Asked by sam
Calculate the net work output of a heat engine following path ABCDA in Figure 14.30, where V1 = 5.0 10-3 m3 and V2 = 20.0 10-3 m3.
Answers
Answered by
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The net work done is the area inside the closed loop in the PV diagram.
W = (1.4 * 3)*(106 N/m2)(10-3 m3) = 4.2*103 J, using area = average height * width.
or
area inside the loop = area of center rectangle + area of to triangle - area of bottom triangle
= (1.4*15 + 0.6*15/2 - 0.4*15/2)*(106 N/m2)(10-3 m3) = 22.5*103 J
W = (1.4 * 3)*(106 N/m2)(10-3 m3) = 4.2*103 J, using area = average height * width.
or
area inside the loop = area of center rectangle + area of to triangle - area of bottom triangle
= (1.4*15 + 0.6*15/2 - 0.4*15/2)*(106 N/m2)(10-3 m3) = 22.5*103 J
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