Chapter 6, Problem 6.19P

Interpretation Introduction

Interpretation:

The total enthalpy and entropy of a liquid-vapor equilibrium mixture at a pressure of 8000 kPa should be deduced.

Concept Introduction:

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

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

Total Enthalpy, H = $\text{1397}\text{.83 kJ/kg}$

Total Entropy, S = $\text{3}\text{.349 kJ/kg-K}$

Given:

Total volume, V^t = 0.15 m³

Equilibrium pressure = 8000 kPa

Explanation:

Based on the steam tables at pressure = 8000 kPa we have:

Specific volume of liquid, v_f = 0.001384 m³/kg

Specific volume of vapor, v_g = 0.023525 m³/kg

Specific enthalpy of liquid, H_f = 1317.1 kJ/kg

Specific enthalpy of vapor, H_g = 2758.7 kJ/kg

Specific entropy of liquid, S_f = 3.2077 kJ/kg-K

Specific entropy of liquid, S_g = 5.7450 kJ/kg-K

Calculation:

Step 1:

Calculate the steam quality x using equation (3)

It is given that in the volumes of the liquid and vapor in the mixture are equal i.e. the total volume is:-

  $\begin{array}{l}{\text{V}}^{\text{t}}{\text{ = V}}^{\text{liq}}{\text{ + V}}^{\text{vap}}\\ {\text{V}}^{\text{t}}{\text{ = m}}_{{}_{\text{liq}}}{\text{×v}}_{\text{f}}{\text{ + m}}_{{}_{\text{vap}}}{\text{×v}}_{\text{g}}\text{ ------(4)}\\ {\text{where, m}}_{\text{liq}}{\text{ and m}}_{\text{vap}}\text{ are mass of the liquid and vapor phases}\\ {\text{v}}_{\text{f}}{\text{ and v}}_{\text{g}}\text{ are the specific volumes of the liquid and vapor}\\ {\text{Since, V}}^{\text{liq}}{\text{ = V}}^{\text{vap}}\\ {\text{V}}^{\text{t}}{\text{ = 2V}}^{\text{liq}}\\ {\text{(or) V}}^{\text{t}}{\text{ = 2(m}}_{{}_{\text{liq}}}{\text{×v}}_{\text{f}}\text{) ----(5)}\\ {\text{m}}_{\text{liq}}\text{ = }\frac{{\text{V}}^{\text{t}}}{2{\text{v}}_{\text{f}}}=\frac{0.15}{2\times 0.001384}=54.19\text{ kg}\\ {\text{Substituting for m}}_{\text{liq}}\text{ in eq(4) we get:}\\ \text{0}\text{.15 = 54}\text{.19×0}{\text{.001384 + m}}_{{}_{\text{vap}}}\text{×0}\text{.023525 }\\ {\text{m}}_{\text{vap}}\text{ = 3}\text{.188 kg}\\ \text{Therefore, based on equation (3)}\\ \text{x = }\frac{3.188}{3.188+54.19}\text{ = 0}\text{.056}\end{array}$

Step 2:

Calculate H and S based on equations 1 and 2

$\begin{array}{l}{\text{H = H}}_{\text{f}}{\text{ + x (H}}_{\text{g}}{\text{-H}}_{\text{f}}\text{)}\\ \text{ = 1317}\text{.1 + 0}\text{.056(2758}\text{.7-1317}\text{.1) = 1397}\text{.83 kJ/kg}\\ {\text{S = S}}_{\text{f}}{\text{ + x (S}}_{\text{g}}{\text{-S}}_{\text{f}}\text{)}\\ \text{ = 3}\text{.2077 + 0}\text{.056(5}\text{.7450-3}\text{.2077) = 3}\text{.349 kJ/kg-K}\end{array}$

Thus the total enthalpy and entropy values are:

Total Enthalpy, H = $\text{1397}\text{.83 kJ/kg}$

Total Entropy, S = $\text{3}\text{.349 kJ/kg-K}$