Question

A closed system consisting of 10 lb of air undergoes a polytropic process from p1 = 75 lbf/in2, v1 = 4 ft3/lb to a final state where p2 = 20 lbf/in2, v2 = 12 ft3/lb. Determine the polytropic exponent, n, and the amount of energy transfer by work, in Btu, for the process.

Answer 1

Answer:

The polytropic exponent (n) is 1.2

The amount of energy transfer by work is 552.45 Btu

Explanation:

P1V1^n = P2V2^n

P1 = 75 lbf/in^2 = 75 lbf/in^2 × 143.3 in^2/ft^2 = 10,747.5 lbf/ft^2

V1 = 4 ft^3/lb × 10 lb = 40 ft^3

P2 = 20 lbf/in^2 = 20 × 143.3 = 2866 lbf/ft^2

V2 = 12 ft^3/lb × 10 lb = 120 ft^3

(V2/V1)^n = P1/P2

(120/40)^n = 10747.5/2866

3^n = 3.75

ln 3^n = ln 3.75

n ln 3 = ln 3.75

n = ln 3.75/ln 3 = 1.2

W = (P2V2 - P1V1)/1 - n = (2866×120 - 10747.5×40)/1-1.2 = (343,920 - 429,900)/-0.2 = -85,980/-0.2 = 429,900 lbf.ft = 429,900/778.17 Btu = 552.45 Btu