To sketch the cycle on PV diagram.

Question

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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, Vol. 3, 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 .

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Students have asked these similar questions

A monatomic ideal gas initially has a temperature of 330k and preassure of 3x105 Pa. The gas expands from a volume of 500cm3 to a volume of 1500cm3. l) Calculate the change in internal energy (in Joules), if the expansion is adiabatic. answer => (150)ln(3)J NOTICE : The work in adiabatic process equals the change in internal energy. So, w = - Δu = - (g/2*n*r*Δt) The answer above is right, and was calculated using integrals. I would like to know how to calculate using Δu = g/2*n*r*Δt, since the work in adiabatic process = -Δu(change in internal energy)

Consider a monatomic ideal gas operating through the Carnot cycle. The initial volume of the gas is V1 = 130 × 10-3 m3. What types of processes are going on for each step in this process? During the isothermal compression step, the volume of gas is reduced by a factor of 4. In the adiabatic heating step, the temperature of the gas is doubled. What is the volume at point 3, in cubic meters?

In an isothermal process, one mole of an ideal monatomic gas at a temperature T is taken from an initial pressure P to a final pressure P/3. Using the convention that work is positive when it is done on the system, what is the work done during the process in terms of R and T?

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, Vol. 3, 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 .

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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, Vol. 3, 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 .

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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, Vol. 3, 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 .

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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, Vol. 3, 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, Vol. 3, 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, Vol. 3, 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.

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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

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