Chapter 11.10, Problem 59P

(a)

To determine

The mass flow rate of the refrigerant through the upper compression cycle.

Answer to Problem 59P

The mass flow rate of the refrigerant through the upper compression cycle is $0.2026\text{kg}/\text{s}$.

Explanation of Solution

Sketch the schematic diagram for the two stage cascade refrigeration system as in Figure (1).

Write the relation between the specific enthalpies at the inlet and exit of throttling process.

$h_{\text{in}}=h_{\text{out}}$ (I)

Here, specific enthalpy at the inlet of throttling is $h_{\text{in}}$, and the specific enthalpy at the exit of throttling is $h_{\text{out}}$.

Write the expression for the isentropic efficiency of the compressor $\left({\eta }_{\text{C}}\right)$.

${\eta }_{\text{C}}=\frac{h_{2s}-h_1}{h_2-h_1}$ (II)

Here, specific enthalpy at the isentropic exit of compressor is $h_{2s}$, and the specific enthalpy at the inlet of compressor is $h_1$.

Write the formula to calculate the dryness fraction at the exit of expansion valve $\left(x_6\right)$ in high pressure cycle.

$x_6=\frac{h_6-h_f}{h_{fg}}$ (III)

Here, specific enthalpy of refrigerant at expansion valve exit is $h_6$, specific enthalpy of saturated liquid is $h_f$, and the specific enthalpy of the saturated liquid vapor mixture is $h_{fg}$.

Write the expression for the mass flow rate of refrigerant $\left(\dot{m}\right)$ from dryness fraction.

$\dot{m}=\frac{{\dot{m}}_7}{1-x_6}$ (IV)

Here, mass flow rate of refrigerant at the inlet of low pressure compressor is ${\dot{m}}_7$ and the dryness fraction at exit of upper pressure compression is $x_6$.

Conclusion:

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure compressor inlet pressure $\left(P_1\right)$ of $200\text{kPa}$ as follows:

$\begin{array}{l}{h}_1=h_g=244.50\text{kJ}/\text{kg}\\ s_1=s_g=0.9379\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, specific enthalpy of the saturated vapor is $h_g$ and the specific entropy of the saturated vapor is $s_g$.

The specific entropy at the end of isentropic compression $\left(s_2\right)$ is calculated as

$s_2=s_1=0.9379\text{kJ}/\text{kg}\cdot \text{K}$

Refer to Table A-13, “Superheated R-134a”, and obtain the values of R-134a at pressure of $\text{450}\text{kPa}$ and specific entropy of $0.9379\text{kJ}/\text{kg}\cdot \text{K}$ by using interpolation method as,

$h_{2s}=261.13\text{kJ}/\text{kg}$

Substitute $0.80$ for ${\eta }_{\text{C}}$, $261.13\text{kJ}/\text{kg}$ for $h_{2s}$, and $244.50\text{kJ}/\text{kg}$ for $h_1$ in Equation (II).

$\begin{array}{l}0.80=\frac{261.13\text{kJ}/\text{kg}-244.50\text{kJ}/\text{kg}}{h_2-244.50\text{kJ}/\text{kg}}\\ h_2=265.28\text{kJ}/\text{kg}\end{array}$

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at throttling inlet pressure $\left(P_3\right)$ of $\text{450}\text{kPa}$ as follows:

$h_3=h_g=257.58\text{kJ}/\text{kg}$

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at second stage compressor inlet pressure $\left(P_5\right)$ of $1200\text{kPa}$ as follows:

$h_5=h_f=117.79\text{kJ}/\text{kg}$

Substitute $117.79\text{kJ}/\text{kg}$ for $h_5$ in Equation (I).

$h_6=117.79\text{kJ}/\text{kg}$

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at second stage expansion inlet pressure $\left(P_7\right)$ of $\text{450}\text{kPa}$ as follows:

$h_7=h_f=68.80\text{kJ}/\text{kg}$

Substitute $68.80\text{kJ}/\text{kg}$ for $h_7$ in Equation (II).

$h_8=68.80\text{kJ}/\text{kg}$

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at pressure $\left(P_6\right)$ of $\text{450}\text{kPa}$ as follows:

$h_g=257.58\text{kJ}/\text{kg}$

$h_f=68.80\text{kJ}/\text{kg}$

Substitute $117.79\text{kJ}/\text{kg}$ for $h_6$, $68.81\text{kJ}/\text{kg}$ for $h_f$, and $188.71\text{kJ}/\text{kg}$ for $h_{fg}$ in Equation (III).

$\begin{array}{c}{x}_6=\frac{117.79-68.81}{188.71}\\ =0.2595\end{array}$

Substitute $0.15\text{kg}/\text{s}$ for ${\dot{m}}_7$, and 0.2595 for $x_6$ in Equation (IV).

