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CANKAYA UNIVERSITY

FACULTY OF ENGINEERING AND ARCHITECTURE

MECHANICAL ENGINEERING DEPARTMENT

ME 212 THERMODYNAMICS II

CAHPTER 9

EXAMPLES SOLUTION

1) An air-standard Otto cycle has a compression ratio of 8.5. At the beginning of

compression, .K300TandkPa100p 11 The heat addition per unit mass of air is 1400

kJ/kg. Determine

a) The net work, in kJ per kg of air.

b) The thermal efficiency of the cycle.

c) The mean effective pressure, in kPa

d) The maximum temperature in the cycle, in K

To investigate the effects of varying compression ratio, plot each of the quantities

calculated in parts (a) through (d) for compression ratios ranging from 1 to 12.

Solution:

Known: An air-standard Otto cycle has a known compression ratio and a specified state

at the beginning of compression. The heat addition per unit mass of air is given.

Find: Determine (a) the net work per unit mass of air, (b) the thermal efficiency, (c) the

mean effective pressure, and (d) the maximum cycle temperature. Plot each of these

quantities versus compression ratio.

Schematic & Given Data:

Assumptions:

1) The air in the piston-cylinder assembly is the closed system.

2) The compression and expansion processes are adiabatic.

3) All processes are internally reversible

4) The air is modeled as an ideal gas

5) Kinetic and potential energy effects are negligible

Analysis: Begin by fixing each principle state of the cycle

State 1:

2.621v,kg/kJ07.214uK300T,kPa100p 1r111

State 2:

For isentropic compression

kg/kJ06.503u,K2.688T,Thus

082.735.8

2.621

V

Vvv

22

1

21r2r

State 3:

The specific internal energy 3u is found by using the energy balance for process 2-3

1

2

4

2

3

2

2

s = c

s = c

v

V1/V2 = 8.5

kg/kJ1400m

Q23

p1 = 100 kPa

T1 = 300 K

air

9192.1v,K3.2231T,Thus

kg

kJ06.1903

kg

kJ06.503

kg

kJ1400u

m

Qu

WQ)uu(m

3r3

2

23

3

0

232323

State 4:

For the isentropic expansion

kg/kJ91.892u,K3.1154T,Finally

3132.16)5.8)(9192.1(V

Vv

V

Vvv

44

2

13r

3

43r4r

(a) To find the net work, note that so,QW cyclecycle

kg

kJ16.721)07.21491.892(1400

)uu(m

Q

m

Q

m

Q

m

W14

234123cycle

(b) The thermal efficiency is

%)5.51(515.0)kg/kJ(1400

)kg/kJ(12.721

m/Q

m/W

23

cycle

(c) The displacement volume is ),vv(mVV 2121 so the mean effective pressure is

given by

)v/v1(v

m/W

vv

m/W

VV

Wmep

121

cycle

21

cycle

21

cycle

Evaluating 1v

kPa2.949

m/N10

kPa1

kJ1

m.N10

5.8

11

kg

m861.0

)kg/kJ12.721(mep

Thus

kg/m861.0kJ1

m.N10

m/N10

kPa1

)kPa100(

)K300(kgK

kJ

97.28

314.8

p

RTv

23

3

3

33

23

1

1

1

Plotting for compression ranging from 1 to 12,

3) The pressure and temperature at the beginning of compression of an air-standard

Diesel cycle are 95 kPa and 290 K, respectively. At the end of the heat addition, the

pressure is 6.5 MPa and the temperature is 2000 K.

Determine

a) The compression ratio.

b) The cut off ratio.

c) The thermal efficiency of the cycle.

d) The mean effective pressure, in kPa.

Solution:

Known: An air-standard Diesel cycle has a specified state at the beginning of

compression and a known pressure and temperature at the end of heat addition.

Find: Determine (a) the compression ratio, (b) the cut off ratio, (c) the thermal

efficiency, and (d) the mean effective pressure

Schematic and Given Data:

Assumptions:



3) All processes are internally reversible.

4) The air is modeled as an ideal gas.

5) Kinetic and potential energy effects are negligible.

Analysis: Begin by fixing each principal state in the cycle.

