Chapter 6, Problem 6.24P

Interpretation Introduction

Interpretation:

The quality of steam in its initial state should be deduced based on the given initial and final conditions of steam.

Concept Introduction:

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

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

• For a process that takes place at constant enthalpy, the change in enthalpy is zero. In other words, the enthalpy in the final state (H₂) is equal to that in the initial state (H₁). The change in enthalpy is given as:

$\begin{array}{l}{\text{ΔH = H}}_{\text{2}}{\text{-H}}_{\text{1}}\text{ -----(4)}\\ \text{When, }\Delta \text{H = 0 }\\ {\text{H}}_{\text{2}}{\text{ = H}}_{\text{1}}\end{array}$

$$

Quality of steam in the initial state = 0.953

Given:

Initial pressure of wet steam, P₁ = 1100 kPa

Final pressure P₂ = 101.33 kPa

Final Temperature of steam = $105\text{C}$

Explanation:

Since this is a constant enthalpy process, H₂ = H₁

The final state enthalpy (H₂) can be deduced by interpolation based on the steam table data for saturated steam and superheated steam at 101.33 kPa. Finally, the initial state steam quality (x) can be deduced based on equation (2) and known values of the initial state H_g and H_ffrom the steam tables.

Calculation:

Step 1:

Calculate the final state enthalpy

Based on the steam tables at the final state pressure = 101.33 kPa we have:

For saturated steam:

Saturation temperature, T = $99.97\text{C}$

Specific enthalpy of vapor, H_g = 2675.6 kJ/kg

For superheated steam:

Saturation temperature, T = $150\text{C}$

Specific enthalpy of vapor, H_g = 2776.6 kJ/kg

Thus the enthalpy at final state T = $105\text{C}$ can be calculated by interpolation:

${\text{H}}_{\text{2}}\text{ = 2675}\text{.6 + }\frac{\text{105-99}\text{.97}}{\text{150-99}\text{.97}}\text{(2776}\text{.6-2675}\text{.6)}=\text{ 2685}\text{.8 kJ/kg}$

Step 2:

Calculate the initial state steam quality

Since, H₂ = H₁ We have, H₁ = 2685.8 kJ/kg

Based on the steam tables at the initial state P = 1100 kPa we have:

Specific enthalpy of vapor, H_g = 2780.7 kJ/kg

Specific enthalpy of liquid, H_f = 781.0 kJ/kg

Based on equation (2) we have for the initial state enthalpy:

$\begin{array}{l}{\text{H}}_1{\text{ = H}}_{\text{f}}{\text{ + x(H}}_{\text{g}}{\text{-H}}_{\text{f}}\text{)}\\ \text{x = }\frac{{\text{H}}_1{\text{-H}}_{\text{f}}}{{\text{H}}_{\text{g}}{\text{-H}}_{\text{f}}}=\frac{2685.8-781.0}{2780.7-781.0}=0.953\end{array}$

The quality of steam in the initial state is, x = 0.953