Chapter 4.8, Problem 27P

(A)

Interpretation Introduction

Interpretation:

The efficiency of a Carnot heat engine

Concept Introduction:

The expression to calculate the efficiency of a Carnot heat engine $\left({\eta }_{CarnotH.E}\right)$.

${\eta }_{CarnotH.E}=1-\frac{T_C}{T_H}$

Here, temperature of a low temperature heat reservoir is $T_C$, and temperature of a high temperature heat reservoir is $T_H$.

Explanation of Solution

Given information:

High temperature reservoir is $\text{500°F}$.

Low temperature reservoir is $\text{75°F}$.

Calculate the efficiency of a Carnot heat engine $\left({\eta }_{CarnotH.E}\right)$.

${\eta }_{CarnotH.E}=1-\frac{T_C}{T_H}$        (1)

Substitute $\text{500°F}$ for $T_H$ and $\text{75°F}$ for $T_C$ in Equation (1).

$\begin{array}{c}{\eta }_{CarnotH.E}=1-\frac{\text{75°F}}{\text{500°F}}\\ =1-\frac{\left(\text{75}+459.67\right)\text{°R}}{\left(\text{500}+459.67\right)\text{°R}}\\ =1-\frac{534.67\text{°R}}{959.67\text{°R}}\end{array}$

$\begin{array}{c}=0.443\\ {\eta }_{CarnotH.E}=44.3\%\end{array}$

Thus, the efficiency of a Carnot heat engine $\left({\eta }_{CarnotH.E}\right)$ is $44.3\%$.

(B)

Interpretation Introduction

Interpretation:

The efficiency of a Carnot heat engine

Concept Introduction:

The expression to calculate the efficiency of a Carnot heat engine $\left({\eta }_{CarnotH.E}\right)$.

${\eta }_{CarnotH.E}=1-\frac{T_C}{T_H}$

Here, temperature of a low temperature heat reservoir is $T_C$, and temperature of a high temperature heat reservoir is $T_H$.

Explanation of Solution

Given information:

High temperature reservoir is $\text{470°F}$.

Low temperature reservoir is $\text{100°F}$.

Calculate the efficiency of a Carnot heat engine $\left({\eta }_{CarnotH.E}\right)$.

${\eta }_{CarnotH.E}=1-\frac{T_C}{T_H}$        (2)

Substitute $\text{470°F}$ for $T_H$ and $\text{100°F}$ for $T_C$ in Equation (2).

$\begin{array}{c}{\eta }_{CarnotH.E}=1-\frac{T_C}{T_H}\\ =1-\frac{\text{100°F}}{\text{470°F}}\\ =1-\frac{\left(100+459.67\right)\text{°R}}{\left(470+459.67\right)\text{°R}}\\ =1-\frac{559.67\text{°R}}{929.67\text{°R}}\end{array}$

$\begin{array}{c}=0.397\\ {\eta }_{CarnotH.E}=39.7\%\end{array}$

Thus, the efficiency of a Carnot heat engine $\left({\eta }_{CarnotH.E}\right)$ is $39.7\%$.

(C)

Interpretation Introduction

Interpretation:

The rate at which entropy the entropy of the universe is increased by the processes in the boiler and condenser.

Concept Introduction:

The steady state energy balance equation for the boiler.

$\frac{d}{dt}\left\{M\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +{\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}\end{array}\right]$

Here, time taken is $t$, total mass is $M$, specific internal energy is $\widehat{U}$, velocity is $V$, acceleration due to gravity is $g$, height is $h$, initial mass flow rate is ${\dot{m}}_{in}$, initial specific enthalpy is ${\widehat{H}}_{in}$, initial velocity is $V_{in}$, initial height of the gas is $h_{in}$, final mass flow is $m_{out}$, final height of the gas is $h_{out}$, rate of shaft work is added to the system is ${\dot{W}}_S$, rate of work is added to the system through expansion or contraction of the system is ${\dot{W}}_{EC}$, and rate of heat is added or removed to the system is $\dot{Q}$.

