Answer to Question #177932 in Mechanical Engineering for MOHAMED SADIQUE

Question #177932

2 m3 of air at 1.1 bar and 20°C is heated in a cylinder at constant pressure until its volume is 4.5 m. It is then compressed according to the law pv#= constant till pressure and volume change to 5.5 bar and 0.6 m calculate change in entropy, work and heat transfer during the processes and the value of n.


1
Expert's answer
2021-04-14T07:45:54-0400

Given that air at initial condition

"P_1=1.1 bar=110 kPa; T_1=20^0 C=20+273 =293 K"

"V_1=2 m^3"

Assume air is ideal gas "R=0.287 kJ\/kg.K"

Now it is heated at constant pressure

"V_2=4.5 m^3 ; (P_1=P_2)=110kPa"

For 1-2

"\\frac{V_1}{T_1}=\\frac{V_2}{T_2} \\implies T_2=\\frac{4.5 \\times 293}{2}=659.25 K"

Now it is compressed to "PV^n=constant" "P_3=5.5 bar =550kPa ; V_3=0.6 m^3"

We have "P_2V_2^n=P_3V_3^n \\implies \\frac{550}{110}=(\\frac{4.5}{0.6})^{n} \\implies n=0.798 =0.8"

"\\frac{T_3}{T_2}=(\\frac{V_2}{V_3})^{n-1} \\implies T_3=659.25 \\times (\\frac{4.5}{0.6})^{-0.2}=440.6 K"

Now consider process 1-2

Q-W = change in internal energy ............(1)

"W_{1-2}=P(V_2-V_1)=110(4.5-2)=275 kJ"

Change in internal energy , "mc_v(T_2-T_1)"

"Q-275 =2.616 \\times 0.718 (659.25-293) \\implies Q_2=962.922 kJ"

Change in entropy, "\\triangle S _{1-2}=m[C_v \\ln \\frac{T_2}{T_1}+R\\ln \\frac{V_2}{V_1}]=2.616[0.718 \\ln \\frac{659.25}{293}+0.287\\ln \\frac{4.5}{2}]=2.132 kJ\/K"

"W_{2-3}= \\frac{P_2V_2-P_3V_3}{n-1}=\\frac{(110 \\times 4.5)-(550 \\times 0.6)}{0.8-1}=-825 kJ"

"Q_{2-3}-W_{2-3}=mc_v(T_3-T_2)"

"Q_{2-3}-(-825)=2.616 \\times0.718(440.6-659.25) \\implies Q_{2-3}=-1235.7kJ"

"\\triangle S _{2-3}=m[C_v \\ln \\frac{T_3}{T_2}+R\\ln \\frac{V_3}{V_2}]=2.616[0.718 \\ln \\frac{440.6}{659.25}+0.287\\ln \\frac{0.6}{4.5}]=-0.8676 kJ\/K"


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