To sketch the cycle on PV diagram.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 19, Problem 83P

(a)

To determine

To sketch the cycle on PV diagram.

(a)

Expert Solution

Explanation of Solution

Introduction:

The Brayton cycle consists of various thermodynamic process, an adiabatic compression, an isobaric expansion, an adiabatic expansion and an isobaric compression.

The figure below shows the PV curve of Brayton cycle,

Physics for Scientists and Engineers, Chapter 19, Problem 83P

Figure (1)

From the range of temperature T1 to T2 , the process called an adiabatic compression, for the temperature range of T2 to T3 , the process is isobaric process, for the temperature T3 to T4 , the process is adiabatic expansion and for the temperature T4 to T1 . The process is isobaric compression.

Conclusion:

Therefore, the diagram of PV curve on Brayton cycle is shown in figure (1).

(b)

To determine

Show that the efficiency of the cycle as ε=1−(T4−T1)(T3−T2)

(b)

Expert Solution

Answer to Problem 83P

The efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

Explanation of Solution

Formula used:

The expression for the efficiency is,

E=1−QcQh

Here, the heat absorbed from the heat reservoir during isobaric expansion is Qh and heat delivered to cold reservoir during isobaric compression is Qc .

The expression for change in enthalpy from first law of thermodynamics is,

Q=dEin+W

The expression for ideal gas law is,

pV=nRT

The expression for work done throught the cycle is,

W=WAB+WBC+WCD+WDA

Calculation:

The heat absorbed from the heat reservoir during isobaric expansion is calculated as,

Qh=Cp(T3−T2)+Ph(V3−V2)=Cp(T3−T2)+PhV3−PhV2

By using the ideal gas law we get,

Qh=Cp(T3−T2)+nRT3−nRT2=Cp(T3−T2)+nR(T3−T2)=(Cp+nR)(T3−T2)

Similarly,

The delivered to cold reservoir during isobaric compression is calculated as

Qc=−[Cp( T 1− T 4)+PL( V 1− V 4)]=−[Cp( T 1− T 4)+PLV1−PLV4]

By using the ideal gas law we get

Qc=−[Cp( T 1− T 4)+nRT1−nRT4]=[Cp+nR](T4−T1)

The efficiency is calculated as,

E=1−[ C p+nR]( T 4 − T 1 )[ C p+nR]( T 3 − T 2 )=1−( T 4 − T 1 )( T 3 − T 2 )

Conclusion:

Therefore, the efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

(c)

To determine

To show the efficiency is ε=1−r(1−γ)/γ .

(c)

Expert Solution

Answer to Problem 83P

The efficiency is ε=1−r(1−γ)/γ .

Explanation of Solution

Formula used:

The expression for adiabatic process is given as,

p1−γTγ=constant

Calculation:

The relation between pressure and temperature for adiabatic process AB is calculated as,

PL1−γT1γ=Ph1−γT2γPL 1−γγT1=Ph 1−γγT2 ........(1)

The relation between pressure and temperature for adiabatic process CD is calculated as,

PL1−γT4γ=Ph1−γT3γPL 1−γγT4=Ph 1−γγT3 ........(2)

Subtracting equation (1) with equation (2) we get,

PL 1−γγT4−PL 1−γγT1=Ph 1−γγT3−Ph 1−γγT2( P h P L ) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )(r) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )

By using efficiency from part (a) we get,

E=1−( T 4 − T 1 )( T 3 − T 2 )E=1−(r) 1−γγ

Conclusion:

Therefore, the efficiency is ε=1−r(1−γ)/γ .

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

How can you tell which vowel is being produced here ( “ee,” “ah,” or “oo”)? Also, how would you be able to tell for the other vowels?

You want to fabricate a soft microfluidic chip like the one below. How would you go about fabricating this chip knowing that you are targeting a channel with a square cross-sectional profile of 200 μm by 200 μm. What materials and steps would you use and why? Disregard the process to form the inlet and outlet. Square Cross Section

1. What are the key steps involved in the fabrication of a semiconductor device. 2. You are hired by a chip manufacturing company, and you are asked to prepare a silicon wafer with the pattern below. Describe the process you would use. High Aspect Ratio Trenches Undoped Si Wafer P-doped Si 3. You would like to deposit material within a high aspect ratio trench. What approach would you use and why? 4. A person is setting up a small clean room space to carry out an outreach activity to educate high school students about patterning using photolithography. They obtained a positive photoresist, a used spin coater, a high energy light lamp for exposure and ordered a plastic transparency mask with a pattern on it to reduce cost. Upon trying this set up multiple times they find that the full resist gets developed, and they are unable to transfer the pattern onto the resist. Help them troubleshoot and find out why pattern of transfer has not been successful. 5. You are given a composite…

Answer 2

Question

Chapter 19, Problem 83P

(a)

To determine

To sketch the cycle on PV diagram.

