The thermal efficiency of the cycle.

Question

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Answer 1

Question

Chapter 9, Problem 180RQ

a)

To determine

The thermal efficiency of the cycle.

a)

Expert Solution

Explanation of Solution

Given:

Maximum pressure in the cycle (P3) is 10 MPa.

Minimum pressure in the cycle (P1) is 30 MPa.

Pressure at which steam is reheated for the first time (P4) is 4 MPa.

Pressure at which steam is reheated for the second time (P6) is 2 MPa.

Net power output of the cycle (W˙net) is 75 MW.

Temperature of steam at all three stages of the turbine (T3/T5/T7) is 550°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 180RQ

The pressures are constant for the process 4 to 5, process 6 to 7 and process 8 to 1.

P4=P5P6=P7P8=P1

The entropies are constant for the process 3 to 4, process 5 to 6 and process 7 to 8.

s3=s4s5=s6s7=s8

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 30 kPa.

h1=hf@30 kPa=289.18 kJ/kgv1=vf@30 kPa=0.001022 m3/kg

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

wp,in=v1(P2−P1)=(0.001022 m3/kg)(10,000 kPa−30 kPa)=(0.001022 m3/kg)(10,000 kPa−30 kPa)(1 kJ1 kPa⋅m3)=10.19kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=289.18 kJ/kg+10.19 kJ/kg=299.37 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 10 MPa and temperature of 550°C.

h3=3500.9 kJ/kgs3=6.7561 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 4 corresponding to the pressure of 4 MPa and specific entropy of 6.7561 kJ/kg⋅K.

h4=3204.9 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 4 MPa and temperature of 550°C.

h5=3559.7 kJ/kgs5=7.2335 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 6 corresponding to the pressure of 2 MPa and specific entropy of 7.2335 kJ/kg⋅K.

h6=3321.1 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 7 corresponding to the pressure of 2 MPa and temperature of 550°C.

h7=3578.4 kJ/kgs7=7.5706 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 30 kPa and specific entropy of 7.5706 kJ/kg⋅K.

hf=289.27 kJ/kghfg=2335.3 kJ/kgsf=0.9441 kJ/kg⋅Ksfg=6.8234 kJ/kg⋅K

Calculate the quality of steam at state 8(s8).

x8=s8−sfsfg=7.5706 kJ/kg⋅K−0.9441 kJ/kg⋅K6.8234 kJ/kg⋅K=0.9711

Calculate the specific enthalpy at state 8(h8).

h8=hf+x8hfg=289.27 kJ/kg+(0.9711)(2335.3 kJ/kg)=2557.1 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=qin−qoutqin=[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)(h3−h2)+(h5−h4)+(h7−h8)

=[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg){(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)}=1548.8 kJ/kg3813.7 kJ/kg=0.4053=40.5%

Thus, the thermal efficiency of the cycle is 40.5%.

b)

To determine

The mass flow rate of the steam.

b)

Expert Solution

Explanation of Solution

Calculate the mass flow rate of the steam (m˙).

m˙=W˙netwnet=W˙netqin−qout=W˙net[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)

=75 MW[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s1548.8 kJ/kg=48.5 kg/s

Thus, the mass flow rate of the steam is 48.5 kg/s.

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Answer 2

Question

Chapter 9, Problem 180RQ

a)

To determine

The thermal efficiency of the cycle.

a)

Expert Solution

Explanation of Solution

Given:

Maximum pressure in the cycle (P3) is 10 MPa.

Minimum pressure in the cycle (P1) is 30 MPa.

Pressure at which steam is reheated for the first time (P4) is 4 MPa.

Pressure at which steam is reheated for the second time (P6) is 2 MPa.

Net power output of the cycle (W˙net) is 75 MW.

Temperature of steam at all three stages of the turbine (T3/T5/T7) is 550°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 180RQ

The pressures are constant for the process 4 to 5, process 6 to 7 and process 8 to 1.

