The power used by the pumps, the power produced by the cycle, the rate of heat transfer in the reheater and the thermal efficiency of the system.

Question

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Answer 1

Question

Chapter 9, Problem 124P

To determine

The power used by the pumps, the power produced by the cycle, the rate of heat transfer in the reheater and the thermal efficiency of the system.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of water at state 3(P3) is 15,000 kPa.

Pressure of water at state 1(P1) is 100 kPa.

Pressure of water at state 5(P5) is 2000 kPa.

Temperature of water at the state 3(T3) is 450°C.

Mass flow rate of the water (m˙) is 1.74 kg/s.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 124P

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 specific enthalpy and specific volume at state 1 corresponding to the pressure of 100 kPa.

h1=hf@100 kPa=417.51 kJ/kgv1=vf@100 kPa=0.001043m3/kg

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

wp,in=v1(P2−P1)=(0.001043m3/kg)(15000 kPa−100 kPa)=(0.001043m3/kg)(15000 kPa−100 kPa)(1 kJ1 kPa⋅m3)=15.54kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=417.51 kJ/kg+15.54 kJ/kg=433.05 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 15,000 kPa and temperature of 450°C.

h3=3157.6kJ/kgs3=6.1434 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 2000 kPa and specific entropy of 6.1434 kJ/kg⋅K.

hf=908.47 kJ/kghfg=1889.8 kJ/kgsf=2.4467 kJ/kg⋅Ksfg=3.8923 kJ/kg⋅K

Calculate the quality of water at state 4(x4).

x4=s4−sfsfg=6.1434 kJ/kg⋅K−2.4467 kJ/kg⋅K3.8923 kJ/kg⋅K=0.9497

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=908.47 kJ/kg+(0.9497)(1889.8 kJ/kg)=2703.3 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 2000 kPa and temperature of 450°C.

h5=3358.2kJ/kgs5=7.2866 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 100 kPa and specific entropy of 7.2866 kJ/kg⋅K.

hf=417.51 kJ/kghfg=2257.5 kJ/kgsf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the quality of water at state 6(x6).

x6=s6−sfsfg=7.2866 kJ/kg⋅K−1.3028 kJ/kg⋅K6.0562 kJ/kg⋅K=0.9880

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=417.51 kJ/kg+(0.9880)(2257.5 kJ/kg)=2648 kJ/kg

Calculate the net power produced by the cycle (W˙net).

W˙net=m˙wnet=m˙(qin−qout)=m˙[(h3−h2)+(h5−h4)−(h6−h1)]

=(1.74 kg/s)[(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)−(2648 kJ/kg−417.51 kJ/kg)]=2000 kW

Thus, the net power produced by the cycle is 2000 kW.

Calculate the rate of heat transfer in the reheater (Q˙reheater).

Q˙reheater=m˙(h5−h4)=(1.74 kg/s)(3358.2 kJ/kg−2703.3 kJ/kg)=1140 kW

Thus, the rate of heat transfer in the reheater is 1140 kW.

Calculate the power used by the pumps (W˙p,in).

W˙p,in=m˙wp,in=(1.74 kg/s)(15.54kJ/kg)=27 kW

Thus, the power used by the pumps is 27 kW.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h6−h1)(h3−h2)+(h5−h4)

=(2648 kJ/kg−417.51 kJ/kg)(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)=2230.5 kJ/kg3379.8 kJ/kg=0.340=34%

Thus, the thermal efficiency of the cycle is 34%.

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Answer 2

Question

Chapter 9, Problem 124P

To determine

The power used by the pumps, the power produced by the cycle, the rate of heat transfer in the reheater and the thermal efficiency of the system.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of water at state 3(P3) is 15,000 kPa.

Pressure of water at state 1(P1) is 100 kPa.

Pressure of water at state 5(P5) is 2000 kPa.

Temperature of water at the state 3(T3) is 450°C.