(a)

Expert Solution

Explanation of Solution

Introduction:

The Brayton cycle consists of various thermodynamic process, an adiabatic compression, an isobaric expansion, an adiabatic expansion and an isobaric compression.

The figure below shows the PV curve of Brayton cycle,

Physics for Scientists and Engineers, Chapter 19, Problem 83P

Figure (1)

From the range of temperature T1 to T2 , the process called an adiabatic compression, for the temperature range of T2 to T3 , the process is isobaric process, for the temperature T3 to T4 , the process is adiabatic expansion and for the temperature T4 to T1 . The process is isobaric compression.

Conclusion:

Therefore, the diagram of PV curve on Brayton cycle is shown in figure (1).

(b)

To determine

Show that the efficiency of the cycle as ε=1−(T4−T1)(T3−T2)

(b)

Expert Solution

Answer to Problem 83P

The efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

Explanation of Solution

Formula used:

The expression for the efficiency is,

E=1−QcQh

Here, the heat absorbed from the heat reservoir during isobaric expansion is Qh and heat delivered to cold reservoir during isobaric compression is Qc .

The expression for change in enthalpy from first law of thermodynamics is,

Q=dEin+W

The expression for ideal gas law is,

pV=nRT

The expression for work done throught the cycle is,

W=WAB+WBC+WCD+WDA

Calculation:

The heat absorbed from the heat reservoir during isobaric expansion is calculated as,

Qh=Cp(T3−T2)+Ph(V3−V2)=Cp(T3−T2)+PhV3−PhV2

By using the ideal gas law we get,

Qh=Cp(T3−T2)+nRT3−nRT2=Cp(T3−T2)+nR(T3−T2)=(Cp+nR)(T3−T2)

Similarly,

The delivered to cold reservoir during isobaric compression is calculated as

Qc=−[Cp( T 1− T 4)+PL( V 1− V 4)]=−[Cp( T 1− T 4)+PLV1−PLV4]

By using the ideal gas law we get

Qc=−[Cp( T 1− T 4)+nRT1−nRT4]=[Cp+nR](T4−T1)

The efficiency is calculated as,

E=1−[ C p+nR]( T 4 − T 1 )[ C p+nR]( T 3 − T 2 )=1−( T 4 − T 1 )( T 3 − T 2 )

Conclusion:

Therefore, the efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

(c)

To determine

To show the efficiency is ε=1−r(1−γ)/γ .

(c)

Expert Solution

Answer to Problem 83P

The efficiency is ε=1−r(1−γ)/γ .

Explanation of Solution

Formula used:

The expression for adiabatic process is given as,

p1−γTγ=constant

Calculation:

The relation between pressure and temperature for adiabatic process AB is calculated as,

PL1−γT1γ=Ph1−γT2γPL 1−γγT1=Ph 1−γγT2 ........(1)

The relation between pressure and temperature for adiabatic process CD is calculated as,

PL1−γT4γ=Ph1−γT3γPL 1−γγT4=Ph 1−γγT3 ........(2)

Subtracting equation (1) with equation (2) we get,

PL 1−γγT4−PL 1−γγT1=Ph 1−γγT3−Ph 1−γγT2( P h P L ) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )(r) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )

By using efficiency from part (a) we get,

E=1−( T 4 − T 1 )( T 3 − T 2 )E=1−(r) 1−γγ

Conclusion:

Therefore, the efficiency is ε=1−r(1−γ)/γ .

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 19, Problem 83P

(a)

To determine

To sketch the cycle on PV diagram.

(a)

Expert Solution

Explanation of Solution

Introduction:

The Brayton cycle consists of various thermodynamic process, an adiabatic compression, an isobaric expansion, an adiabatic expansion and an isobaric compression.

The figure below shows the PV curve of Brayton cycle,

Physics for Scientists and Engineers, Chapter 19, Problem 83P

Figure (1)

From the range of temperature T1 to T2 , the process called an adiabatic compression, for the temperature range of T2 to T3 , the process is isobaric process, for the temperature T3 to T4 , the process is adiabatic expansion and for the temperature T4 to T1 . The process is isobaric compression.

