (turbina)

4-1Rechazo de calor a presin constante

Para el Edo.1

T1 = 30C = 303K

h1 = 303.208 kJ/kg

Pr1 = 1.43556

Para el Edo.2

T2 = 609.2815 K

h2s = 609.2815 kJ/kg

Pr2 = 17.22672

Para el Edo.4

T4 = 500C = 773 K T3 = 1437.345 K

h4s = 792.3825 kJ/kg h3 = 1560.315 kJ/kg

Pr4 = 41.922 Pr3 = 503.064

La razn de trabajo regenerado;

9109 Consider an ideal gas-turbine cycle with two stages of compression and two stagesof expansion. The pressure ratio across each stage of the compressor and turbine is 3.The air enters each stage of the compressor at 300 K and each stage of the turbine at1200 K. Determine the back work ratio and the thermal efficiency of the cycle, assuming

Pr2

P2

Pr1

P1Pr2

Pr1 P2

P1

1200 kPa 1.43556

100 kPa17.22672

Pr4

P4

Pr3

P3Pr3

Pr4 P3

P4

1200 kPa 41.922

100 kPa503.064

wNET wTUB wCOMP T h3 h4sh2s h1

C

wNET 0.88 1560.315 kJ

kg792.3825

kJ

kg

616.7749 kJkg

303.208 kJkg

0.82293.3819

kJ

kg

wTUB 675.7806 kJ

kgwCOMP 382.3986

kJ

kg

rbwwC

wT

382.3980 kJkg

675.7806 kJkg

0.56586

THwNET

Qin

293.3819 kJkg

h3 h4s

293.3819 kJkg

1560.315 kJkg

616.7749 kJkg

0.3109 x 100 % 31.09 %


9/28

an efficiency of 80 percent for each compressor stage and an efficiency of 85 percentfor each turbine stage.Consideraciones:



reversible.


s

T

1200K

300K

4

3

5

10

Qin

2

1

8

7

6

9

Qin

Para el Edo.1T1 = 300 K

h1 = 300.19 kJ/kg = h4

Pr1 = 1.386

T5 = 1200 K

h5 = 1277.79 kJ/kg = h7

Pr5 = 238

h6 = 946.3468 kJ/kg = h8

Pr6 = 79.333

Pr2

P2

Pr1

P1Pr2

Pr1 P2

P13 1.386 4.158

Pr6

P6

Pr5

P5Pr6

Pr5

P6

P5

1

3238 79.333

wNET wTUB wCOMP T 2 h5 h62 h2 h1

C

wNET 0.80 2 1277.79 kJ

kg946.3468

kJ

kg2

411.257 kJkg

300.19 kJkg

0.85268.975

kJ

kg


10/28

9131 A gas-turbine power plant operates on the regenerative Brayton cycle between thepressure limits of 100 and 700 kPa. Air enters the compressor at 30C at a rate of 12.6kg/s and leaves at 260C. It is then heated in a regenerator to 400C by the hotcombustion gases leaving the turbine. A diesel fuel with a heating value of 42,000 kJ/kgis burned in the combustion chamber with a combustion efficiency of 97 percent. Thecombustion gases leave the combustion chamber at 871C and enter the turbine whoseisentropic efficiency is 85 percent. Treating combustion gases as air and using constantspecific heats at 500C, determine (a) the isentropic efficiency of the compressor, (b)the effectiveness of the regenerator, (c) the airfuel ratio in the combustion chamber,(d) the net power output and the back work ratio, (e) the thermal efficiency, and ( f )

the second-law efficiency of the plant. Also determine (g) the second-law (exergetic)efficiencies of the compressor, the turbine, and the regenerator, and (h) the rate of theexergy flow with the combustion gases at the regenerator exit.