$\begin{array}{c}\dot{m}=\frac{0.15\text{kg}/\text{s}}{1-0.2595}\\ =0.2026\text{kg}/\text{s}\end{array}$

The mass flow rate of the refrigerant through the upper compression cycle is $0.2026\text{kg}/\text{s}$.

(b)

To determine

The rate at which heat removed from the refrigerated space.

Answer to Problem 59P

The rate at which heat removed from the refrigerated space is $26.35\text{kW}$.

Explanation of Solution

Write the mass balance equation for the mass transfer in flash chamber.

${\dot{m}}_3=\dot{m}-{\dot{m}}_7$ (V)

Write the energy balance equation for the flash chamber.

$\dot{m}{h}_9={\dot{m}}_7{h}_2+{\dot{m}}_3{h}_3$ (VI)

Here, specific enthalpy at state 9 is $h_9$.

Write the formula to calculate the rate of heat transfer from the refrigerated space $\left({\dot{Q}}_L\right)$.

${\dot{Q}}_L={\dot{m}}_7\left(h_1-h_8\right)$ (VII)

Conclusion:

Refer to Table A-13, “Superheated R-134a”, and obtain the values of R-134a at pressure of $\text{450}\text{kPa}$ and specific enthalpy of $263.28\text{kJ}/\text{kg}$ by using interpolation method as,

$s_9=0.9453\text{kJ}/\text{kg}\cdot \text{K}$

The specific entropy at the end of isentropic compression $\left(s_4\right)$ is calculated as

$s_4=s_9=0.9453\text{kJ}/\text{kg}\cdot \text{K}$

Refer to Table A-13, “Superheated R-134a”, and obtain the values of R-134a at pressure of $\text{1200}\text{kPa}$ and specific entropy of $0.9453\text{kJ}/\text{kg}\cdot \text{K}$ by using interpolation method as,

$h_{4s}=284.32\text{kJ}/\text{kg}$

Substitute $284.32\text{kJ}/\text{kg}\cdot \text{K}$ for $h_{4s}$, $263.28\text{kJ}/\text{kg}\cdot \text{K}$ for $h_9$, and 0.8 for ${\eta }_C$ in Equation (II).

$\begin{array}{l}0.80=\frac{284.32\text{kJ}/\text{kg}\cdot \text{K}-263.28\text{kJ}/\text{kg}\cdot \text{K}}{h_4-263.28\text{kJ}/\text{kg}\cdot \text{K}}\\ h_4=289.57\text{kJ}/\text{kg}\end{array}$

Substitute $0.15\text{kg}/\text{s}$ for ${\dot{m}}_7$, $244.50\text{kJ}/\text{kg}$ for $h_1$, and $68.80\text{kJ}/\text{kg}$ for $h_8$ in Equation (VII).

$\begin{array}{c}{\dot{Q}}_L=\left(0.15\text{kg}/\text{s}\right)\left(244.50-68.80\right)\text{kJ}/\text{kg}\\ =26.35\text{kW}\end{array}$

Thus, the rate at which heat removed from the refrigerated space is $26.35\text{kW}$.

(c)

To determine

The power input required to the two stage cascade refrigeration system.

The coefficient of refrigeration for the two-stage cascade refrigeration system.

Answer to Problem 59P

The power input required to the two stage cascade refrigeration system is $8.443\text{kW}$.

The coefficient of refrigeration for the two-stage cascade refrigeration system is $3.12$.

Explanation of Solution

Write the formula to calculate the total required work input $\left({\dot{W}}_{\text{in}}\right)$ to the compression.

${\dot{W}}_{\text{in}}={\dot{W}}_{\text{comp I,in}}+{\dot{W}}_{\text{comp II,in}}$

${\dot{W}}_{\text{in}}={\dot{m}}_7\left(h_2-h_1\right)+\dot{m}\left(h_4-h_9\right)$ (VIII)

Here, required work input to the first stage compression is ${\dot{W}}_{\text{comp I,in}}$, and required work input to the second stage compression is ${\dot{W}}_{\text{comp II,in}}$.

Write the formula to calculate the COP of the two-stage cascade refrigeration system.

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

Conclusion:

Substitute $0.15\text{kg}/\text{s}$ for ${\dot{m}}_7$, $265.28\text{kJ}/\text{kg}$ for $h_2$, $244.50\text{kJ}/\text{kg}$ for $h_1$, $0.2026\text{kg}/\text{s}$ for $\dot{m}$, $289.57\text{kJ}/\text{kg}$ for $h_4$, and $263.28\text{kJ}/\text{kg}$ for $h_9$ in Equation (VIII).