Conclusion:

Therefore, the diagram of PV curve on Brayton cycle is shown in figure (1).

(b)

To determine

Show that the efficiency of the cycle as ε=1−(T4−T1)(T3−T2)

(b)

Expert Solution

Answer to Problem 83P

The efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

Explanation of Solution

Formula used:

The expression for the efficiency is,

E=1−QcQh

Here, the heat absorbed from the heat reservoir during isobaric expansion is Qh and heat delivered to cold reservoir during isobaric compression is Qc .

The expression for change in enthalpy from first law of thermodynamics is,

Q=dEin+W

The expression for ideal gas law is,

pV=nRT

The expression for work done throught the cycle is,

W=WAB+WBC+WCD+WDA

Calculation:

The heat absorbed from the heat reservoir during isobaric expansion is calculated as,

Qh=Cp(T3−T2)+Ph(V3−V2)=Cp(T3−T2)+PhV3−PhV2

By using the ideal gas law we get,

Qh=Cp(T3−T2)+nRT3−nRT2=Cp(T3−T2)+nR(T3−T2)=(Cp+nR)(T3−T2)

Similarly,

The delivered to cold reservoir during isobaric compression is calculated as

Qc=−[Cp( T 1− T 4)+PL( V 1− V 4)]=−[Cp( T 1− T 4)+PLV1−PLV4]

By using the ideal gas law we get

Qc=−[Cp( T 1− T 4)+nRT1−nRT4]=[Cp+nR](T4−T1)

The efficiency is calculated as,

E=1−[ C p+nR]( T 4 − T 1 )[ C p+nR]( T 3 − T 2 )=1−( T 4 − T 1 )( T 3 − T 2 )

Conclusion:

Therefore, the efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

(c)

To determine

To show the efficiency is ε=1−r(1−γ)/γ .

(c)

Expert Solution

Answer to Problem 83P

The efficiency is ε=1−r(1−γ)/γ .

Explanation of Solution

Formula used:

The expression for adiabatic process is given as,

p1−γTγ=constant

Calculation:

The relation between pressure and temperature for adiabatic process AB is calculated as,

PL1−γT1γ=Ph1−γT2γPL 1−γγT1=Ph 1−γγT2 ........(1)

The relation between pressure and temperature for adiabatic process CD is calculated as,

PL1−γT4γ=Ph1−γT3γPL 1−γγT4=Ph 1−γγT3 ........(2)

Subtracting equation (1) with equation (2) we get,

PL 1−γγT4−PL 1−γγT1=Ph 1−γγT3−Ph 1−γγT2( P h P L ) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )(r) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )

By using efficiency from part (a) we get,

E=1−( T 4 − T 1 )( T 3 − T 2 )E=1−(r) 1−γγ

Conclusion:

Therefore, the efficiency is ε=1−r(1−γ)/γ .

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 19, Problem 83P

(a)

To determine

To sketch the cycle on PV diagram.

(a)

Expert Solution

Explanation of Solution

Introduction:

The Brayton cycle consists of various thermodynamic process, an adiabatic compression, an isobaric expansion, an adiabatic expansion and an isobaric compression.

The figure below shows the PV curve of Brayton cycle,

Physics for Scientists and Engineers, Chapter 19, Problem 83P

Figure (1)

From the range of temperature T1 to T2 , the process called an adiabatic compression, for the temperature range of T2 to T3 , the process is isobaric process, for the temperature T3 to T4 , the process is adiabatic expansion and for the temperature T4 to T1 . The process is isobaric compression.

Conclusion:

Therefore, the diagram of PV curve on Brayton cycle is shown in figure (1).

(b)

To determine

Show that the efficiency of the cycle as ε=1−(T4−T1)(T3−T2)

(b)

Expert Solution

Answer to Problem 83P

The efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

Explanation of Solution

Formula used:

The expression for the efficiency is,

E=1−QcQh

Here, the heat absorbed from the heat reservoir during isobaric expansion is Qh and heat delivered to cold reservoir during isobaric compression is Qc .