Consideraciones:



reversible.


qin h5 h4 h7 h6 1277.79 kJ

kg411.257

kJ

kg1277.79

kJ

kg946.3468

kJ

kg1197.9762

kJ

kg

TH

wNET

Qin

268.975 kJkg

1197.9762 kJkg

0.2245 x 100 % 22.45 %


11/28

s

T

871C

30C

2

1

3

4

Qin

Qout

400C

5Qin

Para los Edo. tenemos;Edo. 1 Edo. 2 Edo. 3 Edo. 4 Edo. 5

T= 30C = 300K T= 260C = 530K T= 871C =1140K

hs=705.438kJ/kg

T=400C=670K

h=300.19 kJ/kg h=533.98 kJ/kg h=1207.57kJ/kg

h=780.758kJ/kg

h=681.14 kJ/kg

P=100 kPa P= 700 kPa P= 700 kPa P= 700 kPa P=100 kPaPr=1.386 Pr=10.37 Pr=193.1 Pr=27.7 Pr=24.46s1=1.70203kJ/kg s2=2.27967kJ/kg

Ch2s h1

h2a h1

523.90157 kJkg

300.19 kJkg

533.98 kJkg

300.19 kJkg

0.956 x 100 % 95.6 %

Pr2

P2

Pr1

P1 Pr2

Pr1 P2

P1

1.386 700 kPa

100 kPa 9.702Pr4

P4

Pr3

P3Pr4

Pr3 P4

P3

193.9 100 kPa

700 kPa27.7

Th3 h4

h3 h4sh4 h3 T h3 h4s 1207.57

kJ

kg0.85 1207.57

kJ

kg705.43

kJ

kg780.758

kJ

kg

h5 h2a

h4a h2a

681.14 kJkg

533.98 kJkg

780.758 kJkg

533.98 kJkg

0.59

wNET h3 h4 h2 h1 1207.57 kJ

kg780.788

kJ

kg533.98

kJ

kg300.19

kJ

kg193.022

kJ

kg

qin h3 h5 1207.57 J

kg681.14

J

kg526.43

J

kg

THw

NETqin

193.022 kJkg

526.43 kJkg

0.37 x 100 % 37 %


12/28

9155 A four-cylinder, four-stroke spark-ignition engine operates on the ideal Otto cyclewith a compression ratio of 11 and a total displacement volume of 1.8 liter. The air is at90 kPa and 50C at the beginning of the compression process. The heat input is 1.5 kJ

per cycle per cylinder. Accounting for the variation of specific heats of air withtemperature, determine (a) the maximum temperature and pressure that occur duringthe cycle, (b) the net work per cycle per cyclinder and the thermal efficiency of thecycle, (c) the mean effective pressure, and (d) the power output for an engine speed of3000 rpm.


13/28

9156 A gas-turbine plant operates on the regenerative Brayton cycle with two stages ofreheating and two-stages of intercooling between the pressure limits of 100 and 1200kPa. The working fluid is air. The air enters the first and the second stages of the

compressor at 300 K and 350 K, respectively, and the first and the second stages of theturbine at 1400 K and 1300 K, respectively. Assuming both the compressor and theturbine have an isentropic efficiency of 80 percent and the regenerator has aneffectiveness of 75 percent and using variable specific heats, determine (a) the backwork ratio and the net work output, (b) the thermal efficiency, and (c) the second-lawefficiency of the cycle. Also determine (d) the exergies at the exits of the combustionchamber (state 6) and the regenerator (state 10) (See Figure 943 in the text).


14/28

1015 A steam power plant operates on a simple ideal Rankine cycle between thepressure limits of 3 MPa and 50 kPa. The temperature of the steam at the turbine inlet is300C, and the mass flow rate of steam through the cycle is 35 kg/s. Show the cycle on

a T-s diagram with respect to saturation lines, and determine (a) the thermal efficiencyof the cycle and (b) the net power output of the power plant.Consideraciones:

Se condiciona el sistema operando en estado estable.

Las energas cintica y potencial son despreciables.

T

2

1

3

4

Qout

Qin

Edo. 1 Edo. 2 Edo. 3 Edo.4h1=340.54 kJ/kg h2=343.5285kJ/kg h3=2994.3kJ/kg h4=2277.3776 kJ/kgv1=0.001030 m3/kg T3=300CP1=50kPa P3=3 MPa P4= 50kPa

