Answer to Question #208357 in Mechanical Engineering for Mohamed sadique

Question

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.

Answer 1

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

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

$P_3V_3= mRT_3$

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

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

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

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