What is pumped steadily out of a flooded basement at a speed of 5.0m/s\nthrough a uniform hose of radius 1.0 cm. The hose passes out through a\nwindow 3.0m above the water line. What is the power of the pump?

Suppose that a mass delta m of water is pumped in time delta t. The pump increases the potential energy of the water by Delta mgh, where h is the vertical distance through which it is lifted, and increases its kinetic energy by 1 /2 delta mv^2 , where v is its final speed. The work it does is
Delta W = delta m gh +1/2 delta mv^2
power P = Delta W / delta t
P = delta m gh +1/2 delta mv^2 / delta t
P = delta m / deta t (gh+1/2v^2)
Now the rate of mass delta m /delta t = rho *A*v where rho is density of water and A is the area of the hose. The area of the hose A = pi*r^2 = 3.14*(0.01)^2 = 3.14*10^-4 m^2
rho*A*v = 1000*3.14x10^-4x5 = 1.57 kg/sec
Thus P = P = delta m / deta t (gh+1/2v^2) = 1.57*[9.8*3 + 5^2 / 2]
P = 66 W
Thanks & Regards,
Nirmal Singh
Askiitians faculty