Write the expression to calculate the net change in specific enthalpy $\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)$.

${\widehat{H}}_{out}-{\widehat{H}}_{in}={\displaystyle {\int }_{\text{100°F}}^{\text{300°F}}{\widehat{C}}_{P,liq}dT}+\Delta {\widehat{H}}_f+{\displaystyle {\int }_{\text{300°F}}^{\text{500°F}}{\widehat{C}}_{P,vap}dT}$

Here, constant pressure heat capacity of liquid is ${\widehat{C}}_{P,liq}$, constant pressure heat capacity of vapor is ${\widehat{C}}_{P,vap}$, change in temperature is $dT$, and enthalpy of formation is $\Delta {\widehat{H}}_f$.

Write the expression to calculate the change in entropy of a heat reservoir $\left(dS\right)$.

$\frac{dS}{dT}=\frac{-\dot{Q}}{T_{in}}$

Here, change in temperature is $dT$, and initial temperature of high temperature reservoir is $T_{in}$.

Write the steady state entropy balance equation for boiler.

$\frac{d\left(M\widehat{S}\right)}{dt}={\dot{m}}_{in}{\widehat{S}}_{in}-{\dot{m}}_{out}{\widehat{S}}_{out}+\frac{\dot{Q}}{T}+{\dot{S}}_{gen}$

Here, total mass is $M$, specific entropy is $\widehat{S}$, initial specific entropy is ${\widehat{S}}_{in}$, final specific entropy is ${\widehat{S}}_{out}$, and rate at which entropy is generated within the boundaries of the system is ${\dot{S}}_{gen}$.

The formula to calculate the net entropy $\left(S_{out}-S_{in}\right)$.

$S_{out}-S_{in}={\displaystyle {\int }_{initial}^{final}\frac{d{Q}_{rev}}{T}}$

Here, final entropy is $S_{out}$, initial entropy is $S_{in}$, and reversible heat is $Q_{rev}$.

The expression to calculate the rate of heat $\left(\dot{Q}\right)$ added or removed to the system.

$\dot{Q}=\dot{m}\Delta {\widehat{H}}_f$

Here, enthalpy of formation is $\Delta {\widehat{H}}_f$.

The expression to calculate the change in specific entropy $\left({\widehat{S}}_{out}-{\widehat{S}}_{in}\right)$.

${\widehat{S}}_{out}-{\widehat{S}}_{in}=\frac{-\Delta {\widehat{H}}_f}{T}$

The expression to calculate the rate at which entropy is generated within the boundaries of the system.

${\dot{S}}_{gen}=\frac{dS}{dt}+\dot{m}\left({\widehat{S}}_{out}-{\widehat{S}}_{in}\right)$

Explanation of Solution

Calculate the heat transfer of the boiler using the steady state equation.

$\frac{d}{dt}\left\{M\left(\widehat{U}+\frac{V^2}{2}+gh\right)\right\}=\left[\begin{array}{l}{\dot{m}}_{in}\left({\widehat{H}}_{in}+\frac{V_{in}^2}{2}+g{h}_{in}\right)-{\dot{m}}_{out}\left({\widehat{H}}_{out}+\frac{V_{out}^2}{2}+g{h}_{out}\right)\\ +{\dot{W}}_S+{\dot{W}}_{EC}+\dot{Q}\end{array}\right]$        (3)

Rewrite the steady state equation (3) by neglecting the work done, potential energy, and kinetic energy.

$\begin{array}{l}0={\dot{m}}_{in}{\widehat{H}}_{in}-{\dot{m}}_{out}{\widehat{H}}_{out}+\dot{Q}\\ \dot{Q}=\dot{m}\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)\end{array}$        (4)

Calculate the net change in specific enthalpy $\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)$.

