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RideAnS [48]
4 years ago
7

Gas contained in a piston-cylinder assembly undergoes a polytropic process for which the relation between pressure and volume is

given by p∀ n = constant. The initial volume is 0.04 m3 and the initial pressure is 2 bar. The final volume of the gas is 0.1 m3 , For each of the following cases, determine the final pressure [bar] and the work for the process [kJ]. Show integration for each case. a. n = 0, b. n = 1, c. n = 1.4.
Engineering
1 answer:
Charra [1.4K]4 years ago
6 0

Answer:

a) P = 2 bar, W =  12 kJ

B)P = 0.8 bar, W =  7.33 kJ

C) P = 0.55 bar, W =  6.14 kJ

Explanation:

pV^n = constant.

The initial volume V_1 = 0.1 m3,

The final volume V_2 = 0.04 m3,

The initial pressure P_1 = 2 bar.

We know that

P_1 V_1^n = P_2 V_2^n,

P_2 = P_1\frac{(V_1}{V_2)^n}

      = 2\frac{(0.04}{0.1)^n}

`a) n = 0, P_2 = 2\frac{(0.04}{0.1)^0}  = 2 bar

W = P_2(V_2 - V_1) = 2*100 kPa * (0.06 m3) = 12 kJ

b) n = 1, P_2 = 2\frac{(0.04}{0.1)^1}  = 0.8 bar

W = P_2 V_2 ln\frac{(V2}{V1} = 7.33 kJ

c)  n = 1.4, P_2 = 2\frac{(0.04}{0.1)^1.3}  = 0.5542 bar

W = \frac{(P_2 V_2 - P_1 V_1)}{(1 - n)} = 6.14 kJ

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