Question

Water is pumped steadily out of a flooded basement at a speed of 5.0 m/s through a uniform hose of radius 1.0 cm. The hose passes out through a window 3.0 m above the waterline. What is the power of the pump. [Hint: Find the rate of mass flow and use the knowledge that the work done (delta W) = (delta m)gh+1/2 (delta m)v^2 ].

Answer 1

Let us take as small mass dm being pumped in small time dt. Doing this increases the potential energy by (dm)gh, increases its kinetic energy by 1 /2 delta mv^2
 dW = (dm)gh +0.5*(dm)v^2
power= P = dW/dt= (dm*(gh)+0.5(dm)v^2) /dt
P = (dm/dt)(gh+1/2v^2)
Rate of mass =dm/dt=rho *A*v 
where rho is density of water 
A is the area of the hose. 
Area= 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
 P = (dm/dt)*(gh+1/2v^2) = 1.57*[9.8*3 + 5^2 / 2]