The quality or temperature (if superheated) of the steam at the turbine exit.

Question

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Answer 1

Question

Chapter 9, Problem 132P

a)

To determine

The quality or temperature (if superheated) of the steam at the turbine exit.

a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at state 3(P3) is 10 MPa.

Pressure of steam at state 1(P1) is 10 kPa.

Temperature of steam at the state 3(T3) is 500°C.

Net power output of the cycle (W˙net) is 80,000 kJ/s.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 132P

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

s3=s4ss5=s6s

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific 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=0.001010m3/kg

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

wp,in=v1(P2−P1)ηP=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95(1 kJ1 kPa⋅m3)=10.09kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.09kJ/kg=201.90 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 500°C.

h3=3375.1kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1 MPa and specific entropy of 6.5995 kJ/kg⋅K.

h4=2783.8 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 1 MPa and temperature of 500°C.

h5=3479.1kJ/kgs5=7.7642 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa and specific entropy of 7.7642 kJ/kg⋅K.

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

Calculate the quality at state 6(x6).

x6=s6−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487=0.949

Thus, the quality of the steam at the turbine exit is 0.949.

b)

To determine

The thermal efficiency of the cycle.

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=191.81 kJ/kg+(0.949)(2392.1 kJ/kg)=2461.2 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=wT,out−wp,in(h3−h2)+(h5−h4)=[(h3−h4)+(h5−h6)]−wp,in(h3−h2)+(h5−h4)=[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg(3375.1 kJ/kg−201.90 kJ/kg)+(3479.1 kJ/kg−2783.7 kJ/kg)

=0.413=41.3%

Thus, the thermal efficiency of the cycle is 41.3%.

c)

To determine

The mass flow rate of the steam.

c)

Expert Solution

Explanation of Solution

Calculation:

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

m˙=W˙netwnet=W˙net[(h3−h4)+(h5−h6)]−wp,in=80,000 kJ/s[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg=50 kg/s

Thus, the mass flow rate of the steam is 50 kg/s.

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Answer 2

Question

Chapter 9, Problem 132P

a)

To determine

The quality or temperature (if superheated) of the steam at the turbine exit.

a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at state 3(P3) is 10 MPa.

Pressure of steam at state 1(P1) is 10 kPa.

Temperature of steam at the state 3(T3) is 500°C.

Net power output of the cycle (W˙net) is 80,000 kJ/s.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 132P

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

s3=s4ss5=s6s

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific 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=0.001010m3/kg

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

wp,in=v1(P2−P1)ηP=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95(1 kJ1 kPa⋅m3)=10.09kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.09kJ/kg=201.90 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 500°C.

h3=3375.1kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1 MPa and specific entropy of 6.5995 kJ/kg⋅K.

h4=2783.8 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 1 MPa and temperature of 500°C.

h5=3479.1kJ/kgs5=7.7642 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa and specific entropy of 7.7642 kJ/kg⋅K.

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

Calculate the quality at state 6(x6).

x6=s6−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487=0.949

Thus, the quality of the steam at the turbine exit is 0.949.

b)

To determine

The thermal efficiency of the cycle.

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=191.81 kJ/kg+(0.949)(2392.1 kJ/kg)=2461.2 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=wT,out−wp,in(h3−h2)+(h5−h4)=[(h3−h4)+(h5−h6)]−wp,in(h3−h2)+(h5−h4)=[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg(3375.1 kJ/kg−201.90 kJ/kg)+(3479.1 kJ/kg−2783.7 kJ/kg)

=0.413=41.3%

Thus, the thermal efficiency of the cycle is 41.3%.

c)

To determine

The mass flow rate of the steam.

c)

Expert Solution

Explanation of Solution

Calculation:

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

m˙=W˙netwnet=W˙net[(h3−h4)+(h5−h6)]−wp,in=80,000 kJ/s[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg=50 kg/s

Thus, the mass flow rate of the steam is 50 kg/s.

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Answer 3

Question

Chapter 9, Problem 132P

a)

To determine

The quality or temperature (if superheated) of the steam at the turbine exit.

a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at state 3(P3) is 10 MPa.

Pressure of steam at state 1(P1) is 10 kPa.

Temperature of steam at the state 3(T3) is 500°C.

Net power output of the cycle (W˙net) is 80,000 kJ/s.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 132P

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

s3=s4ss5=s6s

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific 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=0.001010m3/kg

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

wp,in=v1(P2−P1)ηP=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95(1 kJ1 kPa⋅m3)=10.09kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.09kJ/kg=201.90 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 500°C.

h3=3375.1kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1 MPa and specific entropy of 6.5995 kJ/kg⋅K.

h4=2783.8 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 1 MPa and temperature of 500°C.

h5=3479.1kJ/kgs5=7.7642 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa and specific entropy of 7.7642 kJ/kg⋅K.

