Chapter 6, Problem 6.18P

Interpretation Introduction

Interpretation:

The enthalpy and entropy changes accompanying the change in the state of steam should be calculated based on the steam table data and by considering steam to behave as an ideal gas.

Concept Introduction:

• Enthalpy (H) and entropy (S) are state functions in that they can be defined by two parameters for example, temperature and pressure. When a system changes from state 1 to state 2, the change in enthalpy △H and the change in entropy △S can be given by the difference in the values in the two states, i.e.

$\begin{array}{l}{\text{ΔH = H}}_{\text{2}}{\text{- H}}_{\text{1}}\text{---(1)}\\ {\text{ΔS = S}}_{\text{2}}{\text{- S}}_{\text{1}}\text{----(2)}\end{array}$

• For an ideal gas the corresponding △H and △S can be given as:

$\begin{array}{l}{\text{ΔH = nC}}_{\text{v}}{\text{(T}}_{\text{2}}{\text{-T}}_{\text{1}}\text{) -----(3)}\\ {\text{ΔS = nC}}_{\text{v}}\text{ln}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{ --------(4)}\\ \text{Where, n = number of moles of the gas}\\ {\text{C}}_{\text{v}}\text{ = specific heat at constant volume}\\ {\text{T}}_{\text{1}}{\text{ and T}}_{\text{2}}\text{ = initial and final temperatures}\end{array}$

Based on the steam table data:

△H = $\text{377}\text{.1 Btu/lbm}$

△S = $\text{0}\text{.266 Btu/lbm}\text{.F}$

Based on the ideal gas assumption:

△H = 270.2 Btu/lbm

△S = 0.517 Btu/lbm-F

Given:

Mass of steam, m = 1.0 lbm

Pressure of saturated vapor = 20 psi

Pressure of superheated vapor = 50 psi

Temperature of superheated vapor = $1000\text{F}$

Explanation:

The initial state of steam is saturated vapor and the final state is superheated vapor. Thus the enthalpy and entropy changes can be given based on equations 1 and 2 as:

$\begin{array}{l}{\text{ΔH = H}}_{\text{superheated vapor}}{\text{- H}}_{\text{saturated vapor}}\\ {\text{ΔS = S}}_{\text{superheated vapor}}{\text{- S}}_{\text{saturated vapor}}\end{array}$

These values can be obtained from steam tables

Calculation:

Based on the steam tables, for saturated vapor at P = 20 psi

T_sat = $228\text{F}$

$\begin{array}{l}{\text{Enthalpy of saturated vapor, H}}_{\text{1}}\text{ = 1156}\text{.3 Btu/lbm}\\ {\text{Entropy of saturated vapor, S}}_{\text{1}}\text{ = 1}\text{.732 Btu/lbm}\text{.F}\end{array}$

Again from the steam tables, for the superheated vapor at T = $1000\text{F}$ and P = 50 psi

$\begin{array}{l}{\text{Enthalpy of superheated vapor, H}}_{\text{2}}\text{ = 1533}\text{.4 Btu/lbm}\\ {\text{Entropy of superheated vapor, S}}_{\text{2}}\text{ = 1}\text{.998 Btu/lbm}\text{.F}\end{array}$

Therefore,

$\begin{array}{l}\text{ΔH = 1533}\text{.4 - 1156}\text{.3 = 377}\text{.1 Btu/lbm}\\ \text{ΔS = 1}\text{.998 - 1}\text{.732 = 0}\text{.266 Btu/lbm}\text{.F}\end{array}$

Based on the ideal gas assumptions using equations (3) and (4) we get:

$\begin{array}{l}{\text{ΔH = C}}_{\text{v}}{\text{(T}}_{\text{2}}{\text{-T}}_{\text{1}}\text{) }\\ \text{ = 0}\text{.35Btu/lbm-F}\times \text{(1000-228) F }\\ \text{ = 270}\text{.2 Btu/lbm}\\ {\text{ΔS = C}}_{\text{v}}\text{ln}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\\ \text{ = 0}\text{.35 Btu/lbm-F}\times \text{ ln(}\frac{1000}{228}\text{) }\\ \text{ = 0}\text{.517 Btu/lbm-F}\end{array}$

The values of the change in enthalpy and entropy are as follows:

Based on the steam table data:

△H = $\text{377}\text{.1 Btu/lbm}$

△S = $\text{0}\text{.266 Btu/lbm}\text{.F}$

Based on the ideal gas assumption:

△H = 270.2 Btu/lbm

△S = 0.517 Btu/lbm-F