The temperature at the inlet of each turbine and the thermal efficiency of the cycle.

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Answer 1

Question

Chapter 9, Problem 128P

To determine

The temperature at the inlet of each turbine and the thermal efficiency of the cycle.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 20 kPa.

Pressure of steam at the boiler (P3) is 5000 kPa.

Pressure of steam at the reheat section (P4) is 1200 kPa.

Temperature of water at the turbine inlet (T3) is 450°C.

Quality of steam at the exit of the turbines (x) is 0.96.

Calculation:

Draw the T−s diagram of the cycle as in Figure (1).

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 128P

The pressures are constant for the process 4 to 5 and process 6 to 1.

P4=P5P6=P1

The entropies are constant for the process 3 to 4 and process 5 to 6.

s3=s4s5=s6

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 20 kPa.

h1=hf@20 kPa=251.42 kJ/kgv1=vf@20 kPa=0.0010172 m3/kg

Calculate the work done by the pump during process 1-2(wp,in).

wp,in=v1(P2−P1)=(0.0010172 m3/kg)(5000 kPa−20 kPa)=(0.0010172 m3/kg)(5000 kPa−20 kPa)(1 kJ1 kPa⋅m3)=5.065 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=251.42 kJ/kg+5.065 kJ/kg=256.49 kJ/kg

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1200 kPa and quality of 0.96.

hf=798.33 kJ/kghfg=1985.4 kJ/kgsf=2.2159 kJ/kg⋅Ksfg=4.3058 kJ/kg⋅K

Calculate the enthalpy at state 4(h4).

h4=hf+x4hfg=798.33 kJ/kg+(0.96)(1985.4 kJ/kg)=2704.3 kJ/kg

Calculate the entropy at state 4(s4).

s4=sf+x4sfg=2.2159 kJ/kg⋅K+(0.96)(4.3058 kJ/kg⋅K)=6.3495 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 5000 kPa and entropy of 6.3495 kJ/kg⋅K.

h3=3006.9 kJ/kgT3=327.2°C

Thus, the temperature at the inlet of the high-pressure turbine is 327.2°C.

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 20 kPa and quality of 0.96.

hf=251.42 kJ/kghfg=2357.5 kJ/kgsf=0.8320 kJ/kg⋅Ksfg=7.0752 kJ/kg⋅K

Calculate the enthalpy at state 6(h6).

h6=hf+x6hfg=251.42 kJ/kg+(0.96)(2357.5 kJ/kg)=2514.6 kJ/kg

Calculate the entropy at state 6(s6).

s6=sf+x6sfg=0.8320 kJ/kg⋅K+(0.96)(7.0752 kJ/kg⋅K)=7.6242 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 1200 kPa and entropy of 7.6242 kJ/kg⋅K.

h5=3436.0 kJ/kgT5=481.1°C

Thus, the temperature at the inlet of the low-pressure turbine is 481.1°C.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h6−h1(h3−h2)+(h5−h4)

=1−2514.6 kJ/kg−151.42 kJ/kg(3006.9 kJ/kg−256.49 kJ/kg)+(3436.0 kJ/kg−2704.3 kJ/kg)=1−2263.2 kJ/kg3482.0 kJ/kg=0.35=35%

Thus, the thermal efficiency of the cycle is 35%.

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Answer 2

Question

Chapter 9, Problem 128P

To determine

The temperature at the inlet of each turbine and the thermal efficiency of the cycle.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 20 kPa.

Pressure of steam at the boiler (P3) is 5000 kPa.

Pressure of steam at the reheat section (P4) is 1200 kPa.

Temperature of water at the turbine inlet (T3) is 450°C.

Quality of steam at the exit of the turbines (x) is 0.96.

Calculation:

Draw the T−s diagram of the cycle as in Figure (1).

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 128P

The pressures are constant for the process 4 to 5 and process 6 to 1.

