Horsepower Above Ground

By · Wednesday, July 1st, 2009

Horsepower Above Ground
Need help with the problem that involves physical work and power?

You are asked to design a construction crane that can lift a mass of 5,000 kg (5 tonnes) from the floor to the top of a tall building of 65 meters in 30 seconds. How much power should be specified for the engine raising the load in horsepower? Note 1 HP = 746 watts = 746 joules / second. The average power is defined as work performed by a group divided by the time it took in doing this work, P_avg = W / delta (t) the instantaneous power is the rate at which they are doing the work P = dw / dt = F dx / dt = F * V, where dx, H and V are the infinitesimal displacement vector, the vector of force and the velocity vector, the DT is the infinitesimal time, and the "." Above is the point (or scalar) product of vectors.

P = Power = Work / Time P = mgh / t, where m = Mass adjournment = 5000 kg (given) g = acceleration due to gravity = 9.8 m / s (constant) h = elevation height replace values = 65 mt = 30 sec., Power = (5000) (9.8) (65) / 30 = Power 106,166.7 joules / sec. = 106,166.7 = 106,166.7 Power watts / 746 = 142.3 hp. Assuming a crane efficiency of, say, 75%, then the engine must be 142.3/0.75 = 189.75 hp. NOTE: I would like recommend a 200-engine horsepower for this operation.

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