A Pump supplies oil at $0.002 m^{3} / s e c .$ to a 50 mm diameter double acting cylinder and a rod diameter is 20 mm. If the load is 6000 N bolts in extending and retracting stroke.

Power during extension stroke and return stroke are

6.12 K W, 7.28 K W

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6.12 K W, 12.4 K W

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5.4 K W, 10.2 K W

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10.2 K W, 12.4 K W

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Solution

The correct option is A $6.12 K W, 7.28 K W$
$Cylinder power stroke during extension stroke P_{e x t} = P_{i n} \times Q_{i n} \frac{30.6 \times 10^{6} \times 0.002}{1000} = 6.12 K W Cylinder power stroke during return stroke P_{2} = P_{i n 2} \times Q_{i n 2} = \frac{36.4 \times 10^{6} \times 0.002}{1000} = 7.28 K W .$

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