${\widehat{H}}_{out}-{\widehat{H}}_{in}={\displaystyle {\int }_{\text{100°F}}^{\text{300°F}}{\widehat{C}}_{P,liq}dT}+\Delta {\widehat{H}}_f+{\displaystyle {\int }_{\text{300°F}}^{\text{500°F}}{\widehat{C}}_{P,vap}dT}$        (5)

Substitute $\text{1}\text{.2}\text{BTU}/{\text{lb}}_{\text{m}}$ for ${\widehat{C}}_{P,liq}$, $\text{450}\text{BTU}/{\text{lb}}_{\text{m}}$ for $\Delta {\widehat{H}}_f$, and $1.5+0.003T$ for ${\widehat{C}}_{P,vap}$ in Equation (5).

$\begin{array}{c}{\widehat{H}}_{out}-{\widehat{H}}_{in}={\displaystyle {\int }_{\text{100°F}}^{\text{300°F}}\left(\text{1}\text{.2}\text{BTU}/{\text{lb}}_{\text{m}}\right)dT}+\left(\text{450}\text{BTU}/{\text{lb}}_{\text{m}}\right)+{\displaystyle {\int }_{\text{300°F}}^{\text{500°F}}\left(1.5+0.003T\right)dT}\\ =\left[\begin{array}{c}{\displaystyle {\int }_{\left(100+459.67\right)\text{°R}}^{\left(300+459.67\right)\text{°R}}\left(\text{1}\text{.2}\text{BTU}/{\text{lb}}_{\text{m}}\right)dT}+\left(\text{450}\text{BTU}/{\text{lb}}_{\text{m}}\right)+\\ {\displaystyle {\int }_{\left(300+459.67\right)\text{°R}}^{\left(500+459.67\right)\text{°R}}\left(1.5+0.003T\right)dT}\end{array}\right]\\ ={\displaystyle {\int }_{559.67\text{°R}}^{759.67\text{°R}}\left(\text{1}\text{.2}\text{BTU}/{\text{lb}}_{\text{m}}\right)dT}+\left(\text{450}\text{BTU}/{\text{lb}}_{\text{m}}\right)+{\displaystyle {\int }_{759.67\text{°R}}^{959.67\text{°R}}\left(1.5+0.003T\right)dT}\end{array}$$\begin{array}{c}=\left\{\begin{array}{c}\left[\left(\text{1}\text{.2}\text{BTU}/{\text{lb}}_{\text{m}}\right)\left(759.67\text{°R}-559.67\text{°R}\right)\right]+\left(\text{450}\text{BTU}/{\text{lb}}_{\text{m}}\right)\\ +\left[\left(1.5+0.003\right)\left(959.67\text{°R}-559.67\text{°R}\right)\right]\end{array}\right\}\\ {\widehat{H}}_{out}-{\widehat{H}}_{in}=988.1\text{BTU}/{\text{lb}}_{\text{m}}\end{array}$

Substitute $988.1\text{BTU}/{\text{lb}}_{\text{m}}$ for ${\widehat{H}}_{out}-{\widehat{H}}_{in}$, and $\text{500}{\text{lb}}_{\text{m}}/\text{min}$ for $\dot{m}$ in Equation (4).

$\begin{array}{c}\dot{Q}=\dot{m}\left({\widehat{H}}_{out}-{\widehat{H}}_{in}\right)\\ =\left(\text{500}{\text{lb}}_{\text{m}}/\text{min}\right)\left(988.1\text{BTU}/{\text{lb}}_{\text{m}}\right)\\ =494,000\text{BTU}/\min \end{array}$

Calculate the change in entropy of a heat reservoir $\left(dS\right)$.

$\frac{dS}{dT}=\frac{-\dot{Q}}{T_{in}}$        (6)

Substitute $494,000\text{BTU}/\min$ for $\dot{Q}$, and $\text{500°F}$ for $T_{in}$ in Equation (6).

