Chapter 11.10, Problem 76P

A gas refrigeration system using air as the working fluid has a pressure ratio of 5. Air enters the compressor at 0°C. The high-pressure air is cooled to 35°C by rejecting heat to the surroundings. The refrigerant leaves the turbine at −80°C and then it absorbs heat from the refrigerated space before entering the regenerator. The mass flow rate of air is 0.4 kg/s. Assuming isentropic efficiencies of 80 percent for the compressor and 85 percent for the turbine and using constant specific heats at room temperature, determine (a) the effectiveness of the regenerator, (b) the rate of heat removal from the refrigerated space, and (c) the COP of the cycle. Also, determine (d) the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle. Use the same compressor inlet temperature as given, the same turbine inlet temperature as calculated, and the same compressor and turbine efficiencies.

FIGURE P11–79

To determine

The effectiveness of the regenerator.

Answer to Problem 76P

The effectiveness of the regenerator is $\text{0}\text{.434}$.

Explanation of Solution

Show the T-s diagram as in Figure (1).

Express the temperature at state 2s.

$T_{2s}=T_1{\left[\frac{P_2}{P_1}\right]}^{\left(k-1\right)/k}$ (I)

Here, temperature at state 1 is $T_1$, pressure at state 1 and 2 is $P_1\text{and}{P}_2$ respectively, and heat capacity ratio is $k$.

Express the temperature at state 2 from the isentropic relations.

$\begin{array}{c}{\eta }_C=\frac{h_{2s}-h_1}{h_2-h_1}\\ =\frac{T_{2s}-T_1}{T_2-T_1}\end{array}$ (II)

Here, isentropic efficiency is ${\eta }_C$, enthalpy at state 2s, 1 and 2 is $h_{2s},h_1\text{and}{h}_2$ respectively, temperature at state 2s, 1 and 2 is $T_{2s},T_1\text{and}{T}_2$ respectively.

Express temperature at state 5s.

$T_{5s}=T_4{\left[\frac{P_5}{P_4}\right]}^{\left(k-1\right)/k}$                                                                                   (III)

Here, temperature at state 4 is $T_5$, pressure at state 5 and 4 is $P_5\text{and}{P}_4$ respectively.

Express temperature at state 4.

$\begin{array}{c}{\eta }_T=\frac{h_4-h_5}{h_4-h_{5s}}\\ =\frac{T_4-T_5}{T_4-T_{5s}}\end{array}$ (IV)

Here, thermal efficiency is ${\eta }_T$, enthalpy at state 4, 5 and 5s is $h_4,h_5\text{and}{h}_{5s}$ respectively, temperature at state 4, 5 and 5s is $T_4,T_5\text{and}{T}_{5s}$.

Express the temperature at state 6 using an energy balance.

$\begin{array}{l}\dot{m}{c}_p\left(T_3-T_4\right)=\dot{m}{c}_p\left(T_1-T_6\right)\\ T_3-T_4=T_1-T_6\\ T_6=T_1-T_3+T_4\end{array}$ (V)

Here, mass flow rate is $\dot{m}$, specific heat at constant pressure is $c_p$, temperature at state 1,3,4 and 6 is $T_1,T_3,T_4\text{and}{T}_6$ respectively.

Express the effectiveness of the regenerator.

$\begin{array}{c}{\epsilon }_{\text{regen}}=\frac{h_3-h_4}{h_3-h_6}\\ =\frac{T_3-T_4}{T_3-T_6}\end{array}$                                                                                      (VI)

Here, enthalpy at state 3, 4 and 6 is $h_3,h_4\text{and}{h}_6$ respectively.

Conclusion:

Perform unit conversion of temperature at state 1, 3, and 5 from $°\text{C}\text{to}\text{K}$.

$\begin{array}{c}{T}_1=0\text{°C}\\ =\left(0+273.2\right)\text{K}\\ =273.2\text{K}\end{array}$

$\begin{array}{c}{T}_3=35\text{°C}\\ =\left(35+273.2\right)\text{K}\\ =308.2\text{K}\end{array}$

$\begin{array}{c}{T}_5=-80\text{°C}\\ =\left(-80+273.2\right)\text{K}\\ =193.2\text{K}\end{array}$

Refer Table A-2, “ideal gas specific heats of various common gas”, and write the properties of air.

$\begin{array}{l}{c}_p=1.005\text{kJ}/\text{kg}\cdot \text{K}\\ k=1.4\end{array}$

Substitute $\text{273}\text{.2}\text{K}$ for $T_1$, $\text{5}$ for $P_2/P_1$ and $1.4$ for $k$ in Equation (I).

$\begin{array}{c}{T}_{2s}=\left(\text{273}\text{.2}\text{K}\right){\left(5\right)}^{\left(1.4-1\right)/1.4}\\ =432.4\text{K}\end{array}$

Substitute $0.80$ for ${\eta }_C$, $432.4\text{K}$ for $T_{2s}$ and $\text{273}\text{.2}\text{K}$ for $T_1$ in Equation (II).

