# Power of a Hydraulic Pump and Rate at which Kinetic Energy is imparted to the water

Suppose there is a Horizontal pipe of Length $$L$$ which pumps water through an end with a speed $$v$$, there is no friction. A pump is on the opposite side.

So what will the rate at which of kinetic energy will be imparted to the water which is equal to power of the engine according to this answer on the stack exchange

$$\frac 12\rho AV^3$$ where

$$\rho=$$ density of the liquid and $$A$$ stands for the area of cross section of water.

But according to the traditional formula of power which is equal to

$$\text{force}\times\text{velocity}$$

$$\rho AV^3$$ should be the power of engine

As $$\rho AV^2$$ denoting Force multiplied with Velocity $$v$$

Why there is a difference of factor of $$\frac 12$$? I have researched on the internet and on stack exchange there is a similar question but answers are not very useful as question is closed for some reason.

My take on this, like if we take a column of length $$L$$ of water it is so when half the column flows out of the pipe, space created is filled by a new column so Half of the engine power goes into accelerating that column of water so engine works double but power transmitted to the water column of length $$L$$ is half

• Please after downvoting mention the reason, I think this question is very relevant. Commented Aug 5, 2021 at 10:20
• Power with which Water leaves the Pipe is $\rho AV^2$.This sentence doesn't make sense(to me)
– ACB
Commented Aug 5, 2021 at 10:38
• it is the force required the stop the water by a wall, so multiplying it with the velocity indirectly gives you the power of engine Commented Aug 5, 2021 at 10:41
• You are finding Power. According to the above sentence you've already found the power.
– ACB
Commented Aug 5, 2021 at 10:42
• there is a extra factor of half(1/2) in the answer i have given link to, i want to know the reason behind that factor of half Commented Aug 5, 2021 at 10:44

If the water starts from rest, when doing force × velocity you'll need the average velocity $$\frac{v}{2}$$.
• @ Samardeep singh $P=Fv$ comes from $W=Fd$, work done = force x distance, with both sides divided by a small change in time $dt$, when the water is moving slowly, the force on it only moves through a small distance and the work done during that time is low, but when moving faster the distance is greater and the work done for that time interval is greater. So, if it's a constant acceleration, you can use the average distance moved during a small time interval and use the average $v$ in $P=Fv$ Commented Aug 5, 2021 at 11:22