Chapter 6, Problem 6.25P

Interpretation Introduction

Interpretation:

The temperature of steam in its final state along with the change in entropy should be deduced based on steam table data as well as based on the ideal gas assumption.

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:

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

$$

Based on steam tables:

The final temperature of steam, T = $225\text{C}$

The entropy change, ?S = $\text{1}\text{.2685 kJ/kg-K}$

Based on the ideal gas assumption:

The final temperature of steam, T = $260\text{C}$

The entropy change, ?S = - $1.3032\text{ kJ/kg-K}$

Given:

Initial pressure of steam, P₁ = 2100 kPa

Initial Temperature of steam = $260\text{C}$

Final pressure P₂ = 125 kPa

Explanation:

Since this is a constant enthalpy process, H₁ = H₂

The initial state enthalpy (H₁) and entropy (S₁) can be deduced by interpolation based on the steam table data for superheated steam at 2100 kPa.

The final state temperature and entropy (S₂) can be deduced from steam tables from the calculated initial enthalpy data.

Calculations:

Step 1:

Calculate the initial state enthalpy (H₁) and entropy (S₁) at T = $260\text{C}$

Based on the steam tables at the initial state pressure = 2100 kPa we have:

For superheated steam:

At Saturation temperature, T = $250\text{C}$

Specific enthalpy of vapor, H_g = 2897.9 kJ/kg

Specific entropy of vapor, S_g = 6.5162 kJ/kg-K

At Saturation temperature, T = $275\text{C}$

Specific enthalpy of vapor, H_g = 2961.9 kJ/kg

Specific entropy of vapor, S_g = 6.6356 kJ/kg-K

Thus the enthalpy and entropy at initial state T = $260\text{C}$ can be calculated by interpolation:

${\text{H}}_{\text{1}}\text{ = 2897}\text{.9 + }\frac{\text{260-250}}{\text{275-250}}\text{(2961}\text{.9-2897}\text{.9)}=\text{ 2923}\text{.5 kJ/kg}$

${\text{S}}_{\text{1}}\text{ = 6}\text{.5162 + }\frac{\text{260-250}}{\text{275-250}}\text{(6}\text{.6356-6}\text{.5162)}=\text{ 6}\text{.5639 kJ/kg}$

Step 2:

Calculate the final temperature and ?S

Since, H₂ = H₁

We have, H₂ = 2923.5 kJ/kg

Based on the steam tables the above specific enthalpy corresponds to superheated steam at a pressure P = 125 kPa and T = $225\text{C}$

Thus, the final state temperature, T₂ = $225\text{C}$

Specific entropy of vapor at this final temperature, S₂ = 7.8324 kJ/kg-K

${\text{ΔS = S}}_{\text{2}}{\text{- S}}_{\text{1}}\text{ = 7}\text{.8324 - 6}\text{.5639 = 1}\text{.2685 kJ/kg-K}$

Step 3:

Calculate the final temperature and ?S based on the ideal gas assumption

Enthalpy is a state function and dependent on temperature. Since the process takes place at constant enthalpy, there is will be no change in temperature.

T₂ = T₁ = $260\text{C}$

The entropy change for an ideal gas is:

$\begin{array}{l}{\text{ΔS = C}}_{\text{v}}\text{ln}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{ + Rln}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\\ {\text{Since T}}_{\text{2}}{\text{ = T}}_{\text{1}}\text{; }\\ \text{ΔS = Rln}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}=\frac{0.008314\text{ kJ}{\text{.K}}^{\text{-1}}{\text{.mol}}^{\text{-1}}}{0.018\text{ kg}{\text{.mol}}^{\text{-1}}}\times \ln \frac{125}{2100}=-1.3032\text{ kJ/kg-K}\end{array}$

Thus,

Based on steam tables:

The final temperature of steam, T = $225\text{C}$

The entropy change, ?S = $\text{1}\text{.2685 kJ/kg-K}$

Based on the ideal gas assumption:

The final temperature of steam, T = $260\text{C}$

The entropy change, ?S = - $1.3032\text{ kJ/kg-K}$