Mass flow rate of the water (m˙) is 1.74 kg/s.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 124P

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 specific enthalpy and specific volume at state 1 corresponding to the pressure of 100 kPa.

h1=hf@100 kPa=417.51 kJ/kgv1=vf@100 kPa=0.001043m3/kg

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

wp,in=v1(P2−P1)=(0.001043m3/kg)(15000 kPa−100 kPa)=(0.001043m3/kg)(15000 kPa−100 kPa)(1 kJ1 kPa⋅m3)=15.54kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=417.51 kJ/kg+15.54 kJ/kg=433.05 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 15,000 kPa and temperature of 450°C.

h3=3157.6kJ/kgs3=6.1434 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 2000 kPa and specific entropy of 6.1434 kJ/kg⋅K.

hf=908.47 kJ/kghfg=1889.8 kJ/kgsf=2.4467 kJ/kg⋅Ksfg=3.8923 kJ/kg⋅K

Calculate the quality of water at state 4(x4).

x4=s4−sfsfg=6.1434 kJ/kg⋅K−2.4467 kJ/kg⋅K3.8923 kJ/kg⋅K=0.9497

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=908.47 kJ/kg+(0.9497)(1889.8 kJ/kg)=2703.3 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 2000 kPa and temperature of 450°C.

h5=3358.2kJ/kgs5=7.2866 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 100 kPa and specific entropy of 7.2866 kJ/kg⋅K.

hf=417.51 kJ/kghfg=2257.5 kJ/kgsf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the quality of water at state 6(x6).

x6=s6−sfsfg=7.2866 kJ/kg⋅K−1.3028 kJ/kg⋅K6.0562 kJ/kg⋅K=0.9880

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=417.51 kJ/kg+(0.9880)(2257.5 kJ/kg)=2648 kJ/kg

Calculate the net power produced by the cycle (W˙net).

W˙net=m˙wnet=m˙(qin−qout)=m˙[(h3−h2)+(h5−h4)−(h6−h1)]

=(1.74 kg/s)[(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)−(2648 kJ/kg−417.51 kJ/kg)]=2000 kW

Thus, the net power produced by the cycle is 2000 kW.

Calculate the rate of heat transfer in the reheater (Q˙reheater).

Q˙reheater=m˙(h5−h4)=(1.74 kg/s)(3358.2 kJ/kg−2703.3 kJ/kg)=1140 kW

Thus, the rate of heat transfer in the reheater is 1140 kW.

Calculate the power used by the pumps (W˙p,in).

W˙p,in=m˙wp,in=(1.74 kg/s)(15.54kJ/kg)=27 kW

Thus, the power used by the pumps is 27 kW.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h6−h1)(h3−h2)+(h5−h4)

=(2648 kJ/kg−417.51 kJ/kg)(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)=2230.5 kJ/kg3379.8 kJ/kg=0.340=34%

Thus, the thermal efficiency of the cycle is 34%.

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Answer 3

Question

Chapter 9, Problem 124P

To determine

The power used by the pumps, the power produced by the cycle, the rate of heat transfer in the reheater and the thermal efficiency of the system.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of water at state 3(P3) is 15,000 kPa.

Pressure of water at state 1(P1) is 100 kPa.

Pressure of water at state 5(P5) is 2000 kPa.

Temperature of water at the state 3(T3) is 450°C.

Mass flow rate of the water (m˙) is 1.74 kg/s.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 124P

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 specific enthalpy and specific volume at state 1 corresponding to the pressure of 100 kPa.

h1=hf@100 kPa=417.51 kJ/kgv1=vf@100 kPa=0.001043m3/kg

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

wp,in=v1(P2−P1)=(0.001043m3/kg)(15000 kPa−100 kPa)=(0.001043m3/kg)(15000 kPa−100 kPa)(1 kJ1 kPa⋅m3)=15.54kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=417.51 kJ/kg+15.54 kJ/kg=433.05 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 15,000 kPa and temperature of 450°C.

h3=3157.6kJ/kgs3=6.1434 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 2000 kPa and specific entropy of 6.1434 kJ/kg⋅K.

hf=908.47 kJ/kghfg=1889.8 kJ/kgsf=2.4467 kJ/kg⋅Ksfg=3.8923 kJ/kg⋅K

Calculate the quality of water at state 4(x4).

x4=s4−sfsfg=6.1434 kJ/kg⋅K−2.4467 kJ/kg⋅K3.8923 kJ/kg⋅K=0.9497

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=908.47 kJ/kg+(0.9497)(1889.8 kJ/kg)=2703.3 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 2000 kPa and temperature of 450°C.

h5=3358.2kJ/kgs5=7.2866 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 100 kPa and specific entropy of 7.2866 kJ/kg⋅K.

hf=417.51 kJ/kghfg=2257.5 kJ/kgsf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the quality of water at state 6(x6).

x6=s6−sfsfg=7.2866 kJ/kg⋅K−1.3028 kJ/kg⋅K6.0562 kJ/kg⋅K=0.9880

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=417.51 kJ/kg+(0.9880)(2257.5 kJ/kg)=2648 kJ/kg

Calculate the net power produced by the cycle (W˙net).