The expression for change in enthalpy from first law of thermodynamics is,

Q=dEin+W

The expression for ideal gas law is,

pV=nRT

The expression for work done throught the cycle is,

W=WAB+WBC+WCD+WDA

Calculation:

The heat absorbed from the heat reservoir during isobaric expansion is calculated as,

Qh=Cp(T3−T2)+Ph(V3−V2)=Cp(T3−T2)+PhV3−PhV2

By using the ideal gas law we get,

Qh=Cp(T3−T2)+nRT3−nRT2=Cp(T3−T2)+nR(T3−T2)=(Cp+nR)(T3−T2)

Similarly,

The delivered to cold reservoir during isobaric compression is calculated as

Qc=−[Cp( T 1− T 4)+PL( V 1− V 4)]=−[Cp( T 1− T 4)+PLV1−PLV4]

By using the ideal gas law we get

Qc=−[Cp( T 1− T 4)+nRT1−nRT4]=[Cp+nR](T4−T1)

The efficiency is calculated as,

E=1−[ C p+nR]( T 4 − T 1 )[ C p+nR]( T 3 − T 2 )=1−( T 4 − T 1 )( T 3 − T 2 )

Conclusion:

Therefore, the efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

(c)

To determine

To show the efficiency is ε=1−r(1−γ)/γ .

(c)

Expert Solution

Answer to Problem 83P

The efficiency is ε=1−r(1−γ)/γ .

Explanation of Solution

Formula used:

The expression for adiabatic process is given as,

p1−γTγ=constant

Calculation:

The relation between pressure and temperature for adiabatic process AB is calculated as,

PL1−γT1γ=Ph1−γT2γPL 1−γγT1=Ph 1−γγT2 ........(1)

The relation between pressure and temperature for adiabatic process CD is calculated as,

PL1−γT4γ=Ph1−γT3γPL 1−γγT4=Ph 1−γγT3 ........(2)

Subtracting equation (1) with equation (2) we get,

PL 1−γγT4−PL 1−γγT1=Ph 1−γγT3−Ph 1−γγT2( P h P L ) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )(r) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )

By using efficiency from part (a) we get,

E=1−( T 4 − T 1 )( T 3 − T 2 )E=1−(r) 1−γγ

Conclusion:

Therefore, the efficiency is ε=1−r(1−γ)/γ .

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 19, Problem 83P

(a)

To determine

To sketch the cycle on PV diagram.

(a)

Expert Solution

Explanation of Solution

Introduction:

The Brayton cycle consists of various thermodynamic process, an adiabatic compression, an isobaric expansion, an adiabatic expansion and an isobaric compression.

The figure below shows the PV curve of Brayton cycle,

Physics for Scientists and Engineers, Chapter 19, Problem 83P

Figure (1)

From the range of temperature T1 to T2 , the process called an adiabatic compression, for the temperature range of T2 to T3 , the process is isobaric process, for the temperature T3 to T4 , the process is adiabatic expansion and for the temperature T4 to T1 . The process is isobaric compression.

Conclusion:

Therefore, the diagram of PV curve on Brayton cycle is shown in figure (1).

(b)

To determine

Show that the efficiency of the cycle as ε=1−(T4−T1)(T3−T2)

(b)

Expert Solution

Answer to Problem 83P

The efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

Explanation of Solution

Formula used:

The expression for the efficiency is,

E=1−QcQh

Here, the heat absorbed from the heat reservoir during isobaric expansion is Qh and heat delivered to cold reservoir during isobaric compression is Qc .

The expression for change in enthalpy from first law of thermodynamics is,

Q=dEin+W

The expression for ideal gas law is,

pV=nRT

The expression for work done throught the cycle is,

W=WAB+WBC+WCD+WDA

Calculation:

The heat absorbed from the heat reservoir during isobaric expansion is calculated as,

Qh=Cp(T3−T2)+Ph(V3−V2)=Cp(T3−T2)+PhV3−PhV2

By using the ideal gas law we get,

Qh=Cp(T3−T2)+nRT3−nRT2=Cp(T3−T2)+nR(T3−T2)=(Cp+nR)(T3−T2)

Similarly,

The delivered to cold reservoir during isobaric compression is calculated as

Qc=−[Cp( T 1− T 4)+PL( V 1− V 4)]=−[Cp( T 1− T 4)+PLV1−PLV4]

By using the ideal gas law we get

Qc=−[Cp( T 1− T 4)+nRT1−nRT4]=[Cp+nR](T4−T1)

The efficiency is calculated as,

E=1−[ C p+nR]( T 4 − T 1 )[ C p+nR]( T 3 − T 2 )=1−( T 4 − T 1 )( T 3 − T 2 )

Conclusion:

Therefore, the efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

(c)

To determine

To show the efficiency is ε=1−r(1−γ)/γ .

(c)

Expert Solution

Answer to Problem 83P

The efficiency is ε=1−r(1−γ)/γ .