s3=65412 kJ/kgK s4=s3

wpin v1 P2 P1 0.001030m3

kg3000 kPa 50 kPa

lkJ

1 kPa.m33.0485

kJ

kg

h2 h1 wpin 340.49kJ

kg3.0385

kJ

kg343.5285

kJ

kg

x4s4 sf

sfg

6.5912 kJkg K

1.0912 kJkg K

6.5019 kJkg K

0.8382

h4 hf x4 hfg 340.54 kJ

kg0.8382 2304.7

kJ

kg2277.3776

kJ

kg

qin h3 h2 2994.3kJ

kg

343.5285kJ

kg

2650.7715 kJ

kgqout h4 h1 2277.3776

kJ

kg340.54

kJ

kg1936.8376

kJ

kg

wNET qin qout 2650.7715kJ

kg1936.8376

kJ

kg713.9339

kJ

kg

wNET m wNET 35kg

s713.9339

kJ

kg24987.6865

kJ

s24.98 MW


15/28

1046 A steam power plant operates on an ideal regenerative Rankine cycle with twoopen feedwater heaters. Steam enters the turbine at 10 MPa and 600C and exhausts to

the condenser at 5 kPa. Steam is extracted from the turbine at 0.6 and 0.2 MPa. Waterleaves both feedwater heaters as a saturated liquid. The mass flow rate of steamthrough the boiler is 22 kg/s. Show the cycle on a T-s diagram, and determine (a) thenet power output of the power plant and (b) the thermal efficiency of the cycle.

T

7

10

6

45kPa

0.2MPa

0.6MPa

10MPa

54

32

1

8

9

1-y

1-y-z

Edo.1 Edo. 2 Edo. 3 Edo.4 Edo.5h1=kJ/kg h2=kJ/kg h3=kJ/kg h4=kJ/kg h5=kJ/kgP1=5kPA P3=0.2MPA P5=0.6MPAv1=m3/kg v3=m3/kg v5=m3/kgEdo.6 Edo.7 Edo.8 Edo.9 Edo.10

h6=kJ/kg P7=10MPa P8=0.6MPa P9=0.6MPa P10=5kPah7=kJ/kg s8=s7 s9=s7 s10=s7T7=600C x8=0.9627 h9=kJ/kg h10=kJ/kgs7=kJ/kgK h8=kJ/kg x9= x10=

wpin v1 P2 P1 0.001005m3

kg200 kPa 5 kPa

lkJ

1 kPa.m30.195975

kJ

kg

h2 h1 wpin 137.75kJ

kg0.195975

kJ

kg137.9459

kJ

kg

wpin

v3

P4

P3

0, 001061m3

kg600 kPa 200 kPa

lkJ

1 kPa.m30.4244

kJ

kg

h4 h3 wpin 04.71kJ

kg0.4244

kJ

kg505.1344

kJ

kg

wpin v5 P6 P5 0.001101 m3

kg10000 kPa 600 kPa

lkJ

1 kPa.m310.3494

kJ

kg

h6 h5 wpin 670.38kJ

kg10.3494

kJ

kg680.7294

kJ

kg

x9s9 sf

sfg

6.9045 kJkg K

1.9308 kJkg K

4.8285 kJkg K

0.9699


16/28


kg0.9699 2085.8

kJ

kg2693.4569

kJ

kg

x10 s10 sf

sfg

6.9045 kJ

kg K

0.4762 kJ

kg K

7.9176 kJkg K

0.8119

h10 hf x10 hfg 137.75 kJ

kg0.8119 2423.0

kJ

kg2104.9838

kJ

kg

Ein Eout m8 h8 h2m2 m5 h5 yh8 1 y h4 h5 ym8

m5

yh3 h2

h9 h2

504.71 kJkg

137.9459 kJkg

2693.4569 kJkg

137.9459 kJkg

0.14469

mi hi me he m9 h9 h2m2 m3 h3 zh9 1 y z h2 1 y h3 z m4m5

z h3 h2

h9 h21 y

504.71 kJkg

137.9459 kJkg

2693.4569 kJkg

137.9459 kJkg

1 0.14469 0.12375

qin h7 h6 3625.8kJ

kg680.2794

kJ

kg2945.0706

kJ

kg

qout 1 y z h10 h1 1 0.14469 .12375 2104.9838kJ

kg137.75

kJ

kg1439.144

kJ

kg


kg

1439.144kJ

kg

1505.9265kJ

kg

WNET m wNET 22 kg

s1505.9265

kJ

kg33130.38507kW

TH 1 qout

qin1

1439.144 kJkg

2945.0706 kJkg

0.5113 X100 % 51.13 %


17/28

1050 A steam power plant operates on an ideal reheatregenerative Rankine cycle andhas a net power output of 80 MW. Steam enters the high-pressure turbine at 10 MPa and550C and leaves at 0.8 MPa. Some steam is extracted at this pressure to heat the

