The quality of the steam at the turbine exit.

Question

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Answer 1

Question

Chapter 9, Problem 112P

(a)

To determine

The quality of the steam at the turbine exit.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 10 kPa.

Pressure of steam at the turbine (P3) is 100 MPa.

Temperature of steam at the turbine (T3) is 500°C.

Net power produced by the plant (W˙net) is 210 MW.

Isentropic efficiency of the turbine (ηT) is 0.85.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 112P

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

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 10 kPa.

h1=hf@10 kPa=191.81 kJ/kgv1=vf@10 kPa=0.00101 m3/kg

Refer Table A-6, “Superheated water”, obtain the enthalpy and entropy at state 3 corresponding to the pressure of 10 MPa and the temperature of 500 °C.

h3=3375.1 kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa.

hf=191.81 kJ/kghfg=2392.1 kJ/kgsf=0.6492 kJ/kg⋅Ksfg=7.4996 kJ/kg⋅K

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

wp,in=v1(P2−P1)ηT=(0.001010 m3/kg)(10 MPa−10 kPa)0.85=(0.001010 m3/kg)[(10 MPa×1000 kPa1 MPa)−10 kPa]0.85

=11.87 kPa⋅m3/kg(1 kJ1 kPa⋅m3)=11.87 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+11.87 kJ/kg=203.68 kJ/kg

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

x4=s4−sfsfg=s3−sfsfg=6.5995 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.7934

Thus, the quality of the steam at the turbine exit is 0.7934.

(b)

To determine

The thermal efficiency of the cycle.

(b)

Expert Solution

Explanation of Solution

Calculate the enthalpy at state 4s (h4s).

h4s=hf+x4hfg=191.81 kJ/kg+(0.7934)(2392.1 kJ/kg)=2089.7 kJ/kg

Calculate the enthalpy at state 4(h4).

ηT=h3−h4h3−h4s

h4=h3−(ηT)(h3−h4s)=3375.1 kJ/kg−(0.85)(3375.1 kJ/kg−2089.7 kJ/kg)=2282.5 kJ/kg

Calculate the heat supplied in the boiler (qin).

qin=h3−h2=3375.1 kJ/kg−203.68 kJ/kg=3171.4 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h4−h1)qin=1−2282.5 kJ/kg−191.81 kJ/kg3171.4 kJ/kg

=1−2090.7 kJ/kg3171.4 kJ/kg=0.341=34.1%

Thus, the thermal efficiency of the cycle is 34.1%.

(c)

To determine

The mass flow rate of the steam.

(c)

Expert Solution

Explanation of Solution

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

m˙=W˙netqin−qout=W˙netqin−(h4−h1)=210 MW3171.4 kJ/kg−(2282.5 kJ/kg−191.81 kJ/kg)

=210 MW(1000 kJ/s1 MW)1080.7 kJ/kg=194.3 kg/s

Thus, the mass flow rate of the steam is 194.3 kg/s.

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Answer 2

Question

Chapter 9, Problem 112P

(a)

To determine

The quality of the steam at the turbine exit.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 10 kPa.

Pressure of steam at the turbine (P3) is 100 MPa.

Temperature of steam at the turbine (T3) is 500°C.

Net power produced by the plant (W˙net) is 210 MW.

Isentropic efficiency of the turbine (ηT) is 0.85.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 112P

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

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 10 kPa.

h1=hf@10 kPa=191.81 kJ/kgv1=vf@10 kPa=0.00101 m3/kg

Refer Table A-6, “Superheated water”, obtain the enthalpy and entropy at state 3 corresponding to the pressure of 10 MPa and the temperature of 500 °C.

h3=3375.1 kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa.

hf=191.81 kJ/kghfg=2392.1 kJ/kgsf=0.6492 kJ/kg⋅Ksfg=7.4996 kJ/kg⋅K

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

wp,in=v1(P2−P1)ηT=(0.001010 m3/kg)(10 MPa−10 kPa)0.85=(0.001010 m3/kg)[(10 MPa×1000 kPa1 MPa)−10 kPa]0.85

=11.87 kPa⋅m3/kg(1 kJ1 kPa⋅m3)=11.87 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+11.87 kJ/kg=203.68 kJ/kg

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

x4=s4−sfsfg=s3−sfsfg=6.5995 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.7934

Thus, the quality of the steam at the turbine exit is 0.7934.

