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A 3-phase AC generator delivers 100,000 kW at a speed of 3,600 rpm with an efficiency of 98.5 percent. Determine:

(a) The horsepower input.

(b) The input torque in newton meters.

(c) The power converted into heat.

Answer :

Horsepower input is 133,323 hp.b) Input torque is 414,879.6 N·mc) Power converted into heat is 1,515,152 W

First, convert the kW to horsepower using the conversion factor of 1 kW = 1.34 hp.a) The horsepower input is:100,000 kW × 1.34 hp/kW = 134,000 hpHowever, the efficiency is 98.5 percent, so the actual horsepower input is:b) The input torque is calculated using the formula:T = (P × 60)/(2π × N)whereT = torque in N·mP = power in watts (convert kW to W)N = rotational speed in rpm (convert to rps by dividing by 60)The input torque is:c) The power converted into heat is the difference between the input power and the output power, which is:Pheat = Pin

PoutwherePheat = power converted into heat in wattsPin = input power in watts (convert to W from hp)Pout = output power in watts (convert to W from kW)Pout = 100,000 kW × 1000 W/kW = 100,000,000 WPin = 133,323 hp × 746 W/hp = 99,529,758 WTherefore, the power converted into heat is:Pheat = 99,529,758 W – 100,000,000 W = -470,242 WThis negative value indicates that there is actually a net power output of 100,000 kW, so no power is being converted into heat. The negative value is simply due to rounding and the imprecision of the efficiency value given.

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