P4=P5P6=P7P8=P1

The entropies are constant for the process 3 to 4, process 5 to 6 and process 7 to 8.

s3=s4s5=s6s7=s8

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 30 kPa.

h1=hf@30 kPa=289.18 kJ/kgv1=vf@30 kPa=0.001022 m3/kg

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

wp,in=v1(P2−P1)=(0.001022 m3/kg)(10,000 kPa−30 kPa)=(0.001022 m3/kg)(10,000 kPa−30 kPa)(1 kJ1 kPa⋅m3)=10.19kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=289.18 kJ/kg+10.19 kJ/kg=299.37 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 10 MPa and temperature of 550°C.

h3=3500.9 kJ/kgs3=6.7561 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 4 corresponding to the pressure of 4 MPa and specific entropy of 6.7561 kJ/kg⋅K.

h4=3204.9 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 4 MPa and temperature of 550°C.

h5=3559.7 kJ/kgs5=7.2335 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 6 corresponding to the pressure of 2 MPa and specific entropy of 7.2335 kJ/kg⋅K.

h6=3321.1 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 7 corresponding to the pressure of 2 MPa and temperature of 550°C.

h7=3578.4 kJ/kgs7=7.5706 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 30 kPa and specific entropy of 7.5706 kJ/kg⋅K.

hf=289.27 kJ/kghfg=2335.3 kJ/kgsf=0.9441 kJ/kg⋅Ksfg=6.8234 kJ/kg⋅K

Calculate the quality of steam at state 8(s8).

x8=s8−sfsfg=7.5706 kJ/kg⋅K−0.9441 kJ/kg⋅K6.8234 kJ/kg⋅K=0.9711

Calculate the specific enthalpy at state 8(h8).

h8=hf+x8hfg=289.27 kJ/kg+(0.9711)(2335.3 kJ/kg)=2557.1 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=qin−qoutqin=[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)(h3−h2)+(h5−h4)+(h7−h8)

=[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg){(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)}=1548.8 kJ/kg3813.7 kJ/kg=0.4053=40.5%

Thus, the thermal efficiency of the cycle is 40.5%.

b)

To determine

The mass flow rate of the steam.

b)

Expert Solution

Explanation of Solution

Calculate the mass flow rate of the steam (m˙).

m˙=W˙netwnet=W˙netqin−qout=W˙net[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)

=75 MW[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s1548.8 kJ/kg=48.5 kg/s

Thus, the mass flow rate of the steam is 48.5 kg/s.

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Answer 3

Question

Chapter 9, Problem 180RQ

a)

To determine

The thermal efficiency of the cycle.

a)

Expert Solution

Explanation of Solution

Given:

Maximum pressure in the cycle (P3) is 10 MPa.

Minimum pressure in the cycle (P1) is 30 MPa.

Pressure at which steam is reheated for the first time (P4) is 4 MPa.

Pressure at which steam is reheated for the second time (P6) is 2 MPa.

Net power output of the cycle (W˙net) is 75 MW.

Temperature of steam at all three stages of the turbine (T3/T5/T7) is 550°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 180RQ

The pressures are constant for the process 4 to 5, process 6 to 7 and process 8 to 1.

P4=P5P6=P7P8=P1

The entropies are constant for the process 3 to 4, process 5 to 6 and process 7 to 8.

s3=s4s5=s6s7=s8

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 30 kPa.

h1=hf@30 kPa=289.18 kJ/kgv1=vf@30 kPa=0.001022 m3/kg

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

wp,in=v1(P2−P1)=(0.001022 m3/kg)(10,000 kPa−30 kPa)=(0.001022 m3/kg)(10,000 kPa−30 kPa)(1 kJ1 kPa⋅m3)=10.19kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=289.18 kJ/kg+10.19 kJ/kg=299.37 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 10 MPa and temperature of 550°C.

h3=3500.9 kJ/kgs3=6.7561 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 4 corresponding to the pressure of 4 MPa and specific entropy of 6.7561 kJ/kg⋅K.

h4=3204.9 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 4 MPa and temperature of 550°C.

h5=3559.7 kJ/kgs5=7.2335 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 6 corresponding to the pressure of 2 MPa and specific entropy of 7.2335 kJ/kg⋅K.

h6=3321.1 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 7 corresponding to the pressure of 2 MPa and temperature of 550°C.

h7=3578.4 kJ/kgs7=7.5706 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 30 kPa and specific entropy of 7.5706 kJ/kg⋅K.

hf=289.27 kJ/kghfg=2335.3 kJ/kgsf=0.9441 kJ/kg⋅Ksfg=6.8234 kJ/kg⋅K

Calculate the quality of steam at state 8(s8).

x8=s8−sfsfg=7.5706 kJ/kg⋅K−0.9441 kJ/kg⋅K6.8234 kJ/kg⋅K=0.9711

Calculate the specific enthalpy at state 8(h8).

h8=hf+x8hfg=289.27 kJ/kg+(0.9711)(2335.3 kJ/kg)=2557.1 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=qin−qoutqin=[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)(h3−h2)+(h5−h4)+(h7−h8)

=[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg){(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)}=1548.8 kJ/kg3813.7 kJ/kg=0.4053=40.5%

Thus, the thermal efficiency of the cycle is 40.5%.

b)

To determine

The mass flow rate of the steam.

b)

Expert Solution

Explanation of Solution

Calculate the mass flow rate of the steam (m˙).

m˙=W˙netwnet=W˙netqin−qout=W˙net[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)

=75 MW[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s1548.8 kJ/kg=48.5 kg/s

Thus, the mass flow rate of the steam is 48.5 kg/s.

