Respuesta :
Answer:
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
