Question

A rigid tank of volume 50 m³ contains a pure substance as a saturated liquid-vapour mixture at 400 kPa. Of the total mass of the mixture, 20% mass is liquid and 80% mass is vapour. Properties at 400 kPa are: \( T_{\text{sat}} = 143.61 \, \text{°C}, \, v_f = 0.001084 \, \text{m³/kg}, \, v_g = 0.46242 \, \text{m³/kg} \). The total mass of liquid-vapour mixture in the tank is \(\underline{\hspace{1cm}}\) kg (round off to the nearest integer).

Hint:

The mass of a saturated liquid-vapour mixture can be calculated using the specific volume and the total volume of the tank.

Answer 1

Correct Answer: 134 - 136

The specific volume of the mixture is given by: \[ v_{\text{mix}} = v_f + x \cdot (v_g - v_f), \] where:
- \( v_f = 0.001084 \, \text{m³/kg} \) is the specific volume of the saturated liquid,
- \( v_g = 0.46242 \, \text{m³/kg} \) is the specific volume of the saturated vapour,
- \( x = 0.80 \) is the quality of the mixture (80% vapour).
Thus, the specific volume of the mixture is: \[ v_{\text{mix}} = 0.001084 + 0.80 \cdot (0.46242 - 0.001084) = 0.001084 + 0.369888 = 0.370972 \, \text{m³/kg}. \] Now, the total mass of the mixture is: \[ m = \frac{V}{v_{\text{mix}}} = \frac{50}{0.370972} \approx 134.8 \, \text{kg}. \] Thus, the total mass of the liquid-vapour mixture in the tank is: \[ \boxed{134 \, \text{to} \, 136 \, \text{kg}}. \]