slurry booster pump keeps running

A water pump raises water at a rate of 050m per minute class 11 physics CBSE

A water pump raises water at a rate of 0.50 m per minute from a depth of thirty meters. If the pump is seventy percent, what is the power developed by the engine of the pump.

Hint: Calculate the rate of water per second as it is given in minute. Next, calculate the mass of the water as volume and density is given. Next, the power can be calculated as the work done per unit time. Next, the output power can be obtained from above and input power must be calculated. Efficiency is simply the ratio between the output and input powers.

Formula used:W=m g h m=ρ×v

Complete step-by-step answer:Let’s first calculate the volume of water per unit second, in the question it is given in minutes,60 s=0.50 m⇒1 s=0.0083 m 3Now, let’s calculate the mass of the water,m=ρ×v m=1000×0.0083 m=8.3 k g s−1We know the power of the pump is nothing but the work done per unit time.I.e.,P=W t⇒P=m g h t⇒P=8.3×300⇒P=2490 WNow, we know the output power, we know, the efficiency of the power is equal to the ratio of output and input powers,70 100=2490 P P=3557 WTherefore, the power developed by the engine is obtained.

Additional Information: The pump is powered by an electric motor that drives an impeller, or centrifugal pump. Operating a centrifugal pump, a driven impeller rotates inside the housing to force water through the engine and back to the radiator where it is cooled, and the cycle begins again. Difference with an electric pump is that rather than being driven from a belt, it is driven by an electric motor.

Note: There’s a difference between electric motor and electric pump is that a motor is a device that converts electricity to mechanical energy which results in motion, whereas a pump is a device that is used to transfer a fluid from one place to another without any motion.