Q.4.30 - An Introduction to Thermal Physics

Question

Consider a household refrigerator that uses HFC-134a as the refrigerant, operating between the pressures of $1.0\mathrm{bar}$ and $10\mathrm{bars}$.

(a) The compression stage of the cycle begins with saturated vapor at 1 bar and ends at 10 bars. Assuming that the entropy is constant during compression, find the approximate temperature of the vapor after it is compressed. (You'll have to do an interpolation between the values given in Table 4.4.)

(b) Determine the enthalpy at each of the points 1,2,3 and 4 , and calculate the coefficient of performance. Compare to the COP of a Carnot refrigerator operating between the same extreme temperatures. Does this temperature range seem reasonable for a household refrigerator? Explain briefly.

(c) What fraction of the liquid vaporizes during the throttling step?

Step-by-Step Solution

Answer

a) The approximate temperature of the vapor after it is compressed is $49.17°$

b) The COP of Carnot refrigerator operating between the same reservoir temperatures is $\underset{¯}{3.75}$ which is greater than $\underset{¯}{2.62}$ and this temperature range does not seem reasonable for a household refrigerator in very hot areas.

c)$\text{ The fraction of liquid that vaporizes during the throttling step is }\underset{¯}{0.586}\text{.}$

Part (a) - Step 1: To determine

The approximate temperature of the vapor after it is compressed.

Part (a) - Step 2: Explanation

Given:

A household refrigerator that uses HFC-134a as the refrigerant, operates between the pressures of 1.0 bar and 10 bars.

Compression cycle begins with saturated vapor at 1 bar and ends at 10 bars.

Formula: 

According to table 4.3, the temperature is between$40°$and$50°$ equivalent to a low pressure of 1 bar and a high pressure of 10 bars. Along the compression path, the entropy remains constant, with a value of $0.94 \mathrm{kJ}/\mathrm{K}·\mathrm{kg}$ from table 4.3.

The expression of entropy S along compression path

$S=S_{\mathrm{L}}x+(1-x)S_{\mathrm{H}}$

Here, $S_{\mathrm{L}}=$is entropy at $40°$,

           $S_H =$ is entropy at $50°$ and

           x = is the fraction of liquefaction. 

Rearrange the above expression for x 

$x=\frac{S-S_{\mathrm{H}}}{S_{\mathrm{L}}-S_{\mathrm{H}}}\dots \left(1\right)$

The expression of the temperature of vapor after compression

$T=\left(40°\right)x+(1-x)\left(50°\right)\dots .\text{ (2) }$

Part (a) - Step 3: Calculation

Calculation:

$$\begin{gathered} \text{ Substitute} S_{\mathrm{H}}=0.943 \mathrm{kJ}/\mathrm{K}·\mathrm{kg} \\ {S}_{\mathrm{L}}=0.907 \mathrm{kJ}/\mathrm{K}·\mathrm{kg}\text{ from table }4.4\text{ and} \\ S=0.94 \mathrm{kJ}/\mathrm{K}·\mathrm{kg}\text{ for in equation (1) } \end{gathered}$$

$$\begin{gathered} x=\frac{(0.94 \mathrm{kJ}/\mathrm{K}·\mathrm{kg})-(0.943 \mathrm{kJ}/\mathrm{K}·\mathrm{kg})}{(0.907 \mathrm{kJ}/\mathrm{K}·\mathrm{kg})-(0.943 \mathrm{kJ}/\mathrm{K}·\mathrm{kg})} \\ = 0.083 \end{gathered}$$

$\text{ Substitute x = }0.083\text{ in equation (2) }$

$$\begin{gathered} T=\left(40°\right)(0.083)+(1-0.083)\left(50°\right) \\ =49.17° \end{gathered}$$

$$\begin{gathered} \text{Hence,the approximate temperature of the vapor after it is compressed}\text{is} \\ 49.17°\text{. } \end{gathered}$$

Part (b) - Step 4: To find

The enthalpy at each of the points 1,2,3, and 4 , and the coefficient of performance.

Comparison with the COP of Carnot refrigerator operating between the same reservoir temperatures.

Explanation of whether this temperature range seems reasonable for a household refrigerator.

Part (b) - Step 5: Explanation

Given:

A household refrigerator that uses HFC-134a as the refrigerant, operates between the pressures of 1.0 bar and 10 bars.

