Chapter 6, Problem 6.21P

Interpretation Introduction

Interpretation:

The enthalpy and entropy of a liquid-vapor equilibrium mixture at a temperature of $350\text{F}$ should be deduced.

Concept Introduction:

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

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

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

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

Given:

Total mass, M = 3.0 lbm

Equilibrium temperature = $350\text{F}$

Explanation:

Based on the steam tables at temperature = $350\text{F}$ we have:

Specific volume of liquid, v_f = 0.01799 ft³/lbm

Specific volume of vapor, v_g = 3.3426 ft³/lbm

Specific enthalpy of liquid, H_f = 321.73 Btu/lbm

Specific enthalpy of vapor, H_g = 1192.7 Btu/lbm

Specific entropy of liquid, S_f = 0.50322 Btu/lbm-R

Specific entropy of liquid, S_g = 5.7450 Btu/lbm-R

Calculation:

Step 1:

Calculate the steam quality x using equation (3)

It is given that in the volumes of the vapor is 50 times the volume of liquid. The total volume, V^t is:-

  ${\text{V}}^{\text{t}}{\text{ = V}}^{\text{liq}}{\text{ + V}}^{\text{vap}}$

  ${\text{V}}^{\text{t}}{\text{ = m}}_{{}_{\text{liq}}}{\text{×v}}_{\text{f}}{\text{ + m}}_{{}_{\text{vap}}}{\text{×v}}_{\text{g}}\text{ ------(4)}$

  ${\text{where, m}}_{\text{liq}}{\text{ and m}}_{\text{vap}}\text{ are mass of the liquid and vapor phases}$

  ${\text{v}}_{\text{f}}{\text{ and v}}_{\text{g}}\text{ are the specific volumes of the liquid and vapor}$

  ${\text{Since, V}}^{\text{vap}}{\text{ = 50 V}}^{\text{liq}}$

  ${\text{we can write, V}}^{\text{t}}{\text{ = 51V}}^{\text{liq}}\text{ -----(5)}$

  ${\text{by extension, V}}^{\text{t}}{\text{ = 51(m}}_{{}_{\text{liq}}}{\text{×v}}_{\text{f}}\text{) ------(6)}$

  ${\text{Substituting for V}}^{\text{t}}\text{ in equation (4) we get:}$

  ${\text{51(m}}_{{}_{\text{liq}}}{\text{×v}}_{\text{f}}{\text{) = m}}_{{}_{\text{liq}}}{\text{×v}}_{\text{f}}{\text{ + m}}_{{}_{\text{vap}}}{\text{×v}}_{\text{g}}$

  ${\text{50(m}}_{{}_{\text{liq}}}{\text{×v}}_{\text{f}}{\text{) = m}}_{{}_{\text{vap}}}{\text{×v}}_{\text{g}}$

  $\frac{{\text{m}}_{{}_{\text{vap}}}{\text{×v}}_{\text{g}}}{{\text{m}}_{{}_{\text{liq}}}{\text{×v}}_{\text{f}}}\text{ = }50$

  $\frac{{\text{m}}_{{}_{\text{vap}}}}{{\text{m}}_{{}_{\text{liq}}}}\text{ = 50}\times \frac{{\text{v}}_{\text{f}}}{{\text{v}}_{\text{g}}}$

  ${\text{Substituting the values for v}}_{\text{f}}{\text{ and v}}_{\text{g}},$

  $\frac{{\text{m}}_{{}_{\text{vap}}}}{{\text{m}}_{{}_{\text{liq}}}}\text{ = 50}\times \frac{0.01799}{3.3426}\text{ =0}\text{.269 -------(7)}$

  $\text{Now, the total mass m = 3}\text{.0 lbm , i}\text{.e}\text{.}$

  ${\text{m}}_{{}_{\text{vap}}}+{\text{m}}_{{}_{\text{liq}}}=3$

  $\text{Based on equation 3, we can then write,}$

  ${\text{3x = m}}_{{}_{\text{vap}}}{\text{ and 3(1-x) = m}}_{{}_{\text{liq}}}$

  ${\text{Substituting for m}}_{\text{liq}}{\text{ and m}}_{{}_{\text{vap}}}\text{ in eq(7) we get:}$

  $\frac{\text{x}}{\text{1-x}}\text{ = 0}\text{.269}$

  $\text{x = 0}\text{.212}$

Step 2:

Calculate H and S based on equations 1 and 2

  $\begin{array}{l}{\text{H = H}}_{\text{f}}{\text{ + x (H}}_{\text{g}}{\text{-H}}_{\text{f}}\text{)}\\ \text{ = 321}\text{.73 + 0}\text{.212(1192}\text{.7-321}\text{.73) = 506}\text{.38 Btu/lbm}\end{array}$

  $\begin{array}{l}{\text{S = S}}_{\text{f}}{\text{ + x (S}}_{\text{g}}{\text{-S}}_{\text{f}}\text{)}\\ \text{ = 0}\text{.50322 + 0}\text{.212(1}\text{.5789-0}\text{.50322) = 0}\text{.731 Btu/lbm-R}\end{array}$

Thus the total enthalpy and entropy values are:

Enthalpy, H = $\text{506}\text{.38 Btu/lbm}$

Entropy, S = $\text{0}\text{.731 Btu/lbm-R}$