If 100 m³⁻³ of water is pumped from a reservoir into a tank, 10 m higher than the reservoir, in 20 minutes. If density of water is 1000 kg m⁻³, find

(a) The increase in P.E
(b) The power delivered by pump.

(Ans: (a) 9.8 x 10⁶ J, (b) 8.2 kW)



Given:

Volume of Water = v = 100 m³

Height of tank = h = 10 m

Time used =  t = 20 minutes = 1200 s = 1.2 ｘ10³ s

Density of water = ρ = 1000 kg m⁻³

Value of g = 9.8 m s⁻²


To Find:

(a) Increase in P.E. =  P.E. = ?   
(b) Power of pump =  P = ?   

Solution:

(a) Increase in P.E. =  P.E. = ?   

 We know that

P.E. = m g h
where mass m in term of density ρ and volume V is 

m = ρ V 

P.E. = ρ V g h ----------(1)


by putting the corresponding values in equation (1)

P.E. =  1000 kg m⁻³ ｘ 100 m³ｘ 9.8 m s⁻²ｘ 10 m 

P.E. =  9800000 J

P.E. = 9.8 ｘ10⁶ J -------------Ans. (1)


(b) Power of pump =  P = ?

We know that 

Power = Energy / Time

P = `\frac {P.E}{t}`

Putting the corresponding values

P = `\frac {9.8 ｘ10⁶ J}{1.2 ｘ10³ s}`

P = 8.17 ｘ10⁶⁻³ W

P = 8.17 ｘ10³ W

Or

P = 8.17 kW --------------Ans. (2)




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