Compare the thermal efficiency of the cycle when it is operated such that the saturated liquid enters the pump against that when the sub-cooled liquid enters the pump.

Question

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Answer 1

Question

Chapter 9, Problem 179RQ

To determine

Compare the thermal efficiency of the cycle when it is operated such that the saturated liquid enters the pump against that when the sub-cooled liquid enters the pump.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the boiler (P3) is 6000 kPa.

Pressure of steam at the condenser (P1) is 50 kPa.

Temperature of steam at the turbine inlet (T3) is 600°C.

Temperature of sub-cooled liquid (T) is 11.3°C.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 179RQ

The pressures are constant for the process 2 to 3 and process 4 to 1.

P2=P3P4=P1

The entropies are constant for the process 1 to 2 and process 3 to 4.

s1=s2s3=s4

The thermal efficiency of the cycle when it is operated such that when the saturated liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 50 kPa.

h1=hf@50 kPa=340.54 kJ/kgv1=vf@50 kPa=0.001030m3/kg

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

wp,in=v1(P2−P1)=(0.001030m3/kg)(6000 kPa−50 kPa)=(0.001030m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.13kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=340.54 kJ/kg+6.13 kJ/kg=346.67 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and temperature at state 3 corresponding to the pressure of 6 MPa and temperature of 600°C.

h3=3658.8kJ/kgs3=7.1693 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 50 kPa and specific entropy of 7.1693 kJ/kg⋅K.

hf=340.54 kJ/kghfg=2304.7 kJ/kgsf=1.0912 kJ/kg⋅Ksfg=6.5019 kJ/kg⋅K

Calculate the quality of steam at state 4(s4).

x4=s4−sfsfg=7.1693 kJ/kg⋅K−1.0912 kJ/kg⋅K6.5019 kJ/kg⋅K=0.9348

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=340.54 kJ/kg+(0.9348)(2304.7 kJ/kg)=2495 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−340.64 kJ/kg3658.8 kJ/kg−346.67 kJ/kg=2154.5 kJ/kg3312.1 kJ/kg=0.3495=34.95%

The thermal efficiency of the cycle when it is operated such that when the sub-cooled liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the saturation temperature corresponding to the pressure of 50 kPa.

Tsat@50 kPa=81.3°C

Calculate the temperature at state 1(T1) .

T1=Tsat@50 kPa−T=81.3°C−11.3°C=70°C

Refer Table A-4, “Saturated water-Temperature table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the temperature of 70°C.

h1=hf@70°C=293.07 kJ/kgv1=vf@70°C=0.001023 m3/kg

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

wp,in=v1(P2−P1)=(0.001023 m3/kg)(6000 kPa−50 kPa)=(0.001023 m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.09 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=293.07 kJ/kg+6.09 kJ/kg=299.16 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−293.09 kJ/kg3658.8 kJ/kg−299.16 kJ/kg=2201.9 kJ/kg3359.6 kJ/kg=0.3446=34.46%

By comparing the thermal efficiencies of the cycle, it is found that the thermal efficiency slightly reduces because of the sub-cooling at the inlet of the pump.

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Answer 2

Question

Chapter 9, Problem 179RQ

To determine

Compare the thermal efficiency of the cycle when it is operated such that the saturated liquid enters the pump against that when the sub-cooled liquid enters the pump.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the boiler (P3) is 6000 kPa.

Pressure of steam at the condenser (P1) is 50 kPa.

Temperature of steam at the turbine inlet (T3) is 600°C.

Temperature of sub-cooled liquid (T) is 11.3°C.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 179RQ

The pressures are constant for the process 2 to 3 and process 4 to 1.