W˙net=m˙wnet=m˙(qin−qout)=m˙[(h3−h2)+(h5−h4)−(h6−h1)]

=(1.74 kg/s)[(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)−(2648 kJ/kg−417.51 kJ/kg)]=2000 kW

Thus, the net power produced by the cycle is 2000 kW.

Calculate the rate of heat transfer in the reheater (Q˙reheater).

Q˙reheater=m˙(h5−h4)=(1.74 kg/s)(3358.2 kJ/kg−2703.3 kJ/kg)=1140 kW

Thus, the rate of heat transfer in the reheater is 1140 kW.

Calculate the power used by the pumps (W˙p,in).

W˙p,in=m˙wp,in=(1.74 kg/s)(15.54kJ/kg)=27 kW

Thus, the power used by the pumps is 27 kW.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h6−h1)(h3−h2)+(h5−h4)

=(2648 kJ/kg−417.51 kJ/kg)(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)=2230.5 kJ/kg3379.8 kJ/kg=0.340=34%

Thus, the thermal efficiency of the cycle is 34%.

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Answer 4

Question

Chapter 9, Problem 124P

To determine

The power used by the pumps, the power produced by the cycle, the rate of heat transfer in the reheater and the thermal efficiency of the system.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of water at state 3(P3) is 15,000 kPa.

Pressure of water at state 1(P1) is 100 kPa.

Pressure of water at state 5(P5) is 2000 kPa.

Temperature of water at the state 3(T3) is 450°C.

Mass flow rate of the water (m˙) is 1.74 kg/s.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 124P

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 specific enthalpy and specific volume at state 1 corresponding to the pressure of 100 kPa.

h1=hf@100 kPa=417.51 kJ/kgv1=vf@100 kPa=0.001043m3/kg

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

wp,in=v1(P2−P1)=(0.001043m3/kg)(15000 kPa−100 kPa)=(0.001043m3/kg)(15000 kPa−100 kPa)(1 kJ1 kPa⋅m3)=15.54kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=417.51 kJ/kg+15.54 kJ/kg=433.05 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 15,000 kPa and temperature of 450°C.

h3=3157.6kJ/kgs3=6.1434 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 2000 kPa and specific entropy of 6.1434 kJ/kg⋅K.

hf=908.47 kJ/kghfg=1889.8 kJ/kgsf=2.4467 kJ/kg⋅Ksfg=3.8923 kJ/kg⋅K

Calculate the quality of water at state 4(x4).

x4=s4−sfsfg=6.1434 kJ/kg⋅K−2.4467 kJ/kg⋅K3.8923 kJ/kg⋅K=0.9497

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=908.47 kJ/kg+(0.9497)(1889.8 kJ/kg)=2703.3 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 2000 kPa and temperature of 450°C.

h5=3358.2kJ/kgs5=7.2866 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 100 kPa and specific entropy of 7.2866 kJ/kg⋅K.

hf=417.51 kJ/kghfg=2257.5 kJ/kgsf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the quality of water at state 6(x6).

x6=s6−sfsfg=7.2866 kJ/kg⋅K−1.3028 kJ/kg⋅K6.0562 kJ/kg⋅K=0.9880

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=417.51 kJ/kg+(0.9880)(2257.5 kJ/kg)=2648 kJ/kg

Calculate the net power produced by the cycle (W˙net).

W˙net=m˙wnet=m˙(qin−qout)=m˙[(h3−h2)+(h5−h4)−(h6−h1)]

=(1.74 kg/s)[(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)−(2648 kJ/kg−417.51 kJ/kg)]=2000 kW

Thus, the net power produced by the cycle is 2000 kW.

Calculate the rate of heat transfer in the reheater (Q˙reheater).

Q˙reheater=m˙(h5−h4)=(1.74 kg/s)(3358.2 kJ/kg−2703.3 kJ/kg)=1140 kW

Thus, the rate of heat transfer in the reheater is 1140 kW.

Calculate the power used by the pumps (W˙p,in).

W˙p,in=m˙wp,in=(1.74 kg/s)(15.54kJ/kg)=27 kW

Thus, the power used by the pumps is 27 kW.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h6−h1)(h3−h2)+(h5−h4)

=(2648 kJ/kg−417.51 kJ/kg)(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)=2230.5 kJ/kg3379.8 kJ/kg=0.340=34%

Thus, the thermal efficiency of the cycle is 34%.