(b)

To determine

The thermal efficiency of the cycle.

(b)

Expert Solution

Explanation of Solution

Calculate the enthalpy at state 4s (h4s).

h4s=hf+x4hfg=191.81 kJ/kg+(0.7934)(2392.1 kJ/kg)=2089.7 kJ/kg

Calculate the enthalpy at state 4(h4).

ηT=h3−h4h3−h4s

h4=h3−(ηT)(h3−h4s)=3375.1 kJ/kg−(0.85)(3375.1 kJ/kg−2089.7 kJ/kg)=2282.5 kJ/kg

Calculate the heat supplied in the boiler (qin).

qin=h3−h2=3375.1 kJ/kg−203.68 kJ/kg=3171.4 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h4−h1)qin=1−2282.5 kJ/kg−191.81 kJ/kg3171.4 kJ/kg

=1−2090.7 kJ/kg3171.4 kJ/kg=0.341=34.1%

Thus, the thermal efficiency of the cycle is 34.1%.

(c)

To determine

The mass flow rate of the steam.

(c)

Expert Solution

Explanation of Solution

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

m˙=W˙netqin−qout=W˙netqin−(h4−h1)=210 MW3171.4 kJ/kg−(2282.5 kJ/kg−191.81 kJ/kg)

=210 MW(1000 kJ/s1 MW)1080.7 kJ/kg=194.3 kg/s

Thus, the mass flow rate of the steam is 194.3 kg/s.

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Answer 3

Question

Chapter 9, Problem 112P

(a)

To determine

The quality of the steam at the turbine exit.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 10 kPa.

Pressure of steam at the turbine (P3) is 100 MPa.

Temperature of steam at the turbine (T3) is 500°C.

Net power produced by the plant (W˙net) is 210 MW.

Isentropic efficiency of the turbine (ηT) is 0.85.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 112P

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

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 10 kPa.

h1=hf@10 kPa=191.81 kJ/kgv1=vf@10 kPa=0.00101 m3/kg

Refer Table A-6, “Superheated water”, obtain the enthalpy and entropy at state 3 corresponding to the pressure of 10 MPa and the temperature of 500 °C.

h3=3375.1 kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa.

hf=191.81 kJ/kghfg=2392.1 kJ/kgsf=0.6492 kJ/kg⋅Ksfg=7.4996 kJ/kg⋅K

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

wp,in=v1(P2−P1)ηT=(0.001010 m3/kg)(10 MPa−10 kPa)0.85=(0.001010 m3/kg)[(10 MPa×1000 kPa1 MPa)−10 kPa]0.85

=11.87 kPa⋅m3/kg(1 kJ1 kPa⋅m3)=11.87 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+11.87 kJ/kg=203.68 kJ/kg

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

x4=s4−sfsfg=s3−sfsfg=6.5995 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.7934

Thus, the quality of the steam at the turbine exit is 0.7934.

(b)

To determine

The thermal efficiency of the cycle.

(b)

Expert Solution

Explanation of Solution

Calculate the enthalpy at state 4s (h4s).

h4s=hf+x4hfg=191.81 kJ/kg+(0.7934)(2392.1 kJ/kg)=2089.7 kJ/kg

Calculate the enthalpy at state 4(h4).

ηT=h3−h4h3−h4s

h4=h3−(ηT)(h3−h4s)=3375.1 kJ/kg−(0.85)(3375.1 kJ/kg−2089.7 kJ/kg)=2282.5 kJ/kg

Calculate the heat supplied in the boiler (qin).

qin=h3−h2=3375.1 kJ/kg−203.68 kJ/kg=3171.4 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h4−h1)qin=1−2282.5 kJ/kg−191.81 kJ/kg3171.4 kJ/kg

=1−2090.7 kJ/kg3171.4 kJ/kg=0.341=34.1%

Thus, the thermal efficiency of the cycle is 34.1%.

(c)

To determine

The mass flow rate of the steam.

