Question

A \(4-L\) rigid tank contains 2 kg of saturated liquidvapor mixture of water at \(50^{\circ} \mathrm{C}\). The water is now slowly heated until it exists in a single phase. At the final state, will the water be in the liquid phase or the vapor phase? What would your answer be if the volume of the tank were \(400 \mathrm{L}\) instead of \(4 \mathrm{L} ?\)

Short answer

Answer: The final state for water in the 4L tank will be in the liquid phase, while for the 400L tank, it will be in the vapor phase.

Step-by-step solution

Find the specific volume of the initial mixture

To find the specific volume of the initial mixture, first, we need to find the total volume of the tank and then divide it by the mass of the water contained. For the 4L tank, Total volume = 4 L = 0.004 m³ (1 m³ = 1000 L) Mass = 2 kg Initial specific volume, \(v_{initial}\) = \(\frac{Total \, volume}{Mass}\) = \(\frac{0.004 \, \mathrm{m^3}}{2 \, \mathrm{kg}} =0.002 \, \mathrm{m^3/kg}\).

Determine the saturation properties at 50°C

Determine the specific volume of the saturated liquid and the saturated vapor at the given temperature (50°C) using the saturation property tables for water (we can call these properties \(v_f\) and \(v_g\) respectively). At 50°C, \(v_f \approx 0.001 \, \mathrm{m^3/kg}\) \(v_g \approx 0.015 \, \mathrm{m^3/kg}\).

Compare the initial specific volume to saturation properties for the 4L tank

Determine whether the specific volume of the initial mixture falls between the specific volumes of the liquid and vapor phase at the same temperature. Since \(v_f \lt v_{initial} \lt v_g \Rightarrow 0.001 \, \mathrm{m^3/kg} \lt 0.002 \, \mathrm{m^3/kg} \lt 0.015 \, \mathrm{m^3/kg}\), the water in the 4L tank initially exists in a two-phase region. As the water is slowly heated, the mixture moves towards a single-phase region. Since \(v_{initial}\) is closer to \(v_f\), the final state will be in the liquid phase.

Calculate the initial specific volume for the 400L tank

Repeat step 1 for the 400L tank: Total volume = 400 L = 0.4 m³ (1 m³ = 1000 L) Mass = 2 kg Initial specific volume, \(v_{initial}^{'}\) = \(\frac{Total \, volume}{Mass}\) = \(\frac{0.4 \, \mathrm{m^3}}{2 \, \mathrm{kg}} =0.2 \, \mathrm{m^3/kg}\).

Compare the initial specific volume to saturation properties for the 400L tank

Determine whether the specific volume of the initial mixture falls between the specific volumes of the liquid and vapor phase at the same temperature. Since \(v_g \lt v_{initial}^{'} \Rightarrow 0.015\, \mathrm{m^3/kg} \lt 0.2 \, \mathrm{m^3/kg}\), the water in the 400L tank initially exists in the superheated vapor region. As the water is slowly heated, the mixture remains in the vapor phase. In conclusion, the final state for water in the 4L tank will be in the liquid phase, while for the 400L tank, it will be in the vapor phase.