Question

An insulated rigid closed tank of $2 \, m^3$ contains saturated liquid-vapor mixture of water at $200^\circ C$. Quality of mixture = $0.75$. Find mass of mixture. Data at $200^\circ C$: $v_f = 0.001156 \, m^3/kg$, $v_{fg} = 0.12620 \, m^3/kg$, $v_g = 0.12736 \, m^3/kg$.

Hint:

For mixtures, always use $v = v_f + x v_{fg}$, where $x$ is the dryness fraction.

Answer 1

Correct Answer: 20.5

Step 1: Specific volume of mixture.
\[ v = v_f + x v_{fg} \] where $x = 0.75$. \[ v = 0.001156 + 0.75(0.12620) = 0.001156 + 0.09465 = 0.09581 \, m^3/kg \]
Step 2: Mass of mixture.
\[ m = \frac{V}{v} = \frac{2}{0.09581} = 20.87 \, kg \] Correction check: Using $v_g = 0.12736$ ensures consistency. Correct answer is around $20.9 \, kg$. Final Answer: \[ \boxed{20.9 \, kg} \]