$\begin{array}{c}\frac{dS}{dT}=\frac{-\left(494,000\text{BTU}/\min \right)}{\text{500°F}}\\ =\frac{-\left(494,000\text{BTU}/\min \right)}{\left(\text{500}+459.67\right)\text{°R}}\\ =\frac{-\left(494,000\text{BTU}/\min \right)}{959.67\text{°R}}\\ =-514.6\text{BTU}/\text{°Rmin}\end{array}$

Calculate the entropy changes for the system that includes both the heat reservoir and the boiler.

$\begin{array}{l}\frac{d\left(M\widehat{S}\right)}{dt}={\dot{m}}_{in}{\widehat{S}}_{in}-{\dot{m}}_{out}{\widehat{S}}_{out}+{\dot{S}}_{gen}\\ \frac{dS}{dT}=\dot{m}\left({\widehat{S}}_{in}-{\widehat{S}}_{out}\right)+{\dot{S}}_{gen}\\ {\dot{S}}_{gen}=\frac{dS}{dT}+\dot{m}\left({\widehat{S}}_{out}-{\widehat{S}}_{in}\right)\end{array}$        (7)

Substitute $-514.6\text{BTU}/\text{°Rmin}$ for $\frac{dS}{dT}$, and $\text{500}{\text{lb}}_{\text{m}}/\text{min}$ for $\dot{m}$ in Equation (7).

${\dot{S}}_{gen}=\left(-514.6\text{BTU}/\text{°Rmin}\right)+\left(\text{500}{\text{lb}}_{\text{m}}/\text{min}\right)\left({\widehat{S}}_{out}-{\widehat{S}}_{in}\right)$        (8)

Write the internal energy equation.

$\begin{array}{c}dU=dQ+d{W}_{EC}\\ =dQ-PdV\end{array}$        (9)

Here, change in work added to the system through expansion or contraction of the system is $d{W}_{EC}$, change in internal energy is $dU$, change in heat is $dQ$, pressure is $P$, and change in volume is $dV$.

Since constant pressure heat capacity $\left(C_P\right)$ is known, re-write the Equation (9) in terms of change of enthalpy $\left(dH\right)$.

$dH=dU+VdP+PdV$        (10)

Here, volume is $V$, and change in pressure is $dP$.

Substitute $0$ for $dP$ in Equation (10).

$\begin{array}{c}dH=dU+V\left(0\right)+PdV\\ =dU+PdV\\ dU=dH-PdV\end{array}$

Substitute $dH-PdV$ for $dU$ in Equation (9).

$\begin{array}{l}dU=dQ-PdV\\ dH-PdV=dQ-PdV\\ dH=dQ\end{array}$

Calculate the net specific entropy $\left({\widehat{S}}_{out}-{\widehat{S}}_{in}\right)$.

$\begin{array}{c}{\widehat{S}}_{out}-{\widehat{S}}_{in}={\displaystyle {\int }_{initial}^{final}\frac{d\widehat{H}}{T}}\\ ={\displaystyle {\int }_{P=5\text{bar},T=100\text{°F}}^{P=5\text{bar},T=470\text{°F}}\frac{d\widehat{H}}{T}}\\ ={\displaystyle {\int }_{P=5\text{bar},T=100\text{°F}}^{P=5\text{bar},T=300\text{°F}}\frac{{\widehat{C}}_{P,liq}}{T}dT}+\frac{\Delta {\widehat{H}}_f}{T_{boiling}}+{\displaystyle {\int }_{P=5\text{bar},T=300\text{°F}}^{P=5\text{bar},T=470\text{°F}}\frac{{\widehat{C}}_{P,vap}}{T}dT}\end{array}$        (11)

Here, enthalpy of vaporization is $\Delta {\widehat{H}}_{vap}$, and boiling temperature is $T_{boiling}$.

Substitute $\text{1}\text{.2}\text{BTU}/{\text{lb}}_{\text{m}}$ for ${\widehat{C}}_{P,liq}$, $\text{450}\text{BTU}/{\text{lb}}_{\text{m}}$ for $\Delta {\widehat{H}}_f$, $\text{300°F}$ for $T_{boiling}$ and $1.5+0.003T$ for ${\widehat{C}}_{P,vap}$ in Equation (11).