State 1:

2311.1p,1.676v,kg/kJ91.206ukPa95p,K290T 1r1r111

State 2: For the isentropic compression

v

p2 = p3 = 6.5 MPa = 6500 kPa

T3 = 2000 K

p1 = 95 kPa

T1 = 290 K

1

4

2

Air

3 p

kg/kJ19.962h,K926T,59.31v233.8495

6500)2311.1(

p

ppp 222r

1

21r2r

State 3:

776.2v,kg/kJ1.2252hkPa6500P,K2000T 3r333

State 4:

For the isentropic expansion,

kg/kJ36.734u,K971T48.27vV

Vv

9.92000

926

59.31

1.676

T

T

v

v

T

T

V

V

V

V

V

V

V

V

443r

3

44r

3

2

2r

1r

3

2

2

1

3

2

2

1

3

4

(a) The compression ratio is

4.2159.31

1.676

v

v

V

Vr

2r

1r

2

1

b) The cut off ratio

16.2926

2000

T

T

V

Vr

2

3

2

3

c

c) The thermal efficiency is

%)1.59(591.091.1289

46.762

)19.9621.2252(

)91.20636.734()19.9621.2252(

hh

)uu()hh(

m/Q

m/W

23

1423

23

cycle

d) The mean effective pressure is given as

kPa913

m/N10

kPa1

kJ1

Nm10

4.21

11

kg

m8761.0

)kg/kJ46.762(mep

Thus

kg/m8761.0kJ1

Nm10

m/N10

kPa1

)kPa95(

)K290(K.kg

kJ

97.28

314.8

P

RTv

vevaluating

)v/v1(v

m/W

VV

Wmep

23

3

3

33

23

1

11

1

121

cycle

21

cycle

6) An air-standard dual cycle has a compression ratio of 9. At the beginning of

compression, K300TandkPa100p 11 . The heat addition per unit mass of air

is 1400 kJ/kg, with two thirds added at constant volume and the rest at constant

pressure.

Determine

a) The temperatures at the end of each heat addition process, in K.

b) The net work of the cycle per unit mass of air, in kJ/kg

c) The thermal efficiency.

d) The mean effective pressure, in kPa.

Solution:

Known: An air-standard dual cycle has a known compression ratio and a specified state

at the beginning of compression. The heat additions at constant volume and constant

pressure are also given.

Find: Determine (a) the temperatures at the end of each heat addition process, (b) the net

work per unit mass, (c) the thermal efficiency, and (d) the mean effective pressure.


Assumptions:




4) The air is modeled as an ideal gas.


Analysis: Begin by fixing principal state of the cycle

5

2

3 4

v

Qin/m = 1400 kJ/kg

Q23/m = 933.3 kJ/kg

Q34/m = 466.7 kJ/kg

V1/V2 = 9

p1 = 100 kPa

T1 = 300 K

1

Air

P

State 1:

2.621v,kg/kJ07.214uK300T 1r11

State 2:

For the isentropic compression, 022.69v)V/V(v 1r122r

Thus, kg/kJ4.514uandK7.702T 22

State 3: For the heat addition process from 2 to 3:

kg/kJ7.14474.5143.933um/Qu

or

WQ)uu(m

2233

0

232323

(a) Thus, kg/kJ1952handK5.1758T 33

State 4:

For the heat addition process from 3 to 4:

237.2vandK8.2132T,Thus

kg/kJ7.24187.4661952m/Qhh

or

hhm/Q

4r4

3434

3434

State 5: For the isentropic expansion

5997.16vT

T

V

Vv

V

V

V

Vv)V/V(v 4r

4

3

2

14r

4

3

2

14r455r

kg/kJ24.887uandK8.1147T,Thus 55

(b) For the cycle, Thus.QW cyclecycle

kg/kJ83.726)07.21424.887(1400

uum/Qm/Qm/Qm/Qm/W 15in513423cycle


%)9.51(519.01400

83.726

m/Q

m/W

in

cycle

d) The mean effective pressure is given by

kPa7.949

m/N10

kPa1

kJ1

Nm10

)9/11)(kg/m86096.0(

)kg/kJ83.726(mep

Thus

kg/m86096.0kJ1

Nm10

m/N10

kPa1

)kPa100(

)K300(kgK

kJ

97.28

314.8

P

RTv

vEvaluating

)v/v1(v

m/W

VV

Wmep

23

3

3

33

23

1

11

1

121

cycle

21

cycle

6) Air enters the compressor of an ideal air-standard Brayton cycle at 100 kPa, 300 K,

with a volumetric flow rate of 5 m3/s. The compressor pressure ratio is 10. For turbine

inlet temperatures ranging from 1000 to 1600 K, plot

a) The thermal efficiency of the cycle.

b) The back work ratio.

c) The net power developed, in kW.