closed feedwater heater. The rest of the steam is reheated to 500C and is expanded inthe low-pressure turbine to the condenser pressure of 10 kPa. Show the cycle on a T-sdiagram with respect to saturation lines, and determine (a) the mass flow rate of steamthrough the boiler and (b) the thermal efficiency of the cycle. Assume that thefeedwater leaves the heater at the condensation temperature of the extracted steamand that the extracted steam leaves the heater as a saturated liquid and is pumped tothe line carrying the feedwater.

Consideraciones:

Se condiciona el sistema operando en estado estable.

Las energas cintica y potencial son despreciables.

s

T

2

1

7

8

10MPa

6

5

10

3

49

1-y

0.8MPa

10kPa


18/28

Edo.1 Edo. 2 Edo. 3 Edo.4 Edo.5h1=191.81 kJ/kg h2=201.89

kJ/kgh3=720.87 kJ/kg h4=731.128

kJ/kgh5=3375.1kJ/kg

P1=10kPA P3=0.8MPA h4=h9=h10 P5=10MPAv1=0.001010m3/kg v3=0.001115m3/kg s5=6.5995

kJ/kgK=s6Edo.6 Edo.7 Edo.8P6=0.8MPa P7=0.8MPa P8=10kPah6=2811.9 kJ/kg h7=3481.3

kJ/kgs8=s7

T7=500C x8=0.9627s7=7.8692kJ/kgK

h8=2494.727 kJ/kg

wpin v1 P2 P1 0.001010 m3

kg10000kPa 10 kPa

lkJ

1 kPa.m310.0899

kJ

kg

h2 h1 wpin 191.81 J

kg10.0899

J

kg201.8999

J

kg

wpin v3 P4 P3 0.001115 m

kg10000kPa 800kPa

lkJ

1 kPa.m310.258

kJ

kg

h4 h3 wpin 720.87kJ

kg10.258

kJ

kg731.128

kJ

kg

x8s8 sf

sfg

7.8692 kJkg K

0.6492 kJkg K

7.4996 kJ

kg K

0.9627


kg0.9627 2392.1

kJ

kg2494.727

kJ

kg

Ein Eout m2 h9 h2 m3 h6 h3 1 y h6 h2 y h6 h3 ym3

m4

y h9 h2

h6 h3 h9 h2

731.128 kJkg

201.8999 kJkg

2811.9 kJkg

720.87 kJkg

731.128 kJkg

201.8999 kJkg

0.20197

qin h5 h4 1 y h7 h6 3375.1kJ

kg731.128

kJ

kg1 0.201 3481.3

kJ

kg2811.9

kJ

kg3178.173

qout 1 y h8 h1 1 0.201 2494.727 J

kg198.81

J

kg1837.796

J

kg

wNET qin qout 3178.173 kJ

kg1837.796

kJ

kg1340.377

kJ

kg

mWNET

wNET

80000 kJs

1340.377 kJkg

59.6847kg

s

TH 1 qout

qin1

1837.796 kJkg

3178.173 kJkg

0.4234 X100 % 42.34 %


19/28

1060 Determine the exergy destruction associated with the regenerative cycle describedin Prob.1044. Assume a source temperature of 1500 K and a sink temperature of 290 K.1044 A steam power plant operates on an ideal regenerative Rankine cycle. Steam

enters the turbine at 6 MPa and 450C and is condensed in the condenser at 20 kPa.Steam is extracted from the turbine at 0.4 MPa to heat the feedwater in an openfeedwater heater. Water leaves the feedwater heater as a saturated liquid. Show thecycle on a T-s diagram, and determine (a) the net work output per kilogram of steamflowing through the boiler and (b) the thermal efficiency of the cycle.