P4=P5P6=P1

The entropies are constant for the process 3 to 4 and process 5 to 6.

s3=s4s5=s6

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 20 kPa.

h1=hf@20 kPa=251.42 kJ/kgv1=vf@20 kPa=0.0010172 m3/kg

Calculate the work done by the pump during process 1-2(wp,in).

wp,in=v1(P2−P1)=(0.0010172 m3/kg)(5000 kPa−20 kPa)=(0.0010172 m3/kg)(5000 kPa−20 kPa)(1 kJ1 kPa⋅m3)=5.065 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=251.42 kJ/kg+5.065 kJ/kg=256.49 kJ/kg

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1200 kPa and quality of 0.96.

hf=798.33 kJ/kghfg=1985.4 kJ/kgsf=2.2159 kJ/kg⋅Ksfg=4.3058 kJ/kg⋅K

Calculate the enthalpy at state 4(h4).

h4=hf+x4hfg=798.33 kJ/kg+(0.96)(1985.4 kJ/kg)=2704.3 kJ/kg

Calculate the entropy at state 4(s4).

s4=sf+x4sfg=2.2159 kJ/kg⋅K+(0.96)(4.3058 kJ/kg⋅K)=6.3495 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 5000 kPa and entropy of 6.3495 kJ/kg⋅K.

h3=3006.9 kJ/kgT3=327.2°C

Thus, the temperature at the inlet of the high-pressure turbine is 327.2°C.

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 20 kPa and quality of 0.96.

hf=251.42 kJ/kghfg=2357.5 kJ/kgsf=0.8320 kJ/kg⋅Ksfg=7.0752 kJ/kg⋅K

Calculate the enthalpy at state 6(h6).

h6=hf+x6hfg=251.42 kJ/kg+(0.96)(2357.5 kJ/kg)=2514.6 kJ/kg

Calculate the entropy at state 6(s6).

s6=sf+x6sfg=0.8320 kJ/kg⋅K+(0.96)(7.0752 kJ/kg⋅K)=7.6242 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 1200 kPa and entropy of 7.6242 kJ/kg⋅K.

h5=3436.0 kJ/kgT5=481.1°C

Thus, the temperature at the inlet of the low-pressure turbine is 481.1°C.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h6−h1(h3−h2)+(h5−h4)

=1−2514.6 kJ/kg−151.42 kJ/kg(3006.9 kJ/kg−256.49 kJ/kg)+(3436.0 kJ/kg−2704.3 kJ/kg)=1−2263.2 kJ/kg3482.0 kJ/kg=0.35=35%

Thus, the thermal efficiency of the cycle is 35%.

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Answer 3

Question

Chapter 9, Problem 128P

To determine

The temperature at the inlet of each turbine and the thermal efficiency of the cycle.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 20 kPa.

Pressure of steam at the boiler (P3) is 5000 kPa.

Pressure of steam at the reheat section (P4) is 1200 kPa.

Temperature of water at the turbine inlet (T3) is 450°C.

Quality of steam at the exit of the turbines (x) is 0.96.

Calculation:

Draw the T−s diagram of the cycle as in Figure (1).

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 128P

The pressures are constant for the process 4 to 5 and process 6 to 1.

P4=P5P6=P1

The entropies are constant for the process 3 to 4 and process 5 to 6.

s3=s4s5=s6

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 20 kPa.

h1=hf@20 kPa=251.42 kJ/kgv1=vf@20 kPa=0.0010172 m3/kg

Calculate the work done by the pump during process 1-2(wp,in).

wp,in=v1(P2−P1)=(0.0010172 m3/kg)(5000 kPa−20 kPa)=(0.0010172 m3/kg)(5000 kPa−20 kPa)(1 kJ1 kPa⋅m3)=5.065 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=251.42 kJ/kg+5.065 kJ/kg=256.49 kJ/kg

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1200 kPa and quality of 0.96.

hf=798.33 kJ/kghfg=1985.4 kJ/kgsf=2.2159 kJ/kg⋅Ksfg=4.3058 kJ/kg⋅K

Calculate the enthalpy at state 4(h4).

h4=hf+x4hfg=798.33 kJ/kg+(0.96)(1985.4 kJ/kg)=2704.3 kJ/kg

Calculate the entropy at state 4(s4).

s4=sf+x4sfg=2.2159 kJ/kg⋅K+(0.96)(4.3058 kJ/kg⋅K)=6.3495 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 5000 kPa and entropy of 6.3495 kJ/kg⋅K.

h3=3006.9 kJ/kgT3=327.2°C

Thus, the temperature at the inlet of the high-pressure turbine is 327.2°C.