(c)

Expert Solution

Explanation of Solution

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

m˙=W˙netqin−qout=W˙netqin−(h4−h1)=210 MW3171.4 kJ/kg−(2282.5 kJ/kg−191.81 kJ/kg)

=210 MW(1000 kJ/s1 MW)1080.7 kJ/kg=194.3 kg/s

Thus, the mass flow rate of the steam is 194.3 kg/s.

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Answer 4

Question

Chapter 9, Problem 112P

(a)

To determine

The quality of the steam at the turbine exit.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 10 kPa.

Pressure of steam at the turbine (P3) is 100 MPa.

Temperature of steam at the turbine (T3) is 500°C.

Net power produced by the plant (W˙net) is 210 MW.

Isentropic efficiency of the turbine (ηT) is 0.85.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 112P

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

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 10 kPa.

h1=hf@10 kPa=191.81 kJ/kgv1=vf@10 kPa=0.00101 m3/kg

Refer Table A-6, “Superheated water”, obtain the enthalpy and entropy at state 3 corresponding to the pressure of 10 MPa and the temperature of 500 °C.

h3=3375.1 kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa.

hf=191.81 kJ/kghfg=2392.1 kJ/kgsf=0.6492 kJ/kg⋅Ksfg=7.4996 kJ/kg⋅K

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

wp,in=v1(P2−P1)ηT=(0.001010 m3/kg)(10 MPa−10 kPa)0.85=(0.001010 m3/kg)[(10 MPa×1000 kPa1 MPa)−10 kPa]0.85

=11.87 kPa⋅m3/kg(1 kJ1 kPa⋅m3)=11.87 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+11.87 kJ/kg=203.68 kJ/kg

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

x4=s4−sfsfg=s3−sfsfg=6.5995 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.7934

Thus, the quality of the steam at the turbine exit is 0.7934.

(b)

To determine

The thermal efficiency of the cycle.

(b)

Expert Solution

Explanation of Solution

Calculate the enthalpy at state 4s (h4s).

h4s=hf+x4hfg=191.81 kJ/kg+(0.7934)(2392.1 kJ/kg)=2089.7 kJ/kg

Calculate the enthalpy at state 4(h4).

ηT=h3−h4h3−h4s

h4=h3−(ηT)(h3−h4s)=3375.1 kJ/kg−(0.85)(3375.1 kJ/kg−2089.7 kJ/kg)=2282.5 kJ/kg

Calculate the heat supplied in the boiler (qin).

qin=h3−h2=3375.1 kJ/kg−203.68 kJ/kg=3171.4 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h4−h1)qin=1−2282.5 kJ/kg−191.81 kJ/kg3171.4 kJ/kg

=1−2090.7 kJ/kg3171.4 kJ/kg=0.341=34.1%

Thus, the thermal efficiency of the cycle is 34.1%.

(c)

To determine

The mass flow rate of the steam.

(c)

Expert Solution

Explanation of Solution

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

m˙=W˙netqin−qout=W˙netqin−(h4−h1)=210 MW3171.4 kJ/kg−(2282.5 kJ/kg−191.81 kJ/kg)

=210 MW(1000 kJ/s1 MW)1080.7 kJ/kg=194.3 kg/s

Thus, the mass flow rate of the steam is 194.3 kg/s.

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Answer 5

Chapter 9, Problem 112P

(a)

To determine

The quality of the steam at the turbine exit.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 10 kPa.

Pressure of steam at the turbine (P3) is 100 MPa.

Temperature of steam at the turbine (T3) is 500°C.

Net power produced by the plant (W˙net) is 210 MW.

Isentropic efficiency of the turbine (ηT) is 0.85.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 112P

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

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 10 kPa.

h1=hf@10 kPa=191.81 kJ/kgv1=vf@10 kPa=0.00101 m3/kg

Refer Table A-6, “Superheated water”, obtain the enthalpy and entropy at state 3 corresponding to the pressure of 10 MPa and the temperature of 500 °C.

h3=3375.1 kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa.

hf=191.81 kJ/kghfg=2392.1 kJ/kgsf=0.6492 kJ/kg⋅Ksfg=7.4996 kJ/kg⋅K

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

wp,in=v1(P2−P1)ηT=(0.001010 m3/kg)(10 MPa−10 kPa)0.85=(0.001010 m3/kg)[(10 MPa×1000 kPa1 MPa)−10 kPa]0.85

=11.87 kPa⋅m3/kg(1 kJ1 kPa⋅m3)=11.87 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+11.87 kJ/kg=203.68 kJ/kg

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

x4=s4−sfsfg=s3−sfsfg=6.5995 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.7934

Thus, the quality of the steam at the turbine exit is 0.7934.