Formula:

Since the gas has a pressure of 1.0 bar at point 1, the enthalpy at this point $H_1=231 \mathrm{kJ}/\mathrm{kg}$ from table 4.3, and the liquid has a pressure of 10 bar at point 3, the enthalpy at this point $H_3= 105 \mathrm{kJ}/\mathrm{kg}$ from table 4.3. Because the enthalpy between points 3 and 4 is preserved, the enthalpy at point $4 H_4 =105 \mathrm{kJ}/\mathrm{kg}$

The expression of the enthalpy ${\mathrm{H}}_2$at point 2

$H_2=H_{\mathrm{L}}x+(1-x)H_{\mathrm{H}}$

Where, x is the same fraction in Part (a), $H_{\mathrm{L}}$= is enthalpy at temperature $40°$ and pressure 10 bar and $H_{\mathrm{H}}$= is enthalpy at temperature $50°$ and pressure 10 bar.

Substitute $x= 0.083$

  $H_{\mathrm{L}}=269 \mathrm{kJ}/\mathrm{kg}$ and

  $H_{\mathrm{H}}=280 \mathrm{kJ}/\mathrm{kg}$ for in the above expression

$$\begin{gathered} H_2=(269 \mathrm{kJ}/\mathrm{kg})(0.083)+(1-0.083)(280 \mathrm{kJ}/\mathrm{kg}) \\ =279.08KJ/kg \end{gathered}$$

The expression of COP

$\mathrm{COP}=\frac{H_1-H_3}{H_2-H_1}\dots \text{ (3) }$

Write the expression of the maximum COP for a Carnot refrigerator

${\mathrm{COP}}_{\max }=\frac{T_{\text{cold }}}{T_{\text{bot }}-T_{\text{cold }}}\dots \text{ (4) }$

Part (b) - Step 6: Calculation

Calculation:

$$\begin{gathered} \text{ Substitute }{H}_1=231 \mathrm{kJ}/\mathrm{kg}, \\ H_3=105 \mathrm{kJ}/\mathrm{kg}\text{ and} \\ H_2=279.08 \mathrm{kJ}/\mathrm{kg}\text{ for in equation (3) } \end{gathered}$$

$$\begin{gathered} COP=\frac{(231 \mathrm{kJ}/\mathrm{kg})-(105 \mathrm{kJ}/\mathrm{kg})}{(279.08 \mathrm{kJ}/\mathrm{kg})-(231 \mathrm{kJ}/\mathrm{kg})} \\ = 2.62 \end{gathered}$$

$$\begin{gathered} \text{ Substitute}{T}_{\text{cold }}=-26.4°\text{ and} \\ T_{\text{hot }}=39.4°\text{ for in equation (4) } \end{gathered}$$

$$\begin{gathered} {\mathrm{COP}}_{\max }=\frac{(-26.4+273)\mathrm{K}}{(39.4+273)\mathrm{K}-(-26.4+273)\mathrm{K}} \\ =3.75 \end{gathered}$$

Hence, the enthalpy at point 1 is$231 \mathrm{kJ}/\mathrm{kg}$, the enthalpy at point 2 is $279.08 \mathrm{kJ}/\mathrm{kg}$, the enthalpy at point 3 is $105 \mathrm{kJ}/\mathrm{kg}$, the enthalpy at point 4 is $105 \mathrm{kJ}/\mathrm{kg}$and the coefficient of performance is 2.62.

Part (c) - Step 7: To find

The fraction of liquid that vaporizes during the throttling step.

Part (c) - Step 8: Explanation

Given: A household refrigerator that uses HFC-134a as the refrigerant, operates between the pressures of 1.0 bar and 10 bars.

Formula:

$$\begin{gathered} \text{The expression of the enthalpy at point }4\text{ at temperature }-26.4° \\ \text{ and pressure }1.0\text{ bar } \end{gathered}$$

$H_4=x{H}_{\text{liquid }}+(1-x)H_{\text{gaseous }}$

Here, x = is the fraction of liquid at point 4,

        $H_{\text{liquid }}=$ is the enthalpy of liquid and

       $H_{\text{gaseous }}=$is the enthalpy of the ga at point 4.

Rearrange the above expression for x

$x=\frac{H_4-H_{\text{gaseous }}}{H_{\text{liquid }}-H_{g\text{ aseos }}}\dots \left(5\right)$

Calculation:

$$\begin{gathered} \text{ Substitute}{\mathrm{H}}_4=105 \mathrm{kJ}/\mathrm{kg} \\ H_{\text{gaseous }}=231 \mathrm{kJ}/\mathrm{kg} \\ H_{\text{liquid }}=16 \mathrm{kJ}/\mathrm{kg}\text{ for from table }4.3\text{ in equation (5) } \end{gathered}$$

$$\begin{gathered} x=\frac{(105 \mathrm{kJ}/\mathrm{kg})-(231 \mathrm{kJ}/\mathrm{kg})}{(16 \mathrm{kJ}/\mathrm{kg})-(231 \mathrm{kJ}/\mathrm{kg})} \\ = 0.586 \end{gathered}$$

$\text{Hence the fraction of liquid that vaporizes during the throttling step is }0.586\text{. }$