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).
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Solution and Explanation
- Initial temperature \( T_1 = 200^\circ C \),
- Final temperature \( T_2 = 110^\circ C \),
- Saturated volume of vapour at 200°C \( v_1 = 0.127 \, {m}^3/{kg} \),
- Saturated volume of vapour at 110°C \( v_2 = 1.210 \, {m}^3/{kg} \),
- Saturated volume of liquid at 110°C \( v_f = 0.001 \, {m}^3/{kg} \).
We need to find the ratio of mass of liquid to mass of vapour. We can use the mass balance and specific volumes to find this ratio.
Step 1: The total volume of the system is the sum of the volumes of the liquid and vapour phases at 110°C. The total volume can be written as: \[ V_{{total}} = v_f \cdot m_{{liquid}} + v_2 \cdot m_{{vapour}} \] Let the mass of vapour be \( m_{{vapour}} = m_v \) and the mass of liquid be \( m_{{liquid}} = m_l \). The total mass is the sum of both: \[ m_{{total}} = m_v + m_l \] Thus, the total volume is: \[ V_{{total}} = v_2 \cdot m_v + v_f \cdot m_l \] Step 2: Since the system is closed, we have a relationship between the volumes at two different temperatures using the principle of conservation of mass. After applying the mass balance equations and solving, the ratio of the mass of liquid to the mass of vapour is approximately: \[ \frac{m_l}{m_v} \approx 8.5 \]
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