Chapter 7, Problem 7.56P

(a)

Interpretation Introduction

Interpretation:

To find the downstream temperature and the rate of entropy generation.

Concept Introduction:

$T_1=375K$

$P_1=18bar$

$P_2=1.2bar$

The properties for ethylene are

$\omega =0.087$

$T_c=282.3K$

$P_c=50.40bar$

$A=1.424$

$B={\mathrm{14.394.10}}^{-3}$

$C=-{\mathrm{4.392.10}}^{-6}$

$D=0$

$\Delta H={(C_P^{ig})}_H(T_2-T_1)+H_2^R-H_1^R$ = 0

$\Delta H=0=MCPH(T_1,T_2,A,B,C,D).R(T_2-T_1)\mathrm{...}+R.T_c.HRB\left(\frac{T_2}{T_c},\omega P_{r2},\right)-R.T_c.HRB(T_{r1},\omega P_{r1},)$

  $\Delta S=\left(R\times MCPS(T_1,T_2,A,B,C,D).\ln \left(\frac{T_2}{T_1}\right)-R.\ln \left(\frac{P_2}{P_1}\right)\right)\mathrm{...}+R\times SRB(T_{r2},\omega P_{r2},)-R.SRB(T_{r1},\omega P_{r1},)$

$\Delta S$ = Change in Entropy

Cp = Molar heat capacity

R = Universal Gas Constant

T₁= Initial given temperature = 375 K

P₁ = Initial given Pressure = 18 bar

P₂ = Final Given Pressure = 1.2 bar

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

A = 1.424

B = 14.394 x 10-3 K-1

C = -4.392 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 = 282.3 K

P_c = Critical Pressure = 50.4 bar

  $\omega$ = 0.087

Answer to Problem 7.56P

$T_2=365.474K$

$\Delta S=22.128\frac{J}{mol.K}$

Explanation of Solution

$T_{r1}=\frac{T_1}{T_c}$

$T_{r1}=1.328$

  $T_{r1}=\frac{P_1}{P_c}$

$T_{r1}=0.357$

$$

$P_{r2}=\frac{P_2}{P_c}$

$P_{r2=}0.024$

For throttling process, $\Delta H=0.$ assuming to be adiabatic

$\Delta H={(C_P^{ig})}_H(T_2-T_1)+H_2^R-H_1^R$

Assume that $T_2=T_1$

Given

$0=MCPH(T_1,T_2,A,B,C,D).R(T_2-T_1)\mathrm{...}+R.T_c.HRB\left(\frac{T_2}{T_c},\omega P_{r2},\right)-R.T_c.HRB(T_{r1},\omega P_{r1},)$

Where

$\frac{{(C_p)}_H}{R}=A+\frac{B}{2}{T}_0(\tau +1)+\frac{C}{3}{T}_0^2\left({\tau }^2+\tau +1\right)+\frac{D}{\tau T_0^2}=MCPH(T,T_0,A,B,C,D)$

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

Assuming and substituting variables find T₂(=T in the above equation)

We get T₂= 365.45 K

$T_{r2}=\frac{T_2}{T_c}$

$T_{r2}=1.295$

Entropy is given as

$\Delta S={(C_P^{ig})}_S\ln {\frac{T_2}{T}}_1-R\ln \frac{P_2}{P_1}+S_2^R-S_1^R$

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

$\Delta S=\left(R\times MCPS(T_1,T_2,A,B,C,D).\ln \left(\frac{T_2}{T_1}\right)-R.\ln \left(\frac{P_2}{P_1}\right)\right)\mathrm{...}+R\times SRB(T_{r2},\omega P_{r2},)-R\times SRB(T_{r1},\omega P_{r1})$

Where

$\frac{S^R}{R}=-P_r\left[\frac{0.675}{T_r{}^{2.6}}+\omega \left(\frac{0.722}{T_6{}^{5.2}}\right)\right]=SRB(T_r,\omega P_r)$

$\frac{(C_p^{ig})}{R}=A+\left[B{T}_0+\left(C{T}_0^2+\frac{D}{{\tau }^2{T}_0^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)=MCPS(T,T_0,A,B,C,D)$

Substituting all the values into the above equations,

We get S = 22.13 J/(mol.K)

(b)

Interpretation Introduction

Interpretation

To find the downstream temperature, the rate of entropy generation and power output in kJ/mol.