(b)

To determine

The thermal efficiency of the cycle.

(b)

Expert Solution

Explanation of Solution

Calculate the enthalpy at state 4s (h4s).

h4s=hf+x4hfg=191.81 kJ/kg+(0.7934)(2392.1 kJ/kg)=2089.7 kJ/kg

Calculate the enthalpy at state 4(h4).

ηT=h3−h4h3−h4s

h4=h3−(ηT)(h3−h4s)=3375.1 kJ/kg−(0.85)(3375.1 kJ/kg−2089.7 kJ/kg)=2282.5 kJ/kg

Calculate the heat supplied in the boiler (qin).

qin=h3−h2=3375.1 kJ/kg−203.68 kJ/kg=3171.4 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h4−h1)qin=1−2282.5 kJ/kg−191.81 kJ/kg3171.4 kJ/kg

=1−2090.7 kJ/kg3171.4 kJ/kg=0.341=34.1%

Thus, the thermal efficiency of the cycle is 34.1%.

(c)

To determine

The mass flow rate of the steam.

(c)

Expert Solution

Explanation of Solution

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

m˙=W˙netqin−qout=W˙netqin−(h4−h1)=210 MW3171.4 kJ/kg−(2282.5 kJ/kg−191.81 kJ/kg)

=210 MW(1000 kJ/s1 MW)1080.7 kJ/kg=194.3 kg/s

Thus, the mass flow rate of the steam is 194.3 kg/s.

Answer 6

Chapter 9, Problem 112P

(a)

To determine

The quality of the steam at the turbine exit.

(a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 10 kPa.

Pressure of steam at the turbine (P3) is 100 MPa.

Temperature of steam at the turbine (T3) is 500°C.

Net power produced by the plant (W˙net) is 210 MW.

Isentropic efficiency of the turbine (ηT) is 0.85.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 112P

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

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 10 kPa.

h1=hf@10 kPa=191.81 kJ/kgv1=vf@10 kPa=0.00101 m3/kg

Refer Table A-6, “Superheated water”, obtain the enthalpy and entropy at state 3 corresponding to the pressure of 10 MPa and the temperature of 500 °C.

h3=3375.1 kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa.

hf=191.81 kJ/kghfg=2392.1 kJ/kgsf=0.6492 kJ/kg⋅Ksfg=7.4996 kJ/kg⋅K

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

wp,in=v1(P2−P1)ηT=(0.001010 m3/kg)(10 MPa−10 kPa)0.85=(0.001010 m3/kg)[(10 MPa×1000 kPa1 MPa)−10 kPa]0.85

=11.87 kPa⋅m3/kg(1 kJ1 kPa⋅m3)=11.87 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+11.87 kJ/kg=203.68 kJ/kg

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

x4=s4−sfsfg=s3−sfsfg=6.5995 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.7934

Thus, the quality of the steam at the turbine exit is 0.7934.

Answer 7

Chapter 9, Problem 112P

(a)

To determine

The quality of the steam at the turbine exit.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at the condenser (P1) is 10 kPa.

Pressure of steam at the turbine (P3) is 100 MPa.

Temperature of steam at the turbine (T3) is 500°C.

Net power produced by the plant (W˙net) is 210 MW.

