Computing the Properties of a Two-Phase Mixture Compute the total volume, total enthalpy, and total entropy of 1 kg of water at 100°C, half by weight of which is steam and the remainder liquid water.

Question

Computing the Properties of a Two-Phase Mixture
Compute the total volume, total enthalpy, and total entropy of 1 kg of water at 100°C, half by weight of which is steam and the remainder liquid water.

Accepted Answer

From the saturated steam temperature table at 100°C, the equilibrium pressure is 0.101 35 MPa and

[latex]\begin{aligned}\hat{V}^{\mathrm{L}} &=0.001004 \mathrm{~m}^{3} / \mathrm{kg} & \hat{V}^{\mathrm{V}} &=1.6729 \mathrm{~m}^{3} / \mathrm{kg} \\\hat{H}^{\mathrm{L}} &=419.04 \mathrm{~kJ} / \mathrm{kg} & \hat{H}^{\mathrm{V}} &=2676.1 \mathrm{~kJ} / \mathrm{kg} \\\hat{S}^{\mathrm{L}} &=1.3069 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K}) & \hat{S}^{\mathrm{V}} &=7.3549 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\end{aligned}[/latex]

Using Eq. 7.3-1a on a mass basis gives

[latex]\underline{V}=\omega^{\mathrm{V}} \underline{V}^{\mathrm{V}}+\omega^{\mathrm{L}} \underline{V}^{\mathrm{L}}=\omega^{\mathrm{V}} \underline{V}^{\mathrm{V}}+\left(1-\omega^{\mathrm{V}} ight) \underline{V}^{\mathrm{L}}[/latex] (7.3.1a)

[latex]\hat{V}=0.5 imes 0.001004+0.5 imes 1.6729=0.83645 \mathrm{~m}^{3} / \mathrm{kg}[/latex]

The analogous equation for enthalpy is

[latex]\hat{H}=\omega^{\mathrm{L}} \hat{H}^{\mathrm{L}}+\omega^{\mathrm{V}} \hat{H}^{\mathrm{V}}=0.5 imes 419.04+0.5 imes 2676.1=1547.6 \frac{\mathrm{kJ}}{\mathrm{kg}}[/latex]

and that for entropy is

[latex]\hat{S}=\omega^{\mathrm{L}} \hat{S}^{\mathrm{L}}+\omega^{\mathrm{V}} \hat{S}^{\mathrm{V}}=0.5 imes 1.3069+0.5 imes 7.3549=4.3309 \frac{\mathrm{kJ}}{\mathrm{kg} \mathrm{K}}[/latex]

Question 7.3.1: Computing the Properties of a Two-Phase Mixture Compute the ...

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