2

1 7

6MPa

6

5

3

4

1-y

0.4MPa

20kPa


kJ/kg

h3=604.66 kJ/kg h4=610.7304

kJ/kg

h5=3302.9

kJ/kgP1=20kPA P3=0.4MPa P5=6MPav1=0.001017m3/kg v3=0.001080m3/kg s5=6.7219

kJ/kgK=s6=s7Edo.6 Edo.7 T=450CP6=0.4MPa P7=20kPah6=2665.67 kJ/kg h7=2213.971

kJ/kgx7=0.8324

wpin v1 P2 P1 0.001017m3

kg400 kPa 20 kPa

lkJ

1 kPa.m30.38646

kJ

kg

h2

h1

wpin

251.42kJ

kg0.38646

kJ

kg251.80646

kJ

kg

wpin v3 P4 P3 0.001080 m3

kg6000 kPa 400kPa

lkJ

1 kPa.m36.0704

kJ

kg

h4 h3 wpin 604.66kJ

kg6.0704

kJ

kg610.7304

kJ

kg

x6s6 sf

sfg

6.7219 kJkg K

1.7765 kJkg K

5.1191 kJkg K

0.966

h6 hf x6 hfg 604.66kJ

kg0.966 2133.4

kJ

kg2665.67

kJ

kg


20/28

1065C Consider a cogeneration plant for which the utilization factor is 0.5. Can the

exergy destruction associated with this plant be zero? If yes, under what conditions?

S, si el ciclo no involucra irreversibilidades tales como: desaceleracin, friccin, ytransferencia de calor a travs de una diferencia finita de temperatura.

x7s7 sf

sfg

6.7219 kJkg K

0.8320 kJkg K

7.0752 kJkg K

0.8324


kg0.8324 2357.5

kJ

kg2213.971

kJ

kg

Ein Eout m6 h6 h2m2 m3 h3 yh6 1 y h2 h3 y m6

m3

y h3 h2

h6 h2

604.66 kJkg

251.8064 kJkg

2665.67 kJkg

251.8064 kJkg

0.14617

qin h5 h4 3302.9kJ

kg610.7304

kJ

kg2692.1696

kJ

kg

qout 1 y h7 h1 1 0.14617 2213.971

kJ

kg 251.42

kJ

kg 1675.669

kJ

kg


kg1675.669

kJ

kg1016.5006

kJ

kg

TH 1 qout

qin1

1675.669 kJkg

2692.1696 kJkg

0.3775 X100% 37.75 %

iCICLO T0qOUT

TL

qIN

TH290 K

1675.669 kJkg

290 K

2692.1696 kJkg

1500 K1155.1828

kJ

kg


21/28

1093 Consider a steam power plant that operates on a regenerative Rankine cycle andhas a net power output of 150 MW. Steam enters the turbine at 10 MPa and 500C andthe condenser at 10 kPa. The isentropic efficiency of the turbine is 80 percent, and that

of the pumps is 95 percent. Steam is extracted from the turbine at 0.5 MPa to heat thefeedwater in an open feedwater heater. Water leaves the feedwater heater as asaturated liquid. Show the cycle on a T-s diagram, and determine (a) the mass flow rateof steam through the boiler and (b) the thermal efficiency of the cycle. Also, determinethe exergy destruction associated with the regeneration process. Assume a sourcetemperature of 1300 K and a sink temperature of 303 K.

s

T

2

1 7

10MPa

6s

5

3

4

1-y

0.5MPa

10kPa

7s

6


kJ/kgh3=640.09 kJ/kg h4=651.02

kJ/kgh5=3375.1kJ/kg

P1=10kPA s2=1.8604kJ/kgK P3=0.5MPa P5=10MPav1=0.001010m3/kg P2=10kPa v3=0.001093m3/kg