$\begin{array}{l}0.80=\frac{432.4\text{K}-\text{273}\text{.2}\text{K}}{T_2-\text{273}\text{.2}\text{K}}\\ T_2=472.5\text{K}\end{array}$

Substitute $\frac{1}{5}$ for $P_5/P_4$ and $1.4$ for $k$ in Equation (III).

$T_{5s}=T_4{\left[\frac{1}{5}\right]}^{\left(1.4-1\right)/1.4}$ (VII)

Substitute $0.85$ for ${\eta }_T$ and $\text{193}\text{.2}\text{K}$ for $T_5$ in Equation (IV).

$0.85=\frac{T_4-\text{193}\text{.2}{\text{K}}_5}{T_4-T_{5s}}$ (VIII)

Solve Equations (VII) and (VIII) simultaneously by online calculator to get,

$\begin{array}{l}{T}_4=281.3\text{K}\\ T_{5s}=177.60\text{K}\end{array}$

Substitute $\text{273}\text{.2}\text{K}$ for $T_1$, $\text{308}\text{.2}\text{K}$ for $T_3$ and $281.3\text{K}$ for $T_4$ in Equation (V).

$\begin{array}{c}{T}_6=\text{273}\text{.2}\text{K}-\text{308}\text{.2}\text{K}+281.3\text{K}\\ =\text{246}\text{.3}\text{K}\end{array}$

Substitute $\text{308}\text{.2}\text{K}$ for $T_3$, $281.3\text{K}$ for $T_4$, and $\text{246}\text{.3}\text{K}$ for $T_6$ in Equation (VI).

$\begin{array}{c}{\epsilon }_{\text{regen}}=\frac{\text{308}\text{.2}\text{K}-281.3\text{K}}{\text{308}\text{.2}\text{K}-\text{246}\text{.3}\text{K}}\\ =0.434\end{array}$

Hence, the effectiveness of the regenerator is $\text{0}\text{.434}$.

To determine

The rate of heat removal from the refrigerated space.

Answer to Problem 76P

The rate of heat removal from the refrigerated space is $\text{21}\text{.4}\text{kW}$.

Explanation of Solution

Express the rate of heat removal from the refrigerated space.

${\dot{Q}}_L=\dot{m}{c}_p\left(T_6-T_5\right)$ (IX)

Conclusion:

Substitute $\text{0}\text{.4}\text{kg}/\text{s}$ for $\dot{m}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $\text{246}\text{.3}\text{K}$ for $T_6$ and $\text{193}\text{.2}\text{K}$ for $T_5$ in Equation (IX).

$\begin{array}{c}{\dot{Q}}_L=\left(\text{0}\text{.4}\text{kg}/\text{s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\text{246}\text{.3}\text{K}-\text{193}\text{.2}\text{K}\right)\\ =21.4\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =\text{21}\text{.4}\text{kW}\end{array}$

Hence, the rate of heat removal from the refrigerated space is $\text{21}\text{.4}\text{kW}$.

To determine

The COP of the gas refrigeration cycle.

Answer to Problem 76P

The COP of the gas refrigeration cycle is $\text{0}\text{.478}$.

Explanation of Solution

Express the net work input of the compressor.

${\dot{W}}_{C,\text{in}}=\dot{m}{c}_p\left(T_2-T_1\right)$ (X)

Express the net work output of the turbine.

${\dot{W}}_{T,\text{out}}=\dot{m}{c}_p\left(T_4-T_5\right)$ (XI)

Express the coefficient of performance of the gas refrigeration cycle.

$\text{COP}=\frac{{\dot{Q}}_L}{{\dot{W}}_{C,\text{in}}-{\dot{W}}_{T,\text{out}}}$ (XII)

Conclusion:

Substitute $\text{0}\text{.4}\text{kg}/\text{s}$ for $\dot{m}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $\text{472}\text{.5}\text{K}$ for $T_2$ and $\text{273}\text{.2}\text{K}$ for $T_1$ in Equation (X).

$\begin{array}{c}{\dot{W}}_{C,\text{in}}=\left(\text{0}\text{.4}\text{kg}/\text{s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\text{472}\text{.5}\text{K}-\text{273}\text{.2}\text{K}\right)\\ =80.13\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =\text{80}\text{.13}\text{kW}\end{array}$

Substitute $\text{0}\text{.4}\text{kg}/\text{s}$ for $\dot{m}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $\text{281}\text{.3}\text{K}$ for $T_4$ and $\text{193}\text{.2}\text{K}$ for $T_5$ in Equation (XI).