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Answer 5

Chapter 9, Problem 124P

To determine

The power used by the pumps, the power produced by the cycle, the rate of heat transfer in the reheater and the thermal efficiency of the system.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of water at state 3(P3) is 15,000 kPa.

Pressure of water at state 1(P1) is 100 kPa.

Pressure of water at state 5(P5) is 2000 kPa.

Temperature of water at the state 3(T3) is 450°C.

Mass flow rate of the water (m˙) is 1.74 kg/s.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 124P

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 specific enthalpy and specific volume at state 1 corresponding to the pressure of 100 kPa.

h1=hf@100 kPa=417.51 kJ/kgv1=vf@100 kPa=0.001043m3/kg

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

wp,in=v1(P2−P1)=(0.001043m3/kg)(15000 kPa−100 kPa)=(0.001043m3/kg)(15000 kPa−100 kPa)(1 kJ1 kPa⋅m3)=15.54kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=417.51 kJ/kg+15.54 kJ/kg=433.05 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 15,000 kPa and temperature of 450°C.

h3=3157.6kJ/kgs3=6.1434 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 2000 kPa and specific entropy of 6.1434 kJ/kg⋅K.

hf=908.47 kJ/kghfg=1889.8 kJ/kgsf=2.4467 kJ/kg⋅Ksfg=3.8923 kJ/kg⋅K

Calculate the quality of water at state 4(x4).

x4=s4−sfsfg=6.1434 kJ/kg⋅K−2.4467 kJ/kg⋅K3.8923 kJ/kg⋅K=0.9497

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=908.47 kJ/kg+(0.9497)(1889.8 kJ/kg)=2703.3 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 2000 kPa and temperature of 450°C.

h5=3358.2kJ/kgs5=7.2866 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 100 kPa and specific entropy of 7.2866 kJ/kg⋅K.

hf=417.51 kJ/kghfg=2257.5 kJ/kgsf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the quality of water at state 6(x6).

x6=s6−sfsfg=7.2866 kJ/kg⋅K−1.3028 kJ/kg⋅K6.0562 kJ/kg⋅K=0.9880

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=417.51 kJ/kg+(0.9880)(2257.5 kJ/kg)=2648 kJ/kg

Calculate the net power produced by the cycle (W˙net).

W˙net=m˙wnet=m˙(qin−qout)=m˙[(h3−h2)+(h5−h4)−(h6−h1)]

=(1.74 kg/s)[(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)−(2648 kJ/kg−417.51 kJ/kg)]=2000 kW

Thus, the net power produced by the cycle is 2000 kW.

Calculate the rate of heat transfer in the reheater (Q˙reheater).

Q˙reheater=m˙(h5−h4)=(1.74 kg/s)(3358.2 kJ/kg−2703.3 kJ/kg)=1140 kW

Thus, the rate of heat transfer in the reheater is 1140 kW.

Calculate the power used by the pumps (W˙p,in).

W˙p,in=m˙wp,in=(1.74 kg/s)(15.54kJ/kg)=27 kW

Thus, the power used by the pumps is 27 kW.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h6−h1)(h3−h2)+(h5−h4)

=(2648 kJ/kg−417.51 kJ/kg)(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)=2230.5 kJ/kg3379.8 kJ/kg=0.340=34%

Thus, the thermal efficiency of the cycle is 34%.

Answer 6

Chapter 9, Problem 124P

To determine

The power used by the pumps, the power produced by the cycle, the rate of heat transfer in the reheater and the thermal efficiency of the system.

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of water at state 3(P3) is 15,000 kPa.

Pressure of water at state 1(P1) is 100 kPa.

Pressure of water at state 5(P5) is 2000 kPa.

Temperature of water at the state 3(T3) is 450°C.

Mass flow rate of the water (m˙) is 1.74 kg/s.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 124P

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 specific enthalpy and specific volume at state 1 corresponding to the pressure of 100 kPa.

h1=hf@100 kPa=417.51 kJ/kgv1=vf@100 kPa=0.001043m3/kg

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

wp,in=v1(P2−P1)=(0.001043m3/kg)(15000 kPa−100 kPa)=(0.001043m3/kg)(15000 kPa−100 kPa)(1 kJ1 kPa⋅m3)=15.54kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=417.51 kJ/kg+15.54 kJ/kg=433.05 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 15,000 kPa and temperature of 450°C.

h3=3157.6kJ/kgs3=6.1434 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 2000 kPa and specific entropy of 6.1434 kJ/kg⋅K.