Solution:

Known: Air enters the compressor of an ideal Brayton cycle with known conditions. The

compressor pressure ratio is also known.

Find: Plot for various turbine inlet temperatures (a) the thermal efficiency, (b) the back

work ratio, (c) the net power developed.


Assumptions:

1) Each component is analyzed as a control volume at steady state. The control

volumes are shown on the accompanying sketch by dashed lines.

2) The turbine and compressor processes are isentropic

3) There are no pressure drops for flow through the heat exchangers.


5) The working fluid is air modeled as an ideal gas.

Analysis: Sample calculations are given below for several turbine inlet temperatures.

First, fix each of the principal states.

P1 = 100 kPa

T1 = 300 K

(AV)1 = 5 m3/s

p2/p1 = 10 T3 = 1000 K

T3 = 1200 K

T3 = 1600 K

Compressor

T3’

T3

4

3

4’

3’

2

1

s

T

cycleW

4

3 2

1

outQ

Heat

Exchanger

Turbine

Heat

Exchanger

inQ

State 1: 3860.1pr,kg/kJ19.300hK300T 111

State 2: For the isentropic compression, 860.13prp/pp 1122r

Thus, kg/kJ86.579handK1.574T 22

State 3:

2.791pr,kg/kJ57.1757hK1600

0.238pr,kg/kJ79.1277hK1200

0.114pr,kg/kJ04.1046hK1000

T

33

33

33

3

State 4: For the isentropic expansion, 443344 h,Tprp/ppr

Thus

)K1600(kg/kJ65.945h,K3.911T,12.79

)K1200(kg/kJ84.675h,K665T,8.23

)K1000(kg/kJ45.548h,K9.543T,4.11

pr

44

44

44

4

(a) The thermal efficiency is

23

14

hh

hh1

(b) The back work ratio is

4343

12

hh

67.279

hh

hhbwr

c) The power is

67.279)hh(RT

p)AV(hhhhmW 43

1

111243net

The results are summarized in the plots, which give each of these quantities versus

turbine inlet temperatures ranging from 1000 to 1600K.

7) The compressor and turbine of a simple gas turbine each have isentropic efficiencies of

82%. The compressor pressure ratio is 12. The minimum and maximum temperatures are

290 K and 1400 K, respectively. On the basis of an air-standard analysis, compare the

values of

a) The net work per unit mass of air flowing, in kJ/kg.

b) The heat rejected per unit mass of air flowing, in kJ/kg, and

c) The thermal efficiency to the same quantities evaluated for an ideal cycle.

Solution:

Known: An air-standard gas turbine cycle has a known compressor pressure ratio and

specified minimum and maximum temperatures. The compressor and turbine each have

isentropic efficiencies of 82%.

Find: Determine (a) the net work per unit mass of air flow, (b) the heat rejected per unit

mass of air flow, and (c) the thermal efficiency and compare them to the same quantities

evaluated for an ideal cycle.


Assumptions:

1) Each component is analyzed as a control volume at steady state.

2) The compressor and turbine are adiabatic.




Analysis: First, fix each of the principal states for each cycle using data from Tables.

State 1:

2311.1p,kg/kJ16.290hK290T 1r11

State 2: First, determines2h . For isentropic compression

82.0c

T1 = 290 K

82.0t

T3 = 1400 K p2/p1 = 12

T1 = 290 K 1

2 2s

3

4 4s

T3 = 1400 K

s

T

Compressor

cycleW

4

3 2

1

outQ

Heat

Exchanger

Turbine

Heat

Exchanger

inQ

kg/kJ47.590h7732.14pp/pp s21r12s2r

Using the compressor efficiency; ;hh/hh 121s2c

kg/kJ39.656hh

hhc

1s2

12

State 3;