$\begin{array}{c}{\dot{W}}_{\text{in}}=\left(\begin{array}{l}\left(0.15\text{kg}/\text{s}\right)\left(265.28-244.50\right)\text{kJ}/\text{kg}\\ +\left(0.2026\text{kg}/\text{s}\right)\left(289.57-263.28\right)\text{kJ}/\text{kg}\end{array}\right)\\ =8.443\text{kW}\end{array}$

Thus, the power input required to two stage cascade refrigeration system is $8.443\text{kW}$.

Substitute 26.35 kW for ${\dot{Q}}_L$, and 8.443 kW for ${\dot{W}}_{\text{in}}$ in Equation (IX).

$\begin{array}{c}\text{COP}=\frac{26.35\text{kW}}{8.443\text{kW}}\\ =3.12\end{array}$

Thus, the coefficient of refrigeration for the two-stage cascade refrigeration system is $3.12$.

(d)

To determine

The rate at which heat removed from the refrigerated space.

The coefficient of refrigeration for the two-stage cascade refrigeration system.

Answer to Problem 59P

The rate at which heat removed from the refrigerated space is $25.67\text{kW}$.

The coefficient of refrigeration for the two-stage cascade refrigeration system is $2.71$.

Explanation of Solution

Write the formula to calculate the rate of heat transfer from the refrigerated space $\left({\dot{Q}}_L\right)$.

${\dot{Q}}_L=\dot{m}\left(h_1-h_4\right)$ (X)

Here, the specific enthalpy of refrigerant at the exit of expansion valve is $h_4$.

Write the formula to calculate the required work input $\left({\dot{W}}_{\text{in}}\right)$ to the compression.

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

Conclusion:

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure compressor inlet pressure $\left(P_1\right)$ of $200\text{kPa}$ as follows:

$\begin{array}{l}{h}_1=h_g=244.50\text{kJ}/\text{kg}\\ s_1=s_g=0.9379\text{kJ}/\text{kg}\cdot \text{K}\end{array}$

Here, specific enthalpy of the saturated vapor is $h_g$ and the specific entropy of the saturated vapor is $s_g$.

The specific entropy at the end of isentropic compression $\left(s_2\right)$ is calculated as

$s_2=s_1=0.9379\text{kJ}/\text{kg}\cdot \text{K}$

Refer to Table A-13, “Superheated R-134a”, and obtain the values of R-134a at pressure of $\text{1200}\text{kPa}$ and specific entropy of $0.9379\text{kJ}/\text{kg}\cdot \text{K}$ by using interpolation method as,

$h_{2s}=281.88\text{kJ}/\text{kg}$

Substitute $0.80$ for ${\eta }_{\text{C}}$, $281.88\text{kJ}/\text{kg}$ for $h_{2s}$, and $244.50\text{kJ}/\text{kg}$ for $h_1$ in Equation (II).

$\begin{array}{l}0.80=\frac{281.88\text{kJ}/\text{kg}-244.50\text{kJ}/\text{kg}}{h_2-244.50\text{kJ}/\text{kg}}\\ h_2=291.23\text{kJ}/\text{kg}\end{array}$

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at expansion valve inlet pressure $\left(P_3\right)$ of $1200\text{kPa}$ as follows:

$h_3=h_f=117.79\text{kJ}/\text{kg}$

Substitute $117.79\text{kJ}/\text{kg}$ for $h_3$ in Equation (I).

$h_4=117.79\text{kJ}/\text{kg}$

Substitute $0.2026\text{kg}/\text{s}$ for $\dot{m}$, $244.50\text{kJ}/\text{kg}$ for $h_1$, and $117.79\text{kJ}/\text{kg}$ for $h_4$ in Equation (X).

$\begin{array}{c}{\dot{Q}}_L=\left(0.2026\text{kg}/\text{s}\right)\left(244.50-117.79\right)\text{kJ}/\text{kg}\\ =25.67\text{kW}\end{array}$

Thus, the rate at which heat removed from the refrigerated space is $25.67\text{kW}$.

Substitute $0.2026\text{kg}/\text{s}$ for $\dot{m}$, $291.23\text{kJ}/\text{kg}$ for $h_2$, and $244.50\text{kJ}/\text{kg}$ for $h_1$ in Equation (XI).

$\begin{array}{c}{\dot{W}}_{\text{in}}=\left(0.2026\text{kg}/\text{s}\right)\left(291.23-244.50\right)\text{kJ}/\text{kg}\\ =9.467\text{kW}\end{array}$

Substitute 25.67 kW for ${\dot{Q}}_L$, and 9.467 kW for ${\dot{W}}_{\text{in}}$ in Equation (IX).

$\begin{array}{c}\text{COP}=\frac{25.67\text{kW}}{9.467\text{kW}}\\ =2.71\end{array}$

Thus, the coefficient of refrigeration for the two-stage cascade refrigeration system is $2.71$.