Explanation of Solution

Formula used:

The expression for adiabatic process is given as,

p1−γTγ=constant

Calculation:

The relation between pressure and temperature for adiabatic process AB is calculated as,

PL1−γT1γ=Ph1−γT2γPL 1−γγT1=Ph 1−γγT2 ........(1)

The relation between pressure and temperature for adiabatic process CD is calculated as,

PL1−γT4γ=Ph1−γT3γPL 1−γγT4=Ph 1−γγT3 ........(2)

Subtracting equation (1) with equation (2) we get,

PL 1−γγT4−PL 1−γγT1=Ph 1−γγT3−Ph 1−γγT2( P h P L ) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )(r) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )

By using efficiency from part (a) we get,

E=1−( T 4 − T 1 )( T 3 − T 2 )E=1−(r) 1−γγ

Conclusion:

Therefore, the efficiency is ε=1−r(1−γ)/γ .

Answer 6

Chapter 19, Problem 83P

(a)

To determine

To sketch the cycle on PV diagram.

(a)

Expert Solution

Explanation of Solution

Introduction:

The Brayton cycle consists of various thermodynamic process, an adiabatic compression, an isobaric expansion, an adiabatic expansion and an isobaric compression.

The figure below shows the PV curve of Brayton cycle,

Physics for Scientists and Engineers, Chapter 19, Problem 83P

Figure (1)

From the range of temperature T1 to T2 , the process called an adiabatic compression, for the temperature range of T2 to T3 , the process is isobaric process, for the temperature T3 to T4 , the process is adiabatic expansion and for the temperature T4 to T1 . The process is isobaric compression.

Conclusion:

Therefore, the diagram of PV curve on Brayton cycle is shown in figure (1).

Answer 7

Chapter 19, Problem 83P

(a)

To determine

To sketch the cycle on PV diagram.

Answer 8

(a)

Expert Solution

Explanation of Solution

Introduction:

The Brayton cycle consists of various thermodynamic process, an adiabatic compression, an isobaric expansion, an adiabatic expansion and an isobaric compression.

The figure below shows the PV curve of Brayton cycle,

Physics for Scientists and Engineers, Chapter 19, Problem 83P

Figure (1)

From the range of temperature T1 to T2 , the process called an adiabatic compression, for the temperature range of T2 to T3 , the process is isobaric process, for the temperature T3 to T4 , the process is adiabatic expansion and for the temperature T4 to T1 . The process is isobaric compression.

Conclusion:

Therefore, the diagram of PV curve on Brayton cycle is shown in figure (1).

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

Show that the efficiency of the cycle as ε=1−(T4−T1)(T3−T2)

(b)

Expert Solution

Answer to Problem 83P

The efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

Explanation of Solution

Formula used:

The expression for the efficiency is,

E=1−QcQh

Here, the heat absorbed from the heat reservoir during isobaric expansion is Qh and heat delivered to cold reservoir during isobaric compression is Qc .

The expression for change in enthalpy from first law of thermodynamics is,

Q=dEin+W

The expression for ideal gas law is,

pV=nRT

The expression for work done throught the cycle is,

W=WAB+WBC+WCD+WDA

Calculation:

The heat absorbed from the heat reservoir during isobaric expansion is calculated as,

Qh=Cp(T3−T2)+Ph(V3−V2)=Cp(T3−T2)+PhV3−PhV2

By using the ideal gas law we get,

Qh=Cp(T3−T2)+nRT3−nRT2=Cp(T3−T2)+nR(T3−T2)=(Cp+nR)(T3−T2)

Similarly,

The delivered to cold reservoir during isobaric compression is calculated as

Qc=−[Cp( T 1− T 4)+PL( V 1− V 4)]=−[Cp( T 1− T 4)+PLV1−PLV4]

By using the ideal gas law we get

Qc=−[Cp( T 1− T 4)+nRT1−nRT4]=[Cp+nR](T4−T1)

The efficiency is calculated as,

E=1−[ C p+nR]( T 4 − T 1 )[ C p+nR]( T 3 − T 2 )=1−( T 4 − T 1 )( T 3 − T 2 )

Conclusion:

Therefore, the efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

Answer 13

(b)

To determine

Show that the efficiency of the cycle as ε=1−(T4−T1)(T3−T2)

Answer 14

(b)

Expert Solution

Answer to Problem 83P

The efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

Formula used:

The expression for the efficiency is,

E=1−QcQh

Here, the heat absorbed from the heat reservoir during isobaric expansion is Qh and heat delivered to cold reservoir during isobaric compression is Qc .

