Question

A pump on the ground floor of a building can pump up water to fill a tank of volume $30m^{3}$in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

(Take $g = 10ms^{- 2}$)

• A. 36.5 kW
• B. 44.4 kW
• C. 52.5 kW
• D. 60.5 kW

Answer 1

Answer: (B)

44.4 kW

Answer 2

Here,

Volume of water $= 30m^{3}$

$t = 15\min = 15 \times 60s = 900s$

Height, h = 40m

Efficiency, $\eta = 30\%$

Density of water = $10^{3}kgm^{- 3}$

$\therefore Mass$of water pumped = Volume × Density

$= (30m^{3})(10^{3}kgm^{- 3}) = 3 \times 10^{4}kg$

$P_{output} = \frac{W}{t} = \frac{mgh}{t} = \frac{(3 \times 10^{4}kg)(10ms^{- 2})(40m)}{900s}$

$= \frac{4}{3} \times 10^{4}W$

Efficiency, $\eta = \frac{P_{output}}{P_{input}}\eta = \frac{P_{ifj.kke}}{P_{ifj.kke}}$

$P_{input} = \frac{P_{Output}}{\eta} = \frac{4 \times 10^{4}}{3\frac{30}{100}} = \frac{4}{9} \times 10^{5}$

$= 44.4 \times 10^{3}W = 44.4KW$