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A four-cylinder four-stroke SI engine develops an output of 44 kW. If the pumping work is 5% of the indicated work and mechanical loss is an additional 7%, then the power consumed in pumping work is : (a) 50 kW (b) 25 kW (c) 5.0 kW (d) 2.5 kW |
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Answer» (d) 2.5 kW bp = ip – 0.05ip – 0.07ip = ip – 0.12ip = 0.88ip \(\therefore\) ip = \(\cfrac{bp}{0.88}\) = \(\cfrac{44}{0.88}\) = 50 kW \(\therefore\) Pumping work = 50 × 0.05 = 2.5 KW |
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