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 20 kPa and quality of 0.96.

hf=251.42 kJ/kghfg=2357.5 kJ/kgsf=0.8320 kJ/kg⋅Ksfg=7.0752 kJ/kg⋅K

Calculate the enthalpy at state 6(h6).

h6=hf+x6hfg=251.42 kJ/kg+(0.96)(2357.5 kJ/kg)=2514.6 kJ/kg

Calculate the entropy at state 6(s6).

s6=sf+x6sfg=0.8320 kJ/kg⋅K+(0.96)(7.0752 kJ/kg⋅K)=7.6242 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 1200 kPa and entropy of 7.6242 kJ/kg⋅K.

h5=3436.0 kJ/kgT5=481.1°C

Thus, the temperature at the inlet of the low-pressure turbine is 481.1°C.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h6−h1(h3−h2)+(h5−h4)

=1−2514.6 kJ/kg−151.42 kJ/kg(3006.9 kJ/kg−256.49 kJ/kg)+(3436.0 kJ/kg−2704.3 kJ/kg)=1−2263.2 kJ/kg3482.0 kJ/kg=0.35=35%

Thus, the thermal efficiency of the cycle is 35%.

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Answer 4

Question

Chapter 9, Problem 128P

To determine

The temperature at the inlet of each turbine and the thermal efficiency of the cycle.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 20 kPa.

Pressure of steam at the boiler (P3) is 5000 kPa.

Pressure of steam at the reheat section (P4) is 1200 kPa.

Temperature of water at the turbine inlet (T3) is 450°C.

Quality of steam at the exit of the turbines (x) is 0.96.

Calculation:

Draw the T−s diagram of the cycle as in Figure (1).

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 128P

The pressures are constant for the process 4 to 5 and process 6 to 1.

P4=P5P6=P1

The entropies are constant for the process 3 to 4 and process 5 to 6.

s3=s4s5=s6

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 20 kPa.

h1=hf@20 kPa=251.42 kJ/kgv1=vf@20 kPa=0.0010172 m3/kg

Calculate the work done by the pump during process 1-2(wp,in).

wp,in=v1(P2−P1)=(0.0010172 m3/kg)(5000 kPa−20 kPa)=(0.0010172 m3/kg)(5000 kPa−20 kPa)(1 kJ1 kPa⋅m3)=5.065 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=251.42 kJ/kg+5.065 kJ/kg=256.49 kJ/kg

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1200 kPa and quality of 0.96.

hf=798.33 kJ/kghfg=1985.4 kJ/kgsf=2.2159 kJ/kg⋅Ksfg=4.3058 kJ/kg⋅K

Calculate the enthalpy at state 4(h4).

h4=hf+x4hfg=798.33 kJ/kg+(0.96)(1985.4 kJ/kg)=2704.3 kJ/kg

Calculate the entropy at state 4(s4).

s4=sf+x4sfg=2.2159 kJ/kg⋅K+(0.96)(4.3058 kJ/kg⋅K)=6.3495 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 5000 kPa and entropy of 6.3495 kJ/kg⋅K.

h3=3006.9 kJ/kgT3=327.2°C

Thus, the temperature at the inlet of the high-pressure turbine is 327.2°C.

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 20 kPa and quality of 0.96.

hf=251.42 kJ/kghfg=2357.5 kJ/kgsf=0.8320 kJ/kg⋅Ksfg=7.0752 kJ/kg⋅K

Calculate the enthalpy at state 6(h6).

h6=hf+x6hfg=251.42 kJ/kg+(0.96)(2357.5 kJ/kg)=2514.6 kJ/kg

Calculate the entropy at state 6(s6).

s6=sf+x6sfg=0.8320 kJ/kg⋅K+(0.96)(7.0752 kJ/kg⋅K)=7.6242 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 1200 kPa and entropy of 7.6242 kJ/kg⋅K.

h5=3436.0 kJ/kgT5=481.1°C

Thus, the temperature at the inlet of the low-pressure turbine is 481.1°C.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h6−h1(h3−h2)+(h5−h4)

=1−2514.6 kJ/kg−151.42 kJ/kg(3006.9 kJ/kg−256.49 kJ/kg)+(3436.0 kJ/kg−2704.3 kJ/kg)=1−2263.2 kJ/kg3482.0 kJ/kg=0.35=35%

Thus, the thermal efficiency of the cycle is 35%.

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Answer 5

Chapter 9, Problem 128P

To determine

The temperature at the inlet of each turbine and the thermal efficiency of the cycle.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 20 kPa.

Pressure of steam at the boiler (P3) is 5000 kPa.

Pressure of steam at the reheat section (P4) is 1200 kPa.

Temperature of water at the turbine inlet (T3) is 450°C.

