Answer to Question #158974 in Mechanical Engineering for loy

Question #158974

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


1
Expert's answer
2021-02-01T09:26:44-0500


Since Wcycle = 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

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


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