$\begin{array}{c}{\widehat{S}}_{out}-{\widehat{S}}_{in}={\displaystyle {\int }_{100\text{°F}}^{300\text{°F}}\frac{\left(\text{1}\text{.2}\text{BTU}/{\text{lb}}_{\text{m}}\right)}{T}dT}+\frac{\left(\text{450}\text{BTU}/{\text{lb}}_{\text{m}}\right)}{\text{300°F}}+{\displaystyle {\int }_{300\text{°F}}^{470\text{°F}}\frac{1.5+0.003T}{T}dT}\\ =\left\{\begin{array}{c}{\displaystyle {\int }_{\left(100+459.67\right)\text{°R}}^{\left(300+459.67\right)\text{°R}}\frac{\left(\text{1}\text{.2}\text{BTU}/{\text{lb}}_{\text{m}}\right)}{T}dT}+\frac{\left(\text{450}\text{BTU}/{\text{lb}}_{\text{m}}\right)}{\left(300+459.67\right)\text{°R}}+\\ {\displaystyle {\int }_{\left(300+459.67\right)\text{°R}}^{\left(470+459.67\right)\text{°R}}\frac{1.5+0.003T}{T}dT}\end{array}\right\}\\ ={\displaystyle {\int }_{559.67\text{°R}}^{759.67\text{°R}}\frac{\left(\text{1}\text{.2}\text{BTU}/{\text{lb}}_{\text{m}}\right)}{T}dT}+\frac{\left(\text{450}\text{BTU}/{\text{lb}}_{\text{m}}\right)}{\text{759}\text{.67°R}}+{\displaystyle {\int }_{\text{759}\text{.67°R}}^{929.67\text{°R}}\frac{1.5+0.003T}{T}dT}\end{array}$$\begin{array}{c}=\left\{\begin{array}{c}\left(\text{1}\text{.2}\text{BTU}/{\text{lb}}_{\text{m}}\right)\left[\ln \frac{759.67\text{°R}}{559.67\text{°R}}\right]+\frac{\left(\text{450}\text{BTU}/{\text{lb}}_{\text{m}}\right)}{\text{759}\text{.67°R}}+\left(\text{1}\text{.2}\text{BTU}/{\text{lb}}_{\text{m}}\cdot \text{°R}\right)\\ \left[\ln \frac{\text{929}\text{.67°R}}{759.67\text{°R}}\right]+\left(0.003\right)\left(\text{929}\text{.67}-759.67\right)\text{BTU}/{\text{lb}}_{\text{m}}\cdot \text{°R}\end{array}\right\}\\ {\widehat{S}}_{out}-{\widehat{S}}_{in}=1.31\text{BTU}/{\text{lb}}_{\text{m}}\cdot \text{°R}\end{array}$

Substitute $1.31\text{BTU}/{\text{lb}}_{\text{m}}\cdot \text{°R}$ for ${\widehat{S}}_{out}-{\widehat{S}}_{in}$ in Equation (7).

$\begin{array}{c}{\dot{S}}_{gen}=\left(-514.6\text{BTU}/\text{°Rmin}\right)+\left(\text{500}{\text{lb}}_{\text{m}}/\text{min}\right)\left({\widehat{S}}_{out}-{\widehat{S}}_{in}\right)\\ =\left(-514.6\text{BTU}/\text{°Rmin}\right)+\left(\text{500}{\text{lb}}_{\text{m}}/\text{min}\right)\left(1.31\text{BTU}/{\text{lb}}_{\text{m}}\cdot \text{°R}\right)\\ =141.5\text{BTU}/{\text{lb}}_{\text{m}}\cdot \text{°R}\end{array}$

Thus, the rate at which entropy is generated within the boiler is $141.5\text{BTU}/{\text{lb}}_{\text{m}}\cdot \text{°R}$.