5.450p,kg/kJ42.1515hK1400T 3r33

State 4: First, determine s4h . For isentropic expansion

kg/kJ38.768h542.37Pp/pp s43r34s4r

Using the turbine efficiency; ;hh/hh s4343t

kg/kJ85.902h)hh(hh 4s43t34

(a) The net work per unit mass of air flowing is

kg/kJ3.246

kg

kJ)16.29039.656(

kg

kJ)85.9024.1515(hhhh

m

W1243

cycle

For the ideal cycle

kg/kJ7.446hhhhm

W1s2s43

ideal

cycle

Thus, irreversibilities reduce the net work by nearly one half

(b) The heat rejected per unit mass of air flowing is

kg/kJ7.612kg

kJ16.290

kg

kJ85.902hh

m

Q14

out

For the ideal cycle

kg/kJ2.478hhm

Q1s4

ideal

out

Thus, less heat is rejected for the ideal cycle.


%)7.28(287.039.6564.1515

7.6121

hh

hh1

23

14

For the ideal cycle

%)3.48(483.047.59042.1515

2.4781

hh

hh1

s23

1s4

Thus, irreversibilities cause a substantial decrease in thermal efficiency.

8) Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric

flow rate 5 m3/s. The compressor pressure ratio is 10 and its isentropic efficiency is 85%.

At the inlet to the turbine, the pressure is 950 kPa, and temperature is 1400 K. The

turbine has an isentropic efficiency of 88% and the exit pressure is 100 kPa. On the basis

of an air-standard analysis, evaluate all availability inputs, destructions, and losses. Let T0

= 300 K, p0 = 100 kPa.

Solution:

Known: A simple gas turbine is analyzed on an air-standard basis from an availability

viewpoint. Data are known at various locations.

Find: Determine all availability inputs, destructions, and losses.

Schematic and given data:

p1 = 100 kPa

T1 = 300 K

s/m5AV 3

1

p4 = 100 kPa

p = 1000 kPa

p = 100 kPa

85.0c

88.0t

T3 = 1400 K

p3 = 950 kPa p2/p1 = 10

950 K

1

2 2s

3

4 4s

s

T

Compressor

cycleW

4

3 2

1

outQ

Heat

Exchanger

Turbine

Heat

Exchanger

inQ

Assumptions:


2) The compressor and turbine are adiabatic.



5) Let kPa100pandK300T 00

Analysis: Data are obtained for each principal state from Tables

State T(K) P(kPa) h(kJ/kg) 0s (kJ/kg.K)

1 300 100 300.19 1.70203

2 621.1 1000 629.21 2.44539

3 1400 950 1515.42 3.36200

4 873.9 100 903.72 2.81558

The increase in flow availability of the air passing through the heat exchanger is taken as

the net input of availability to the gas turbine.

Input: 23

0

2

0

30232f3f p/plnRss(T)hh(m)ee(m

Evaluating :m

s/kg807.5Nm10

kJ1

kPa1

m/N10

)K300(K.kg97.28

kJ314.8

)kPa100)(s/m5(

RT

p)AV(m

3

233

1

11

)Input(kW3524)ee(m

s/kJ1

kW1

kgK

kJ

1000

950ln

97.28

314.844539.236200.3)K300(

kg

kJ)21.62942.1515(

s

kg807.5)ee(m

2f3f

2f3f

Availability is destroyed due to irreversibilities in the compressor and turbine. Thus

Destructions:

kW6.173p/plnR)ss(mTI

kW8.143p/plnR)ss(mT)ss(mTI

34

0

3

0

40turb

12

0

1

0

20120comp

The net power developed by the cycle represents the output of availability from the cycle,

or

)Output(kW5.1641)hh()hh(mW 1243cycle

Finally, the air enters the gas turbine at00 Tandp . Thus, the change in flow availability

from inlet to exit represents the loss due to the hot air being discharged. Thus,

kW8.1564)p/plnRss(T)hh(m)ee(m:Loss 14

0

1

0

40141f4f

Summarizing

Input: 3524 kW

Disposition

Output: 1641.5 kW

Destroyed: 317.4 Kw

Loss: 1564.8 kW

3523.7 kW

9) A regenerative gas turbine power plant is shown in Figure. Air enters the compressor

at 1 bar, 270C and is compressed to 4 bars. The isentropic efficiency of the compressor is

80%, and the regenerator effectiveness is 90%. All the power developed by the higher-

pressure turbine is used to run the compressor and the lower-pressure turbine provides the

net power output of 97 kW. Each turbine has an isentropic efficiency of 87% and the

temperature at the inlet to the high-pressure turbine is 1200 K. Determine

a) The mass flow rate of air into the compressor, in kg/s.

b) The thermal efficiency

c) The temperature of the air at the exit of the regenerator, in K.

