Chapter 6, Problem 6.20P

Interpretation Introduction

Interpretation:

The enthalpy and entropy of a liquid-vapor equilibrium mixture at a pressure of 1000 kPa should be deduced.

Concept Introduction:

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

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

Enthalpy, H = $\text{788}\text{.7 kJ/kg}$

Entropy, S = $\text{2}\text{.196 kJ/kg-K}$

Given:

Total mass, M = 1.0 kg

Equilibrium pressure = 1000 kPa

Explanation:

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

Specific volume of liquid, v_f = 0.001127 m³/kg

Specific volume of vapor, v_g = 0.19436 m³/kg

Specific enthalpy of liquid, H_f = 762.51 kJ/kg

Specific enthalpy of vapor, H_g = 2777.1 kJ/kg

Specific entropy of liquid, S_f = 2.1381 kJ/kg-K

Specific entropy of liquid, S_g = 6.5850 kJ/kg-K

Calculation:

Step 1:

Calculate the steam quality x using equation (3)

It is given that in the volumes of the vapor is 70% of the total volume in the mixture. The total volume, V^t is:-

  $\begin{array}{l}{\text{V}}^{\text{t}}{\text{ = V}}^{\text{liq}}{\text{ + V}}^{\text{vap}}\\ {\text{V}}^{\text{t}}{\text{ = m}}_{{}_{\text{liq}}}{\text{×v}}_{\text{f}}{\text{ + m}}_{{}_{\text{vap}}}{\text{×v}}_{\text{g}}\text{ ------(4)}\\ {\text{where, m}}_{\text{liq}}{\text{ and m}}_{\text{vap}}\text{ are mass of the liquid and vapor phases}\\ {\text{v}}_{\text{f}}{\text{ and v}}_{\text{g}}\text{ are the specific volumes of the liquid and vapor}\\ {\text{Since, V}}^{\text{vap}}\text{ = 0}{\text{.7 V}}^{\text{t}}\\ {\text{we can write, m}}_{{}_{\text{vap}}}{\text{×v}}_{\text{g}}\text{ = 0}{\text{.7 V}}^{\text{t}}\text{ -----(5)}\\ {\text{by extension, m}}_{{}_{\text{liq}}}{\text{×v}}_{\text{f}}=\text{0}{\text{.3 V}}^{\text{t}}\text{ ------(6)}\\ \frac{{\text{m}}_{{}_{\text{vap}}}{\text{×v}}_{\text{g}}}{{\text{m}}_{{}_{\text{liq}}}{\text{×v}}_{\text{f}}}\text{ = }\frac{\text{0}{\text{.7 V}}^{\text{t}}}{\text{0}{\text{.3 V}}^{\text{t}}}\\ \frac{{\text{m}}_{{}_{\text{vap}}}}{{\text{m}}_{{}_{\text{liq}}}}\text{ = 2}\text{.333}\times \frac{{\text{v}}_{\text{f}}}{{\text{v}}_{\text{g}}}\\ {\text{Substituting the values for v}}_{\text{f}}{\text{ and v}}_{\text{g}},\\ \frac{{\text{m}}_{{}_{\text{vap}}}}{{\text{m}}_{{}_{\text{liq}}}}\text{ = 2}\text{.333}\times \frac{0.001127}{0.19436}\text{ =0}\text{.0135 -------(7)}\\ \text{Now, the total mass m = 1}\text{.0 kg, i}\text{.e}\text{.}\\ {\text{m}}_{{}_{\text{vap}}}+{\text{m}}_{{}_{\text{liq}}}=1\\ \text{Based on equation 3, we can then write,}\\ {\text{x = m}}_{{}_{\text{vap}}}{\text{ and (1-x) = m}}_{{}_{\text{liq}}}\\ {\text{Substituting for m}}_{\text{liq}}{\text{ and m}}_{{}_{\text{vap}}}\text{ in eq(7) we get:}\\ \frac{\text{x}}{\text{1-x}}\text{ = 0}\text{.0135}\\ \text{x = 0}\text{.013}\end{array}$

Step 2:

Calculate H and S based on equations 1 and 2

$\begin{array}{l}{\text{H = H}}_{\text{f}}{\text{ + x (H}}_{\text{g}}{\text{-H}}_{\text{f}}\text{)}\\ \text{ = 762}\text{.51 + 0}\text{.013(2777}\text{.1-762}\text{.51) = 788}\text{.7 kJ/kg}\end{array}$

$\begin{array}{l}{\text{S = S}}_{\text{f}}{\text{ + x (S}}_{\text{g}}{\text{-S}}_{\text{f}}\text{)}\\ \text{ = 2}\text{.1381 + 0}\text{.013(6}\text{.5850-2}\text{.1381) = 2}\text{.196 kJ/kg-K}\end{array}$

Thus the total enthalpy and entropy values are:

Enthalpy, H = $\text{788}\text{.7 kJ/kg}$

Entropy, S = $\text{2}\text{.196 kJ/kg-K}$