Chapter 19, Problem 84P

(a)

To determine

The PV diagram for the cycle.

Answer to Problem 84P

The required PV diagram is shown in Figure 1.

Explanation of Solution

Formula:

The range of the isobaric expansion cycle is given by,

  $T_1\to T_4$

The range of the adiabatic compression cycle is given by,

  $T_4\to T_3$

The range for the isobaric compression cycle is given by,

  $T_3\to T_2$

The range of the adiabatic expansion cycle is given by,

  $T_2\to T_1$

Calculation:

The state functions A for the PV graph are given by,

  $\left(P_L,V_1,T_1\right)$

The state functions B for the PV graph are given by,

  $\left(P_L,V_4,T_4\right)$

The state functions C for the PV graph are given by,

  $\left(P_h,V_3,T_3\right)$

The state functions D for the PV graph are given by,

  $\left(P_h,V_2,T_2\right)$

The required PV diagram is shown in Figure 1

  

Figure 1

Conclusion:

Therefore, the required PV diagram is shown in Figure 1

(b)

To determine

The expression for the coefficient of performance.

Answer to Problem 84P

The value of ${\text{COP}}_{\text{B}}$ is $\frac{T_4-T_1}{T_3-T_2-T_4-T_1}$ .n

Explanation of Solution

Formula Used:

The expression for the coefficient of performance is given by,

  $\text{COP}=\frac{Q}{W}$

The expression for $\frac{Q}{W}$ is given by,

  $\frac{Q_C}{W}=\frac{Q_c}{Q_h-Q_c}$

The expression for the value of $Q$ is given by,

  $Q=d{E}_{\mathrm{int}}+W$

The expression for the work done during the cycle is given by,

  $W=P_L\left(V_4-V_1\right)$

The expression for the heat released in the entire cycle is given by,

  $d{E}_{\mathrm{int}}=C_P\left(T_4-T_1\right)$

The expression to determine the value of $Q_{\text{h}}$ is given by,

  $Q_{\text{h}}=\left[{\text{C}}_{\text{p}}+nR\right]\left[T_3-T_2\right]$

The expression to determine the value of ${\text{COP}}_{\text{B}}$ is given by,

  ${\text{COP}}_{\text{B}}=\frac{Q_C}{Q_h-Q_C}$

Calculation:

The expression for the value of the $Q$ is evaluated as,

  $\begin{array}{l}Q=d{E}_{\mathrm{int}}+W\\ Q_C=C_P\left(T_4-T_1\right)+nR{T}_4-nR{T}_1\\ Q_C=\left(T_4-T_1\right)\left(C_p+nR\right)\end{array}$

The difference between the value of $Q_{\text{h}}$ and $Q_C$ is evaluated as,

  $\begin{array}{c}{Q}_{\text{h}}-Q_C=\left[{\text{C}}_{\text{p}}+nR\right]\left[T_3-T_2\right]-\left(T_4-T_1\right)\left(C_p+nR\right)\\ =\left[{\text{C}}_{\text{p}}+nR\right]\left[T_3-T_2-T_4-T_1\right]\end{array}$

The value of ${\text{COP}}_{\text{B}}$ is evaluated as,

  $\begin{array}{c}{\text{COP}}_{\text{B}}=\frac{Q_C}{Q_h-Q_C}\\ =\frac{\left(T_4-T_1\right)\left(C_p+nR\right)}{\left[{\text{C}}_{\text{p}}+nR\right]\left[T_3-T_2-T_4-T_1\right]}\\ =\frac{T_4-T_1}{T_3-T_2-T_4-T_1}\end{array}$

Conclusion:

Therefore, the value of ${\text{COP}}_{\text{B}}$ is $\frac{T_4-T_1}{T_3-T_2-T_4-T_1}$ .

(c)

To determine

The coefficient of performance of the refrigerator.

Answer to Problem 84P

The value of value of ${\text{COP}}_{\text{B}}$ is $1.12$ .

Explanation of Solution

Given:

The volume $V_1$ is $60\text{mL}$ .

The initial pressure $P_1$ is $1.0\text{atm}$ .

The volume $V_4$ after expansion is $75\text{mL}$ .

The temperature $T_4$ after expansion is $-25°\text{C}$ .

The pressure ratio $r$ is $5$ .

Formula:

The expression for the temperature and the pressure after adiabatic process BC is given by,

  $P_r^{\frac{1-r}{r}}{T}_4=P_{\text{h}}^{\frac{1-r}{r}}{T}_3$ ….. (1)

The expression for the temperature and the pressure after adiabatic process DA is given by,

  $P_L^{\frac{1-r}{r}}{T}_1=P_L^{\frac{1-r}{r}}{T}_2$

From above and from equation (1) the value of $r^{\frac{r-1}{r}}$ is evaluated as,

  $r^{\frac{r-1}{r}}=\frac{\left(T_3-T_2\right)}{T_4-T_1}$

The expression for ${\text{COP}}_{\text{B}}$ is given by,

  ${\text{COP}}_{\text{B}}=\frac{1}{r^{\frac{r-1}{r}}-1}$

Calculation:

The value of ${\text{COP}}_{\text{B}}$ is evaluated as,

  $\begin{array}{c}{\text{COP}}_{\text{B}}=\frac{1}{r^{\frac{r-1}{r}}-1}\\ =\frac{1}{5^{\frac{1.66-1}{1.66}}-1}\\ =1.12\end{array}$

Conclusion:

Therefore, the value of ${\text{COP}}_{\text{B}}$ is $1.12$ .

(d)

To determine

The electrical energy that must be supplied to the motor of rotor is given by,

Answer to Problem 84P

The electrical energy that must be supplied is $107.55\text{W}$ .

Explanation of Solution

Given:

The rate $Q_C$ at which heat is transferred is $120\text{W}$ .

Formula:

The expression to determine the electrical energy that must be supplied is given by,

  $W=\frac{Q_C}{\text{COP}}$

Calculation:

The electrical energy that must be supplied is calculated as,

  $\begin{array}{c}W=\frac{Q_C}{\text{COP}}\\ =\frac{120\text{W}}{1.12}\\ =107.55\text{W}\end{array}$

Conclusion:

Therefore, the electrical energy that must be supplied is $107.55\text{W}$ .

(e)

To determine

The amount added to the monthly electricity bill.

Answer to Problem 84P

The electrical energy that must be supplied is $193.59\text{cents}$ .

Explanation of Solution

Given:

The number of days $d$ are $30$ .

The time for which the motor runs is $4\text{h}$ .

The charge $x$ for $1\text{kWh}$ of energy is $15\text{cents}$ .

Formula:

The amount of electrical energy utilized by the refrigerator per month is given by,

  $E=Wtdx$

Calculation:

The electrical energy utilized by the refrigerator per month is calculated as,

  $\begin{array}{c}E=Wtdx\\ =\left(107.55\text{W}\right)\left(4\right)\left(30\right)\left(15\text{cents}\right)\\ =193.59\text{cents}\end{array}$

Conclusion:

Therefore, the electrical energy that must be supplied is $193.59\text{cents}$ .