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The approximate temperature of the vapor after it is compressed.

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

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

The expression of the temperature of vapor after compression

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

Calculation:

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

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

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

role="math" localid="1648684064666" $$\begin{gathered} T=\left(40°\right)(0.083)+(1-0.083)\left(50°\right) \\ =49.17° \end{gathered}$$

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

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.

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 $4H_4=105\mathrm{kJ}/\mathrm{kg}$

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$x=0.083$

role="math" localid="1648684832923" $H_{\mathrm{L}}=269\mathrm{kJ}/\mathrm{kg}$and

role="math" localid="1648684857165" $H_{\mathrm{H}}=280\mathrm{kJ}/\mathrm{kg}$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

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

Write the expression of the maximum COP for a Carnot refrigerator

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

Calculation:

$$\begin{gathered} \text{Substitute}{H}_1=231\mathrm{kJ}/\mathrm{kg}, \\ {H}_3=105\mathrm{kJ}/\mathrm{kg}\text{and} \\ {H}_2=279.08\mathrm{kJ}/\mathrm{kg}\text{forin equation (3)} \end{gathered}$$

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

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

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

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.

The fraction of liquid that vaporizes during the throttling step.

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

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

Calculation:

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

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

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