Solution:

Known: In a regenerative gas turbine power plant, a high pressure turbine runs the

compressor and the net power output is provided by a low pressure turbine. Data are

known at various locations.

Find: Determine (a) the mass flow of air into the compressor, (b) the thermal efficiency,

and (c) the temperature of the air at the exit of the regenerator.


1 bar, 270C

inQ

T4 = 1200 K

netW Low pressure turbine

High pressure

turbine

Combustor

Compressor

P2 = 4 bar

1 bar

Regenerator

6

3

4

5 1

2

7

Assumptions:


2) The compressors, turbines, and regenerator are adiabatic.




Analysis: First, fix each of the principal states.

kW97Wnet 87.0t

8.0c

9.0reg

1 bar, 270C

inQ

T4 = 1200 K

Low pressure turbine

High pressure

turbine

Combustor

Compressor

p2 = 4 bar

1 bar Regenerator

6

3

4

5 1

2

7

4 bars

T1 = 300 K

6s 6

5s 5

4

3 2

1

2s

s

T

7

T4 = 1200 K

State 1:

3860.1pr,kg/kJ19.300h,K300T 111

State 2:

For an isentropic compression, kg/kJ49.446h544.5pr)p/p(pr s2112s2

Using the compressor efficiency,

kg/kJ06.483hh

hhhh

hh

c

1s2

12

12

1s2

c

State 4:

238pr,kg/kJ79.1277hK1200T 444

State 5:

The work of the compressor and the high pressure turbine are equal. Thus,

kg/kJ92.1094hhhhh 55412

From the turbine efficiency; kg/kJ59.1067h)hh(hh s5s54t54

Thus, bars065.2pp/ppand86.122p 44rs5r5s5r

From 714.134pkg/kJ92.1094h 5r5

State 6: For isentropic expansion through the low pressure turbine

kg/kJ40.896h237.65)714.134)(065.2/1(pp/pp s65r56s6r

Using the turbine efficiency

kg/kJ21.922)hh(hh s65t56

State 3: Now using the regenerator effectiveness

kg/kJ3.878hhhhhh

hh226reg3

26

23

reg

a) the mass flow rate is the same for each component. Thus, for the low pressure

turbine

s/kg562.0kW1

s/kJ1

kg/kJ)21.92292.1094(

kW97

)hh(

Wm

or

)hh(mW

65

net

65net

b) The thermal efficiency is

%)2.43(432.0)3.87879.1277)(562.0(

97

)hh(m

W

Q

W

34

net

in

net

c) The specific enthalpy 7h is found from an energy balance for the regenerator,

kg/kJ97.526kg

kJ21.922

kg

kJ3.878

kg

kJ06.483hhhh

or

)hh(hh0

6327

7632

Thus, from Table; K2.523T7

10) Air enters the turbine of a gas turbine at 1200 kPa, 1200 K, and expands to 100 kPa

in two stages. Between the stages, the air is reheated at a constant pressure of 350 kPa to

1200 K. The expansion through each turbine stage is isentropic. Determine, in kJ per kg

of air flowing

a) The work developed by each stage

b) The heat transfer for the reheat process

c) The increase in net work as compared to a single stage of expansion with no

reheats.

Solution:

Known: Air expands in two stages through a turbine with reheat between the stages. The

states are specified at the inlet and exit of each component.

Find: Determine per unit mass of air flowing, (a) the work developed by each stage, (b)

the heat transfer for reheat, and (c) the increase in net work compared to a single stage of

expansion with no reheat.


Assumptions:

1) Each component volume is at steady state.

2) The turbines operate isentropically.