P2=P3P4=P1

The entropies are constant for the process 1 to 2 and process 3 to 4.

s1=s2s3=s4

The thermal efficiency of the cycle when it is operated such that when the saturated liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 50 kPa.

h1=hf@50 kPa=340.54 kJ/kgv1=vf@50 kPa=0.001030m3/kg

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

wp,in=v1(P2−P1)=(0.001030m3/kg)(6000 kPa−50 kPa)=(0.001030m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.13kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=340.54 kJ/kg+6.13 kJ/kg=346.67 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and temperature at state 3 corresponding to the pressure of 6 MPa and temperature of 600°C.

h3=3658.8kJ/kgs3=7.1693 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 50 kPa and specific entropy of 7.1693 kJ/kg⋅K.

hf=340.54 kJ/kghfg=2304.7 kJ/kgsf=1.0912 kJ/kg⋅Ksfg=6.5019 kJ/kg⋅K

Calculate the quality of steam at state 4(s4).

x4=s4−sfsfg=7.1693 kJ/kg⋅K−1.0912 kJ/kg⋅K6.5019 kJ/kg⋅K=0.9348

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=340.54 kJ/kg+(0.9348)(2304.7 kJ/kg)=2495 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−340.64 kJ/kg3658.8 kJ/kg−346.67 kJ/kg=2154.5 kJ/kg3312.1 kJ/kg=0.3495=34.95%

The thermal efficiency of the cycle when it is operated such that when the sub-cooled liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the saturation temperature corresponding to the pressure of 50 kPa.

Tsat@50 kPa=81.3°C

Calculate the temperature at state 1(T1) .

T1=Tsat@50 kPa−T=81.3°C−11.3°C=70°C

Refer Table A-4, “Saturated water-Temperature table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the temperature of 70°C.

h1=hf@70°C=293.07 kJ/kgv1=vf@70°C=0.001023 m3/kg

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

wp,in=v1(P2−P1)=(0.001023 m3/kg)(6000 kPa−50 kPa)=(0.001023 m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.09 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=293.07 kJ/kg+6.09 kJ/kg=299.16 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−293.09 kJ/kg3658.8 kJ/kg−299.16 kJ/kg=2201.9 kJ/kg3359.6 kJ/kg=0.3446=34.46%

By comparing the thermal efficiencies of the cycle, it is found that the thermal efficiency slightly reduces because of the sub-cooling at the inlet of the pump.

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Answer 3

Question

Chapter 9, Problem 179RQ

To determine

Compare the thermal efficiency of the cycle when it is operated such that the saturated liquid enters the pump against that when the sub-cooled liquid enters the pump.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the boiler (P3) is 6000 kPa.

Pressure of steam at the condenser (P1) is 50 kPa.

Temperature of steam at the turbine inlet (T3) is 600°C.

Temperature of sub-cooled liquid (T) is 11.3°C.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 179RQ

The pressures are constant for the process 2 to 3 and process 4 to 1.

P2=P3P4=P1

The entropies are constant for the process 1 to 2 and process 3 to 4.

s1=s2s3=s4

The thermal efficiency of the cycle when it is operated such that when the saturated liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 50 kPa.

h1=hf@50 kPa=340.54 kJ/kgv1=vf@50 kPa=0.001030m3/kg

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

wp,in=v1(P2−P1)=(0.001030m3/kg)(6000 kPa−50 kPa)=(0.001030m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.13kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=340.54 kJ/kg+6.13 kJ/kg=346.67 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and temperature at state 3 corresponding to the pressure of 6 MPa and temperature of 600°C.

h3=3658.8kJ/kgs3=7.1693 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 50 kPa and specific entropy of 7.1693 kJ/kg⋅K.

hf=340.54 kJ/kghfg=2304.7 kJ/kgsf=1.0912 kJ/kg⋅Ksfg=6.5019 kJ/kg⋅K

Calculate the quality of steam at state 4(s4).

x4=s4−sfsfg=7.1693 kJ/kg⋅K−1.0912 kJ/kg⋅K6.5019 kJ/kg⋅K=0.9348

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=340.54 kJ/kg+(0.9348)(2304.7 kJ/kg)=2495 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−340.64 kJ/kg3658.8 kJ/kg−346.67 kJ/kg=2154.5 kJ/kg3312.1 kJ/kg=0.3495=34.95%

The thermal efficiency of the cycle when it is operated such that when the sub-cooled liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the saturation temperature corresponding to the pressure of 50 kPa.

Tsat@50 kPa=81.3°C

Calculate the temperature at state 1(T1) .