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

Calculate the quality at state 6(x6).

x6=s6−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487=0.949

Thus, the quality of the steam at the turbine exit is 0.949.

b)

To determine

The thermal efficiency of the cycle.

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=191.81 kJ/kg+(0.949)(2392.1 kJ/kg)=2461.2 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=wT,out−wp,in(h3−h2)+(h5−h4)=[(h3−h4)+(h5−h6)]−wp,in(h3−h2)+(h5−h4)=[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg(3375.1 kJ/kg−201.90 kJ/kg)+(3479.1 kJ/kg−2783.7 kJ/kg)

=0.413=41.3%

Thus, the thermal efficiency of the cycle is 41.3%.

c)

To determine

The mass flow rate of the steam.

c)

Expert Solution

Explanation of Solution

Calculation:

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

m˙=W˙netwnet=W˙net[(h3−h4)+(h5−h6)]−wp,in=80,000 kJ/s[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg=50 kg/s

Thus, the mass flow rate of the steam is 50 kg/s.

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Answer 4

Question

Chapter 9, Problem 132P

a)

To determine

The quality or temperature (if superheated) of the steam at the turbine exit.

a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at state 3(P3) is 10 MPa.

Pressure of steam at state 1(P1) is 10 kPa.

Temperature of steam at the state 3(T3) is 500°C.

Net power output of the cycle (W˙net) is 80,000 kJ/s.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 132P

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

s3=s4ss5=s6s

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific 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=0.001010m3/kg

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

wp,in=v1(P2−P1)ηP=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95(1 kJ1 kPa⋅m3)=10.09kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.09kJ/kg=201.90 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 500°C.

h3=3375.1kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1 MPa and specific entropy of 6.5995 kJ/kg⋅K.

h4=2783.8 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 1 MPa and temperature of 500°C.

h5=3479.1kJ/kgs5=7.7642 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa and specific entropy of 7.7642 kJ/kg⋅K.

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

Calculate the quality at state 6(x6).

x6=s6−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487=0.949

Thus, the quality of the steam at the turbine exit is 0.949.

b)

To determine

The thermal efficiency of the cycle.

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=191.81 kJ/kg+(0.949)(2392.1 kJ/kg)=2461.2 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=wT,out−wp,in(h3−h2)+(h5−h4)=[(h3−h4)+(h5−h6)]−wp,in(h3−h2)+(h5−h4)=[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg(3375.1 kJ/kg−201.90 kJ/kg)+(3479.1 kJ/kg−2783.7 kJ/kg)

=0.413=41.3%

Thus, the thermal efficiency of the cycle is 41.3%.

c)

To determine

The mass flow rate of the steam.

c)

Expert Solution

Explanation of Solution

Calculation:

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

m˙=W˙netwnet=W˙net[(h3−h4)+(h5−h6)]−wp,in=80,000 kJ/s[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg=50 kg/s

Thus, the mass flow rate of the steam is 50 kg/s.

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Answer 5

Chapter 9, Problem 132P

a)

To determine

The quality or temperature (if superheated) of the steam at the turbine exit.

a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at state 3(P3) is 10 MPa.

Pressure of steam at state 1(P1) is 10 kPa.

Temperature of steam at the state 3(T3) is 500°C.

Net power output of the cycle (W˙net) is 80,000 kJ/s.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 132P

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

s3=s4ss5=s6s

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific 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=0.001010m3/kg

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

wp,in=v1(P2−P1)ηP=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95(1 kJ1 kPa⋅m3)=10.09kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.09kJ/kg=201.90 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 500°C.

h3=3375.1kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1 MPa and specific entropy of 6.5995 kJ/kg⋅K.

h4=2783.8 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 1 MPa and temperature of 500°C.

h5=3479.1kJ/kgs5=7.7642 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa and specific entropy of 7.7642 kJ/kg⋅K.

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

Calculate the quality at state 6(x6).

x6=s6−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487=0.949

Thus, the quality of the steam at the turbine exit is 0.949.

b)

To determine

The thermal efficiency of the cycle.

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=191.81 kJ/kg+(0.949)(2392.1 kJ/kg)=2461.2 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=wT,out−wp,in(h3−h2)+(h5−h4)=[(h3−h4)+(h5−h6)]−wp,in(h3−h2)+(h5−h4)=[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg(3375.1 kJ/kg−201.90 kJ/kg)+(3479.1 kJ/kg−2783.7 kJ/kg)

=0.413=41.3%

Thus, the thermal efficiency of the cycle is 41.3%.

c)

To determine

The mass flow rate of the steam.

c)

Expert Solution

Explanation of Solution

Calculation:

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

m˙=W˙netwnet=W˙net[(h3−h4)+(h5−h6)]−wp,in=80,000 kJ/s[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg=50 kg/s

Thus, the mass flow rate of the steam is 50 kg/s.