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Answer 4

Question

Chapter 9, Problem 180RQ

a)

To determine

The thermal efficiency of the cycle.

a)

Expert Solution

Explanation of Solution

Given:

Maximum pressure in the cycle (P3) is 10 MPa.

Minimum pressure in the cycle (P1) is 30 MPa.

Pressure at which steam is reheated for the first time (P4) is 4 MPa.

Pressure at which steam is reheated for the second time (P6) is 2 MPa.

Net power output of the cycle (W˙net) is 75 MW.

Temperature of steam at all three stages of the turbine (T3/T5/T7) is 550°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 180RQ

The pressures are constant for the process 4 to 5, process 6 to 7 and process 8 to 1.

P4=P5P6=P7P8=P1

The entropies are constant for the process 3 to 4, process 5 to 6 and process 7 to 8.

s3=s4s5=s6s7=s8

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 30 kPa.

h1=hf@30 kPa=289.18 kJ/kgv1=vf@30 kPa=0.001022 m3/kg

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

wp,in=v1(P2−P1)=(0.001022 m3/kg)(10,000 kPa−30 kPa)=(0.001022 m3/kg)(10,000 kPa−30 kPa)(1 kJ1 kPa⋅m3)=10.19kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=289.18 kJ/kg+10.19 kJ/kg=299.37 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 10 MPa and temperature of 550°C.

h3=3500.9 kJ/kgs3=6.7561 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 4 corresponding to the pressure of 4 MPa and specific entropy of 6.7561 kJ/kg⋅K.

h4=3204.9 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 4 MPa and temperature of 550°C.

h5=3559.7 kJ/kgs5=7.2335 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 6 corresponding to the pressure of 2 MPa and specific entropy of 7.2335 kJ/kg⋅K.

h6=3321.1 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 7 corresponding to the pressure of 2 MPa and temperature of 550°C.

h7=3578.4 kJ/kgs7=7.5706 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 30 kPa and specific entropy of 7.5706 kJ/kg⋅K.

hf=289.27 kJ/kghfg=2335.3 kJ/kgsf=0.9441 kJ/kg⋅Ksfg=6.8234 kJ/kg⋅K

Calculate the quality of steam at state 8(s8).

x8=s8−sfsfg=7.5706 kJ/kg⋅K−0.9441 kJ/kg⋅K6.8234 kJ/kg⋅K=0.9711

Calculate the specific enthalpy at state 8(h8).

h8=hf+x8hfg=289.27 kJ/kg+(0.9711)(2335.3 kJ/kg)=2557.1 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=qin−qoutqin=[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)(h3−h2)+(h5−h4)+(h7−h8)

=[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg){(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)}=1548.8 kJ/kg3813.7 kJ/kg=0.4053=40.5%

Thus, the thermal efficiency of the cycle is 40.5%.

b)

To determine

The mass flow rate of the steam.

b)

Expert Solution

Explanation of Solution

Calculate the mass flow rate of the steam (m˙).

m˙=W˙netwnet=W˙netqin−qout=W˙net[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)

=75 MW[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s1548.8 kJ/kg=48.5 kg/s

Thus, the mass flow rate of the steam is 48.5 kg/s.

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Answer 5

Chapter 9, Problem 180RQ

a)

To determine

The thermal efficiency of the cycle.

a)

Expert Solution

Explanation of Solution

Given:

Maximum pressure in the cycle (P3) is 10 MPa.

Minimum pressure in the cycle (P1) is 30 MPa.

Pressure at which steam is reheated for the first time (P4) is 4 MPa.

Pressure at which steam is reheated for the second time (P6) is 2 MPa.

Net power output of the cycle (W˙net) is 75 MW.

Temperature of steam at all three stages of the turbine (T3/T5/T7) is 550°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 180RQ

The pressures are constant for the process 4 to 5, process 6 to 7 and process 8 to 1.