T1=Tsat@50 kPa−T=81.3°C−11.3°C=70°C

Refer Table A-4, “Saturated water-Temperature table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the temperature of 70°C.

h1=hf@70°C=293.07 kJ/kgv1=vf@70°C=0.001023 m3/kg

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

wp,in=v1(P2−P1)=(0.001023 m3/kg)(6000 kPa−50 kPa)=(0.001023 m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.09 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=293.07 kJ/kg+6.09 kJ/kg=299.16 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−293.09 kJ/kg3658.8 kJ/kg−299.16 kJ/kg=2201.9 kJ/kg3359.6 kJ/kg=0.3446=34.46%

By comparing the thermal efficiencies of the cycle, it is found that the thermal efficiency slightly reduces because of the sub-cooling at the inlet of the pump.

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Answer 4

Question

Chapter 9, Problem 179RQ

To determine

Compare the thermal efficiency of the cycle when it is operated such that the saturated liquid enters the pump against that when the sub-cooled liquid enters the pump.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the boiler (P3) is 6000 kPa.

Pressure of steam at the condenser (P1) is 50 kPa.

Temperature of steam at the turbine inlet (T3) is 600°C.

Temperature of sub-cooled liquid (T) is 11.3°C.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 179RQ

The pressures are constant for the process 2 to 3 and process 4 to 1.

P2=P3P4=P1

The entropies are constant for the process 1 to 2 and process 3 to 4.

s1=s2s3=s4

The thermal efficiency of the cycle when it is operated such that when the saturated liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 50 kPa.

h1=hf@50 kPa=340.54 kJ/kgv1=vf@50 kPa=0.001030m3/kg

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

wp,in=v1(P2−P1)=(0.001030m3/kg)(6000 kPa−50 kPa)=(0.001030m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.13kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=340.54 kJ/kg+6.13 kJ/kg=346.67 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and temperature at state 3 corresponding to the pressure of 6 MPa and temperature of 600°C.

h3=3658.8kJ/kgs3=7.1693 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 50 kPa and specific entropy of 7.1693 kJ/kg⋅K.

hf=340.54 kJ/kghfg=2304.7 kJ/kgsf=1.0912 kJ/kg⋅Ksfg=6.5019 kJ/kg⋅K

Calculate the quality of steam at state 4(s4).

x4=s4−sfsfg=7.1693 kJ/kg⋅K−1.0912 kJ/kg⋅K6.5019 kJ/kg⋅K=0.9348

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=340.54 kJ/kg+(0.9348)(2304.7 kJ/kg)=2495 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−340.64 kJ/kg3658.8 kJ/kg−346.67 kJ/kg=2154.5 kJ/kg3312.1 kJ/kg=0.3495=34.95%

The thermal efficiency of the cycle when it is operated such that when the sub-cooled liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the saturation temperature corresponding to the pressure of 50 kPa.

Tsat@50 kPa=81.3°C

Calculate the temperature at state 1(T1) .

T1=Tsat@50 kPa−T=81.3°C−11.3°C=70°C

Refer Table A-4, “Saturated water-Temperature table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the temperature of 70°C.

h1=hf@70°C=293.07 kJ/kgv1=vf@70°C=0.001023 m3/kg

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

wp,in=v1(P2−P1)=(0.001023 m3/kg)(6000 kPa−50 kPa)=(0.001023 m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.09 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=293.07 kJ/kg+6.09 kJ/kg=299.16 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−293.09 kJ/kg3658.8 kJ/kg−299.16 kJ/kg=2201.9 kJ/kg3359.6 kJ/kg=0.3446=34.46%

By comparing the thermal efficiencies of the cycle, it is found that the thermal efficiency slightly reduces because of the sub-cooling at the inlet of the pump.

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Answer 5

Chapter 9, Problem 179RQ

To determine

Compare the thermal efficiency of the cycle when it is operated such that the saturated liquid enters the pump against that when the sub-cooled liquid enters the pump.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the boiler (P3) is 6000 kPa.

Pressure of steam at the condenser (P1) is 50 kPa.

Temperature of steam at the turbine inlet (T3) is 600°C.

Temperature of sub-cooled liquid (T) is 11.3°C.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 179RQ

The pressures are constant for the process 2 to 3 and process 4 to 1.

