W 2 3 pdv pv p 0m 3 0 kj w31 is found using

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Unformatted text preview: − 2 − W1− 2 Q1− 2 = W1− 2 = −85 kJ For W2 ­3 it is noticed in the diagram that volume is constant during this step. W 2− 3 = € ∫ pdV = pΔV = p * 0m 3 = 0 kJ W3 ­1 is found using the same equation above, where volume here is changing at constant pressure. € W 3 −1 = ∫ pdV = pΔV = 150kPa[0.5 − 0.1]m 3 = 60 kJ (b) Is this a refrigeration or power cycle? € Wnet for the cycle will give us this answer since Wnet > 0 for a power cycle and...
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