91Ó°ÊÓ

The argument of whether and by how much, if any, the change will affect the COP of a refrigerator.and the reason why real refrigerators use a throttle instead of a turbine.

The refrigerator runs on HFC-234a at pressures ranging from 1 to 10 bar and temperatures ranging from $-26.4°$to $39.4°$. Along the throttling path, entropy is conserved.

The expression of the entropy at point 4

$S=x{S}_1+(1-x)S_{\mathrm{g}}$

Where, x is the fraction of liquid,

$S_1$is the entropy of liquid and

$S_{\mathrm{g}}$is the entropy of the gas.

Rearrange the above expression

$x=\frac{S-S_2}{S_1-S_2}$

Substitute role="math" localid="1648669923161" $S=0.384\mathrm{kJ}/\mathrm{K}·\mathrm{kg}$

$S_1=0.068\mathrm{kJ}/\mathrm{K}·\mathrm{kg}$

$S_{\mathrm{g}}=0.940\mathrm{kJ}/\mathrm{K}·\mathrm{kg}$ from table 4.3 in the above expression.

$$\begin{gathered} x=\frac{(0.384\mathrm{kJ}/\mathrm{K}·\mathrm{kg})-(0.940\mathrm{kJ}/\mathrm{K}·\mathrm{kg})}{(0.068\mathrm{kJ}/\mathrm{K}·\mathrm{kg})-(0.940\mathrm{kJ}/\mathrm{K}·\mathrm{kg})} \\ =0.6376 \end{gathered}$$

Now,

The expression for the enthalpy ${\mathrm{H}}_4$at point 4

$H_4=x{H}_1+(1-x)H_g$

Here, $H_1$= is enthalpy of liquid and

$H_{\mathrm{g}}$= is enthalpy of gas.

$$\begin{gathered} \text{Substitute}{H}_1=16\mathrm{kJ}/\mathrm{kg} \\ {H}_{\mathrm{g}}=231\mathrm{kJ}/\mathrm{kg} \\ x=0.6376\text{from table in the above expression.} \end{gathered}$$

$$\begin{gathered} H_4=(0.6376)(16\mathrm{kJ}/\mathrm{kg})+(1-0.6376)(231\mathrm{kJ}/\mathrm{kg}) \\ =93.92\mathrm{kJ}/\mathrm{kg} \end{gathered}$$

The expression of the COP

$\mathrm{COP}=\frac{H_1-H_4}{H_2-H_3-\left(H_1-H_4\right)}$

Here,$H_1=$enthalpy at point 1,

$H_2$= is enthalpy at point 2 and

localid="1648670937721" $H_3=$is enthalpy at point 3 .

$$\begin{gathered} \text{Substitute}{H}_1=231\mathrm{kJ}/\mathrm{kg} \\ {\mathrm{H}}_2=279.08\mathrm{kJ}/\mathrm{kg}, \\ {\mathrm{H}}_3=105\mathrm{kJ}/\mathrm{kg} \\ {\mathrm{H}}_4=93.92\mathrm{kJ}/kg\text{in above expression} \end{gathered}$$

$$\begin{gathered} \mathrm{COP}=\frac{(231\mathrm{kJ}/\mathrm{kg})-(93.92\mathrm{kJ}/\mathrm{kg})}{(279.08\mathrm{kJ}/\mathrm{kg})-(105\mathrm{kJ}/\mathrm{kg})-\left(\right(231\mathrm{kJ}/\mathrm{kg})-(93.92\mathrm{kJ}/\mathrm{kg}\left)\right)} \\ =3.70 \end{gathered}$$

$\text{The value of increment in COP is}$

$$\begin{gathered} =(3.70-2.62) \\ =1.08 \end{gathered}$$

Hence The COP is enhanced by 1.08, and practical refrigerators use throttle rather than the turbine to reduce cost and complexity.