Quality of steam at the exit of the turbines (x) is 0.96.

Calculation:

Draw the T−s diagram of the cycle as in Figure (1).

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 128P

The pressures are constant for the process 4 to 5 and process 6 to 1.

P4=P5P6=P1

The entropies are constant for the process 3 to 4 and process 5 to 6.

s3=s4s5=s6

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 20 kPa.

h1=hf@20 kPa=251.42 kJ/kgv1=vf@20 kPa=0.0010172 m3/kg

Calculate the work done by the pump during process 1-2(wp,in).

wp,in=v1(P2−P1)=(0.0010172 m3/kg)(5000 kPa−20 kPa)=(0.0010172 m3/kg)(5000 kPa−20 kPa)(1 kJ1 kPa⋅m3)=5.065 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=251.42 kJ/kg+5.065 kJ/kg=256.49 kJ/kg

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1200 kPa and quality of 0.96.

hf=798.33 kJ/kghfg=1985.4 kJ/kgsf=2.2159 kJ/kg⋅Ksfg=4.3058 kJ/kg⋅K

Calculate the enthalpy at state 4(h4).

h4=hf+x4hfg=798.33 kJ/kg+(0.96)(1985.4 kJ/kg)=2704.3 kJ/kg

Calculate the entropy at state 4(s4).

s4=sf+x4sfg=2.2159 kJ/kg⋅K+(0.96)(4.3058 kJ/kg⋅K)=6.3495 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 5000 kPa and entropy of 6.3495 kJ/kg⋅K.

h3=3006.9 kJ/kgT3=327.2°C

Thus, the temperature at the inlet of the high-pressure turbine is 327.2°C.

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 20 kPa and quality of 0.96.

hf=251.42 kJ/kghfg=2357.5 kJ/kgsf=0.8320 kJ/kg⋅Ksfg=7.0752 kJ/kg⋅K

Calculate the enthalpy at state 6(h6).

h6=hf+x6hfg=251.42 kJ/kg+(0.96)(2357.5 kJ/kg)=2514.6 kJ/kg

Calculate the entropy at state 6(s6).

s6=sf+x6sfg=0.8320 kJ/kg⋅K+(0.96)(7.0752 kJ/kg⋅K)=7.6242 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 1200 kPa and entropy of 7.6242 kJ/kg⋅K.

h5=3436.0 kJ/kgT5=481.1°C

Thus, the temperature at the inlet of the low-pressure turbine is 481.1°C.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h6−h1(h3−h2)+(h5−h4)

=1−2514.6 kJ/kg−151.42 kJ/kg(3006.9 kJ/kg−256.49 kJ/kg)+(3436.0 kJ/kg−2704.3 kJ/kg)=1−2263.2 kJ/kg3482.0 kJ/kg=0.35=35%

Thus, the thermal efficiency of the cycle is 35%.

Answer 6

Chapter 9, Problem 128P

To determine

The temperature at the inlet of each turbine and the thermal efficiency of the cycle.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 20 kPa.

Pressure of steam at the boiler (P3) is 5000 kPa.

Pressure of steam at the reheat section (P4) is 1200 kPa.

Temperature of water at the turbine inlet (T3) is 450°C.

Quality of steam at the exit of the turbines (x) is 0.96.

Calculation:

Draw the T−s diagram of the cycle as in Figure (1).

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 128P

The pressures are constant for the process 4 to 5 and process 6 to 1.

P4=P5P6=P1

The entropies are constant for the process 3 to 4 and process 5 to 6.

s3=s4s5=s6

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 20 kPa.

h1=hf@20 kPa=251.42 kJ/kgv1=vf@20 kPa=0.0010172 m3/kg

Calculate the work done by the pump during process 1-2(wp,in).

wp,in=v1(P2−P1)=(0.0010172 m3/kg)(5000 kPa−20 kPa)=(0.0010172 m3/kg)(5000 kPa−20 kPa)(1 kJ1 kPa⋅m3)=5.065 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=251.42 kJ/kg+5.065 kJ/kg=256.49 kJ/kg

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1200 kPa and quality of 0.96.

hf=798.33 kJ/kghfg=1985.4 kJ/kgsf=2.2159 kJ/kg⋅Ksfg=4.3058 kJ/kg⋅K

Calculate the enthalpy at state 4(h4).

h4=hf+x4hfg=798.33 kJ/kg+(0.96)(1985.4 kJ/kg)=2704.3 kJ/kg

Calculate the entropy at state 4(s4).

s4=sf+x4sfg=2.2159 kJ/kg⋅K+(0.96)(4.3058 kJ/kg⋅K)=6.3495 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 5000 kPa and entropy of 6.3495 kJ/kg⋅K.

h3=3006.9 kJ/kgT3=327.2°C

Thus, the temperature at the inlet of the high-pressure turbine is 327.2°C.

