Chapter 6, Problem 6.23P

Interpretation Introduction

Interpretation:

The final volume and mass of liquid water should be deduced based on the given initial and final condition of steam

Concept Introduction:

• For a two-phase liquid-vapor equilibrium mixture the specific volume (V), enthalpy (H) and entropy (S) are given as:

$\begin{array}{l}{\text{V = V}}_{\text{f}}{\text{ + x(V}}_{\text{g}}{\text{-V}}_{\text{f}}\text{) -----(1)}\\ {\text{H = H}}_{\text{f}}{\text{ + x(H}}_{\text{g}}{\text{-H}}_{\text{f}}\text{)-----(2)}\\ {\text{S = S}}_{\text{f}}{\text{ + x(S}}_{\text{g}}{\text{-S}}_{\text{f}}\text{)--------(3)}\\ {\text{where, V}}_{\text{g}}{\text{ and V}}_{\text{f}}\text{ are the specific volumes in the vapor and liquid phases respectively }\\ {\text{H}}_{\text{g}}{\text{ and H}}_{\text{f}}\text{ are the specific enthalpies in the vapor and liquid phases respectively}\\ {\text{S}}_{\text{g}}{\text{ and S}}_{\text{f}}\text{ are the specific entropies in the vapor and liquid phases respectively}\\ \text{x = steam quality }\end{array}$

$$

Final mass of liquid water = 0.377 kg

Final volume of liquid water = 0.00378 m³

Given:

Volume of vessel V = 0.15 m³

Initial Temperature of saturated steam = $150\text{C}$

Final temperature = $30\text{C}$

Explanation:

Based on the steam tables at Temperature = $150\text{C}$ we have:

Specific volume of liquid, v_f = 0.001091 m³/kg

Specific volume of vapor, v_g = 0.39248 m³/kg

Again from the steam tables, at Temperature = $30\text{C}$ we have:

Specific volume of liquid, v_f = 0.001004 m³/kg

Specific volume of vapor, v_g = 32.879 m³/kg

Calculation:

Step 1:

Calculate the final mass of liquid water

In the initial state, the total mass is equivalent to the mass of the vapor

The total volume V = 0.15 m³

Specific volume of vapor, v_g = 0.39248 m³/kg

  $\begin{array}{l}{\text{Total Mass, m}}_{\text{total}}\text{ = }\frac{\text{V}}{{\text{v}}_{\text{g}}}=\frac{0.15{\text{ m}}^3}{\text{0}{\text{.39248 m}}^{\text{3}}\text{/kg}}=0.382\text{ kg}\\ \text{Now,}\\ {\text{V}}_{\text{total}}{\text{ = m}}_{\text{total}}{\text{×v}}_{\text{f}}{\text{+ m}}_{\text{vap}}{\text{×v}}_{\text{g}}\\ {\text{where, m}}_{\text{vap}}\text{ = mass of vapor phase in the final state}\\ 0.15\text{ = 0}\text{.382×}0.001004{\text{+ m}}_{\text{vap}}\text{×}32.879\\ {\text{m}}_{\text{vap}}\text{ = 0}\text{.00455 kg}\\ {\text{Thus, m}}_{\text{liq}}{\text{ = m}}_{\text{total}}{\text{ - m}}_{\text{vap}}\text{ = 0}\text{.382 - 0}\text{.00455 = 0}\text{.377 kg}\end{array}$

Step 2:

Calculate final volume of liquid water

$\begin{array}{l}\text{Volume of liquid water is given as:-}\\ {\text{V}}_{\text{liq }}{\text{= m}}_{\text{liq}}\times {\text{v}}_{\text{f}}\\ \text{ = 0}\text{.377}\times 0.001004\text{ = 0}{\text{.000378 m}}^3\end{array}$

Thus,

The final mass of liquid water = 0.377 kg

The final volume of liquid water = 0.00378 m³