Question

A steam boiler contains saturated water vapour at 200 °C. After a certain period, the temperature of the boiler drops to 110 °C. Assume that all the valves of the boiler are closed and the energy is lost as heat to the surroundings. The ratio of mass of liquid to the mass of vapour is ............. (rounded off to two decimal places). Data: - \(v_g(200^\circ C) = 0.127 \, m^3/kg\) - \(v_g(110^\circ C) = 1.210 \, m^3/kg\) - \(v_f(110^\circ C) = 0.001 \, m^3/kg\)

Hint:

For rigid vessels, specific volume remains constant. Quality is obtained from mixing equation: \[ v = (1-x)v_f + x v_g \] and mass ratio \(m_f/m_g = (1-x)/x\).

Answer 1

Correct Answer: 8.1

Step 1: Initial condition.
At 200 °C, saturated vapour, so total specific volume = \(v = v_g = 0.127 \, m^3/kg\).

Step 2: Final condition (mixture at 110 °C).
For mixture: \[ v = (1-x) v_f + x v_g \] where \(x\) = quality (mass fraction of vapour). \[ 0.127 = (1-x)(0.001) + x (1.210) \] \[ 0.127 = 0.001 - 0.001x + 1.210x \] \[ 0.126 = 1.209x \Rightarrow x = 0.104 \]

Step 3: Mass ratio liquid to vapour.
\[ \frac{m_f}{m_g} = \frac{1-x}{x} = \frac{0.896}{0.104} = 8.62 \] Final Answer:
\[ \boxed{8.62} \]