s3=0.6492kJ/kgKs5=6.5995kJ/kgK=s6=s7s

Edo.6 Edo.7 T=500CP6=0.4MPa P7s=10kPah6=2798.288 kJ/kg h7=2346.81

kJ/kgx6=0.9554 x7=0.7934


22/28

h2 h1 wpin 191.81kJ

kg0.5209

kJ

kg192.3304

kJ

kg

wpin v1 P2 P1 0.001010m3

kg500kPa 10 kPa

lkJ

1 kPa.m30.5209

kJ

kg

wpinv3 P4 P3

th

0.001093 m3

kg

10000 kPa 500 kPa

.95

lkJ

1 kPa.m310.93

kJ

kg

h4 h3 wpin 640.09kJ

kg10.93

kJ

kg651.02

kJ

kg

x6s6 sf

sfg

6.5995 kJkg K

1.8604 kJkg K

4.9603 kJkg K

0.9554


kg0.9554 2108.0

kJ

kg2798.2884

kJ

kg

x7ss7s sf

sfg

66.5995 kJkg K

0.6492 kJkg K

7.4996 kJkg K

0.7934

th5 h7

h5 h7sh7 h5 t h5 h7s 3375.1

kJ

kg0.8 3375.1

kJ

kg2089.739

kJ

kg5 2346.8116

kJ

kg

Ein Eout m6 h6 h2m2 m3 h3 yh6 1 y h2 h3 ym6

m3

y h3 h2

h6 h2

640.09 kJkg

192.3304

2654.0856 kJkg

192.3304 kJkg

0.1818

qin h5 h4 3375.1kJ

kg651.02

kJ

kg2724.08

kJ

kg

qout 1 y h7 h1 1 0.1818 2346.8116

kJ

kg 191.81

kJ

kg 1763.036

kJ

kg


kg1763.036

kJ

kg961.0437

kJ

kg

TH 1 qout

qin1

1763.036 kJkg

2724.08 kJkg

0.3528 X100 % 35.28 %

m WNET

wNET

150000 kJs

961.0437 kJkg

156.08kg

s

iregen T0 me se mi siqalr

TL

T0 s3 ys6 1 y s2

iregen 303 K 1.8604 kJ

kgK0.1818 6.9308

kJ

kgK1 0.1818 0.6492

kJ

kgK120.8048

kJ

kg


23/28

1095 Consider an ideal reheatregenerative Rankine cycle with one open feedwaterheater. The boiler pressure is 10 MPa, the condenser pressure is 15 kPa, the reheaterpressure is 1 MPa, and the feedwater pressure is 0.6 MPa. Steam enters both the high-

and low-pressure turbines at 500C. Show the cycle on a T-s diagram with respect tosaturation lines, and determine (a) the fraction of steam extracted for regeneration and(b) the thermal efficiency of the cycle.

s

2

1 9

10MPa

5

3

40.6MPa

15kPa

6

7

8

1MPa


kJ/kgh3=670.38 kJ/kg h4=680.7294

kJ/kgh5=3375.1kJ/kg

P1=15kPA P3=0.6MPa P5=10MPav1=0.001014m3/kg

v3=0.001101m3/kg

s5=6.5995kJ/kgK=s6=s7s

Edo.6 Edo.7 Edo. 8 Edo. 9 T=500CP6=1MPa P7=1MPa P8=0.6MPa P7s=15kPah6=2783.8124kJ/kg

h7=3479.1kJ/kg

h7=3310.1569kJ/kg

h7=2518.7838 kJ/kg

s7=7.7642m3/kg

s8=s7=s9 x9=0.966506

wpin v1 P2 P1 0.001014m3

kg600 kPa 15 kPa

lkJ

1 kPa.m30.59319

kJ

kg

h2 h1 wpin 225.94kJ

kg0.59319

kJ

kg226.53319

kJ

kg

wpinv3 P4 P3

th

0.001101 mkg

10000 kPa 600 kPa lkJ1 kPa.m3

10.3494 kJkg

h4 h3 wpin 670.38kJ

kg10.3494

kJ

kg680.7294

kJ

kg

x9s9 sf

sfg

7.7642 kJkg K

0.7549 kJkg K

7.2522 kJkg K

0.966506

h9 hf x9 hfg 225.94kJ

kg0.9665064 2372.3

kJ

kg2518.7838

kJ

kg


24/28

1112 A refrigerator uses refrigerant-134a as the working fluid and operates on an idealvapor-compression refrigeration cycle between 0.12 and 0.7 MPa. The mass flow rate ofthe refrigerant is 0.05 kg/s. Show the cycle on a T-s diagram with respect to saturationlines. Determine (a) the rate of heat removal from the refrigerated space and the powerinput to the compressor, (b) the rate of heat rejection to the environment, and (c) thecoefficient of performance.

s

T

2

1

0.7MPa3

40.12MPa

4sQL.

QH.

Win.