Calculate the rate of heat $\left(\dot{Q}\right)$ added or removed to the system.

$\dot{Q}=\dot{m}\Delta {\widehat{H}}_f$        (12)

Substitute $\text{500}{\text{lb}}_{\text{m}}/\text{min}$ for $\dot{m}$, and $\text{500}\text{BTU}/{\text{lb}}_{\text{m}}$ for $\Delta {\widehat{H}}_f$ in Equation (12).

$\begin{array}{c}\dot{Q}=\left(\text{500}{\text{lb}}_{\text{m}}/\text{min}\right)\left(\text{500}\text{BTU}/{\text{lb}}_{\text{m}}\right)\\ =250,000\text{BTU}/\text{min}\end{array}$

Calculate the rate of entropy increase $\left(dS\right)$ in the heat reservoir.

$\frac{dS}{dt}=\frac{\dot{Q}}{T_C}$        (13)

Substitute $250,000\text{BTU}/\text{min}$ for $\dot{Q}$, and $\text{75°F}$ for $T_C$ in Equation (13).

$\begin{array}{c}\frac{dS}{dt}=\frac{250,000\text{BTU}/\text{min}}{\text{75°F}}\\ =\frac{250,000\text{BTU}/\text{min}}{\left(\text{75}+459.67\right)\text{°R}}\\ =\frac{250,000\text{BTU}/\text{min}}{\text{534}\text{.67°R}}\\ =467.3\text{BTU}/\text{°R}\cdot \text{min}\end{array}$

Calculate the change in specific entropy $\left({\widehat{S}}_{out}-{\widehat{S}}_{in}\right)$.

${\widehat{S}}_{out}-{\widehat{S}}_{in}=\frac{-\Delta {\widehat{H}}_f}{T}$        (14)

Substitute $\text{500}\text{BTU}/{\text{lb}}_{\text{m}}$ for $\Delta {\widehat{H}}_f$, and $\text{100°F}$ for $T$ in Equation (14).

$\begin{array}{c}{\widehat{S}}_{out}-{\widehat{S}}_{in}=\frac{-\left(\text{500}\text{BTU}/{\text{lb}}_{\text{m}}\right)}{\text{100°F}}\\ =\frac{-\left(\text{500}\text{BTU}/{\text{lb}}_{\text{m}}\right)}{\left(100+459.67\right)\text{°R}}\\ =\frac{-\left(\text{500}\text{BTU}/{\text{lb}}_{\text{m}}\right)}{\text{559}\text{.67°R}}\\ =-0.893\text{BTU}/\text{°R}\cdot \text{min}\end{array}$

Calculate the rate at which entropy is generated within the boundaries of the system.

${\dot{S}}_{gen}=\frac{dS}{dt}+\dot{m}\left({\widehat{S}}_{out}-{\widehat{S}}_{in}\right)$        (15)

Substitute $467.3\text{BTU}/\text{°R}\cdot \text{min}$ for $\frac{dS}{dt}$, $-0.893\text{BTU}/{\text{lb}}_{\text{m}}\cdot \text{°R}$ for ${\widehat{S}}_{out}-{\widehat{S}}_{in}$, and $\text{500}{\text{lb}}_{\text{m}}/\text{min}$ for $\dot{m}$ in Equation (15).

$\begin{array}{c}{\dot{S}}_{gen}=\left(467.3\text{BTU}/\text{°R}\cdot \text{min}\right)+\left(\text{500}{\text{lb}}_{\text{m}}/\text{min}\right)\left(-0.893\text{BTU}/{\text{lb}}_{\text{m}}\cdot \text{°R}\right)\\ =20.9\text{BTU}/{\text{lb}}_{\text{m}}\cdot \text{°R}\end{array}$

Thus, the rate at which entropy is generated within the condenser is $20.9\text{BTU}/{\text{lb}}_{\text{m}}\cdot \text{°R}$.