Energy/Power Question for pumping water out of a well

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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.
  • #1
Physics_Amazing
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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=?
 
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  • #2
How much does each liter of water weigh? And then how much energy does it take to lift each liter of water 20 meters? ...
 
  • #3
berkeman said:
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
??
 
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  • #4
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...

Physics_Amazing said:
E=1[kg](9.8[m/s^2])(20[m])
 
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  • #5
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?
 
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  • #6
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 )
 
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  • #8
Lnewqban said:
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.

Please, see:
https://processengineering.co.uk/article/2019233/maximum-suction
Well pumps aren’t suction side pumps. They are lowered into the well so NPSH will not be an issue.

Generally speaking though, you are correct. The location of a pump in a particular application can have consequences. If you tried to suck it up from the top of the well thinking it’s a pump maintenance time saver, there is going to be a problem!
 
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FAQ: Energy/Power Question for pumping water out of a well

1. How much energy is required to pump water out of a well?

The amount of energy required to pump water out of a well depends on several factors, such as the depth of the well, the amount of water being pumped, and the efficiency of the pump. In general, it takes more energy to pump water from deeper wells and to pump larger volumes of water.

2. What type of energy is used to pump water out of a well?

The most common type of energy used to pump water out of a well is electricity. However, some well pumps can also be powered by other sources of energy such as solar power, wind power, or even human or animal power.

3. How can energy efficiency be improved when pumping water out of a well?

There are several ways to improve energy efficiency when pumping water out of a well. These include using a properly sized pump, maintaining the pump regularly, minimizing friction losses in the piping system, and using renewable energy sources if possible.

4. Is it possible to use gravity to pump water out of a well?

Yes, it is possible to use gravity to pump water out of a well. This is known as a gravity-fed or gravity-powered system. In this system, the water flows down from a higher source to the well, eliminating the need for a pump. However, this method is only feasible if there is a significant elevation difference between the water source and the well.

5. How can the power needed to pump water out of a well be calculated?

The power needed to pump water out of a well can be calculated using the following formula: Power (Watts) = Flow rate (liters per second) x Head (meters) x Specific gravity (unitless) x Pump efficiency (decimal). The specific gravity refers to the density of the fluid being pumped, while the pump efficiency takes into account the efficiency of the pump itself.

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