Q.4.32 - An Introduction to Thermal Physics

Question

Suppose you are told to design a household air conditioner using 

HFC-134a as its working substance. Over what range of pressures would you have it operate? Explain your reasoning. Calculate the COP for your design, and compare to the COP of an ideal Carnot refrigerator operating between the same extreme temperatures.

Step-by-Step Solution

Answer

The household air conditioner will be operated between pressures of 4 and 12 bar, with a COP of 5.67 for the design and 7.54 for an ideal Carnot refrigerator running between the same reservoir temperatures, which is higher than the design's COP.

Step 1: To find

The range of pressures over which the household air conditioner will be operated and the COP for the design.Comparison of COP for design to the COP for an ideal Carnot refrigerator operating between the same reservoir temperatures.

Step 2: Explanation

Given:

HFC-134a is used as a working substance to design a household air conditioner.

Formula: Tables 4.3 and 4.4 can be used to select the high temperature, which must be higher than the temperature outside the room, and the low temperature, which must be lower than the temperature outside the room, as $46.3°$and $8.9°$, respectively. Table 4.3 shows that the pressure at high temperature is 12 bar and the pressure at low temperature is 4 bar.

The enthalpy at the low temperature is $252 \mathrm{kJ}/\mathrm{kg}$ and the enthalpy at the high temperature is $116 \mathrm{kJ}/\mathrm{kg}$ from table 4.3.

Since the process is assumed to be adiabatic along the compression path, the entropy at the beginning and end is the same. From table 4.3, it has a value of $0.915 \mathrm{kJ}/\mathrm{K}·\mathrm{kg}$, which corresponds to the temperature $50°$from table 4.4. The corresponding enthalpy can be calculated as $276 \mathrm{kJ}/\mathrm{kg}$.

The expression of COP for the design

$\mathrm{COP}=\frac{H_1-H_3}{H_2-H_1}\dots \left(1\right)$

Where, $H_1$is enthalpy at low temperature,

            $H_3$ is enthalpy at high temperature and

           $H_2$is enthalpy at temperature $50°$ for the compression path.

The expression of the maximum COP for a Carnot air conditioner

${\mathrm{COP}}_{\max }=\frac{T_{\text{cold }}}{T_{\text{boc }}-T_{\text{cold }}}\dots ..\left(2\right)$

$$\begin{gathered} \text{ Where, }{T}_{\text{cold }}\text{ is the temperature of the cold reservoir and} \\ T_{\text{hot }}\text{ is the temperature of hot reservoir. } \end{gathered}$$

Step 3: Calculation

Calculation:

$$\begin{gathered} \text{ Substitute } {\mathrm{H}}_1=252 \mathrm{kJ}/\mathrm{kg} \\ {\mathrm{H}}_2=276 \mathrm{kJ}/\mathrm{kg} \\ {\mathrm{H}}_3=116 \mathrm{kJ}/\mathrm{kg}\text{ in equation (1) } \end{gathered}$$

$$\begin{gathered} \mathrm{COP}=\frac{(252 \mathrm{kJ}/\mathrm{kg})-(116 \mathrm{kJ}/\mathrm{kg})}{(276 \mathrm{kJ}/\mathrm{kg})-(252 \mathrm{kJ}/\mathrm{kg})} \\ = 5.67 \end{gathered}$$

$$\begin{gathered} \text{ Substitute }{T}_{\text{cold }}=8.9\text{ and } \\ T_{\text{hot }}=46.3°\text{ for in equation (2) } \end{gathered}$$

$$\begin{gathered} {\mathrm{COP}}_{\max }=\frac{\left(\right(8.9+273\left)\mathrm{K}\right)}{\left(\right(46.3+273\left)\mathrm{K}\right)-\left(\right(8.9+273\left)\mathrm{K}\right)} \\ = 7.54 \end{gathered}$$

Hence The household air conditioner will be operated between pressures of 4 and 12 bar, with a COP of 5.67 for the design and 7.54 for an ideal Carnot refrigerator running between the same reservoir temperatures, which is higher than the design's COP.