# Energy/Power Question for pumping water out of a well

• Physics_Amazing
In summary, the conversation discusses the formula P = E/t and how to solve for v. It also includes the information that the density of water is 1.0 kg/L and the formula E=mgh is used to calculate the energy needed to lift water. The conversation also mentions the importance of carrying units in calculations and converting to a per unit volume basis. The conversation ends with a discussion about pump power and the location of the pump in relation to the water source.f

#### Physics_Amazing

Homework Statement
A water pump has a power rating of 1000. W. If the kinetic energy change is negligible, how many liters of water per second can it pump from a 20.0 m deep well?
Relevant Equations
P= E/t
Ek= (1/2)mv^2
Eg= mgh
E = Fd
Im guessing you start off with P=E/t but I'm unsure of what I need to sub in, in order to solve for v.

P = 1000W
h = 20m
v=?

How much does each liter of water weigh? And then how much energy does it take to lift each liter of water 20 meters? ...

How much does each liter of water weigh? And then how much energy does it take to lift each liter of water 20 meters? ...
density of water is 1.0 kg/L

So... E=mgh
E=1(9.8)(20)
E=196J
then sub that in P=E/t
??

• berkeman
Looks good so far. Be sure to keep carrying units along in your calculations like you are doing so far...

But maybe include units in some of your intermediate calculations too...

E=1[kg](9.8[m/s^2])(20[m])

• erobz
I think it helps significantly to think about it on a per unit volume basis.

The energy the pump added was(as you found):

$$E = mgh$$

You can use the density of water to look at it on a per unit volume basis as follows:

$$E = m g h = \rho V\llap{-} g h \implies \frac{E}{ V\llap{-} } = \rho g h$$

Now, what do you get if you multiply the Energy per unit Volume by the Volume per unit Time?

Last edited:
Relevant Equations:
P= E/t
Ek= (1/2)mv^2
Eg= mgh
E = Fd

On the other hand, the derivation can be performed before putting in values.

starting with E = Fd

the pump has to supply that much work, so W = Fd.

Multiply the right hand side by 1, using an area m2 of a volume that the force acts upon,
W = ( F / m2 ) (d m2 )

giving W = p V , where p is pressure, V is volume

take the derivative of both sides,
dW/dt = p dV/dt, assuming pressure p is constant,

giving, the general equation.
pump power P = p Q , or power = pressure( N/ m2 ) times flow rate ( m3/ sec )

• berkeman
I would say zero, unless the pump is located inside the well and not higher than around 7 meters from the free surface of the water.

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