State 1: 0.238p,kg/kJ79.1277hK1200T 1r11

State 1: kg/kJ11.912h417.69p)p/p(p 21r122r

State 3: 0.238pp,kg/kJ79.1277hhK1200T 1r3r133

State 4: kg/kJ85.906h68p)p/p(p 43r344r

a) The work developed by each stage is

4

a

2

3 1

s

T

1200 K

100 kPa

350 kPa

1200 kPa

inQ

4 p4 = 100 kPa

netW

p2 = p3 = 350 kPa

T3 = 1200 K

3 2 1

reheater

Turbine

2 Turbine

1

p1 = 1200 kPa

T1 = 1200 K

kg/kJ94.370hhm/W

kg/kJ68.365hhm/W

432t

211t

b) For the reheater

kg/kJ7.365)11.91279.1277(hhm

Q23

in

c) To determine the work for a single stage of expansion, determineah , as follows.

%2.151002.639

21.639)94.37068.365(increase%

and

kg/kJ21.639)hh(m/WThus

kg/kJ58.638h833.19PrP/PPr

a1

a11aa

11) A two-stage air compressor operates at steady state, compressing 10m3/min of air

from 100 kPa, 300 K, to 1200 kPa. An intercooler between the two stages cools the air to

300 K at a constant pressure of 350 kPa. The compression processes are isentropic.

Calculate the power required to run the compressor, in kW, and compare the result to the

power required for isentropic compression from the same inlet state to the same final

pressure.

Solution:

Known: Air is compressed into a two-stage compressor with intercooling between the

stages. Operating pressures and temperatures are given.

Find: Determine the power to run the compressor and compare this to the power required

for isentropic compression from the same inlet state to the same final pressure.


Assumptions:

T1 = 300 K 1

c d

3 2 p

v

p1 = 100 kPa

pi = 350 kPa

p2 = 1200 kPa

cW

p2 = 1200 kPa

2

(AV)1 = 10 m3/min

T1 = 300 K

p1 = 100 kPa

Intercooler

outQ

Td = 300 K

d c 1

Compressor

stage 1

Compressor

stage 2

1) The compressor stages and intercooler are analyzed as control volumes at steady state.

The control volumes are shown on the accompanying sketch by dashed lines.

2) The compression processes are isentropic.

3) There is no pressure drop for flow through the intercooler. Kinetic and potential

energy effects are negligible. The air is modeled as an ideal gas.


State 1: 3860.1p,kg/kJ19.300hK300T 1r11

State c: kg/kJ77.429h851.4p)p/p(p c1r1crc

State d: 3860.1p,kg/kJ19.300hK300TT rdd1d

State 2: kg/kJ21.427h752.4p)p/p(p 2rdd22r

The mass flow rate is

s/kg1936.0Nm10

kJ1

kPa1

m/N10

s60

min1

)K300(kgK

kJ

97.28

314.8

)kPa100min)(/m10(

RT

p)AV(m

3

233

1

11

The compressor power is

kW68.49W

)19.30021.427)(1936.0(s/kJ1

kW1

kg

kJ)19.30077.429(

s

kg1936.0W

)hh(m)hh(mWWW

c

c

d21c2c1cc

To find the power a single stage of compression, determine 3h as follows:

kg/kJ65.610h632.16pp/pp 31r123r

Thus, kW11.60)hh(mW 13

The decrease in power with two-stage, intercooled compression is

%35.1710011.60

68.4911.60decrease%

12) Air at 22 kPa, and 220 K, and 250 m/s enters a turbojet engine in flight at an altitude

of 10,000 m. The pressure ratio across the compressor is 12. The turbine inlet

temperature is 1400 K, and the pressure at the nozzle exit is 22 kPa. The diffuser and

nozzle processes are isentropic efficiencies of 85 and 88%, respectively, and there is no

pressure drop for flow through the combustor. On the basis of an air-standard analysis,

determine

a) The pressures and temperatures at each principal state, in kPa and K, respectively.

b) The velocity at the nozzle exit, in m/s

Neglect kinetic energy except at the diffuser inlet and the nozzle exit.

Solution:

Known: A turbojet engine is analyzed as an air standard basis. Data are known at various

locations.

Find: Determine (a) the pressures and temperatures at each principal state and (b) the

velocity at the nozzle exit.


Assumptions:

1) Each component is analyzed as a control volume at steady state. The control

volumes are shown on the accompanying sketch by dashed lines.

2) There is no pressure drop for flow through the combustor.