Answer 6

Chapter 9, Problem 132P

a)

To determine

The quality or temperature (if superheated) of the steam at the turbine exit.

a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at state 3(P3) is 10 MPa.

Pressure of steam at state 1(P1) is 10 kPa.

Temperature of steam at the state 3(T3) is 500°C.

Net power output of the cycle (W˙net) is 80,000 kJ/s.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 132P

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

s3=s4ss5=s6s

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific 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=0.001010m3/kg

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

wp,in=v1(P2−P1)ηP=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95(1 kJ1 kPa⋅m3)=10.09kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.09kJ/kg=201.90 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 500°C.

h3=3375.1kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1 MPa and specific entropy of 6.5995 kJ/kg⋅K.

h4=2783.8 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 1 MPa and temperature of 500°C.

h5=3479.1kJ/kgs5=7.7642 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa and specific entropy of 7.7642 kJ/kg⋅K.

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

Calculate the quality at state 6(x6).

x6=s6−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487=0.949

Thus, the quality of the steam at the turbine exit is 0.949.

Answer 7

Chapter 9, Problem 132P

a)

To determine

The quality or temperature (if superheated) of the steam at the turbine exit.

Answer 8

a)

Expert Solution

Explanation of Solution

Given:

Pressure of steam at state 3(P3) is 10 MPa.

Pressure of steam at state 1(P1) is 10 kPa.

Temperature of steam at the state 3(T3) is 500°C.

Net power output of the cycle (W˙net) is 80,000 kJ/s.

Calculation:

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

Fundamentals of Thermal-Fluid Sciences, Chapter 9, Problem 132P

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

s3=s4ss5=s6s

Refer Table A-5, “Saturated water-Pressure table”, obtain the specific 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=0.001010m3/kg

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

wp,in=v1(P2−P1)ηP=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95=(0.001010m3/kg)(10,000 kPa−10 kPa)0.95(1 kJ1 kPa⋅m3)=10.09kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.09kJ/kg=201.90 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 500°C.

h3=3375.1kJ/kgs3=6.5995 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 1 MPa and specific entropy of 6.5995 kJ/kg⋅K.

h4=2783.8 kJ/kg

Refer Table A-6, “Superheated water”, obtain the specific enthalpy and specific entropy at state 5 corresponding to the pressure of 1 MPa and temperature of 500°C.

h5=3479.1kJ/kgs5=7.7642 kJ/kg⋅K

Refer Table A-5, “Saturated water-Pressure table”, obtain the following properties corresponding to the pressure of 10 kPa and specific entropy of 7.7642 kJ/kg⋅K.

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

Calculate the quality at state 6(x6).

x6=s6−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487=0.949

Thus, the quality of the steam at the turbine exit is 0.949.

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

Calculation:

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=191.81 kJ/kg+(0.949)(2392.1 kJ/kg)=2461.2 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=wT,out−wp,in(h3−h2)+(h5−h4)=[(h3−h4)+(h5−h6)]−wp,in(h3−h2)+(h5−h4)=[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg(3375.1 kJ/kg−201.90 kJ/kg)+(3479.1 kJ/kg−2783.7 kJ/kg)

=0.413=41.3%

Thus, the thermal efficiency of the cycle is 41.3%.

Answer 13

b)

To determine

The thermal efficiency of the cycle.

Answer 14

b)

Expert Solution

Explanation of Solution

Calculation:

Calculate the specific enthalpy at state 6(h6).

h6=hf+x6hfg=191.81 kJ/kg+(0.949)(2392.1 kJ/kg)=2461.2 kJ/kg

Calculate the thermal efficiency of the cycle (ηth).

ηth=wnetqin=wT,out−wp,in(h3−h2)+(h5−h4)=[(h3−h4)+(h5−h6)]−wp,in(h3−h2)+(h5−h4)=[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg(3375.1 kJ/kg−201.90 kJ/kg)+(3479.1 kJ/kg−2783.7 kJ/kg)

=0.413=41.3%

Thus, the thermal efficiency of the cycle is 41.3%.

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

Calculation:

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

m˙=W˙netwnet=W˙net[(h3−h4)+(h5−h6)]−wp,in=80,000 kJ/s[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg=50 kg/s

Thus, the mass flow rate of the steam is 50 kg/s.

Answer 19

c)

To determine

The mass flow rate of the steam.

Answer 20

c)

Expert Solution

Explanation of Solution

Calculation:

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

m˙=W˙netwnet=W˙net[(h3−h4)+(h5−h6)]−wp,in=80,000 kJ/s[(3375.1 kJ/kg−2783.7 kJ/kg)+(3479.1 kJ/kg−2461.2 kJ/kg)]−10.09 kJ/kg=50 kg/s

Thus, the mass flow rate of the steam is 50 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 or temperature (if superheated) of the steam at the turbine exit.

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The quality or temperature (if superheated) 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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