Chapter 6.11, Problem 141RP

To determine

The ratio of the maximum to minimum pressure.

Answer to Problem 141RP

The ratio of the maximum to minimum pressure is $\underset{\_}{5.67}$.

Explanation of Solution

Determine the COP of a reversible heat pump depends on the temperature limits in the cycle only.

${\text{COP}}_{\text{HP,rev}}=\frac{1}{1-\left(T_L/T_H\right)}$ (I)

Here, the temperature of higher temperature body is $T_H$ and the temperature of lower temperature body is $T_L$.

Determine rate of heat reject by the heat pump minimum.

$\begin{array}{l}{\dot{W}}_{\text{in}}=\frac{{\dot{Q}}_H}{{\text{COP}}_{\text{HP}}}\\ {\dot{Q}}_H={\text{COP}}_{\text{HP}}\times {\dot{W}}_{\text{in}}\end{array}$ (II)

Here, the power input required to operate the heat pump is ${\dot{W}}_{\text{in}}$.

Determine the amount of heat supplied per kg of Carnot heat pump cycle.

$q_H=\frac{{\dot{Q}}_H}{\dot{m}}$ (IV)

Here, the mass flow rate of the working fluid is $\dot{m}$.

Conclusion:

Substitute $1.2T_L$ for $T_H$ in Equation (I).

$\begin{array}{c}{\text{COP}}_{\text{HP,rev}}=\frac{1}{1-\left(1/1.2\right)}\\ =6.0\end{array}$

Substitute 6.0 for ${\text{COP}}_{\text{HP}}$ and 5 kW for ${\dot{W}}_{\text{in}}$ in Equation (II).

$\begin{array}{c}{\dot{Q}}_H=\left(6.0\right)\times \left(5\text{kW}\right)\\ =30.0\text{kW}\\ =30.0\text{kW}\times \left(\frac{1\text{kJ}/\text{s}}{1\text{kW}}\right)\\ =30.0\text{kJ}/\text{s}\end{array}$

Substitute $30.0\text{kJ}/\text{s}$ for ${\dot{Q}}_H$ and $0.18\text{kg}/\text{s}$ for $\dot{m}$ in Equation (III).

$\begin{array}{c}{q}_H=\frac{\left(30.0\text{kJ}/\text{s}\right)}{\left(0.18\text{kg}/\text{s}\right)}\\ =166.67\text{kJ}/\text{kg}\end{array}$

Refer to Table A-4, “Saturated water”, obtain the below properties at the enthalpy of vaporization of $166.67\text{kJ}/\text{kg}$ using interpolation method of two variables.

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y are enthalpy of vaporization and temperature.

Show the temperature at $36°\text{C}$ and $38°\text{C}$ as in Table (1).

S. No	enthalpy of vaporization $\text{kJ}/\text{kg}$ $\left(x\right)$	Temperature, $°\text{C}$ $\left(y\right)$
1	$167.19\text{kJ}/\text{kg}$	$36°\text{C}$
2	$166.67\text{kJ}/\text{kg}$	$y_2=?$
3	$165.13\text{kJ}/\text{kg}$	$38°\text{C}$

Calculate final temperature at enthalpy of vaporization of $166.67\text{kJ}/\text{kg}$ for liquid phase using interpolation method.

Substitute $167.19\text{kJ}/\text{kg}$ for $x_1$, $166.67\text{kJ}/\text{kg}$ for $x_2$, $165.13\text{kJ}/\text{kg}$ for $x_3$, $36°\text{C}$ for $y_1$, and $38°\text{C}$ for $y_3$ in Equation (V).

$\begin{array}{c}{y}_2=\frac{\left(166.67\text{kJ}/\text{kg}-167.19\text{kJ}/\text{kg}\right)\left(38°\text{C}-36°\text{C}\right)}{\left(165.13\text{kJ}/\text{kg}-167.19\text{kJ}/\text{kg}\right)}+36°\text{C}\\ =36.5°\text{C}\\ =36.5°\text{C+273}\text{.15}\\ =\text{309}\text{.6}\text{K}\end{array}$

From above calculation the temperature of the working fluid is $36.5°\text{C}$ at enthalpy of vaporization of $166.67\text{kJ}/\text{kg}$.

Repeat the above Equation (IV) for saturated maximum pressure at $36.5°\text{C}$ temperature is 925.3 kPa.

From this cycle maximum absolute temperature in the cycle is 1.2 time the minimum absolute temperature.

$\begin{array}{l}{T}_H=1.2T_L\\ T_L=\frac{T_H}{1.2}\end{array}$ (V)

Substitute 309.6 K for $T_H$ in Equation (V).

$\begin{array}{c}{T}_L=\frac{\left(309.6\text{K}\right)}{1.2}\\ =258\text{K}\\ =258\text{K}-\text{273}\text{.15}\\ =\text{-15}\text{.15}°\text{C}\end{array}$

Repeat the above Equation (IV) for saturated maximum pressure at $-15.15°\text{C}$ temperature is 163.1 kPa.

Calculate the ratio of the maximum to minimum pressure.

$\text{Ratio}=\frac{P_{\max }}{P_{\min }}$ (VI).

Here, the minimum pressure is $P_{\min }$ and the maximum pressure is $P_{\max }$.

Substitute 925.3 kPa for $P_{\max }$ and 163.1 kPa for $P_{\min }$ in Equation (VI).

$\begin{array}{c}\text{Ratio}=\frac{925.3\text{kPa}}{163.1\text{kPa}}\\ =5.67\end{array}$

Thus, the ratio of the maximum to minimum pressure is $\underset{\_}{5.67}$.