A closed, rigid tank contains a two-phase liquid-vapor mixture of Refrigerant 22 initially at - 20°C 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 °C. If the final state is in the superheated vapor region, at what temperature, in °C, does the tank contain only saturated vapor?

T_2 = 43.75 C

T = 0 C

Explanation:

Given:

- State 1:

Quality of liquid-vapor mixture x = 0.5036

T_1 = - 20 C

- State 2:

P_2 = 6 bar

Find:

- Final Temperature T_2

- If final state is super-heated, the temperature T the tank. Does the tank only contains saturated vapor.

Solution:

- A constant volume process is applicable on a rigid tank. The system is in equilibrium.

State 1:

Table A-7 property Table:

-v_f = 0.7427*10^-3 m^3 / kg

-v_g = 0.0926 m^3 / kg

v_1 = v_f + x * (v_g - v_g)

v_1 = 0.7427*10^-3 + 0.5036 * (0.0926 - 0.7427*10^-3)

v_1 = 0.047 m^3 / kg

State 2:

Table A-8 property Table:

-v_f = 0.3927*10^-3 m^3 / kg

-v_g = 0.0392 m^3 / kg

-v_2 = v_1 = 0.047 m^3 / kg

v_2 > v_g (Hence, in super-heated region)

-Refer to Table A-9 @ P_2 = 6 bar and v_2 = 0.047 m^3 / kg

T_a = 40 C

v_a = 0.04628 m^3 / kg

T_b = 45 C

v_b = 0.04724 m^3 / kg

Interpolation: T_2 = T_a - (T_a - T_b) * (v_a - v_2) / (v_a - v_b)

T_2 = 40 - (-5) * (0.04628 - 0.047) / (0.04628-0.04724)

T_2 = 43.75 C

- If v_2 = v_g = 0.047 m^3 / kg, the corresponding T can be looked from Table A-7:

T = 0C, P = 4.9811 bar