Chapter 7, Problem 7.35P

Interpretation Introduction

Interpretation:

To find final temperature, T

To find work required, W

Concept Introduction:

Methane:

T_c= 190.6 K

P_C= 45.99 bar

$\omega =0.012$

$T_0=308.15$ K

$P_0=3500$ kPa

$P=5500$ kPa

For isenthalpic process, $\Delta S=0$

For the heat/capacity of Methane:

$A=1.702$

$B=9.081\times {10}^{-3}{K}^{-1}$

$C=-2.164\times {10}^{-6}.K^{-2}$

Use generalized second-virial correlation:

The entropy change is given by

$\Delta S={\displaystyle {\int }_{T_1}^{T_2}{C}_P^{ig}}\frac{dT}{T}-R.\ln (\frac{P_2}{P_1})+S_2^R-S_1^R$ Combined with

  ${\displaystyle {\int }_{T_0}^T\frac{C_P}{R}}\frac{dT}{T}=A.\ln \tau +[B{T}_0+(C{T}_0^2+\frac{D}{{\tau }^2{T}_0^2})(\frac{\tau +1}{\tau })](\tau -1)$ ; D = 0;

We know that

  $\Delta S=R.\left[A.\ln (\tau )+\left[B.T_{{}_0}+C.T_0^2.\left(\frac{\tau +1}{2}\right)\right].(\tau -1)-\ln \frac{P}{P_0}\mathrm{...}\right]+SRB\left(\frac{\tau -T_0}{T_C},\omega P_r\right)-SRB(T_{r0},\omega P_{r0})$

$\begin{array}{l}\Delta H=R.\left[A.T_0.(\tau -1)+\frac{B}{2}.T_0{}^2.({\tau }^2-1)+\frac{C}{3}{T}_0{}^3.({\tau }^3-1)\mathrm{...}+T_c.\left(HRB\left(\frac{\tau .T_0}{T_c},\omega P_{r,}\right)-HRB(T_{r0},\omega P_{r0},)\right)\right]\\ \end{array}$

$\Delta H=R.\left[A.T_0(\tau -1)+\frac{B}{2}{T}_0^2({\tau }^2-1)+\frac{C}{3}{T}_0^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\right]=R\times ICPH(T_0,T,A,B,C,D)$

$\frac{S^R}{R}=-P_r\left[\frac{0.675}{T_r{}^{2.6}}+\omega \left(\frac{0.722}{T_r{}^{5.2}}\right)\right]=SRB\left(T_{r,}\omega P_r\right)$

$\frac{H^R}{R{T}_C}=P_r\left[0.083-\frac{1.097}{T_r{}^{1.6}}+\omega \left(0.139-\frac{0.894}{T_r{}^{4.2}}\right)\right]=HRB(T_r\omega P_r)$

Where,

$\Delta S$ = Change in Entropy

Cp = Molar heat capacity

R = Universal Gas Constant

T₀= Initial given temperature = 308.15 K

P₀ = Initial given Pressure = 3500 kPa

P₂ = Final Given Pressure = 5500 kPa

$\eta$ = Given efficiency = 0.78

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

A = 1.702

B = 9.081 x 10-3 K-1

C = -2.164 x 10-6 K-2

D = 0

$\Delta H$ = Actual Change in Enthalpy

$\Delta H_{ig}$ = Ideal Gas Change in Enthalpy

W = Power required of the compressor

T_c= Critical Temperature = 190.6 K

P_c = Critical Pressure = 45.99 bar

  $\omega$ = 0.012

m = 1500 mol/sec