Chapter 3.8, Problem 95P

Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft³/lbm. Determine the temperature of the refrigerant based on (a) the ideal-gas equation, (b) the van der Waals equation, and (c) the refrigerant tables.

To determine

The temperature of the refrigerant using the ideal gas equation.

Answer to Problem 95P

The temperature of the refrigerant using the ideal gas equation is $\underset{\_}{435\text{R}}$.

Explanation of Solution

Determine the temperature of the refrigerant using the ideal gas equation.

$T=\frac{Pv}{R}$ (I)

Here, the pressure of the refrigerant is $P$, the specific volume of the refrigerant is $v$, the universal gas constant is $R$ and the pressure of the refrigerant is $T$.

Conclusion:

Refer to Table A-1E to find the gas constant, the critical pressure, and the critical temperature of refrigerant-134a as $\text{0}\text{.1052}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$, $588.7\text{psia}$, and $673.6\text{R}$.

Substitue $\text{0}\text{.1052}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for $R$, $400\text{psia}$ for $P$, and $0.1144{\text{ft}}^{\text{3}}/\text{lbm}$ for $v$ in Equation (I).

$\begin{array}{c}T=\frac{\left(400\text{psia}\right)\left(0.1144{\text{ft}}^{\text{3}}/\text{lbm}\right)}{\left(\text{0}\text{.1052}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)}\\ =\frac{\left(45.76\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\right)}{\left(\text{0}\text{.1052}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)}\\ =434.98\text{R}\\ \cong 435\text{R}\end{array}$

Thus, the temperature of the refrigerant using the ideal gas equation is $\underset{\_}{435\text{R}}$.

To determine

The temperature of the refrigerant using the van der Waals.

Answer to Problem 95P

The temperature of the refrigerant using the van der Waals is $\underset{\_}{637.5\text{K}}$.

Explanation of Solution

Determine the temperature of the refrigerant using the van der Waals.

$\begin{array}{c}T=\frac{1}{R}\left(P+\frac{a}{v^2}\right)\left(v-b\right)\\ =\frac{1}{R}\left(P+\frac{\left(\frac{27R^2{T}_{\text{cr}}^2}{64P_{\text{cr}}}\right)}{v^2}\right)\left(v-\left(\frac{R{T}_{\text{cr}}}{8P_{\text{cr}}}\right)\right)\end{array}$ (II)

Here, the critical temperature is $T_{\text{cr}}$, the critical pressure is $P_{\text{cr}}$.

Conclusion:

Substitute $\text{0}\text{.1052}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}$ for $R$, $400\text{psia}$ for $P$, $0.1144{\text{ft}}^{\text{3}}/\text{lbm}$ for $v$, $673.6\text{R}$ for $T_{\text{cr}}$, and $588.7\text{psia}$ for $P_{\text{cr}}$ in Equation (II).

$\begin{array}{c}T=\left[\begin{array}{l}\frac{1}{\left(\text{0}\text{.1052}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)}\\ \times \left(\left(400\text{psia}\right)+\frac{\left(\frac{27{\left(\text{0}\text{.1052}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)}^2{\left(673.6\text{R}\right)}^2}{64\left(588.7\text{psia}\right)}\right)}{{\left(0.1144{\text{ft}}^{\text{3}}/\text{lbm}\right)}^2}\right)\\ \times \left(\left(0.1144{\text{ft}}^{\text{3}}/\text{lbm}\right)-\left(\frac{\left(\text{0}\text{.1052}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)\left(673.6\text{R}\right)}{8\left(588.7\text{psia}\right)}\right)\right)\end{array}\right]\\ =\left[\begin{array}{l}\frac{1}{\left(\text{0}\text{.1052}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)}\left(\left(400\text{psia}\right)+\frac{\left(3.598\text{psia}\cdot {\text{ft}}^{\text{6}}/{\text{lbm}}^2\right)}{{\left(0.1144{\text{ft}}^{\text{3}}/\text{lbm}\right)}^2}\right)\\ \times \left(\left(0.1144{\text{ft}}^{\text{3}}/\text{lbm}\right)-\left(0.015046{\text{ft}}^{\text{3}}/\text{lbm}\right)\right)\end{array}\right]\\ =\left[\begin{array}{l}\frac{1}{\left(\text{0}\text{.1052}\text{psia}\cdot {\text{ft}}^{\text{3}}/\text{lbm}\cdot \text{R}\right)}\left(674.96\text{psia}\right)\\ \times \left(0.099353{\text{ft}}^{\text{3}}/\text{lbm}\right)\end{array}\right]\\ =637.45\text{R}\end{array}$

                $\cong 637.5\text{R}$

Thus, the temperature of the refrigerant using the van der Waals is $\underset{\_}{637.5\text{K}}$.

To determine

The temperature of the refrigerant using the refrigerant table R-134.

Answer to Problem 95P

The temperature of the refrigerant using the refrigerant table R-134 is $\underset{\_}{660\text{R}}$.

Explanation of Solution

From the Table A-13E, “Superheated refrigrenat-134a” to obtain the value of the temperature of the refrigerant at $400\text{psia}$ of pressure and $0.1144{\text{ft}}^{\text{3}}/\text{lbm}$ of specific volume of the refrigerant as $200°\text{F}$.

Unit conversion temperature from $°\text{F}$ to R.

$\begin{array}{c}T=200°\text{F}\\ =200+460\text{R}\\ =660\text{R}\end{array}$

Thus, the temperature of the refrigerant using the refrigerant table R-134 is $\underset{\_}{660\text{R}}$.