Question

An electric motor is required to raise 400kg of water through a vertical height of 120m in 2min
What will be electric power required) If the overall efficiency is 80%
a) 3.2 KW
b) 5 KW
c) 32 KW
d) 300 KW

Answer 1

Answer:

5kw

Explanation:

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Answer 2

Answer:

Option A.

Explanation:

Given:

M = 400 kg

g. = 10/s^2

H. = 120 m

T. = 2 min = 2 x 60 sec = 120 sec

Solution:

Energy = M x g x H

= 400 x 10 x 120

= 480000 J

Power = Energy/sec

= 480000 / 120 = 4000 J/s

Output at 80% efficiency = 4000 x 0.8

= 3200 J/s = 3.2KW