P4=P5P6=P7P8=P1

The entropies are constant for the process 3 to 4, process 5 to 6 and process 7 to 8.

s3=s4s5=s6s7=s8

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 30 kPa.

h1=hf@30 kPa=289.18 kJ/kgv1=vf@30 kPa=0.001022 m3/kg

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

wp,in=v1(P2−P1)=(0.001022 m3/kg)(10,000 kPa−30 kPa)=(0.001022 m3/kg)(10,000 kPa−30 kPa)(1 kJ1 kPa⋅m3)=10.19kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=289.18 kJ/kg+10.19 kJ/kg=299.37 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 10 MPa and temperature of 550°C.

h3=3500.9 kJ/kgs3=6.7561 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 4 corresponding to the pressure of 4 MPa and specific entropy of 6.7561 kJ/kg⋅K.

h4=3204.9 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 4 MPa and temperature of 550°C.

h5=3559.7 kJ/kgs5=7.2335 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 6 corresponding to the pressure of 2 MPa and specific entropy of 7.2335 kJ/kg⋅K.

h6=3321.1 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 7 corresponding to the pressure of 2 MPa and temperature of 550°C.

h7=3578.4 kJ/kgs7=7.5706 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 30 kPa and specific entropy of 7.5706 kJ/kg⋅K.

hf=289.27 kJ/kghfg=2335.3 kJ/kgsf=0.9441 kJ/kg⋅Ksfg=6.8234 kJ/kg⋅K

Calculate the quality of steam at state 8(s8).

x8=s8−sfsfg=7.5706 kJ/kg⋅K−0.9441 kJ/kg⋅K6.8234 kJ/kg⋅K=0.9711

Calculate the specific enthalpy at state 8(h8).

h8=hf+x8hfg=289.27 kJ/kg+(0.9711)(2335.3 kJ/kg)=2557.1 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=qin−qoutqin=[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)(h3−h2)+(h5−h4)+(h7−h8)

=[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg){(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)}=1548.8 kJ/kg3813.7 kJ/kg=0.4053=40.5%

Thus, the thermal efficiency of the cycle is 40.5%.

b)

To determine

The mass flow rate of the steam.

b)

Expert Solution

Explanation of Solution

Calculate the mass flow rate of the steam (m˙).

m˙=W˙netwnet=W˙netqin−qout=W˙net[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)

=75 MW[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s1548.8 kJ/kg=48.5 kg/s

Thus, the mass flow rate of the steam is 48.5 kg/s.

Answer 6

Chapter 9, Problem 180RQ

a)

To determine

The thermal efficiency of the cycle.

a)

Expert Solution

Explanation of Solution

Given:

Maximum pressure in the cycle (P3) is 10 MPa.

Minimum pressure in the cycle (P1) is 30 MPa.

Pressure at which steam is reheated for the first time (P4) is 4 MPa.

Pressure at which steam is reheated for the second time (P6) is 2 MPa.

Net power output of the cycle (W˙net) is 75 MW.

Temperature of steam at all three stages of the turbine (T3/T5/T7) is 550°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 180RQ

The pressures are constant for the process 4 to 5, process 6 to 7 and process 8 to 1.

P4=P5P6=P7P8=P1

The entropies are constant for the process 3 to 4, process 5 to 6 and process 7 to 8.

s3=s4s5=s6s7=s8

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 30 kPa.

h1=hf@30 kPa=289.18 kJ/kgv1=vf@30 kPa=0.001022 m3/kg

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

wp,in=v1(P2−P1)=(0.001022 m3/kg)(10,000 kPa−30 kPa)=(0.001022 m3/kg)(10,000 kPa−30 kPa)(1 kJ1 kPa⋅m3)=10.19kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=289.18 kJ/kg+10.19 kJ/kg=299.37 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 10 MPa and temperature of 550°C.

h3=3500.9 kJ/kgs3=6.7561 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 4 corresponding to the pressure of 4 MPa and specific entropy of 6.7561 kJ/kg⋅K.

h4=3204.9 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 4 MPa and temperature of 550°C.

h5=3559.7 kJ/kgs5=7.2335 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 6 corresponding to the pressure of 2 MPa and specific entropy of 7.2335 kJ/kg⋅K.

h6=3321.1 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 7 corresponding to the pressure of 2 MPa and temperature of 550°C.

h7=3578.4 kJ/kgs7=7.5706 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 30 kPa and specific entropy of 7.5706 kJ/kg⋅K.

hf=289.27 kJ/kghfg=2335.3 kJ/kgsf=0.9441 kJ/kg⋅Ksfg=6.8234 kJ/kg⋅K

Calculate the quality of steam at state 8(s8).

x8=s8−sfsfg=7.5706 kJ/kg⋅K−0.9441 kJ/kg⋅K6.8234 kJ/kg⋅K=0.9711

Calculate the specific enthalpy at state 8(h8).

h8=hf+x8hfg=289.27 kJ/kg+(0.9711)(2335.3 kJ/kg)=2557.1 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=qin−qoutqin=[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)(h3−h2)+(h5−h4)+(h7−h8)

=[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg){(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)}=1548.8 kJ/kg3813.7 kJ/kg=0.4053=40.5%

Thus, the thermal efficiency of the cycle is 40.5%.

