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25 December, 10:57

Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a) the ideal-gas equation, (b) the van der Waals equation, (c) the refrigerant tables.

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  1. 25 December, 14:50
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    a) Using Ideal gas Equation, T = 434.98°R = 435°R

    b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

    c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

    Explanation:

    a) Ideal gas Equation

    PV = mRT

    T = PV/mR

    P = pressure = 400 psia

    V/m = specific volume = 0.1144 ft³/lbm

    R = gas constant = 0.1052 psia. ft³/lbm.°R

    T = 400 * 0.1144/0.1052 = 434.98 °R

    b) Van Der Waal's Equation

    T = (1/R) (P + (a/v²)) (v - b)

    a = Van Der Waal's constant = (27R² (T꜀ᵣ) ²) / (64P꜀ᵣ)

    R = 0.1052 psia. ft³/lbm.°R

    T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

    P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

    a = (27 * 0.1052² * 673.6²) / (64 * 588.7)

    a = 3.596 ft⁶. psia/lbm²

    b = (RT꜀ᵣ) / 8P꜀ᵣ

    b = (0.1052 * 673.6) / (8 * 588.7) = 0.01504 ft³/lbm

    T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

    c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

    T = 100°F = 559.67°R
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