Chapter 18, Problem 55P

(a)

To determine

Rate at which air conditioner remove energy.

Answer to Problem 55P

The energy is $\underset{\_}{8.48\text{kW}}$.

Explanation of Solution

Air conditioner removes its energy in the form of heat from the cold reservoir.

Write the equation for coefficient of performance.

    $\text{COP}=\left(\frac{T_c}{T_h-T_c}\right)$        (I)

Here $T_c$ is the temperature of the cold reservoir and $T_h$ is the temperature of the hot reservoir.

The efficiency is given as $40\%$. Then actual coefficient of performance would be

    $0.400\left(\text{COP}\right)=\frac{\left|Q_c\right|}{\left|Q_h-Q_c\right|}$        (II)

Here $Q_c$ is the heat in the cold reservoir and $Q_h$ is the heat in the hot reservoir.

Divide the numerator and denominator of (II) with $\Delta t$

    $0.400\left(\text{COP}\right)=\frac{\left|Q_c\right|/\Delta t}{\left|Q_h\right|/\Delta t-\left|Q_c\right|/\Delta t}$        (III)

Rewrite (III) in terms of $Q_c/\Delta t$

    $\left|Q_c\right|/\Delta t=\frac{\left|Q_h\right|/\Delta t}{\left(1+\frac{1}{0.400\left(\text{COP}\right)}\right)}$        (IV)

Conclusion:

Substitute $7°\text{C}$ for $T_c$ and $27°\text{C}$ for $T_h$ in (I)

    $\begin{array}{c}\text{COP}=\left(\frac{7°\text{C}+273\text{K}}{27°\text{C+273}\text{K}-7°\text{C}+273\text{K}}\right)\\ =\frac{280\text{K}}{20\text{K}}\\ =14\end{array}$

Substitute $14$ for $\text{COP}$ and $10\text{kW}$ for $\left|Q_h\right|/\Delta t$.

    $\begin{array}{c}\left|Q_c\right|/\Delta t=\frac{10\text{kW}}{\left(1+\frac{1}{0.400\left(14\right)}\right)}\\ =8.48\text{kW}\end{array}$

The energy is $\underset{\_}{8.48\text{kW}}$.

(b)

To determine

Power required for input work.

Answer to Problem 55P

The input work is $\underset{\_}{1.52\text{kW}}$.

Explanation of Solution

Write the equation for work done by the engine

    $W_{\text{eng}}=Q_h-Q_c$        (V)

Divide (V) with $\Delta t$

    $\begin{array}{c}\frac{W_{\text{eng}}}{\Delta t}=P\\ =\frac{Q_h}{\Delta t}-\frac{Q_c}{\Delta t}\end{array}$        (VI)

Conclusion:

Substitute $10\text{kW}$ for $\left|Q_h\right|/\Delta t$ and $8.48\text{kW}$ for $\left|Q_c\right|/\Delta t$ in (VI)

    $\begin{array}{c}P=10\text{kW}-8.48\text{kW}\\ \text{=1}\text{.52}\text{kW}\end{array}$

The input work is $\underset{\_}{1.52\text{kW}}$.

(c)

To determine

The change in entropy.

Answer to Problem 55P

The net entropy change is $\underset{\_}{1.09\times {10}^4\text{J}/\text{K}}$.

Explanation of Solution

The hot reservoir increases in entropy by $\left|Q_h\right|/T_h$.

The cold reservoir decreases its entropy by $-\left|Q_c\right|/T_c$.

Write the equation for change in entropy

    $\Delta S=\frac{Q_h}{T_h}-\frac{\left|Q_c\right|}{T_c}$        (VII)

Conclusion:

Substitute $10\times {10}^3\text{J}/\text{h}$ for $Q_h$, $27°\text{C}$ for $T_h$, $8.48\times {10}^3\text{J}/\text{h}$ for $Q_c$ and $7°\text{C}$ for $T_c$ in (VII)

    $\begin{array}{c}\Delta S=\frac{\left(10\times {10}^3\text{J}/\text{h}\left(\frac{1\text{h}}{3600\text{s}}\right)\right)}{27°\text{C+273}\text{K}}-\frac{\left|\left(8.48\times {10}^3\text{J}/\text{h}\left(\frac{1\text{h}}{3600\text{s}}\right)\right)\right|}{7°\text{C+273}\text{K}}\\ =\frac{\left(10\times {10}^3\text{J}/\text{s}\left(3600\text{s}\right)\right)}{\text{300}\text{K}}-\frac{\left|\left(8.48\times {10}^3\text{J}/\text{s}\left(3600\text{s}\right)\right)\right|}{\text{280}\text{K}}\\ =-1.09\times {10}^4\text{J}/\text{K}\end{array}$

The net entropy change is $\underset{\_}{1.09\times {10}^4\text{J}/\text{K}}$.

(d)

To determine

Fractional change in $\text{COP}$ of the air conditioner.

Answer to Problem 55P

The fractional change is to drop by $\underset{\_}{20\%}$.

Explanation of Solution

Using (I) find the $\text{COP}$ for the given temperature

    $\text{COP}=\left(\frac{T_c}{T_h-T_c}\right)$

Then the actual $\text{COP}$ would be $\text{COP}\text{'}=0.400\left(\text{COP}\right)$

Take the ratio of $\text{COP}$ for $27°\text{C}$ and $\text{CO}{\text{P}}^{\prime }$

    $\frac{0.400\left(\text{COP}\right)}{0.400\left(14\right)}=\frac{\text{COP}}{14}$        (VIII)

Conclusion:

Substitute $7°\text{C}$ for $T_c$ and $32°\text{C}$ for $T_h$ in (I)

    $\begin{array}{c}\text{COP}=\left(\frac{7°\text{C}+273\text{K}}{32°\text{C+273}\text{K}-7°\text{C}+273\text{K}}\right)\\ =\frac{280\text{K}}{25\text{K}}\\ =11.2\end{array}$

Substitute $11.2$ for $\text{COP}$ in(VIII)

    $\frac{11.2}{14}=0.800$

Therefore, the fractional change is to drop by $\underset{\_}{20\%}$