Question

An insulated rigid container is divided into two parts by a thin partition. One part of the container contains 6 kg of saturated liquid-vapor mixture with a dryness fraction of 0.7 at 0.3 MPa. The other part contains 12 kg of saturated liquid at 0.6 MPa of the same substance. When the partition is removed and the system attains equilibrium, the final specific volume of the mixture is _________ m$^3$/kg (round off to 2 decimal places).

Hint:

When partitioned mixtures reach equilibrium, the total volume is the sum of individual volumes, and the total mass is the sum of individual masses.

Answer 1

Correct Answer: 0.13

For the first part (saturated mixture at 0.3 MPa): Dryness fraction: \[ x_1 = 0.7 \] Specific volume of mixture: \[ v_1 = v_f + x_1(v_g - v_f) \] At 0.3 MPa: \[ v_f = 0.001073~\text{m}^3/\text{kg}, \quad v_g = 0.60582~\text{m}^3/\text{kg} \] \[ v_1 = 0.001073 + 0.7(0.60582 - 0.001073) = 0.001073 + 0.7(0.604747) = 0.001073 + 0.423323 = 0.424396~\text{m}^3/\text{kg} \] For the second part (saturated liquid at 0.6 MPa): At 0.6 MPa: \[ v_f = 0.001101~\text{m}^3/\text{kg}, \quad v_g = 0.31560~\text{m}^3/\text{kg} \] Since it's saturated liquid: \[ v_2 = v_f = 0.001101~\text{m}^3/\text{kg} \] Now, total volume: \[ V = 6 \times v_1 + 12 \times v_2 \] \[ V = 6 \times 0.424396 + 12 \times 0.001101 = 2.546376 + 0.013212 = 2.559588~\text{m}^3 \] Total mass: \[ m = 6 + 12 = 18~\text{kg} \] Final specific volume: \[ v_f = \frac{V}{m} = \frac{2.559588}{18} = 0.1422~\text{m}^3/\text{kg} \] \[ \boxed{0.14~\text{m}^3/\text{kg}} \]