Chapter 7, Problem 7.32P

Interpretation Introduction

Interpretation

For the given conditions i.e. adiabatic ideal gas, the discharge stream temperature and power required of the compressor is to be found out for any one of the set of operating conditions ( in this case (a)).

Concept Introduction:

Assume ideal gas for given conditions, $\Delta S=0$

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

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

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

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

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 = 25^oC = 298.15K

P₀ = Initial given Pressure = 101.33 kPa

P₂ = Final Given Pressure = 375 kPa

n = Given Molar flow rate = 100 mol / s

$\eta$ = Given efficiency = 0.75

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

A = 3.355

B = 0.575 x 10-3 K-1

C = 0

D = -0.016 x 10⁵K²$\Delta H$ = Actual Change in Enthalpy

$\Delta \overrightarrow{H}$ = Ideal Change in Enthalpy

W = Power required of the compressor