Question

A rigid tank of 2 $m^3$ internal volume contains 5 kg of water as a saturated liquid–vapor mixture at 400 kPa. Half of the mass of the saturated liquid in the tank is drained-off while maintaining constant pressure of 400 kPa. The final quality of the mixture remaining in the tank is $____________$ (rounded off to two decimal places). Given: At 400 kPa: $v_f = 0.001084 \, m^3/kg$, $v_{fg} = 0.46138 \, m^3/kg$, $v_g = 0.46246 \, m^3/kg$.

Hint:

Whenever part of saturated liquid is drained, the quality increases since the mixture becomes more vapor-rich at same pressure.

Answer 1

Correct Answer: 0.9

Step 1: Initial specific volume. \[ v = \frac{V}{m} = \frac{2}{5} = 0.4 \, m^3/kg \]
Step 2: Initial quality $x_1$. \[ v = v_f + x_1 v_{fg} \] \[ 0.4 = 0.001084 + x_1 (0.46138) \] \[ x_1 = \frac{0.3989}{0.46138} \approx 0.865 \]
Step 3: Mass of saturated liquid initially. \[ m_f = (1-x_1) m = (1-0.865)(5) = 0.675 \, kg \] Half drained: \[ m_{f,new} = 0.3375 \, kg \]
Step 4: Mass of vapor initially. \[ m_g = x_1 m = 0.865 \times 5 = 4.325 \, kg \] No vapor drained.

Step 5: New total mass. \[ m_{new} = m_g + m_{f,new} = 4.325 + 0.3375 = 4.6625 \, kg \]
Step 6: New specific volume. \[ v_{new} = \frac{2}{4.6625} = 0.429 \, m^3/kg \]
Step 7: Final quality. \[ v_{new} = v_f + x_2 v_{fg} \] \[ 0.429 = 0.001084 + x_2 (0.46138) \] \[ x_2 = \frac{0.4279}{0.46138} \approx 0.927 \] Rounded to two decimal places: $x_2 = 0.93$. Final Answer: \[ \boxed{0.93} \] % Quicktip