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 131P

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.

Isentropic efficiency of the boiler (ηT) is 0.80.

Isentropic efficiency of the pump (ηP) is 0.95.

Calculation:

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

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 131P

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)=10.62kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.62kJ/kg=202.43 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.

h4s=2783.8 kJ/kg

Calculate the specific enthalpy at state 4(h4).

h4=h3−ηT(h3−h4s)=3375.1 kJ/kg−(0.80)(3375.1 kJ/kg−2783.7 kJ/kg)=2902.0 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 6s(x6s).

x6s=s6s−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487

Calculate the specific enthalpy at state 6s(h6s).

h6s=h5−ηT(h5−h6s)=3479.1 kJ/kg−(0.80)(3479.1 kJ/kg−2461.2 kJ/kg)=2664.8 kJ/kg>hg

Since the specific enthalpy at state 6s is greater than hg, the state is superheated.

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

T6=88.1°C

Thus, the temperature of the steam at the turbine exit is 88.1°C.

b)

To determine

The thermal efficiency of the cycle.

b)

Expert Solution

Explanation of Solution

Calculation:

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg(3375.1 kJ/kg−202.43 kJ/kg)+(3479.1 kJ/kg−2902.0 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

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg=62.7 kg/s

Thus, the mass flow rate of the steam is 62.7 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

Water is supplied at 150 ft³/s and 70 psi to a hydraulic turbine through a 3-ft inside-diameter inlet pipe as indicated in the figure below. The turbine discharge pipe has a 4.8-ft inside diameter. The static pressure at section (2), 10 ft below the turbine inlet, is 10 in. Hg vacuum. If the turbine develops 2400 hp, determine the rate of loss of available energy between sections (1) and (2). Section (1) P₁ =70psi Q=150ft³/s D₁ = 3 ft 10 ft Turbine power loss = i P₂ = 10 in. Hg vacuum D₂ =4.8ft Section (2) de hp

This problem studies the response of two single degree of freedom bridge systems shown in Figure 1 under three loading cases. The problem has two parts. Part A and Part B use the same loading cases but the system is modified. Assume the following three loading cases in both Part A and Part B: (a) Harmonic wind load acting on the bridge deck pw(t) = powsin(ωwt) with amplitude pow and forcing circular frequency ωw. (b) Harmonic displacement base excitation acting at the base of the bridge pier ug(t) = ugosin(ωgt) with amplitude ugo and displacement circular frequency ωg. (c) Rectangular pulse load acting on the bridge deck with amplitude pop and pulse duration td. Part A The system includes part of a bridge deck and a bridge pier shown in Figure 1(a). For each loading case find the symbolic expression of the peak shear force in the bridge pier assuming the following: • The bridge deck is rigid and it has a mass m. • The bridge deck is rigidly connected with the bridge pier (i.e.,…

specific speed P #2 Q.2. A Pelton wheel turbine of 1.9 m diameter works under a head of 50 m at 150 rpm. The buckets are exposed to water jet which delivers from a nozzle of 20 cm in diameter. Find the overall efficiency power produced by the wheel if the buckets deflects the jet through an angle of 163°. coefficient of velocity as 0.98 [50 Marks] ·licosply Y and no Take the

Answer 2

Question

Chapter 9, Problem 131P

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.

Isentropic efficiency of the boiler (ηT) is 0.80.

Isentropic efficiency of the pump (ηP) is 0.95.

Calculation:

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

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 131P

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)=10.62kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.62kJ/kg=202.43 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.

h4s=2783.8 kJ/kg

Calculate the specific enthalpy at state 4(h4).

h4=h3−ηT(h3−h4s)=3375.1 kJ/kg−(0.80)(3375.1 kJ/kg−2783.7 kJ/kg)=2902.0 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 6s(x6s).

x6s=s6s−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487

Calculate the specific enthalpy at state 6s(h6s).

h6s=h5−ηT(h5−h6s)=3479.1 kJ/kg−(0.80)(3479.1 kJ/kg−2461.2 kJ/kg)=2664.8 kJ/kg>hg

Since the specific enthalpy at state 6s is greater than hg, the state is superheated.

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

T6=88.1°C

Thus, the temperature of the steam at the turbine exit is 88.1°C.

b)

To determine

The thermal efficiency of the cycle.

b)

Expert Solution

Explanation of Solution

Calculation:

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg(3375.1 kJ/kg−202.43 kJ/kg)+(3479.1 kJ/kg−2902.0 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

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg=62.7 kg/s

Thus, the mass flow rate of the steam is 62.7 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 131P

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.

Isentropic efficiency of the boiler (ηT) is 0.80.

Isentropic efficiency of the pump (ηP) is 0.95.

Calculation:

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

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 131P

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)=10.62kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.62kJ/kg=202.43 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.

h4s=2783.8 kJ/kg

Calculate the specific enthalpy at state 4(h4).

h4=h3−ηT(h3−h4s)=3375.1 kJ/kg−(0.80)(3375.1 kJ/kg−2783.7 kJ/kg)=2902.0 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 6s(x6s).

x6s=s6s−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487

Calculate the specific enthalpy at state 6s(h6s).

h6s=h5−ηT(h5−h6s)=3479.1 kJ/kg−(0.80)(3479.1 kJ/kg−2461.2 kJ/kg)=2664.8 kJ/kg>hg

Since the specific enthalpy at state 6s is greater than hg, the state is superheated.

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

T6=88.1°C

Thus, the temperature of the steam at the turbine exit is 88.1°C.

b)

To determine

The thermal efficiency of the cycle.

b)

Expert Solution

Explanation of Solution

Calculation:

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg(3375.1 kJ/kg−202.43 kJ/kg)+(3479.1 kJ/kg−2902.0 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

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg=62.7 kg/s

Thus, the mass flow rate of the steam is 62.7 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 131P

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.

