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

Question

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

1) In each of the following scenarios, based on the plane of impact (shown with an (n, t)) and the motion of mass 1, draw the direction of motion of mass 2 after the impact. Note that in all scenarios, mass 2 is initially at rest. What can you say about the nature of the motion of mass 2 regardless of the scenario? m1 15 <+ m2 2) y "L χ m1 m2 m1 בז m2 F

8. In the following check to see if the set S is a vector subspace of the corresponding Rn. If it is not, explain why not. If it is, then find a basis and the dimension. X1 (a) S = X2 {[2], n ≤ n } c X1 X2 CR² X1 (b) S X2 = X3 X4 x1 + x2 x3 = 0

2) Suppose that two unequal masses m₁ and m₂ are moving with initial velocities V₁ and V₂, respectively. The masses hit each other and have a coefficient of restitution e. After the impact, mass 1 and 2 head to their respective gaps at angles a and ẞ, respectively. Derive expressions for each of the angles in terms of the initial velocities and the coefficient of restitution. m1 m2 8 m1 ↑ บา m2 ñ В

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Videos

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

Want to see more full solutions like this?

Chapter 9 Solutions