Chapter 6, Problem 6.27P

Interpretation Introduction

Interpretation:

The final enthalpy of superheated steam as it expands under isentropic conditions 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:

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

• 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:

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

$$

Final enthalpy of steam = 2597.5 kJ/kg

Given:

Initial pressure of steam, P₁ = 500 kPa

Initial Temperature of steam = $300\text{C}$

Final pressure P₂ = 50 kPa

Explanation:

Since this is a constant entropy process, S₁ = S₂

The initial state enthalpy (H₁) and entropy (S₁) can be deduced based on the steam table data for superheated steam at 500 kPa and $300\text{C}$

The final state temperature and enthalpy (H₂) can be deduced from steam tables from the calculated initial entropy data.

Calculations:

Step 1:

Calculate the initial state enthalpy (H₁) and entropy (S₁) at T = $300\text{C}$

Based on the steam tables at the initial state pressure = 500 kPa we have:

For superheated steam:

At Saturation temperature, T = $300\text{C}$

Specific enthalpy of vapor, H_g = H₁ = 2925.7 kJ/kg

Specific entropy of vapor, S_g = S₁= 7.4598 kJ/kg-K

Step 2:

Calculate the final enthalpy (H₂)

Since, S₂ = S₁ We have, S₂ = 7.4598 kJ/kg-K

This is wet steam. Hence, for a final pressure P₂ = 50 kPa for saturated steam

Specific entropy of vapor, S_g = 7.5939 kJ/kg-K

Specific entropy of liquid, S_f = 1.0910 kJ/kg-K

The steam quality x can be deduced from equation (3) as:

$\begin{array}{l}{\text{S}}_2{\text{ = S}}_{\text{f}}{\text{ + x(S}}_{\text{g}}{\text{-S}}_{\text{f}}\text{)}\\ \text{x = }\frac{{\text{S}}_2{\text{-S}}_{\text{f}}}{{\text{S}}_{\text{g}}{\text{-S}}_{\text{f}}}=\frac{7.4598-1.0910}{7.5939-1.0910}=0.979\end{array}$

At 50 kPa,

Specific enthalpy of vapor, H_g = 2645.9 kJ/kg

Specific enthalpy of liquid, H_f = 340.47 kJ/kg

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

$\begin{array}{l}{\text{H}}_2{\text{ = H}}_{\text{f}}{\text{ + x(H}}_{\text{g}}{\text{-H}}_{\text{f}}\text{)}\\ {\text{H}}_2\text{ = 340}\text{.47 + 0}\text{.979(2645}\text{.9-340}\text{.47)=2597}\text{.5 kJ/kg}\end{array}$

Thus, final enthalpy = 2597.5 kJ/kg