Answer to Question #156508 in Mechanical Engineering for Rodel

Question

Determine the work done by a 1 Kg_m fluid system as it expand slowly in a closed system from P₁ = Mpa, V₁ = 0.08 m³ to V₂ = 0.22 m³ in accordance to the defining relations: (a) p = c, (b) pV = c, (c) pV³ = c, (d) pV (lnV) = c and (e) pV^1.4 = c. Express your answer in calories.

Answer 1

(a) p = c

p₁ = 1 MPa = 10⁶ Pa

V₁ = 0.08 m³ 

V₂ = 0.22 m³ 

W = $\smallint$ pdV = p$\smallint$ dV = p(V₂ - V₁)

W = 10⁶ x (0.22 - 0.08) = 0.14 x 10⁶ J = 140 kJ = 33461 cal

(b) pV = c

W = $\smallint$ pdV = $\smallint$ p₁V₁/V dV = p₁V₁$\smallint$ dV/V = p₁V₁lnV₂/V₁

W = 10⁶ Pa x 0.08 m³ x ln (0.22 m³/0.08 m³) = 81 kJ = 19360 cal

(c) pV³ = c

W =  $\smallint$ pdV = p₁V₁ⁿ$\smallint$ dV/Vⁿ = (p₂V₂ - p₁V₁) / (1 - n)

c = p₁V₁³ = 10⁶ Pa x (0.08 m³)³ =  512 Pa/m⁹

p₂ = c/V₂³ = 512 / (0.22)³ = 48 084 Pa = 48 kPa

W = (48 084 x 0.22 - 10⁶ x 0.08) / (1 - 3) = 34 711 J = 34.7 kJ = 8296 cal

(e)pV^1.4 = c

W =  $\smallint$ pdV = p₁V₁ⁿ$\smallint$ dV/Vⁿ = (p₂V₂ - p₁V₁) / (1 - n)

c = p₁V₁^1.4 = 10⁶ Pa x (0.08 m³)^1.4 =  29 129 Pa/m⁹

p₂ = c/V₂³ = 29 129 / (0.22)^1.4 = 242 742 Pa = 243 kPa

W = (242 742 x 0.22 - 10⁶ x 0.08) / (1 - 1.4) = 66 492 J = 66.5 kJ = 15892 cal