Chapter 7, Problem 7.36P

Interpretation Introduction

Interpretation

To Find S_G, W_lost and ${\eta }_t$ ( Thermodynamic efficiency)

Concept Introduction:

  $\frac{\Delta S}{R}={\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}\frac{dT}{T}-\ln (\frac{P}{P_0})$

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}\frac{dT}{T}=A.\ln \tau +[B{T}_0+(C{T}_0^2+\frac{D}{{\tau }^2{T}_0^2})(\frac{\tau +1}{\tau })](\tau -1)$

$\Delta H={\displaystyle {\int }_{T_1}^{T_2}{C}_P}dT$

${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}dT=A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$

Consider operating conditions (a)

$W=(m.\Delta H)$

$\tau =\frac{T}{T_0}$

$\Delta H=\frac{\overrightarrow{\Delta H}}{\eta }$

Where,

$\Delta S$ = Change in Entropy

Cp = Molar heat capacity

R = Universal Gas Constant

T₀= Initial given temperature = 293.15 K

T = Final given temperature = 428.65 K

P₀ = Initial given Pressure = 140 kPa

P = Final Given Pressure = 560 kPa

$\eta$ = Given efficiency

A, B, C, D = Constants for heat capacity of air

A = 1.702

B = 9.081 x 10-3

C = -2.164 x 10-6

D = 0

$\Delta H$ = Actual Change in Enthalpy

W = Work