Answer to Question #181522 in Mechanical Engineering for Mohamed sadique

Question

A mixture of gases has the following composition by mass : N2 = 60 %, O2 = 30 % CO, = 10 %. 1 kg of mixture at 1 bar and 40°C is compressed to 5 bar. The index of compression may be taken as 1.2. Find the work done, heat transfer and change in entropy during the process. Also, find the molecular weight, gas constant and specific heat at constant pressure for the gas. Take Cp, for constituent gases as 1.04, 0.924 and 0.85 kJ/kgK for nitrogen, oxygen and carbon-dioxide respectively.

Answer 1

Given data is

mass of the mixture (m) = 1 kg

Pressure of the mixture $P_1 = 1 bar ; 40^0C$

$P_2 = 5 bar$

Compression index n = 1.2

Specific heat at constant pressure of nitrogen , Cp = 1.04 kJ/kgK

Oxygen, Cp = 0.924 kJ/kgK

Carbon Dioxide, Cp = 0.85 kJ/Kgk

molecular weight M = mass/no of moles

from equation of state of gases, PV = mRT

$\frac{P_{N_2}V}{T}=m_{N_2}R_{N_2}........(1)$

$R_{N_2}=\frac{\bar{R}}{M_{N_2}}$ , Here M= molecular weight and $\bar{R}$ is the universal gas constant

Similarly for the oxygen and carbon dioxide

$P_{O_2}V=m_{O_2}R_{O_2}T........(2)$

$P_{CO_2}V=m_{CO_2}R_{CO_2}T........(3)$

From equations 1, 2 and 3

$\frac{V}{T}(P_{N_2}+P_{O_2}+P_{CO_2})= \frac{m_{N_2}\bar{R}}{M_{N_2}}+\frac{m_{O_2}\bar{R}}{M_{O_2}}+\frac{m_{CO_2}\bar{R}}{M_{CO_2}}$

$\frac{V}{T}P= [\frac{m_{N_2}}{M_{N_2}}+\frac{m_{O_2}}{M_{O_2}}+\frac{m_{CO_2}}{M_{CO_2}}]\bar{R}........(4)$

Where $P=P_{N_2}+P_{O_2}+P_{CO_2}$ pressure of mixture

$R= [\frac{0.6}{M_{N_2}}+\frac{0.3}{M_{O_2}}+\frac{0.1}{M_{CO_2}}]\bar{R}$

$R= [\frac{0.6}{25}+\frac{0.3}{32}+\frac{0.1}{44}] \times 8314$

$R= 274.86J/kgK$

Gas constant, $R= 274.86J/kgK$

Molecular weight of mixture

M $= M_{N_2}+M_{O_2}+M_{CO_2}$

$M= \frac{m_{N_2}}{n_{N_2}}+\frac{m_{O_2}}{n_{O_2}}+\frac{m_{CO_2}}{n_{CO_2}}$

$M= \frac{0.6}{2}+ \frac{0.3}{2}+ \frac{0.1}{3}$

$M= 0.483$

Work done of the mixture at 40^o C

$\int^0_1pdv \implies (\frac{P_1}{P_2})^{\frac{n-1}{n}}=\frac{T_1}{T_2}$

$(\frac{1}{5})^{\frac{1.2-1}{1.2}}=\frac{40+273}{T_2}$

$T_2=409.298 K$

$W=\int^2_1Pdv=P(v_2-v_1)=mR(T_2-T_1)$

$W=0.483 \times 274.80(409.298-313)=12.784kJ$

From first law of thermodynamics

$Q= \triangle U+W$

$\triangle U=C_vdT$

However, $C_v=\frac{R}{n-1}=\frac{278.86}{1.2-1}=1.374 kJ/kgK$

$\triangle U=1.374(409.298-313)=132.313 kJ$

$Q= \triangle U+W=132.313+12.784=145.1kJ$

Change in entropy , $\triangle s=C_p \ln{\frac{T_2}{T_1}}-R \ln{\frac{P_2}{P_1}}$

However, $C_p=\frac{nR}{n-1}=\frac{1.2 \times 278.86}{1.2-1}=1.649 kJ/kgK$

$\triangle s=1.649 \ln{\frac{409.298}{313}}-0.27486 \ln{\frac{5}{1}}=0.44kJ/k$