Question

A mixture of saturated water and saturated steam at a temperature of 300ᵒC is contained in a closed vessel of 0.1 m3 capacity. If the mass of saturated water is 1kg find the mass of steam in the vessel. Also find the pressure, specific volume, dryness fraction and enthalpy of the mixture

Answer 1

Answer on Question #73338, Physics / Molecular Physics | Thermodynamics

A mixture of saturated water and saturated steam at a temperature of $300{}^{\circ}\mathrm{C}$ is contained in a closed vessel of $0.1\mathrm{m}^3$ capacity. If the mass of saturated water is $1\mathrm{kg}$ find the mass of steam in the vessel. Also, find the pressure, specific volume, dryness fraction and enthalpy of the mixture.

Solution:

From the Steam tables corresponding to $300{}^{\circ}\mathrm{C}$,

$$\begin{array}{l} v_{f} = v_{1} = 0.001403 \, \mathrm{m}^{3} / \mathrm{kg} \\ v_{g} = v_{s} = 0.022 \, \mathrm{m}^{3} / \mathrm{kg} \\ p = 85.81 \, \text{bar} \end{array}$$

Total volume occupied by the liquid,

$$V_{1} = m_{1} \times v_{1} = 1 \times 0.001403 = 0.001403 \, \mathrm{m}^{3}.$$

Total volume of the vessel,

$$\begin{array}{l} V = \text{Volume of liquid} + \text{Volume of steam} = V_{1} + V_{s} \\ 0.1 = 0.001403 + V_{s} \\ V_{s} = 0.1 - 0.001403 = 0.098597 \, \mathrm{m}^{3} \end{array}$$

Mass of steam,

$$m_{s} = V_{S} / v_{s} = 0.098597 / 0.022 = 4.48 \, \mathrm{kg}.$$

Mass of mixture of liquid and steam,

$$m = m_{1} + m_{s} = 1 + 4.48 = 5.48 \, \mathrm{kg}.$$

Total specific volume of the mixture,

$$v = 0.1 / 5.48 = 0.01825 \, \mathrm{m}^{3} / \mathrm{kg}.$$

Dryness fraction can be expressed:

$$\begin{array}{l} \zeta = m_{s} / (m_{w} + m_{s}) \\ \zeta = 4.48 / 5.48 = 0.8 \end{array}$$ $$\begin{array}{l} v = v_{f} + x \, v_{fg} \\ 0.01825 = 0.001403 + x (0.022 - 0.001403) \\ x = 0.81 \end{array}$$

From Steam table corresponding to $300{}^{\circ}\mathrm{C}$,

$$\begin{array}{l} h_{f} = 1344 \, \mathrm{KJ} / \mathrm{kg} \\ h_{fg} = 1405 \, \mathrm{KJ} / \mathrm{kg} \end{array}$$

Enthalpy of mixture,

$$\begin{array}{l} h = h_{f} + x \, h_{fg} = 1344 + 0.81 \times 1405 = 2482.05 \, \mathrm{KJ} / \mathrm{kg} \\ \text{Answer: } 85.81 \, \text{bar; } 4.48 \, \text{kg; } 0.01825 \, \text{m}^{3} / \text{kg; } 0.8; 2482.05 \, \text{KJ} / \text{kg} \end{array}$$

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