3) The turbine work output is just sufficient to drive the compressor.

4) Except at the inlet and exit of the engine, kinetic energy effects can be ignored.

Potential energy effects are negligible throughout.



State a: 4690.0p,kg/kJ97.219hkPa22p,K220T raaaa

State 1: An energy balance for the diffuser gives; .h2

Vh0 1

2

a

a Thus,

7454.0pkg/kJ22.251Nm10

kJ1

s/kgm1

N1

2

s/m250

kg

kJ97.219

2

Vhh 1r32

2222

a

a1

Since the process is isentropic, kPa97.34pp/pp ara1r1

State 2: For isentropic compression, 9453.8pp/pp 1r122r

.kg/kJ9.511h s2 With the compressor efficiency

kPa64.419p,kPa12p,kg/kJ9.557hh

hh 12

c

1s2

12

State 3: 5.450p,kg/kJ42.1515hpp,K1400T 3r3233

State 4: For a turbojet, m/Wm/W tc . Thus

kg/kJ92.1166h)hh()hh( s4s43t12

Thus, kPa62.158)p/p(ppand28.170p 3rs4r34s4r

And

81.193pkg/kJ74.1208hhhh 4rs43t34

State 5:

kg/kJ48.699h;881.26p/ppp 5454r5r

b) An energy balance on the nozzle gives

s/m1009

N1

s/kgm1

kJ1

Nm10

kg

kJ)48.69974.1208(2)hh(2V

or

2

Vhh0

23

545

2

5

54

13) A combined gas turbine-vapor power plant has a net power output of 10 MW. Air

enters the compressor of the gas turbine at 100 kPa, 300 K, and is compressed to 1200

kPa. The isentropic efficiency of the compressor is 84%. The conditions at the inlet to the

turbine are 1200 kPa and 1400 K. Air expands through the turbine, which has an

isentropic efficiency of 88%, to a pressure of 100 kPa. The air then passes through the

interconnecting heat exchanger, and is finally discharged at 480 K. Steam enters the

turbine of the vapor power cycle at 8 MPa, 4000C, and expands to the condenser pressure

of 8 kPa. Water enters the pump as saturated liquid at 8 kPa. The turbine and pump have

isentropic efficiencies of 90 and 80%, respectively. Determine

a) The mass flow rates of air and water, each in kg/s.

b) The rate of heat transfer to the combined cycle, in MW.

c) The thermal efficiency of the combined cycle.

Solution:

Known: A combined gas turbine-vapor power plant has a known net power output. Data

are known at various locations in both cycles.

Find: Determine (a) The mass flow rates of water and air, (b) the rate of heat transfer to

the combined cycle, and (c) the overall thermal efficiency.


Assumptions:


2) The gas turbine is analyzed on an air-standard basis.

3) The turbines, compressor, pump, and interconnecting heat exchanger operate

adiabatically.


5) The working fluid for the gas turbine is air modeled as an ideal gas.

Analysis: First, fix each of the principal states. For the gas turbine cycle:

State 1: 3860.1p,kg/kJ19.300hK300T 1r11

State 2: For isentropic compression,

kg/kJ65.610h632.16pp/pp s21r122r

Using the compressor efficiency,

kg/kJ78.669hh

hhc

1s2

12

State 3: 5.450p,kg/kJ42.1515hK1400T 3r33

State 4: kg/kJ858h88.0,kg/kJ38.768h542.37pp/pp 4ts43r344r

State 5: kg/kJ49.482hK480T 55

For the vapor cycle;

State a: kgK/kJ3634.6s,kg/kJ3.3138hMPa8p,C400T aaa

0

a

State b: kg/kJ9.1989h,7557.0xss,kPa8p bsbsabsb

Thus, with the turbine efficiency, kg/kJ74.2104)hh(hh bsatab

State c: ,kPa8p c saturated liquid kg/kJ88.173hc

State d:

