Question #181522

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.


1
Expert's answer
2021-04-21T07:05:55-0400

Given data is

mass of the mixture (m) = 1 kg

Pressure of the mixture P1=1bar;400CP_1 = 1 bar ; 40^0C

P2=5barP_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

PN2VT=mN2RN2........(1)\frac{P_{N_2}V}{T}=m_{N_2}R_{N_2}........(1)

RN2=RˉMN2R_{N_2}=\frac{\bar{R}}{M_{N_2}} , Here M= molecular weight and Rˉ\bar{R} is the universal gas constant

Similarly for the oxygen and carbon dioxide

PO2V=mO2RO2T........(2)P_{O_2}V=m_{O_2}R_{O_2}T........(2)

PCO2V=mCO2RCO2T........(3)P_{CO_2}V=m_{CO_2}R_{CO_2}T........(3)

From equations 1, 2 and 3

VT(PN2+PO2+PCO2)=mN2RˉMN2+mO2RˉMO2+mCO2RˉMCO2\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}}

VTP=[mN2MN2+mO2MO2+mCO2MCO2]Rˉ........(4)\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=PN2+PO2+PCO2P=P_{N_2}+P_{O_2}+P_{CO_2} pressure of mixture

R=[0.6MN2+0.3MO2+0.1MCO2]RˉR= [\frac{0.6}{M_{N_2}}+\frac{0.3}{M_{O_2}}+\frac{0.1}{M_{CO_2}}]\bar{R}

R=[0.625+0.332+0.144]×8314R= [\frac{0.6}{25}+\frac{0.3}{32}+\frac{0.1}{44}] \times 8314

R=274.86J/kgKR= 274.86J/kgK

Gas constant, R=274.86J/kgKR= 274.86J/kgK

Molecular weight of mixture

M =MN2+MO2+MCO2= M_{N_2}+M_{O_2}+M_{CO_2}

M=mN2nN2+mO2nO2+mCO2nCO2M= \frac{m_{N_2}}{n_{N_2}}+\frac{m_{O_2}}{n_{O_2}}+\frac{m_{CO_2}}{n_{CO_2}}

M=0.62+0.32+0.13M= \frac{0.6}{2}+ \frac{0.3}{2}+ \frac{0.1}{3}

M=0.483M= 0.483

Work done of the mixture at 40o C

10pdv    (P1P2)n1n=T1T2\int^0_1pdv \implies (\frac{P_1}{P_2})^{\frac{n-1}{n}}=\frac{T_1}{T_2}

(15)1.211.2=40+273T2(\frac{1}{5})^{\frac{1.2-1}{1.2}}=\frac{40+273}{T_2}

T2=409.298KT_2=409.298 K

W=12Pdv=P(v2v1)=mR(T2T1)W=\int^2_1Pdv=P(v_2-v_1)=mR(T_2-T_1)

W=0.483×274.80(409.298313)=12.784kJW=0.483 \times 274.80(409.298-313)=12.784kJ

From first law of thermodynamics

Q=U+WQ= \triangle U+W

U=CvdT\triangle U=C_vdT

However, Cv=Rn1=278.861.21=1.374kJ/kgKC_v=\frac{R}{n-1}=\frac{278.86}{1.2-1}=1.374 kJ/kgK

U=1.374(409.298313)=132.313kJ\triangle U=1.374(409.298-313)=132.313 kJ

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

Change in entropy , s=CplnT2T1RlnP2P1\triangle s=C_p \ln{\frac{T_2}{T_1}}-R \ln{\frac{P_2}{P_1}}

However, Cp=nRn1=1.2×278.861.21=1.649kJ/kgKC_p=\frac{nR}{n-1}=\frac{1.2 \times 278.86}{1.2-1}=1.649 kJ/kgK

s=1.649ln409.2983130.27486ln51=0.44kJ/k\triangle s=1.649 \ln{\frac{409.298}{313}}-0.27486 \ln{\frac{5}{1}}=0.44kJ/k


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