A pump lifts a liquid of density to a height h and accelerates it from rest to a

Question

A pump lifts a liquid of density to a height h and accelerates it from rest to a final velocity v.

What power P does the pump deliver to the liquid, if the liquid is being pumped through a pipe with a cross sectional area A?

My professor didn't go much into detail of this subject, so if you could throw in the step by step process it would be much appreciated!

a v p g h a v3 rho/2 not enough information is provided. A v2 rho g h/2 zero, of course a liquid will naturally flow in that way without any pumping. A v rho (g h+1/2 v2) a v rho (gh-1/2 v2) a v rho + rho g h a v rho (1/2 v2) -g h a v rho

Explanation / Answer

a)

Energy = kinetic energy + potential energy

E = 1/2 m v^2 + m g h

==> E = m (1/2 v^2 + g h)

==> E = rho V (1/2 v^2 + g h)

==> E = rho A y (1/2 v^2 + g h)

P = dE/dt = d(rho A y (1/2 v^2 + g h))/dt = (rho A (1/2 v^2 + g h)) (dy/dt)
dy/dt = v

==> P = (rho A (1/2 v^2 + g h)) (v)

==> P = A v rho (gh + 1/2 v^2)

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