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Power & Efficiency









Power is by definition the rate of working.

Since work = force x distance moved, it follows that :



work theory #01






A military tank of mass 20 metric tonnes moves up a 30o hill at a uniform speed of 5 ms-1 .

If all the frictional forces opposing motion total 5000N, what is the power delivered by the engine?
(g = 10ms-2 , answer in kW)



If the tank is moving at constant speed then the forces forwards are balanced by the forces backwards.


m is the tank's mass, then mgsin30o is the component of the weight down the hill
R is the total of resistive forces down the hill
T is the tractive force forwards up the hill


mgsin30o + R = T


T = (20,000 x 10 x 0.5) + 5000 = 105,000N


power = force x speed


power of tank engine = 105,000 x 5 = 525,000W



Ans. 525 kW



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Efficiency is the ratio of useful work out divided by total work done, expressed as a percentage.



definition of efficiency






A pump running at an efficiency of 70% delivers oil at a rate of 4 kgs-1 with a speed of 3 ms-1to an oil heater .

If the vertical distance moved by the oil is 10 m, what is the power consumption of the pump?
(g = 10 ms-2, answer to 1 d.p.)



Ef = 70%,  m=4kg,  v=3 ms-1 ,  h=10 m,  g=10 ms-2


work/sec. to raise oil 8 m high = mgh = 4x10x10 = 400 J/s

work/sec. to produce discharge speed = 0.5x4x3x3 = 18 J/s

total work/sec. = 400 + 18 = 418 W


418 W represents 70% of the power supplied,


therefore total power consumption of pump =


efficiency problem #02



Ans. power consumption of pump is 597.1W (1.d.p.)




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