Question

A closed, rigid tank contains a two-phase liquid–vapor mixture of Refrigerant 22 initially at 2208C with a quality of 50.36%. Energy transfer by heat into the tank occurs until the refrigerant is at a final pressure of 6 bar. Determine the final temperature, in 8C. If the final state is in the superheated vapor region, at what temperature, in 8C, does the tank contain only saturated vapor?

Answer 1

Answer:

1. The final temperature, when the pressure is 6 bar, is 43.75 C. And it is indeed in the superheated vapor region.
2. The tank only would contain saturated vapor at 0 C.

Explanation:

1. It is needed property tables for Refrigerant 22.
2. It must be noted, that if the tank is rigid, then the specific volume would be constant in each state of the refrigerant.
3. Then knowing the initial state is at -20 C and 50.36% quality, it is possible to calculate the specific volume at the initial state.
4. The specific volume would be calculated using $v_{1}=v_{1f}+X_1*(v_{1g}-v_{1f})$ where: $v_{1}$ is the specific volume at the initial state, $v_{1f}$  is the specific volume of the saturated liquid state at the temperature of the initial state, $(v_{1g}$ is the specific volume of the saturated vapor state at the same temperature of the initial state, and $X_1$ is the quality at the initial state.
5. If $v_{1f}$ at -20 C is $0.7427*10^{-3}\frac{m^3}{kg}$, and $v_1g$ at -20 C is $0.0926\frac{m^3}{kg}$, then: $v_1=0.7427*10^{-3}\frac{m^3}{kg}+0.5036(0.0926\frac{m3}{kg}-0.7427*10^{-3}\frac{m^3}{kg})=0.047\frac{m^3}{kg}$
6. As the specific volume stays constant, then the specific volume would be compared with the specific volume at saturated vapor state at 6 bar. At this pressure $v_g=0.0392\frac{m^3}{kg}$. Then, as this value is lower than the specific volume previously calculated ($0.047\frac{m^3}{kg}$), it means that the final state is in the superheated vapor region.
7. Looking for the specific volume at 6 bar pressure, it is found that is between 45 C ($0.04724\frac{m^3}{kg}$) and 40 C ($0.04623\frac{m^3}{kg}$. Interpolation gives that with a specific volume of $0.047 \frac{m^3}{kg}$ and a pressure of 6 bar, the temperature would be 43.75 C.
8. To found the temperature when the tank only contains saturated vapor, only is required to look when the saturated vapor specific volume is equal to the specific volume calculated for the process ($0.047\frac{m^3}{kg}$). This is found to be at 0 C.