Chapter 7, Problem 121RQ

To determine

The ratio of maximum pressure to minimum pressure.

Explanation of Solution

Given:

The absolute temperature reservoir $\left(T_H\right)$ is $1.2T_L$.

The rate of heat rejected by heat pump $\left({\dot{W}}_{\text{net,in}}\right)$ is $5\text{kW}$.

The mass flow rate $\left(\dot{m}\right)$ is $0.22\text{kg/s}$.

Calculation:

Write the coefficient of performance of reversible heat pump $\left({\text{COP}}_{\text{HP,rev}}\right)$.

  ${\text{COP}}_{\text{HP,rev}}=\frac{1}{1-\frac{T_L}{T_H}}$

  $\begin{array}{c}{\text{COP}}_{\text{HP,rev}}=\frac{1}{1-\frac{1}{1.2}}\\ =6\end{array}$

Write the coefficient of performance of a Carnot heat pump.

  ${\text{COP}}_{\text{HP}}=\frac{{\dot{Q}}_H}{{\dot{W}}_{\text{net,in}}}$        (I)

Rewrite Equation (I) to calculate the heat gained by the pump.

  ${\dot{Q}}_H={\dot{W}}_{\text{net,in}}\times {\text{COP}}_{\text{HP}}$

  $\begin{array}{c}{\dot{Q}}_H=5\text{kW}\times \text{6}\\ \text{=30}\text{kW}\end{array}$

Write the heat transfer from the Carnot heat pump cycle $\left(q_H\right)$.

  $q_H=\frac{{\dot{Q}}_H}{\dot{m}}$

  $\begin{array}{c}{q}_H=\frac{\text{30}\text{kW}}{0.22\text{kg/s}}\\ =\frac{\text{30}\text{kW}\times \frac{\text{kJ/s}}{\text{kW}}}{0.22\text{kg/s}}\\ =136.36\text{ kJ/kg}\end{array}$

Obtain the temperature and pressure using the above calculated enthalpy of vaporization.

Refer to Table A-12, “Saturated pressure table” enthalpy of vaporization is $136.36\text{ kJ/kg}$ obtain the following properties using interpolation method.

Show the temperature and constant pressure at temperature of $548\text{ K}$ as in Table (1).

Temperature, $°\text{C}$	Enthalpy $\text{kJ}/\text{kg}$	Pressure kPa
57.88	141.96	1600
$y_2$	$136.36$	$z_2$
62.87	135.14	1800

Write the formula of interpolation method of two variables.

  $y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$        (II)

Here, the variables denoted by x and y are enthalpy and entropy.

Substitute $x_1=141.96\text{ kJ/kg}$, $x_2=136.36\text{ kJ/kg}$, $x_3=135.14\text{ kJ/kg}$, $y_1=57.88°\text{C}$, and $y_3=62.87°\text{C}$ in Equation (II).

  $\begin{array}{c}{y}_2=\frac{\left(136.36-141.96\right)\left(62.87-57.88\right)}{\left(135.14-141.96\right)}+57.88\\ =\frac{-5.6\times 4.99}{-6.82}+57.88\\ =62°\text{C}\end{array}$

Substitute $x_1=141.96\text{ kJ/kg}$, $x_2=136.36\text{ kJ/kg}$, $x_3=135.14\text{ kJ/kg}$, $y_1=1600\text{}\text{ kPa}$, and  $y_3=1600\text{}\text{ kPa}$ in Equation (II).

  $\begin{array}{c}{y}_2=\frac{\left(136.36-141.96\right)\left(1800-1600\right)}{\left(135.14-141.96\right)}+1600\\ =\frac{-5.6\times 200}{-6.82}+1600\\ =1764\text{kPa}\end{array}$

Calculate the lower temperature of the reservoir.

  $\begin{array}{c}{T}_L=\frac{T_H}{1.2}\\ =\frac{62°\text{C}}{1.2}\\ =\frac{\left(62+273.2\right)\text{K}}{1.2}\\ =279.3333\text{ K}\\ \simeq 280\text{K}\end{array}$

      $\begin{array}{c}=\left(280-273.2\right)°\text{C}\\ =\text{6}\text{.8}°\text{C}\end{array}$

Refer table A-11, “Saturated refrigerant-134a temperature table”, obtain pressure value at the temperature of $\text{6}\text{.8}°\text{C}$ is $P_{\min }=372.55\text{ kPa}$ (using interpolation method).

Calculate the ratio of maximum pressure to minimum pressure.

  $\begin{array}{c}\frac{P_{\max }}{P_{\min }}=\frac{1764\text{ kPa}}{372.55\text{ kPa}}\\ =4.735\end{array}$

Thus, the ratio of maximum pressure to minimum pressure is s.