Question
A pump lifts up 200 kg of water per minute from a depth of 3 m. The pump's efficiency is 40%. (g = 10)
a) What is the pump's useful power output?
b) What is the pump's power consumption from the electrical grid?
c) Assuming the same efficiency, how much water per minute can the same pump lift up from a depth of 5 m?
Answer
The water of mass m = 200 kg is being raised by the electrical pump of efficiency `\eta` = 40 % = 40 / 100 = 0.4 in a time of t = 1 min = 60 s from a depth of h = 3 m.
(c) If the pump can lift the water of mass M from a depth of H = 5 m in time t = 1 min = 60 s, then :
`\frac{MgH/t}{P_(text{in}` = `\eta`.
Hence, M = `\frac{\etatP_(text{in})}{gH}`= ( 0.4 x 60 s x 250 W ) / ( 10 m/`s^2` x 5 m ) = 120 kg.
A pump lifts up 200 kg of water per minute from a depth of 3 m. The pump's efficiency is 40%. (g = 10)
a) What is the pump's useful power output?
b) What is the pump's power consumption from the electrical grid?
c) Assuming the same efficiency, how much water per minute can the same pump lift up from a depth of 5 m?
Answer
The water of mass m = 200 kg is being raised by the electrical pump of efficiency `\eta` = 40 % = 40 / 100 = 0.4 in a time of t = 1 min = 60 s from a depth of h = 3 m.
(a) Hence, its power output = `P_(out)`
= `\frac{text{The work done by it against gravity}}{time}`
=`\frac{mgh}{t}`
= 200 kg x 10 m/`s^2` x 3 m / 60 s
= 100 W.
(b) The pump's power consumption from the electrical grid is :
`P_(text{in})` = `\frac{P_(out)}{\eta}` = `\frac{100 W}{0.4}` = 250 W.
= `\frac{text{The work done by it against gravity}}{time}`
=`\frac{mgh}{t}`
= 200 kg x 10 m/`s^2` x 3 m / 60 s
= 100 W.
(b) The pump's power consumption from the electrical grid is :
`P_(text{in})` = `\frac{P_(out)}{\eta}` = `\frac{100 W}{0.4}` = 250 W.
(c) If the pump can lift the water of mass M from a depth of H = 5 m in time t = 1 min = 60 s, then :
`\frac{MgH/t}{P_(text{in}` = `\eta`.
Hence, M = `\frac{\etatP_(text{in})}{gH}`= ( 0.4 x 60 s x 250 W ) / ( 10 m/`s^2` x 5 m ) = 120 kg.
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