Question

A centrifugal pump deliver water at the rate of 50 litres against a total head of 40 meters. Then the power required to drive the pump is

Answer 1

I have any other information in the best regards and thanks again and will in fact that I want a little more time and I will be a good day please

Answer 2

The power required to drive the pump is 20kN.

Explanation:

The power is given as,

$P=\frac{mgh}{t}$.....(1)

Where,

P=power required

m=mass of the body

g=acceleration due to gravity=10m/s²

h=height to which the body is raised

t=time required

Mass in terms of density can be written as,

$m=\rho \times V$.......(2)

ρ=density of the water=1000kg/m³

V=volume occupied by the water

By substituting equation (2) in the equation (1) we get;

$P=\frac{\rho \times V \times g\times h}{t}$ .....(3)

From the question we have;

$\frac{V}{t}=50L/s=50\times 10^{-3}m^3 s^{-1}$  ($1L=1\times 10^{-2}m^2$)

$h=40m$

By substituting the required values in equation (3) we get;

$P=1000\times 50\times 10^{-3} \times 10\times 40=50 \times 10 \times 40 = 20000N = 20kN$

Hence, the power required to drive the pump is $20kN$.

#SPJ3