The expression for change in enthalpy from first law of thermodynamics is,

Q=dEin+W

The expression for ideal gas law is,

pV=nRT

The expression for work done throught the cycle is,

W=WAB+WBC+WCD+WDA

Calculation:

The heat absorbed from the heat reservoir during isobaric expansion is calculated as,

Qh=Cp(T3−T2)+Ph(V3−V2)=Cp(T3−T2)+PhV3−PhV2

By using the ideal gas law we get,

Qh=Cp(T3−T2)+nRT3−nRT2=Cp(T3−T2)+nR(T3−T2)=(Cp+nR)(T3−T2)

Similarly,

The delivered to cold reservoir during isobaric compression is calculated as

Qc=−[Cp( T 1− T 4)+PL( V 1− V 4)]=−[Cp( T 1− T 4)+PLV1−PLV4]

By using the ideal gas law we get

Qc=−[Cp( T 1− T 4)+nRT1−nRT4]=[Cp+nR](T4−T1)

The efficiency is calculated as,

E=1−[ C p+nR]( T 4 − T 1 )[ C p+nR]( T 3 − T 2 )=1−( T 4 − T 1 )( T 3 − T 2 )

Conclusion:

Therefore, the efficiency of the cycle as ε=1−(T4−T1)(T3−T2) .

Answer 19

(c)

To determine

To show the efficiency is ε=1−r(1−γ)/γ .

(c)

Expert Solution

Answer to Problem 83P

The efficiency is ε=1−r(1−γ)/γ .

Explanation of Solution

Formula used:

The expression for adiabatic process is given as,

p1−γTγ=constant

Calculation:

The relation between pressure and temperature for adiabatic process AB is calculated as,

PL1−γT1γ=Ph1−γT2γPL 1−γγT1=Ph 1−γγT2 ........(1)

The relation between pressure and temperature for adiabatic process CD is calculated as,

PL1−γT4γ=Ph1−γT3γPL 1−γγT4=Ph 1−γγT3 ........(2)

Subtracting equation (1) with equation (2) we get,

PL 1−γγT4−PL 1−γγT1=Ph 1−γγT3−Ph 1−γγT2( P h P L ) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )(r) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )

By using efficiency from part (a) we get,

E=1−( T 4 − T 1 )( T 3 − T 2 )E=1−(r) 1−γγ

Conclusion:

Therefore, the efficiency is ε=1−r(1−γ)/γ .

Answer 20

(c)

To determine

To show the efficiency is ε=1−r(1−γ)/γ .

Answer 21

(c)

Expert Solution

Answer to Problem 83P

The efficiency is ε=1−r(1−γ)/γ .

Answer 22

(c)

Expert Solution

Answer 23

(c)

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

Formula used:

The expression for adiabatic process is given as,

p1−γTγ=constant

Calculation:

The relation between pressure and temperature for adiabatic process AB is calculated as,

PL1−γT1γ=Ph1−γT2γPL 1−γγT1=Ph 1−γγT2 ........(1)

The relation between pressure and temperature for adiabatic process CD is calculated as,

PL1−γT4γ=Ph1−γT3γPL 1−γγT4=Ph 1−γγT3 ........(2)

Subtracting equation (1) with equation (2) we get,

PL 1−γγT4−PL 1−γγT1=Ph 1−γγT3−Ph 1−γγT2( P h P L ) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )(r) 1−γγ=( T 4 − T 1 )( T 3 − T 2 )

By using efficiency from part (a) we get,

E=1−( T 4 − T 1 )( T 3 − T 2 )E=1−(r) 1−γγ

Conclusion:

Therefore, the efficiency is ε=1−r(1−γ)/γ .

Answer 26

Ch. 19 - Prob. 1P Ch. 19 - Prob. 2P Ch. 19 - Prob. 3P Ch. 19 - Prob. 4P Ch. 19 - Prob. 5P Ch. 19 - Prob. 6P Ch. 19 - Prob. 7P Ch. 19 - Prob. 8P Ch. 19 - Prob. 9P Ch. 19 - Prob. 10P

To sketch the cycle on PV diagram.

Videos

To sketch the cycle on PV diagram.

Explanation of Solution

Show that the efficiency of the cycle as ε=1−(T4−T1)(T3−T2)

Answer to Problem 83P

Explanation of Solution

To show the efficiency is ε=1−r(1−γ)/γ .

Answer to Problem 83P

Explanation of Solution

Want to see more full solutions like this?

Chapter 19 Solutions