Chapter 6.11, Problem 108P

(a)

To determine

The actual coefficient of performance of the refrigerator.

Answer to Problem 108P

The actual coefficient of performance of the refrigerator is $4.33$.

Explanation of Solution

Write the formula to calculate the mass flow rate of a refrigerant $\left({\dot{m}}_R\right)$.

${\dot{m}}_R=\frac{{\dot{v}}_1}{v_1}$ (I)

Here, volume flow rate at inlet 1 is ${\dot{v}}_1$.

Write the formula to calculate the power consumption of the compressor $\left({\dot{W}}_{\text{in}}\right)$.

${\dot{W}}_{\text{in}}={\dot{m}}_R\left(h_2-h_1\right)$ (II)

Write the formula to calculate the refrigeration load $\left({\dot{Q}}_L\right)$.

${\dot{Q}}_L={\dot{Q}}_{\text{heat}}+{\dot{Q}}_{\text{equipment}}$ (III)

Here, heating load rate is ${\dot{Q}}_{\text{heat}}$ and other equipment load rate is ${\dot{Q}}_{\text{equipment}}$.

Write the formula to calculate the actual coefficient of performance.

${\text{COP}}_{\text{actual}}=\frac{{\dot{Q}}_L}{{\dot{W}}_{\text{in}}}$ (IV)

Conclusion:

Refer Table A-12, “Saturated refrigerant-134a: Pressure table”, obtain the properties of refrigerant R-134a at initial pressure $\left(P_1\right)$ of $\text{400 kPa}$ and quality 1.

$\begin{array}{l}{h}_1=255.61\text{kJ}/\text{kg}\\ v_1=0.05127{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Refer Table A-13, “Superheated refrigerant-134a”, obtain the properties of refrigerant R-134a at exit pressure $\left(P_2\right)$ of $\text{1}\text{.2 MPa}$ and temperature $\left(T_2\right)$ of $70°\text{C}$.

$h_2=300.63\text{kJ}/\text{kg}$

Substitute $0.05127{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ and $80\text{L}/\min$ for ${\dot{v}}_1$ in Equation (I).

$\begin{array}{c}{\dot{m}}_R=\frac{80\text{L}/\min }{0.05127{\text{m}}^{\text{3}}/\text{kg}}\\ =\left(\frac{80\text{L}/\min \left(\frac{1{\text{m}}^{\text{3}}}{1000\text{L}}\right)\left(\frac{1\min }{60\text{s}}\right)}{0.05127{\text{m}}^{\text{3}}/\text{kg}}\right)\\ =0.02601\text{kg}/\text{s}\end{array}$

Substitute $0.02601\text{kg}/\text{s}$ for ${\dot{m}}_R$, $255.61\text{kJ}/\text{kg}$ for $h_1$, and $300.63\text{kJ}/\text{kg}$ for $h_2$ in Equation (II).

$\begin{array}{c}{\dot{W}}_{\text{in}}=0.02601\text{kg}/\text{s}\left(300.63\text{kJ}/\text{kg}-255.61\text{kJ}/\text{kg}\right)\\ =1.171\text{kW}\end{array}$

Substitute $250\text{kJ}/\min$ for ${\dot{Q}}_{\text{heat}}$ and 0.9 kW for ${\dot{Q}}_{\text{equipment}}$ in Equation (III).

$\begin{array}{c}{\dot{Q}}_L=250\text{kJ}/\min +0.9\text{kW}\\ =250\text{kJ}/\min \left(\frac{1\min }{60\text{}\text{s}}\right)+0.9\text{kW}\\ =\text{5}\text{.067}\text{kW}\end{array}$

Substitute $1.171\text{kW}$ for ${\dot{W}}_{\text{in}}$ and 5.067 kW for ${\dot{Q}}_L$ in Equation (IV).

$\begin{array}{c}{\text{COP}}_{\text{actual}}=\frac{5.067\text{kW}}{1.171\text{kW}}\\ =4.33\end{array}$

Thus, the actual coefficient of performance of the refrigerator is $4.33$.

(b)

To determine

The maximum coefficient of performance of a reversible refrigerator.

Answer to Problem 108P

The maximum coefficient of performance of a reversible refrigerator is $26.91$.

Explanation of Solution

Write the formula to calculate the maximum coefficient of performance of a reversible refrigerator.

${\text{COP}}_{\text{max}}=\frac{1}{\frac{T_H}{T_L}-1}$ (V)

Here, absolute temperature of high and low temperature reservoir is $T_H\text{and}{T}_L$ respectively.

Conclusion:

Convert the unit of temperature $T_L$ from $°\text{C}\text{to}\text{K}$.

$\begin{array}{c}{T}_L=23°\text{C}\\ =\left(23+273\right)K\\ =296\text{K}\end{array}$

Convert the unit of temperature $T_H$ from $°\text{C}\text{to}\text{K}$.

$\begin{array}{c}{T}_H=34°\text{C}\\ =\left(34+273\right)K\\ =307\text{K}\end{array}$

Substitute 296 K for $T_L$ and 307 K for $T_H$ in Equation (V).

$\begin{array}{c}{\text{COP}}_{\text{max}}=\frac{1}{\frac{307\text{K}}{296\text{K}}-1}\\ =26.91\end{array}$

Thus, the maximum coefficient of performance of a reversible refrigerator is $26.91$.

(c)

To determine

The minimum volume flow rate at the compressor inlet.

Answer to Problem 108P

The minimum volume flow rate at the compressor inlet is $12.9\text{L}/\min$.

Explanation of Solution

Write the formula to calculate the minimum power input to the condenser for the same refrigeration load $\left({\dot{W}}_{\text{in,min}}\right)$.

${\dot{W}}_{\text{in,min}}=\frac{{\dot{Q}}_L}{{\text{COP}}_{\text{max}}}$ (VI)

Write the formula to calculate the minimum mass flow rate $\left({\dot{m}}_{R,\min }\right)$.

${\dot{m}}_{R,\min }=\frac{{\dot{W}}_{\text{in,min}}}{h_2-h_1}$ (VII)

Write the formula to calculate the minimum volume flow rate at the compressor inlet.$\left({\dot{v}}_{\min ,1}\right)$.

${\dot{v}}_{\min ,1}={\dot{m}}_{R,\min }{v}_1$ (VIII)

Conclusion:

Substitute 26.91 for ${\text{COP}}_{\text{max}}$ and 5.067 kW for ${\dot{Q}}_L$ in Equation (VI).

$\begin{array}{c}{\dot{W}}_{\text{in,min}}=\frac{5.067\text{kW}}{26.91}\\ =0.1883\text{kW}\end{array}$

Substitute $0.1883\text{kW}$ for ${\dot{W}}_{\text{in,min}}$, $255.61\text{kJ}/\text{kg}$ for $h_1$, and $300.63\text{kJ}/\text{kg}$ for $h_2$ in Equation (VII).

$\begin{array}{c}{\dot{m}}_{R,\min }=\frac{0.1883\text{kW}}{300.63\text{kJ}/\text{kg}-255.61\text{kJ}/\text{kg}}\\ =0.004182\text{kg}/\text{s}\end{array}$

Substitute $0.004182\text{kg}/\text{s}$ for ${\dot{m}}_{R,\min }$ and $0.05127{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (VIII).

$\begin{array}{c}{\dot{v}}_{\min ,1}=\left(0.004182\text{kg}/\text{s}\right)\left(0.05127{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.0002144{\text{m}}^{\text{3}}/\text{s}\\ =12.9\text{L}/\min \end{array}$

Thus, the minimum volume flow rate at the compressor inlet is $12.9\text{L}/\min$.