Isentropic efficiency of the boiler (ηT) is 0.80.

Isentropic efficiency of the pump (ηP) is 0.95.

Calculation:

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

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 131P

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)=10.62kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.62kJ/kg=202.43 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.

h4s=2783.8 kJ/kg

Calculate the specific enthalpy at state 4(h4).

h4=h3−ηT(h3−h4s)=3375.1 kJ/kg−(0.80)(3375.1 kJ/kg−2783.7 kJ/kg)=2902.0 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 6s(x6s).

x6s=s6s−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487

Calculate the specific enthalpy at state 6s(h6s).

h6s=h5−ηT(h5−h6s)=3479.1 kJ/kg−(0.80)(3479.1 kJ/kg−2461.2 kJ/kg)=2664.8 kJ/kg>hg

Since the specific enthalpy at state 6s is greater than hg, the state is superheated.

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

T6=88.1°C

Thus, the temperature of the steam at the turbine exit is 88.1°C.

b)

To determine

The thermal efficiency of the cycle.

b)

Expert Solution

Explanation of Solution

Calculation:

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg(3375.1 kJ/kg−202.43 kJ/kg)+(3479.1 kJ/kg−2902.0 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

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg=62.7 kg/s

Thus, the mass flow rate of the steam is 62.7 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 131P

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.

Isentropic efficiency of the boiler (ηT) is 0.80.

Isentropic efficiency of the pump (ηP) is 0.95.

Calculation:

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

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 131P

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)=10.62kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.62kJ/kg=202.43 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.

h4s=2783.8 kJ/kg

Calculate the specific enthalpy at state 4(h4).

h4=h3−ηT(h3−h4s)=3375.1 kJ/kg−(0.80)(3375.1 kJ/kg−2783.7 kJ/kg)=2902.0 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 6s(x6s).

x6s=s6s−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487

Calculate the specific enthalpy at state 6s(h6s).

h6s=h5−ηT(h5−h6s)=3479.1 kJ/kg−(0.80)(3479.1 kJ/kg−2461.2 kJ/kg)=2664.8 kJ/kg>hg

Since the specific enthalpy at state 6s is greater than hg, the state is superheated.

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

T6=88.1°C

Thus, the temperature of the steam at the turbine exit is 88.1°C.

b)

To determine

The thermal efficiency of the cycle.

b)

Expert Solution

Explanation of Solution

Calculation:

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg(3375.1 kJ/kg−202.43 kJ/kg)+(3479.1 kJ/kg−2902.0 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

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg=62.7 kg/s

Thus, the mass flow rate of the steam is 62.7 kg/s.

Answer 6

Chapter 9, Problem 131P

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.

Isentropic efficiency of the boiler (ηT) is 0.80.

Isentropic efficiency of the pump (ηP) is 0.95.

Calculation:

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

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 131P

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)=10.62kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.62kJ/kg=202.43 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.

h4s=2783.8 kJ/kg

Calculate the specific enthalpy at state 4(h4).

h4=h3−ηT(h3−h4s)=3375.1 kJ/kg−(0.80)(3375.1 kJ/kg−2783.7 kJ/kg)=2902.0 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 6s(x6s).

x6s=s6s−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487

Calculate the specific enthalpy at state 6s(h6s).

h6s=h5−ηT(h5−h6s)=3479.1 kJ/kg−(0.80)(3479.1 kJ/kg−2461.2 kJ/kg)=2664.8 kJ/kg>hg

Since the specific enthalpy at state 6s is greater than hg, the state is superheated.

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

T6=88.1°C

Thus, the temperature of the steam at the turbine exit is 88.1°C.

Answer 7

Chapter 9, Problem 131P

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.

Isentropic efficiency of the boiler (ηT) is 0.80.

Isentropic efficiency of the pump (ηP) is 0.95.

Calculation:

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

EBK FUNDAMENTALS OF THERMAL-FLUID SCIEN, Chapter 9, Problem 131P

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)=10.62kJ/kg

Calculate the specific enthalpy at state 2(h2).

h2=h1+wp,in=191.81 kJ/kg+10.62kJ/kg=202.43 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.

h4s=2783.8 kJ/kg

Calculate the specific enthalpy at state 4(h4).

h4=h3−ηT(h3−h4s)=3375.1 kJ/kg−(0.80)(3375.1 kJ/kg−2783.7 kJ/kg)=2902.0 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 6s(x6s).

x6s=s6s−sfsfg=7.7642 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.9487

Calculate the specific enthalpy at state 6s(h6s).

h6s=h5−ηT(h5−h6s)=3479.1 kJ/kg−(0.80)(3479.1 kJ/kg−2461.2 kJ/kg)=2664.8 kJ/kg>hg

Since the specific enthalpy at state 6s is greater than hg, the state is superheated.

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

T6=88.1°C

Thus, the temperature of the steam at the turbine exit is 88.1°C.

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 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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg(3375.1 kJ/kg−202.43 kJ/kg)+(3479.1 kJ/kg−2902.0 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

Calculation:

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg(3375.1 kJ/kg−202.43 kJ/kg)+(3479.1 kJ/kg−2902.0 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

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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg=62.7 kg/s

Thus, the mass flow rate of the steam is 62.7 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−2902 kJ/kg)+(3479.1 kJ/kg−2664.8 kJ/kg)]−10.62 kJ/kg=62.7 kg/s

Thus, the mass flow rate of the steam is 62.7 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