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 20 kPa and quality of 0.96.

hf=251.42 kJ/kghfg=2357.5 kJ/kgsf=0.8320 kJ/kg⋅Ksfg=7.0752 kJ/kg⋅K

Calculate the enthalpy at state 6(h6).

h6=hf+x6hfg=251.42 kJ/kg+(0.96)(2357.5 kJ/kg)=2514.6 kJ/kg

Calculate the entropy at state 6(s6).

s6=sf+x6sfg=0.8320 kJ/kg⋅K+(0.96)(7.0752 kJ/kg⋅K)=7.6242 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 1200 kPa and entropy of 7.6242 kJ/kg⋅K.

h5=3436.0 kJ/kgT5=481.1°C

Thus, the temperature at the inlet of the low-pressure turbine is 481.1°C.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h6−h1(h3−h2)+(h5−h4)

=1−2514.6 kJ/kg−151.42 kJ/kg(3006.9 kJ/kg−256.49 kJ/kg)+(3436.0 kJ/kg−2704.3 kJ/kg)=1−2263.2 kJ/kg3482.0 kJ/kg=0.35=35%

Thus, the thermal efficiency of the cycle is 35%.

Answer 7

Chapter 9, Problem 128P

To determine

The temperature at the inlet of each turbine and the thermal efficiency of the cycle.

Answer 8

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 20 kPa.

Pressure of steam at the boiler (P3) is 5000 kPa.

Pressure of steam at the reheat section (P4) is 1200 kPa.

Temperature of water at the turbine inlet (T3) is 450°C.

Quality of steam at the exit of the turbines (x) is 0.96.

Calculation:

Draw the T−s diagram of the cycle as in Figure (1).

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 128P

The pressures are constant for the process 4 to 5 and process 6 to 1.

P4=P5P6=P1

The entropies are constant for the process 3 to 4 and process 5 to 6.

s3=s4s5=s6

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 20 kPa.

h1=hf@20 kPa=251.42 kJ/kgv1=vf@20 kPa=0.0010172 m3/kg

Calculate the work done by the pump during process 1-2(wp,in).

wp,in=v1(P2−P1)=(0.0010172 m3/kg)(5000 kPa−20 kPa)=(0.0010172 m3/kg)(5000 kPa−20 kPa)(1 kJ1 kPa⋅m3)=5.065 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=251.42 kJ/kg+5.065 kJ/kg=256.49 kJ/kg

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1200 kPa and quality of 0.96.

hf=798.33 kJ/kghfg=1985.4 kJ/kgsf=2.2159 kJ/kg⋅Ksfg=4.3058 kJ/kg⋅K

Calculate the enthalpy at state 4(h4).

h4=hf+x4hfg=798.33 kJ/kg+(0.96)(1985.4 kJ/kg)=2704.3 kJ/kg

Calculate the entropy at state 4(s4).

s4=sf+x4sfg=2.2159 kJ/kg⋅K+(0.96)(4.3058 kJ/kg⋅K)=6.3495 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 5000 kPa and entropy of 6.3495 kJ/kg⋅K.

h3=3006.9 kJ/kgT3=327.2°C

Thus, the temperature at the inlet of the high-pressure turbine is 327.2°C.

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 20 kPa and quality of 0.96.

hf=251.42 kJ/kghfg=2357.5 kJ/kgsf=0.8320 kJ/kg⋅Ksfg=7.0752 kJ/kg⋅K

Calculate the enthalpy at state 6(h6).

h6=hf+x6hfg=251.42 kJ/kg+(0.96)(2357.5 kJ/kg)=2514.6 kJ/kg

Calculate the entropy at state 6(s6).

s6=sf+x6sfg=0.8320 kJ/kg⋅K+(0.96)(7.0752 kJ/kg⋅K)=7.6242 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the enthalpy and temperature at state 3 corresponding to the pressure of 1200 kPa and entropy of 7.6242 kJ/kg⋅K.

h5=3436.0 kJ/kgT5=481.1°C