hf=908.47 kJ/kghfg=1889.8 kJ/kgsf=2.4467 kJ/kg⋅Ksfg=3.8923 kJ/kg⋅K

Calculate the quality of water at state 4(x4).

x4=s4−sfsfg=6.1434 kJ/kg⋅K−2.4467 kJ/kg⋅K3.8923 kJ/kg⋅K=0.9497

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=908.47 kJ/kg+(0.9497)(1889.8 kJ/kg)=2703.3 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 2000 kPa and temperature of 450°C.

h5=3358.2kJ/kgs5=7.2866 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 100 kPa and specific entropy of 7.2866 kJ/kg⋅K.

hf=417.51 kJ/kghfg=2257.5 kJ/kgsf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K

Calculate the quality of water at state 6(x6).

x6=s6−sfsfg=7.2866 kJ/kg⋅K−1.3028 kJ/kg⋅K6.0562 kJ/kg⋅K=0.9880

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=417.51 kJ/kg+(0.9880)(2257.5 kJ/kg)=2648 kJ/kg

Calculate the net power produced by the cycle (W˙net).

W˙net=m˙wnet=m˙(qin−qout)=m˙[(h3−h2)+(h5−h4)−(h6−h1)]

=(1.74 kg/s)[(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)−(2648 kJ/kg−417.51 kJ/kg)]=2000 kW

Thus, the net power produced by the cycle is 2000 kW.

Calculate the rate of heat transfer in the reheater (Q˙reheater).

Q˙reheater=m˙(h5−h4)=(1.74 kg/s)(3358.2 kJ/kg−2703.3 kJ/kg)=1140 kW

Thus, the rate of heat transfer in the reheater is 1140 kW.

Calculate the power used by the pumps (W˙p,in).

W˙p,in=m˙wp,in=(1.74 kg/s)(15.54kJ/kg)=27 kW

Thus, the power used by the pumps is 27 kW.

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h6−h1)(h3−h2)+(h5−h4)

=(2648 kJ/kg−417.51 kJ/kg)(3157.9 kJ/kg−433.03 kJ/kg)+(3358.2 kJ/kg−2703.3 kJ/kg)=2230.5 kJ/kg3379.8 kJ/kg=0.340=34%

Thus, the thermal efficiency of the cycle is 34%.

Answer 7

Chapter 9, Problem 124P

To determine

The power used by the pumps, the power produced by the cycle, the rate of heat transfer in the reheater and the thermal efficiency of the system.

Answer 8

Expert Solution & Answer

Explanation of Solution

Given:

Pressure of water at state 3(P3) is 15,000 kPa.

Pressure of water at state 1(P1) is 100 kPa.

Pressure of water at state 5(P5) is 2000 kPa.

Temperature of water at the state 3(T3) is 450°C.

Mass flow rate of the water (m˙) is 1.74 kg/s.

Calculation:

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

Fundamentals Of Thermal-fluid Sciences In Si Units, Chapter 9, Problem 124P

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 specific enthalpy and specific volume at state 1 corresponding to the pressure of 100 kPa.

h1=hf@100 kPa=417.51 kJ/kgv1=vf@100 kPa=0.001043m3/kg

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

wp,in=v1(P2−P1)=(0.001043m3/kg)(15000 kPa−100 kPa)=(0.001043m3/kg)(15000 kPa−100 kPa)(1 kJ1 kPa⋅m3)=15.54kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=417.51 kJ/kg+15.54 kJ/kg=433.05 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 15,000 kPa and temperature of 450°C.

h3=3157.6kJ/kgs3=6.1434 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 2000 kPa and specific entropy of 6.1434 kJ/kg⋅K.

hf=908.47 kJ/kghfg=1889.8 kJ/kgsf=2.4467 kJ/kg⋅Ksfg=3.8923 kJ/kg⋅K

Calculate the quality of water at state 4(x4).

x4=s4−sfsfg=6.1434 kJ/kg⋅K−2.4467 kJ/kg⋅K3.8923 kJ/kg⋅K=0.9497

Calculate the specific enthalpy at state 4(h4).

h4=hf+x4hfg=908.47 kJ/kg+(0.9497)(1889.8 kJ/kg)=2703.3 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 2000 kPa and temperature of 450°C.

h5=3358.2kJ/kgs5=7.2866 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 100 kPa and specific entropy of 7.2866 kJ/kg⋅K.

hf=417.51 kJ/kghfg=2257.5 kJ/kgsf=1.3028 kJ/kg⋅Ksfg=6.0562 kJ/kg⋅K