Chapter 6, Problem 6.22P

Interpretation Introduction

Interpretation:

The steam quality, enthalpy and entropy of wet steam at a temperature of $230\text{C}$ 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:

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

Steam quality, x = 0.867

Enthalpy, H = $\text{2572}\text{.9 kJ/kg}$

Entropy, S = $\text{5}\text{.636 kJ/kg-K}$

Given:

Density of wet steam, D = 0.025 g/cm³

Temperature = $230\text{C}$

Explanation:

Based on the steam tables at Temperature = $230\text{C}$ we have:

Specific volume of liquid, v_f = 0.001252 m³/kg

Specific volume of vapor, v_g = 0.045941 m³/kg

Specific enthalpy of liquid, H_f = 1085.7 kJ/kg

Specific enthalpy of vapor, H_g = 2801.0 kJ/kg

Specific entropy of liquid, S_f = 2.7933 kJ/kg-K

Specific entropy of liquid, S_g = 6.0721 kJ/kg-K

Calculation:

Step 1:

Calculate the steam quality x

It is given that:

Density of steam, D = 0.025 g/cm³ = 25 kg/m³

$\begin{array}{l}\text{The specific volume V is given as:}\\ \text{V = }\frac{\text{1}}{\text{D}}\text{ = }\frac{\text{1}}{\text{25}}\text{ = 0}{\text{.04 m}}^{\text{3}}\text{/kg}\\ \text{Based on equation (1), the steam quality x is:-}\\ \text{x = }\frac{{\text{V - V}}_{\text{f}}}{{\text{V}}_{\text{g}}{\text{ - V}}_{\text{f}}}=\frac{\text{0}\text{.04 - 0}\text{.001252}}{\text{0}\text{.045941 - 0}\text{.001252}}=0.867\end{array}$

Step 2:

Calculate H and S based on equations 2 and 3

$\begin{array}{l}{\text{H = H}}_{\text{f}}{\text{ + x (H}}_{\text{g}}{\text{-H}}_{\text{f}}\text{)}\\ \text{ = 1085}\text{.7 + 0}\text{.867(2801}\text{.0-1085}\text{.7) = 2572}\text{.9 kJ/kg}\end{array}$

$\begin{array}{l}{\text{S = S}}_{\text{f}}{\text{ + x (S}}_{\text{g}}{\text{-S}}_{\text{f}}\text{)}\\ \text{ = 2}\text{.7933 + 0}\text{.867(6}\text{.0721-2}\text{.7933) = 5}\text{.636 kJ/kg-K}\end{array}$

Thus the quality, total enthalpy and entropy values are:

x = 0.867

Enthalpy, H = $\text{2572}\text{.9 kJ/kg}$

Entropy, S = $\text{5}\text{.636 kJ/kg-K}$