)pp(vhh cdccds

kg/kJ96.183hh

hh

kg/kJ94.181h

Nm10

kJ1

bar1

m/N10bars)08.080(

kg

m100084.1

kg

kJ88.173h

p

cds

cd

ds

3

2533

ds

a) To find the mass flow rates of water stm and air airm , begin with energy and mass

balances on the interconnecting heat exchanger

1217.0)kg/kJ(96.183)kg/kJ(3.3138

)kg/kJ(49.482)kg/kJ(858

hh

hh

m

m

or

)hh(m)hh(m0

da

54

air

st

adst54air

For the gas turbine cycle

)kg/kJ83.287(mhhhhmW air1243airgas

and for the vapor cycle

s/kg041.3m)1271.0(m

s/kg93.23

kg

kJ)5.1023)(1271.0(

kg

kJ)83.287(

s/kJ10000m

)5.1023(m

m)83.287(mW,Thus

)5.1023(mhhhhmW

airst

air

air

st

airnet

stcdbastvap

b) The rate of heat addition to the combined cycle is

MW24.20Q

s/kJ10

MW1

kg

kJ)78.66942.1515(

s

kg93.23hhmQ

in

323airin

c) The overall thermal efficiency is

%)4.49(494.024.20

10

Q

W

in

net

14) Air enters the compressor of an Ericsson cycle at 300 K, 1 bar, with a mass flow rate

of 5 kg/s. The pressure and temperature at the inlet to the turbine are 10 bar and 1400 K,

respectively. Determine

a) The net power developed, in kW.

b) The thermal efficiency.

c) The back work ratio.

Solution:

Known: Air is the working fluid in an Ericsson cycle with data known at various

locations.

Find: Determine a) the net power developed, b) the thermal efficiency, c) the backwork

ratio


Assumptions:



3) The compression and expansion processes are isothermal.


5) The air behaves as an ideal gas.

Analysis: a) The turbine power is evaluated using

kW4626s/kJ1

kW1

10

1ln)K1400(

kgK

kJ

97.28

314.8)s/kg5(W

)p/pln(RTmvdpmW

t

121

2

1t

and for the compressor

kW2.991s/kJ1

kW1

1

10ln)K300(

kgK

kJ

97.28

314.8)s/kg5(W

)p/pln(RTmvdpmW

c

343

4

3c

kW3635WWW ctcycle

b) The thermal efficiency is

%)6.78(786.01400

3001

T

T1

1

3

Alternatively, from an energy balance on the turbine;

tin WQ

Thus,

786.04626

3635

Q

W

in

cycle

c) The backwork ratio is

214.04626

2.991

W

Wbwr

t

c

15) Consider an ideal gas-turbine cycle with two stages of compression and two stages of

expansion. The pressure ratio across each stage of compressor and turbine is 3. The air

enters each stage of the compressor at 300 K and each stage of the turbine at 1200 K.

Determine the back work ratio and the thermal efficiency of the cycle, assuming

a) no regenerator is used and

b) a regenerator with 75 percent effectiveness is used. Use constant specific heats at

room temperature

Solution:

Assumptions:

1) The air standard assumptions are applicable.

2) Air is an ideal gas with variable specific heats.

3) Kinetic and potential energy changes are negligible.

Analysis: a) The work inputs to each stage of compressor are identical, so are the work

outputs of each stage of the turbine since this is an ideal cycle. Then

kg/kJ26.411hh)386.1)(3(pp

pp

386.1p

kg/kJ19.300hK300T

421r

1

22r

1r

11

238p

kg/kJ79.1277hhK1200T

5r

755

1200 K

300 K

T

8

7

6

5

4

9

3

2

1

s

%8.36kg/kJ96.1197

kg/kJ72.440

q

W

kg/kJ72.44014.22286.662WWW

kg/kJ96.1197)36.94679.1277()26.41179.1277()hh()hh(q

%5.33kg/kJ86.662

kg/kJ14.222

W

Wr

,Thus

kg/kJ86.662)36.94679.1277(2)hh(2W

kg/kJ14.222)19.30026.411(2)hh(2W

kg/kJ36.946hh33.79)238(3

1p

p

pp

in

net

th

in,cToutnet

6745in

out,T

in,c

bw

65out,T

12in,c

865r

5

6

6r

b) When a regenerator is used, bwr remains the same. The thermal efficiency in this case

becomes

%3.55kg/kJ63.796

kg/kJ72.440

q

W

kg/kJ63.796kg

kJ33.401

kg

kJ96.1197qqq

kg/kJ33.401kg

kJ)26.41136.946)(75.0()hh(q

in

net

th

regenold,inin

49regregen

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