Pros and cons of throttling valve and adiabatic expander

Concept Introduction:

$T_1=375K$

$P_1=18bar$

$P_2=1.2bar$

The properties for ethylene are

$\omega =0.087$

$T_c=282.3K$

$P_c=50.40bar$

$A=1.424$

$B={\mathrm{14.394.10}}^{-3}$

$C=-{\mathrm{4.392.10}}^{-6}$

$D=0$

$\Delta H={(C_P^{ig})}_H(T_2-T_1)+H_2^R-H_1^R$ = 0

$0.\frac{J}{mol}=MCPH(T_1,T_2,A,B,C,D).R(T_2-T_1)\mathrm{...}+R.T_c.HRB\left(\frac{T_2}{T_c},\omega P_{r2},\right)-R.T_c.HRB(T_{r1},\omega P_{r1},)$

$\Delta S=\left(R\times MCPS(T_1,T_2,A,B,C,D).\ln \left(\frac{T_2}{T_1}\right)-R.\ln \left(\frac{P_2}{P_1}\right)\right)\mathrm{...}+R\times SRB(T_{r2},\omega P_{r2}-R\times SRB(T_{r1},\omega P_{r1}))$

$\Delta S$ = Change in Entropy

Cp = Molar heat capacity

R = Universal Gas Constant

T₁= Initial given temperature = 375 K

P₁ = Initial given Pressure = 18 bar

P₂ = Final Given Pressure = 1.2 bar

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

A = 1.424

B = 14.394 x 10-3 K-1

C = -4.392 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 = 282.3 K

P_c = Critical Pressure = 50.4 bar

  $\omega$ = 0.087

$\eta$ = Given efficiency = 0.80

Answer to Problem 7.56P

$T_2=270K$

$\Delta S=7.768\frac{J}{mol.K}$

$P=-3.147\frac{kJ}{mol}$

Pros of expander − Can be used to generate power

Cons of expander − High cost and maintenance

Pros of throttling valve − Temperature of outlet gas is high and can be used for various purposes

Cons of throttling valve − Consumes power

Explanation of Solution

$T_{r1}=\frac{T_1}{T_c}$

$T_{r1}=1.328$

  $T_{r1}=\frac{P_1}{P_c}$

$T_{r1}=0.357$

$$

$P_{r2}=\frac{P_2}{P_c}$

$P_{r2=}0.024$

$\Delta S={(C_P^{ig})}_S\ln {\frac{T_2}{T}}_1-R\ln \frac{P_2}{P_1}+S_2^R-S_1^R$

With $\Delta S=0$ for isentropic expansion

Guess: $T_2=T_1$

Given

  $\Delta S=0=\left(R\times MCPS(T_1,T_2,A,B,C,D).\ln \left(\frac{T_2}{T_1}\right)-R.\ln \left(\frac{P_2}{P_1}\right)\right)\mathrm{...}+SRB\left(\frac{T_2}{T_c},\omega P_{r2},\right).R-SRB(T_{rl},\omega P_{rl},).R$

$T_2=Find(T_2)$

$T_2=219.793K$

$T_{r2}=\frac{T_2}{T_c}$

$T_{r2}=0.779$

Now calculate the isentropic enthalpy change, $\Delta H_s$

$\begin{array}{l}H{R}_2=HRB\left(T_{r2},\omega P_{r2},\right).R.T_c\\ \end{array}$

  $\Delta H_S=\left[R\times MCPH\left(T_1,T_2,A,B,C,D\right).\left(T_2-T_1\right)\right]\mathrm{...}+HRB(T_{r2},\omega P_{r2}).R.T_c-HRB(T_{rl},\omega P_{rl},).R.T_c$

$\Delta H_S=-6.423*{10}^3\frac{J}{mol}$

Calculate actual enthalpy change using the expander efficiency.

$\Delta H=\eta \Delta$

$H_S$

$\Delta H=-4.496*{10}^3\frac{J}{mol}$

Assume T₂to satisfy the above value of $\Delta H$ by trial and error method,

Given

$\eta \Delta H_S=MCPH(T_1,T_2,A,B,C,D).R(T_2-T_1)\mathrm{...}+R.T_c.HRB\left(\frac{T_2}{T_c},\omega P_{r2},\right)-R.T_c.HRB(T_{rl},\omega P_{rl},)$

We get T₂= 270 K

$\Delta S$ at calculated $T_2$ can be found using

$\Delta S=\left(R\times MCPS(T_1,T_2,A,B,C,D).\ln \left(\frac{T_2}{T_1}\right)-R.\ln \left(\frac{P_2}{P_1}\right)\right)\mathrm{...}+R\times SRB(T_{r2},\omega P_{r2},)-R\times SRB(T_{rl},\omega P_{rl},)$

Substituting the values in the above equation, we get

$\Delta S$ =7.768 J/mol.K

Power is given as, $P=\eta \Delta H$

$$ = 0.7*(-4.496) = -3.147 KJ/mol

Pros of expander − Can be used to generate power

Cons of expander − High cost and maintenance

Pros of throttling valve − Temperature of outlet gas is high and can be used for various purposes

Cons of throttling valve − Consumes power