$\begin{array}{c}{\dot{W}}_{T,\text{out}}=\left(\text{0}\text{.4}\text{kg}/\text{s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\text{281}\text{.3}\text{K}-\text{193}\text{.2}\text{K}\right)\\ =35.43\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =\text{35}\text{.43}\text{kW}\end{array}$

Substitute $\text{21}\text{.36}\text{kW}$ for ${\dot{Q}}_L$, $\text{80}\text{.13}\text{kW}$ for ${\dot{W}}_{C,\text{in}}$ and $\text{35}\text{.43}\text{kW}$ for ${\dot{W}}_{T,\text{out}}$ in Equation (XII).

$\begin{array}{c}\text{COP}=\frac{\text{21}\text{.36}\text{kW}}{\text{80}\text{.13}\text{kW}-\text{35}\text{.43}\text{kW}}\\ =0.478\end{array}$

Hence, the COP of the gas refrigeration cycle is $\text{0}\text{.478}$.

To determine

The refrigeration load and the COP of the system.

Answer to Problem 76P

The refrigeration load is $\text{24}\text{.74}\text{kW}$ and the COP of the system is $\text{0}\text{.599}$.

Explanation of Solution

Show the T-s diagram as in Figure (2).

Express temperature at state 4s.

$T_{4s}=T_3{\left[\frac{P_4}{P_3}\right]}^{\left(k-1\right)/k}$ (XIII)

Here, temperature at state 3 is $T_3$, pressure at state 3 and 4 is $P_3\text{and}{P}_4$ respectively.

Express temperature at state 4.

${\eta }_T=\frac{T_3-T_4}{T_3-T_{4s}}$ (XIV)

Express the refrigeration load.

${\dot{Q}}_L=\dot{m}{c}_p\left(T_1-T_4\right)$ (XV)

Express the net work input.

${\dot{W}}_{\text{net,in}}=\dot{m}{c}_p\left(T_2-T_1\right)-\dot{m}{c}_p\left(T_3-T_4\right)$ (XVI)

Express the coefficient of performance of the system.

$\text{COP}=\frac{{\dot{Q}}_L}{{\dot{W}}_{\text{net,in}}}$ (XVII)

Conclusion:

Substitute $\text{308}\text{.2}\text{K}$ for $T_3$, $\frac{\text{1}}{5}$ for $P_4/P_3$ and $1.4$ for $k$ in Equation (XIII).

$\begin{array}{c}{T}_{4s}=\left(\text{308}\text{.2}\text{K}\right){\left[\frac{\text{1}}{5}\right]}^{\left(1.4-1\right)/1.4}\\ =194.6\text{K}\end{array}$

Substitute $0.85$ for ${\eta }_T$,  $\text{308}\text{.2}\text{K}$ for $T_3$ and $194.6\text{K}$ for $T_{4s}$ in Equation (XIV).

$\begin{array}{l}0.85=\frac{\text{308}\text{.2}\text{K}-T_4}{\text{308}\text{.2}\text{K}-194.6\text{K}}\\ T_4=211.6\text{K}\end{array}$

Substitute $\text{0}\text{.4}\text{kg}/\text{s}$ for $\dot{m}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $\text{211}\text{.6}\text{K}$ for $T_4$ and $\text{273}\text{.2}\text{K}$ for $T_1$ in Equation (XV).

$\begin{array}{c}{\dot{Q}}_L=\left(\text{0}\text{.4}\text{kg}/\text{s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\text{273}\text{.2}\text{K}-\text{211}\text{.6}\text{K}\right)\\ =24.74\text{kJ}/\text{s}\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =\text{24}\text{.74}\text{kW}\end{array}$

Hence, the refrigeration load is $\text{24}\text{.74}\text{kW}$.

Substitute $\text{0}\text{.4}\text{kg}/\text{s}$ for $\dot{m}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_p$, $\text{472}\text{.5}\text{K}$ for $T_2$, $\text{273}\text{.2}\text{K}$ for $T_1$, $\text{308}\text{.2}\text{K}$ for $T_3$ and $\text{211}\text{.6}\text{K}$ for $T_4$ in Equation (XVI).

$\begin{array}{c}{\dot{W}}_{\text{net,in}}=\left[\begin{array}{l}\left(\text{0}\text{.4}\text{kg}/\text{s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\text{472}\text{.5}\text{K}-\text{273}\text{.2}\text{K}\right)\\ -\left(\text{0}\text{.4}\text{kg}/\text{s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\left(\text{308}\text{.2}\text{K}-\text{211}\text{.6}\text{K}\right)\end{array}\right]\\ =41.32\left[\frac{\text{kW}}{\text{kJ}/\text{s}}\right]\\ =\text{41}\text{.32}\text{kW}\end{array}$

Substitute $\text{24}\text{.74}\text{kW}$ for ${\dot{Q}}_L$ and $\text{41}\text{.32}\text{kW}$ for ${\dot{W}}_{\text{net,in}}$ in Equation (XVII).

$\begin{array}{c}\text{COP}=\frac{\text{24}\text{.74}\text{kW}}{\text{41}\text{.32}\text{kW}}\\ =0.599\end{array}$

Hence, the coefficient of performance of the system is $0.599$.