P2=P3P4=P1

The entropies are constant for the process 1 to 2 and process 3 to 4.

s1=s2s3=s4

The thermal efficiency of the cycle when it is operated such that when the saturated liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 50 kPa.

h1=hf@50 kPa=340.54 kJ/kgv1=vf@50 kPa=0.001030m3/kg

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

wp,in=v1(P2−P1)=(0.001030m3/kg)(6000 kPa−50 kPa)=(0.001030m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.13kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=340.54 kJ/kg+6.13 kJ/kg=346.67 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and temperature at state 3 corresponding to the pressure of 6 MPa and temperature of 600°C.

h3=3658.8kJ/kgs3=7.1693 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 50 kPa and specific entropy of 7.1693 kJ/kg⋅K.

hf=340.54 kJ/kghfg=2304.7 kJ/kgsf=1.0912 kJ/kg⋅Ksfg=6.5019 kJ/kg⋅K

Calculate the quality of steam at state 4(s4).

x4=s4−sfsfg=7.1693 kJ/kg⋅K−1.0912 kJ/kg⋅K6.5019 kJ/kg⋅K=0.9348

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=340.54 kJ/kg+(0.9348)(2304.7 kJ/kg)=2495 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−340.64 kJ/kg3658.8 kJ/kg−346.67 kJ/kg=2154.5 kJ/kg3312.1 kJ/kg=0.3495=34.95%

The thermal efficiency of the cycle when it is operated such that when the sub-cooled liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the saturation temperature corresponding to the pressure of 50 kPa.

Tsat@50 kPa=81.3°C

Calculate the temperature at state 1(T1) .

T1=Tsat@50 kPa−T=81.3°C−11.3°C=70°C

Refer Table A-4, “Saturated water-Temperature table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the temperature of 70°C.

h1=hf@70°C=293.07 kJ/kgv1=vf@70°C=0.001023 m3/kg

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

wp,in=v1(P2−P1)=(0.001023 m3/kg)(6000 kPa−50 kPa)=(0.001023 m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.09 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=293.07 kJ/kg+6.09 kJ/kg=299.16 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−293.09 kJ/kg3658.8 kJ/kg−299.16 kJ/kg=2201.9 kJ/kg3359.6 kJ/kg=0.3446=34.46%

By comparing the thermal efficiencies of the cycle, it is found that the thermal efficiency slightly reduces because of the sub-cooling at the inlet of the pump.

Answer 6

Chapter 9, Problem 179RQ

To determine

Compare the thermal efficiency of the cycle when it is operated such that the saturated liquid enters the pump against that when the sub-cooled liquid enters the pump.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the boiler (P3) is 6000 kPa.

Pressure of steam at the condenser (P1) is 50 kPa.

Temperature of steam at the turbine inlet (T3) is 600°C.

Temperature of sub-cooled liquid (T) is 11.3°C.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 179RQ

The pressures are constant for the process 2 to 3 and process 4 to 1.

P2=P3P4=P1

The entropies are constant for the process 1 to 2 and process 3 to 4.

s1=s2s3=s4

The thermal efficiency of the cycle when it is operated such that when the saturated liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 50 kPa.

h1=hf@50 kPa=340.54 kJ/kgv1=vf@50 kPa=0.001030m3/kg

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

wp,in=v1(P2−P1)=(0.001030m3/kg)(6000 kPa−50 kPa)=(0.001030m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.13kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=340.54 kJ/kg+6.13 kJ/kg=346.67 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and temperature at state 3 corresponding to the pressure of 6 MPa and temperature of 600°C.

h3=3658.8kJ/kgs3=7.1693 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 50 kPa and specific entropy of 7.1693 kJ/kg⋅K.

hf=340.54 kJ/kghfg=2304.7 kJ/kgsf=1.0912 kJ/kg⋅Ksfg=6.5019 kJ/kg⋅K

Calculate the quality of steam at state 4(s4).

x4=s4−sfsfg=7.1693 kJ/kg⋅K−1.0912 kJ/kg⋅K6.5019 kJ/kg⋅K=0.9348

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=340.54 kJ/kg+(0.9348)(2304.7 kJ/kg)=2495 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−340.64 kJ/kg3658.8 kJ/kg−346.67 kJ/kg=2154.5 kJ/kg3312.1 kJ/kg=0.3495=34.95%

The thermal efficiency of the cycle when it is operated such that when the sub-cooled liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the saturation temperature corresponding to the pressure of 50 kPa.

