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1 April, 07:35

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?

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  1. 1 April, 08:20
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    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
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