Question

A 1.2 m$^{3}$ rigid vessel contains 8 kg of saturated liquid–vapor mixture at 150 kPa. The specific enthalpy of this mixture is _________ kJ/kg (round off to 2 decimal places). At 150 kPa:
$v_f = 0.001053$ m$^{3}$/kg,
$v_g = 1.1594$ m$^{3}$/kg
$h_f = 467.13$ kJ/kg,
$h_g = 2693.1$ kJ/kg

Hint:

Use mixture relations: $v = v_f + x(v_g - v_f)$ and $h = h_f + x(h_g - h_f)$.

Answer 1

Correct Answer: 750

Total specific volume of mixture: \[ v = \frac{V}{m} = \frac{1.2}{8} = 0.15~\text{m}^3/\text{kg} \] Quality: \[ v = v_f + x(v_g - v_f) \] \[ 0.15 = 0.001053 + x(1.1594 - 0.001053) \] \[ 0.15 - 0.001053 = x(1.158347) \] \[ x = \frac{0.148947}{1.158347} = 0.1286 \] Specific enthalpy: \[ h = h_f + x(h_g - h_f) \] \[ h = 467.13 + 0.1286(2693.1 - 467.13) \] \[ h = 467.13 + 0.1286(2225.97) \] \[ h = 467.13 + 286.4 = 753.53~\text{kJ/kg} \] \[ \boxed{753.53~\text{kJ/kg}} \]