Tsat@50 kPa=81.3°C

Calculate the temperature at state 1(T1) .

T1=Tsat@50 kPa−T=81.3°C−11.3°C=70°C

Refer Table A-4, “Saturated water-Temperature table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the temperature of 70°C.

h1=hf@70°C=293.07 kJ/kgv1=vf@70°C=0.001023 m3/kg

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

wp,in=v1(P2−P1)=(0.001023 m3/kg)(6000 kPa−50 kPa)=(0.001023 m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.09 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=293.07 kJ/kg+6.09 kJ/kg=299.16 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−293.09 kJ/kg3658.8 kJ/kg−299.16 kJ/kg=2201.9 kJ/kg3359.6 kJ/kg=0.3446=34.46%

By comparing the thermal efficiencies of the cycle, it is found that the thermal efficiency slightly reduces because of the sub-cooling at the inlet of the pump.

Answer 7

Chapter 9, Problem 179RQ

To determine

Compare the thermal efficiency of the cycle when it is operated such that the saturated liquid enters the pump against that when the sub-cooled liquid enters the pump.

Answer 8

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of steam at the boiler (P3) is 6000 kPa.

Pressure of steam at the condenser (P1) is 50 kPa.

Temperature of steam at the turbine inlet (T3) is 600°C.

Temperature of sub-cooled liquid (T) is 11.3°C.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 179RQ

The pressures are constant for the process 2 to 3 and process 4 to 1.

P2=P3P4=P1

The entropies are constant for the process 1 to 2 and process 3 to 4.

s1=s2s3=s4

The thermal efficiency of the cycle when it is operated such that when the saturated liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 50 kPa.

h1=hf@50 kPa=340.54 kJ/kgv1=vf@50 kPa=0.001030m3/kg

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

wp,in=v1(P2−P1)=(0.001030m3/kg)(6000 kPa−50 kPa)=(0.001030m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.13kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=340.54 kJ/kg+6.13 kJ/kg=346.67 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and temperature at state 3 corresponding to the pressure of 6 MPa and temperature of 600°C.

h3=3658.8kJ/kgs3=7.1693 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 50 kPa and specific entropy of 7.1693 kJ/kg⋅K.

hf=340.54 kJ/kghfg=2304.7 kJ/kgsf=1.0912 kJ/kg⋅Ksfg=6.5019 kJ/kg⋅K

Calculate the quality of steam at state 4(s4).

x4=s4−sfsfg=7.1693 kJ/kg⋅K−1.0912 kJ/kg⋅K6.5019 kJ/kg⋅K=0.9348

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=340.54 kJ/kg+(0.9348)(2304.7 kJ/kg)=2495 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−340.64 kJ/kg3658.8 kJ/kg−346.67 kJ/kg=2154.5 kJ/kg3312.1 kJ/kg=0.3495=34.95%

The thermal efficiency of the cycle when it is operated such that when the sub-cooled liquid enters the pump:

Refer Table A-5, “Saturated water-Pressure table”, obtain the saturation temperature corresponding to the pressure of 50 kPa.

Tsat@50 kPa=81.3°C

Calculate the temperature at state 1(T1) .

T1=Tsat@50 kPa−T=81.3°C−11.3°C=70°C

Refer Table A-4, “Saturated water-Temperature table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the temperature of 70°C.

h1=hf@70°C=293.07 kJ/kgv1=vf@70°C=0.001023 m3/kg

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

wp,in=v1(P2−P1)=(0.001023 m3/kg)(6000 kPa−50 kPa)=(0.001023 m3/kg)(6000 kPa−50 kPa)(1 kJ1 kPa⋅m3)=6.09 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=293.07 kJ/kg+6.09 kJ/kg=299.16 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−h4−h1h3−h2

=2495 kJ/kg−293.09 kJ/kg3658.8 kJ/kg−299.16 kJ/kg=2201.9 kJ/kg3359.6 kJ/kg=0.3446=34.46%

By comparing the thermal efficiencies of the cycle, it is found that the thermal efficiency slightly reduces because of the sub-cooling at the inlet of the pump.