Chapter 8, Problem 8.21P

(a)

Interpretation Introduction

Interpretation:

The air-standard cycles are to be sketched on PT diagrams for the Carnot cycle and also explain why a PT diagram would not be helpful for depicting power cycles involving liquid/vapor phase changes.

Concept Introduction :

Initially a general treatment is given of paths on a PT diagram for an ideal gas with constant heat capacities undergoing reversible polytropic processes. Equation 3.35c may be written as:

  $P=K{T}^{\delta /\delta -1}$

  $\ln P=\ln K+\frac{\delta }{\delta -1}\ln T$

  $\frac{dP}{P}=\frac{\delta }{\delta -1}\frac{dT}{T}$

  $\frac{dP}{dT}=\frac{\delta }{\delta -1}\frac{P}{T}$ ......(A) Sign of $dP/dT$ is that of $\delta -1,i.e.,+$

Special cases { $\begin{array}{l}\delta =0\to dP/dT=0\\ \delta =1\to dP/dT=\infty \end{array}$

Where P and T are constant

According to equation A,

  $\frac{d^{2}P}{d{T}^2}=\frac{\delta }{\delta -1}(\frac{1}{T}\frac{dP}{dT}-\frac{P}{T^2})=\frac{\delta }{\delta -1}\frac{1}{T}(\frac{\delta }{\delta -1}\frac{P}{T}-\frac{P}{T})$

  $\frac{d^{2}P}{d{T}^2}=\frac{\delta }{\delta -1}\frac{P}{T^2}$ .....eq B Sign of $d^{2}P/d{T}^2$ is that of $\delta ,i.e.,+$

(b)

Interpretation Introduction

Interpretation:

The air-standard cycles are to be sketched on PT diagrams for the Otto Cycle and also explain why a PT diagram would not be helpful for depicting power cycles involving liquid/vapor phase changes.

Concept Introduction :

Initially a general treatment is given of paths on a PT diagram for an ideal gas with constant heat capacities undergoing reversible polytropic processes. Equation 3.35c may be written as:

  $P=K{T}^{\delta /\delta -1}$

  $\ln P=\ln K+\frac{\delta }{\delta -1}\ln T$

  $\frac{dP}{P}=\frac{\delta }{\delta -1}\frac{dT}{T}$

  $\frac{dP}{dT}=\frac{\delta }{\delta -1}\frac{P}{T}$ ......(A) Sign of $dP/dT$ is that of $\delta -1,i.e.,+$

Special cases { $\begin{array}{l}\delta =0\to dP/dT=0\\ \delta =1\to dP/dT=\infty \end{array}$

Where P and T are constant

According to equation A,

  $\frac{d^{2}P}{d{T}^2}=\frac{\delta }{\delta -1}(\frac{1}{T}\frac{dP}{dT}-\frac{P}{T^2})=\frac{\delta }{\delta -1}\frac{1}{T}(\frac{\delta }{\delta -1}\frac{P}{T}-\frac{P}{T})$

  $\frac{d^{2}P}{d{T}^2}=\frac{\delta }{\delta -1}\frac{P}{T^2}$ .....eq B Sign of $d^{2}P/d{T}^2$ is that of $\delta ,i.e.,+$

(c)

Interpretation Introduction

Interpretation:

The air-standard cycles are to be sketched on PT diagrams for the Diesel cycle and also explain why a PT diagram would not be helpful for depicting power cycles involving liquid/vapor phase changes.

Concept Introduction :

Initially a general treatment is given of paths on a PT diagram for an ideal gas with constant heat capacities undergoing reversible polytropic processes. Equation 3.35c may be written as:

  $P=K{T}^{\delta /\delta -1}$

  $\ln P=\ln K+\frac{\delta }{\delta -1}\ln T$

  $\frac{dP}{P}=\frac{\delta }{\delta -1}\frac{dT}{T}$

  $\frac{dP}{dT}=\frac{\delta }{\delta -1}\frac{P}{T}$ ......(A) Sign of $dP/dT$ is that of $\delta -1,i.e.,+$

Special cases { $\begin{array}{l}\delta =0\to dP/dT=0\\ \delta =1\to dP/dT=\infty \end{array}$

Where P and T are constant

According to equation A,

  $\frac{d^{2}P}{d{T}^2}=\frac{\delta }{\delta -1}(\frac{1}{T}\frac{dP}{dT}-\frac{P}{T^2})=\frac{\delta }{\delta -1}\frac{1}{T}(\frac{\delta }{\delta -1}\frac{P}{T}-\frac{P}{T})$

  $\frac{d^{2}P}{d{T}^2}=\frac{\delta }{\delta -1}\frac{P}{T^2}$ .....eq B Sign of $d^{2}P/d{T}^2$ is that of $\delta ,i.e.,+$

(d)

Interpretation Introduction

Interpretation:

The air-standard cycles are to be sketched on PT diagrams for the Brayton cycle and also explain why a PT diagram would not be helpful for depicting power cycles involving liquid/vapor phase changes.

Concept Introduction:

Initially a general treatment is given of paths on a PT diagram for an ideal gas with constant heat capacities undergoing reversible polytropic processes. Equation 3.35c may be written as:

  $P=K{T}^{\delta /\delta -1}$

  $\ln P=\ln K+\frac{\delta }{\delta -1}\ln T$

  $\frac{dP}{P}=\frac{\delta }{\delta -1}\frac{dT}{T}$

  $\frac{dP}{dT}=\frac{\delta }{\delta -1}\frac{P}{T}$ ......(A) Sign of $dP/dT$ is that of $\delta -1,i.e.,+$

Special cases { $\begin{array}{l}\delta =0\to dP/dT=0\\ \delta =1\to dP/dT=\infty \end{array}$

Where P and T are constant

According to equation A,

  $\frac{d^{2}P}{d{T}^2}=\frac{\delta }{\delta -1}(\frac{1}{T}\frac{dP}{dT}-\frac{P}{T^2})=\frac{\delta }{\delta -1}\frac{1}{T}(\frac{\delta }{\delta -1}\frac{P}{T}-\frac{P}{T})$

  $\frac{d^{2}P}{d{T}^2}=\frac{\delta }{\delta -1}\frac{P}{T^2}$ .....eq B Sign of $d^{2}P/d{T}^2$ is that of $\delta ,i.e.,+$