Answer to Question #158974 in Mechanical Engineering for loy

Question

Problem 1 : 5 kg of air experience the three processes cycle as shown in figure. Calculate the network take R= 0.287 KJ/K.g.k

Answer 1

Since W_cycle = W₁₂ + W₂₃ + W₃₁, we will work our way around the cycle and calculate each work term along the way.

Step 1-2 is isobaric, therefore, the definition of boundary work becomes:

W₁₂ = $\smallint$ pdV = p₁ (V₂ - V₁)

P₂ = P₁ and pV = nRT

W₁₂ = p₂V₂ - p₁V₁ = nR(T₂ - T₁)

W₁₂ = 5 kg x 0.287 kJ/(kg K) x (873 K - 373 K) = 717.5 kJ

Because the volume is constant in step 2-3: W₂₃ = 0 kJ

Step 3-1 is isothermal, therefore, the definition of boundary work becomes:

W₃₁ = $\smallint$ pdV = nRT₁ln(V₁/V₃) = nRT₁ln(p₃/p₁)

V₃ = V₂ = nRT₂ / p₂ = nRT₂ / p₁ = 5 kg x 0.287 kJ/(kg K) x 873 K / 200 kPa = 6.263 m³

V₁ = nRT₁ / p₁ = 5 kg x 0.287 kJ/(kg K) x 373 K / 200 kPa = 2.676 m³

W₃₁ = 5 kg x 0.287 kJ/(kg K) x 373 K x ln(2.676 m³/6.263 m³) = -455.15 kJ

W_cycle = 717.5 kJ + 0 kJ - 455.15 kJ = 262.35 kJ