Answer to Question #208357 in Mechanical Engineering for Mohamed sadique

Question #208357

A certain mass of air is initially at 267°C and 7 bar occupies 0.21 m3. The air is expanded at. constant pressure such that volume becomes three times the initial volume. A polytropic process with n =1.3 is then carried out, followed by an isothermal process which completes the cycle. Considering all the processes reversible, find : (i) The heat rejected and received during each process. (ii) Net work done during the cycle. iii) Sketch the cycle on p-V and T-s diagram.


1
Expert's answer
2021-06-21T05:47:32-0400

"P_1V_1= mRT_1 \\implies m =\\frac{P_1V_1}{RT_1}=0.948 kg"

"P_2V_2= mRT_2\\implies T_1 =\\frac{P_1V_1}{mR}= 1620.87 K"

"P_3V_3= mRT_3"

"\\frac{T_2}{T_3}=(\\frac{P_2}{P_3})^{\\frac{\\gamma-1}{\\gamma}}"

"\\frac{1620.87}{540}=(\\frac{7}{P_3})^{\\frac{1.4-1}{1.4}} \\implies P_3=0.149 bar"

"P_3V_3= mRT_3 \\implies v_3 = 9.86 m^3"

Part i

Hea received in cycle

Applying the first law at P=C (1-2)

"Q=\\triangle U +W"

"W= \\int_1^2Pdv=P(v_2-v_1)=294000J =294 kJ"

"Q=mc_v(T_2-T_1)+294 = 1029.7 kJ"

Process (2-3)

"Q=\\triangle U +W"

"W= \\frac{P_2V_2-P_1V_1}{n-1}= \\frac{mR(T_2-T_1)}{n-1}=980.262 kJ"

"Q=mc_v(T_3-T_2)+W=244.55 kJ"

Total heat received in the cycle = 1029.7 +244.55=1274.25 kJ

Heat rejected in the cycle

The first law in process 3-1

"Q=\\triangle U +W"

"W=P_3V_3\\ln(\\frac{V_1}{V_3})=-5.65 kJ"

Part ii

"W_{net}=W_{1-2}+W_{2-3}+W_{3-1}=294+980.262-5.65 = 1268.12kJ"

Part iii


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