# 10 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determi

10 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed.

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Troy Lesure

Step 1
First, we calculate the total specific volume of the mixture:
${\alpha }_{\to t}=\frac{V}{m}$
$=\frac{0.014{m}^{3}}{10kg}$
$=0.0014\frac{{m}^{3}}{kg}$
Next, we look at table A-12 to determine the values of the specific volumes of the liquid and gas parts and their enthalpies. Since we don't have values for 300 kPa, we have to interpolate from the values for 280 and 320 kPa. Now, we calculate the quality:
${q}_{1}=\frac{{\alpha }_{\text{tot}}-{\alpha }_{\text{liq}}}{{\alpha }_{\text{vap}}-{\alpha }_{\text{liq}}}$
$=\frac{\left(0.0014-0.0007734\right)\frac{{m}^{3}}{kg}}{\left(0.0680575-0.0007734\right)\frac{{m}^{3}}{kg}}$
$=0.009$
The enthalpy is calculated from the values given in the table A-13, again using interpolations:
${h}_{1}={h}_{\text{liq}}+{q}_{1}{h}_{\text{evap}}$
$=52.65\frac{\text{kJ}}{\text{kg}}+0.009\cdot 198.195\frac{\text{kJ}}{\text{kg}}$
$=\begin{array}{|c|}\hline 54.43\frac{\text{kJ}}{\text{kg}}\\ \hline\end{array}$
The temperature is determined from the given table values also using interpolation:
${T}_{1}=\begin{array}{|c|}\hline {0.61}^{\circ }C\\ \hline\end{array}$
The temperature at 600 kPa, determined from A-12, is:
${T}_{2}=\begin{array}{|c|}\hline {21.55}^{\circ }C\\ \hline\end{array}$
Step 2
The quality of the mixture this pressure is:
${q}_{1}=\frac{{\alpha }_{\text{tot}}-{\alpha }_{\text{liq}}}{{\alpha }_{\text{vap}}-{\alpha }_{\text{liq}}}$
$\frac{\left(0.0014-0.0008198\right)\frac{{m}^{3}}{kg}}{\left(0.034335-0.0008198\right)\frac{{m}^{3}}{kg}}$
$=0.017$
The final enthalpy is:
${h}_{2}={h}_{\text{liq}}+{q}_{2}{h}_{\text{evap}}$
$=81.5\frac{kJ}{kg}+0.017\cdot 180.95\frac{kJ}{kg}$
$=\begin{array}{|c|}\hline 84.58\frac{kJ}{kg}\\ \hline\end{array}$
Result
${T}_{1}={0.61}^{\circ }C,{h}_{1}=54.43\frac{kJ}{kg}$
${T}_{2}={21.55}^{\circ }C,{h}_{2}=84.58\frac{kJ}{kg}$