Isentropic efficiency of the turbine (ηT) is 0.85.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 112P

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

Refer Table A-5, “Saturated water-Pressure table”, obtain the enthalpy and specific volume at state 1 corresponding to the pressure of 10 kPa.

h1=hf@10 kPa=191.81 kJ/kgv1=vf@10 kPa=0.00101 m3/kg

Refer Table A-6, “Superheated water”, obtain the enthalpy and entropy at state 3 corresponding to the pressure of 10 MPa and the temperature of 500 °C.

h3=3375.1 kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa.

hf=191.81 kJ/kghfg=2392.1 kJ/kgsf=0.6492 kJ/kg⋅Ksfg=7.4996 kJ/kg⋅K

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

wp,in=v1(P2−P1)ηT=(0.001010 m3/kg)(10 MPa−10 kPa)0.85=(0.001010 m3/kg)[(10 MPa×1000 kPa1 MPa)−10 kPa]0.85

=11.87 kPa⋅m3/kg(1 kJ1 kPa⋅m3)=11.87 kJ/kg

Calculate the enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+11.87 kJ/kg=203.68 kJ/kg

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

x4=s4−sfsfg=s3−sfsfg=6.5995 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.7934

Thus, the quality of the steam at the turbine exit is 0.7934.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

The thermal efficiency of the cycle.

(b)

Expert Solution

Explanation of Solution

Calculate the enthalpy at state 4s (h4s).

h4s=hf+x4hfg=191.81 kJ/kg+(0.7934)(2392.1 kJ/kg)=2089.7 kJ/kg

Calculate the enthalpy at state 4(h4).

ηT=h3−h4h3−h4s

h4=h3−(ηT)(h3−h4s)=3375.1 kJ/kg−(0.85)(3375.1 kJ/kg−2089.7 kJ/kg)=2282.5 kJ/kg

Calculate the heat supplied in the boiler (qin).

qin=h3−h2=3375.1 kJ/kg−203.68 kJ/kg=3171.4 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h4−h1)qin=1−2282.5 kJ/kg−191.81 kJ/kg3171.4 kJ/kg

=1−2090.7 kJ/kg3171.4 kJ/kg=0.341=34.1%

Thus, the thermal efficiency of the cycle is 34.1%.

Answer 13

(b)

To determine

The thermal efficiency of the cycle.

Answer 14

(b)

Expert Solution

Explanation of Solution

Calculate the enthalpy at state 4s (h4s).

h4s=hf+x4hfg=191.81 kJ/kg+(0.7934)(2392.1 kJ/kg)=2089.7 kJ/kg

Calculate the enthalpy at state 4(h4).

ηT=h3−h4h3−h4s

h4=h3−(ηT)(h3−h4s)=3375.1 kJ/kg−(0.85)(3375.1 kJ/kg−2089.7 kJ/kg)=2282.5 kJ/kg

Calculate the heat supplied in the boiler (qin).

qin=h3−h2=3375.1 kJ/kg−203.68 kJ/kg=3171.4 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=1−qoutqin=1−(h4−h1)qin=1−2282.5 kJ/kg−191.81 kJ/kg3171.4 kJ/kg

=1−2090.7 kJ/kg3171.4 kJ/kg=0.341=34.1%

Thus, the thermal efficiency of the cycle is 34.1%.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

The mass flow rate of the steam.

(c)

Expert Solution

Explanation of Solution

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

m˙=W˙netqin−qout=W˙netqin−(h4−h1)=210 MW3171.4 kJ/kg−(2282.5 kJ/kg−191.81 kJ/kg)

=210 MW(1000 kJ/s1 MW)1080.7 kJ/kg=194.3 kg/s

Thus, the mass flow rate of the steam is 194.3 kg/s.

Answer 19

(c)

To determine

The mass flow rate of the steam.

Answer 20

(c)

Expert Solution

Explanation of Solution

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

m˙=W˙netqin−qout=W˙netqin−(h4−h1)=210 MW3171.4 kJ/kg−(2282.5 kJ/kg−191.81 kJ/kg)

=210 MW(1000 kJ/s1 MW)1080.7 kJ/kg=194.3 kg/s

Thus, the mass flow rate of the steam is 194.3 kg/s.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Ch. 9 - Prob. 1P Ch. 9 - What is the difference between air-standard...Ch. 9 - Prob. 3P Ch. 9 - Prob. 4P Ch. 9 - Prob. 5P Ch. 9 - Prob. 6P Ch. 9 - Prob. 7P Ch. 9 - Prob. 8P Ch. 9 - Prob. 9P Ch. 9 - Prob. 10P

The quality of the steam at the turbine exit.

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The quality of the steam at the turbine exit.

Explanation of Solution

The thermal efficiency of the cycle.

Explanation of Solution

The mass flow rate of the steam.

Explanation of Solution

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