Answer to Question #173964 in Mechanical Engineering for Randal Rodriguez

Question

Note: Draw the P-V, T-S diagrams for all problems

13. The work required to compress a gas reversibly according to PV 1.3 =C is 68KJ. if there is no flow

determine the change in internal energy, enthalpy and heat if the gas is CO 2 .

14. Air is compressed polytropically so that the quantity PV 1.4 is constant. If 0.02 m 3 of air at atmospheric pressure of 101.3 KPA and 4C are compressed to a gage pressure of 405KPAg, determine the final temperature of air in C.

Answer 1

13. pV^1.3 = C

Q = 0

∆U = Q - W = -W

∆H = ∆U = -68 000 J

14. pV^1.4 = const

V₁ = 0.02 m³; p₁ = 101.3 kPa; T₁ = 4°C = 4 + 273 = 277 K;

p₂ = 101.3 + 405 = 506.3 kPa

p₁V₁ⁿ = p₂V₂ⁿ

p₁/p₂ = (V₂/V₁)ⁿ

101.3/506.3 = (V₂/0.02)^1.4

V₂ = 0.006 m³

Substance is ideal gas pV = RT

T = pV/R = 506.3 x 10³ Pa x 0.006 m³ / 8.31 m³ Pa K^-1 = 365 K