Answer 7

Chapter 9, Problem 180RQ

a)

To determine

The thermal efficiency of the cycle.

Answer 8

a)

Expert Solution

Explanation of Solution

Given:

Maximum pressure in the cycle (P3) is 10 MPa.

Minimum pressure in the cycle (P1) is 30 MPa.

Pressure at which steam is reheated for the first time (P4) is 4 MPa.

Pressure at which steam is reheated for the second time (P6) is 2 MPa.

Net power output of the cycle (W˙net) is 75 MW.

Temperature of steam at all three stages of the turbine (T3/T5/T7) is 550°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 180RQ

The pressures are constant for the process 4 to 5, process 6 to 7 and process 8 to 1.

P4=P5P6=P7P8=P1

The entropies are constant for the process 3 to 4, process 5 to 6 and process 7 to 8.

s3=s4s5=s6s7=s8

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific enthalpy and specific volume at state 1 corresponding to the pressure of 30 kPa.

h1=hf@30 kPa=289.18 kJ/kgv1=vf@30 kPa=0.001022 m3/kg

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

wp,in=v1(P2−P1)=(0.001022 m3/kg)(10,000 kPa−30 kPa)=(0.001022 m3/kg)(10,000 kPa−30 kPa)(1 kJ1 kPa⋅m3)=10.19kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=289.18 kJ/kg+10.19 kJ/kg=299.37 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 3 corresponding to the pressure of 10 MPa and temperature of 550°C.

h3=3500.9 kJ/kgs3=6.7561 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 4 corresponding to the pressure of 4 MPa and specific entropy of 6.7561 kJ/kg⋅K.

h4=3204.9 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 4 MPa and temperature of 550°C.

h5=3559.7 kJ/kgs5=7.2335 kJ/kg⋅K

Refer Table A-6, “Superheated water”, obtain the specific enthalpy at state 6 corresponding to the pressure of 2 MPa and specific entropy of 7.2335 kJ/kg⋅K.

h6=3321.1 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 7 corresponding to the pressure of 2 MPa and temperature of 550°C.

h7=3578.4 kJ/kgs7=7.5706 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 30 kPa and specific entropy of 7.5706 kJ/kg⋅K.

hf=289.27 kJ/kghfg=2335.3 kJ/kgsf=0.9441 kJ/kg⋅Ksfg=6.8234 kJ/kg⋅K

Calculate the quality of steam at state 8(s8).

x8=s8−sfsfg=7.5706 kJ/kg⋅K−0.9441 kJ/kg⋅K6.8234 kJ/kg⋅K=0.9711

Calculate the specific enthalpy at state 8(h8).

h8=hf+x8hfg=289.27 kJ/kg+(0.9711)(2335.3 kJ/kg)=2557.1 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=qin−qoutqin=[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)(h3−h2)+(h5−h4)+(h7−h8)

=[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg){(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)}=1548.8 kJ/kg3813.7 kJ/kg=0.4053=40.5%

Thus, the thermal efficiency of the cycle is 40.5%.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

b)

To determine

The mass flow rate of the steam.

b)

Expert Solution

Explanation of Solution

Calculate the mass flow rate of the steam (m˙).

m˙=W˙netwnet=W˙netqin−qout=W˙net[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)

=75 MW[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s1548.8 kJ/kg=48.5 kg/s

Thus, the mass flow rate of the steam is 48.5 kg/s.

Answer 13

b)

To determine

The mass flow rate of the steam.

Answer 14

b)

Expert Solution

Explanation of Solution

Calculate the mass flow rate of the steam (m˙).

m˙=W˙netwnet=W˙netqin−qout=W˙net[(h3−h2)+(h5−h4)+(h7−h8)]−(h8−h1)

=75 MW[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s[(3500.9 kJ/kg−299.37 kJ/kg)+(3559.7 kJ/kg−3204.9 kJ/kg)+(3758.4 kJ/kg−3321.1 kJ/kg)]−(2557.1 kJ/kg−289.19 kJ/kg)=75,000 kJ/s1548.8 kJ/kg=48.5 kg/s