Edo.1 Edo. 2 Edo. 3 Edo.4

h1=236.97 kJ/kg h2=273.5377 kJ/kg h3=88.82 kJ/kg h4h3

P1=0.12MPA P2=0.7MPA P3=0.7MPa

s1=0.94779 kJ/kgK s2=s1

Ein Eout m8 h8 h2m2 m3 h3 yh8 1 y h2 h3 y m8

m3

y h3 h2

h8 h2

670.38

kJ

kg 226.53319

kJ

kg

3310.1569 kJkg

226.53319 kJkg

0.143936

qin h5 h4 h7 h6 3375.1kJ

kg680.7299

kJ

kg3479.1

kJ

kg2783.8124

kJ

kg3389.6577

kJ

kg

qout 1 y h9 h1 1 0.143936 2518.7838 kJ

kg225.94

kJ

kg1962.81428

kJ

kg

TH 1 qout

qin1

1962.81428 kJkg

3389.6577 kJkg

0.4209 X100 % 42.09 %

QL m h1 h4 0.05 kg

s236.97

kJ

kg88.82

kJ

kg7.4075 kW

Win m h2 h1 0.05kg

s273.5377

kJ

kg236.97

kJ

kg1.828 kW


25/28

QH QL Win 7.4075kW 1.828kW 9.2355kW

COPR

QL

Win

7.4075 kW

1.828 kW 4.0522


26/28

1132 A heat pump using refrigerant-134a heats a house by using underground water at8C as the heat source. The house is losing heat at a rate of 60,000 kJ/h. Therefrigerant enters the compressor at 280 kPa and 0C, and it leaves at 1 MPa and 60C.

The refrigerant exits the condenser at 30C. Determine (a) the power input to the heatpump, (b) the rate of heat absorption from the water, and (c) the increase in electricpower input if an electric resistance heater is used instead of a heat pump.

T

2

1

1MPa3

40.28MPa

4sQL.

QH.

Win.

30C

60C

casa

0C

Edo.1 Edo. 2 Edo. 3 Edo.4

h1=250.83 kJ/kg h2=293.38 kJ/kg h3=93.58 kJ/kg h4h3

P1=280kPA P2=1MPA P3=1MPa

T1=0C T2=60C T3=30C

mR QHqH

60000

3600

kJ

S

293.38 kJkg

93.58 kJkg

0.0834 kgs

Win m h2 h1 0.0834kg

s293.38

kJ

kg250.83

kJ

kg3.54938 kW

QL m h1 h4 0.0834kg

s250.83

kJ

kg93.58

kJ

kg13.1172 kW

We QH3600

J

S16.6666 kW

Wincrem We Win 16.6666 k 3.54938 kW 13.11728kW


27/28

1196 Consider a two-stage compression refrigeration system operating between thepressure limits of 0.8 and 0.14 MPa. The working fluid is refrigerant-134a. The

refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamberoperating at 0.4 MPa. Part of the refrigerant evaporates during this flashing process, andthis vapor is mixed with the refrigerant leaving the low-pressure compressor. Themixture is then compressed to the condenser pressure by the high-pressure compressor.The liquid in the flash chamber is throttled to the evaporator pressure, and it cools therefrigerated space as it vaporizes in the evaporator. Assuming the refrigerant leaves theevaporator as saturated vapor and both compressors are isentropic, determine (a) thefraction of the refrigerant that evaporates as it is throttled to the flash chamber, (b) theamount of heat removed from the refrigerated space and the compressor work per unitmass of refrigerant flowing through the condenser, and (c) the coefficient ofperformance.

s

4

1

0.8MPa5

80.14MPa

qL

0.4MPa76

2

9

h1=239.16 kJ/kg h2=256.58 kJ/kg h3=255.55 kJ/kg h5h695.47kJ/kg

h7=63.94 kJ/kgh8 h9=256.4105 kJ/kg

x6h6 hf

hfg

95.47 kJkg

63.94 kJkg

191.62 kJkg

0.16454

Ein Eout me he mi hi

h9 x6 h3 1 x6 h2 0.16454 255.55 kJ

kg1 0.16454 256.58

kJ

kg256.4105

k

k

qL 1 x6 h1 h8 1 0.1645 239.16 kJ

kg63.94

kJ

kg146.389

kJ

kg


28/28

win 1 0.1645 256.58kJ

kg239.16

kJ

kg279.3185

kJ

kg256.4105

kJ

kg37.4617

kJ

kg

COPR qL

win

146.389 J

kg

37.4617 kJkg

3.9076

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