Answer:
At 300 kPa: The temperature is 0.529 ºC and the total enthalpy, 2016.77kJ.
At 600 kPa: The temperature is 21.45 ºC and the total enthalpy, 2317.00 kJ.
Explanation:
1. Get the container volume in cubic meters: [tex]1m^{3}=1000L[/tex]
[tex]V=14L*\frac{1m^{3}}{1000L} =0.014m^{3}[/tex]
2. Find the specific volume of R-134a:
[tex]v=\frac{V}{m}=\frac{0.014m^{3}}{9.96kg}=0.001416\frac{m^{3}}{kg}[/tex]
3. Find the phase of R-134a, for this, look at the steam tables (Note: I am using the table B.5.1 from van Wylen 6th Edition.) Look for a pressure of 300 kPa at the table; I found this values:
T(ºC) P(kPa) vf([tex]\frac{m^{3}}{kg}[/tex] vg([tex]\frac{m^{3}}{kg}[/tex]
0 294.0 0.000773 0.06919
5 350.9 0.000783 0.05833
It is possible to predict that the properties of saturated R134-a at 300 kPa would be so closely to that of 294 kPa. From that data we find that the specific volume of our R134-a is between vf and vg, so we conclude that it is a vapor liquid mixture.
4. Find the quality (x)
From a linear interpolation for the pressure, it is possible to know the saturation data for 300 KPa:
T(ºC) P(kPa) vf([tex]\frac{m^{3}}{kg}[/tex] vg([tex]\frac{m^{3}}{kg}[/tex]
0.529 300 0.0007405 0.06796
Applying the quality relation for specific volume:
[tex]v=v_{f}+xv_{fg}\\x=\frac{v-v_{f}}{v_{fg}}\\x=0.009989[/tex]
The total enthalpy is calculated in the same way:
[tex]h=h_{f}+xh_{fg}\\h=200.71+0.00898*197.96=202.48 \frac{kJ}{kg}[/tex]
[tex]H=202.48*9.96=2016.77kJ[/tex]
The temperature is 0.529 ºC and the total enthalpy, 2016.77kJ.
When the substance is heated, the volume remains constant because the container is rigid, so the calculation requires to do the same process again with 600 kPa.
T(ºC) P(kPa) vf([tex]\frac{m^{3}}{kg}[/tex] vg([tex]\frac{m^{3}}{kg}[/tex]
21.45 600 0.000820 0.03458
hf([tex]\frac{kJ}{kg}[/tex] hg([tex]\frac{kJ}{kg}[/tex]
229.56 406.13
[tex]x=0.0173[/tex]
[tex]h=232.63 \frac{kJ}{kg}[/tex]
[tex]H=m*h=2317.00kJ[/tex]
The temperature is 21.45 ºC and the total enthalpy, 2317.00 kJ.
