Difference in Potential Energy between two Reservoirs

[catsandtrees]

Homework Statement

3. Water is pumped from one reservoir to another 100m away. The water level in the second reservoir is 20m above the water level of the first reservoir. What is the increase in specific potential energy of the water in J/kg?

Homework Equations

The relevant equation I've been given in my notes is

P = mgh

This problem is for a renewable energy course. In class I only received a definition of potential energy and the above equation, the teacher didn't do any problems with potential energy in class either. The given textbook doesn't touch on the quantitative side of renewable energies. I've been trying to look through old physics textbooks but can't find anything too helpful yet.

The Attempt at a Solution

I have been away from physics for almost 8 years, so physics problems aren't very intuitive for me anymore.

So far I've assumed I'm calculating the change in P. Setting up the equation as :
ΔP = (mgh)2nd reservoir - (mgh)1st reservoir

with
Δ (mass*g) = (20m* ? * density of water)*9.8m^2/sec
Δh = 100m

I can't figure out how to get the Δ in the volume of water, which I need to get the mass. Or since the teacher wants the answer in J/kg, should I be assuming that I should set up the equation as:
ΔP/m = Δ(g*h) ?

Any help would be great.
Thanks,
Alyssa

 

[TSny]

Hello and welcome to PF!

So far I've assumed I'm calculating the change in P. Setting up the equation as :
ΔP = (mgh)2nd reservoir - (mgh)1st reservoir

Yes. This is very good.

Note that you can write this as ΔP = mgh₂ - mgh₁ = mg(h₂ - h₁) = mgΔh.

You can imagine that you skim a mass of m = 1 kg of water off the surface of the first reservoir and move it to the surface of the second reservoir. You just need to calculate ΔP for this situation.

 

[catsandtrees]

I attempted the question again keeping in mind I need the specific potential energy.
This would give me:

ΔP/Δm = gΔh = 9.80 * 100m = 980 m²/s² = 980 J/kg.

However this doesn't use the 20m water level difference. At this point in the course I don't think the teacher would give us useless info. Does this mean I need to use a trig function to get actual difference in height? Or should I be incorporating the 20m into the Δm?

 

[TSny]

In the equation P = mgh, h is the vertical height. So, Δh is the change in vertical height, or the change in water level of the two reservoirs. There is no change in gravitational potential energy when a mass is moved horizontally. You can think of moving 1 kg of mass 100 m horizontally and then 20 m vertically. Only the 20 m vertical change contributes to the change in potential energy.

 

[Nickie]

Hi Alyssa,

Thank you for reaching out for help with your physics problem. It's great that you are taking a renewable energy course and trying to understand the quantitative side of it.

To solve this problem, we can use the equation P = mgh, where P is the potential energy, m is the mass of the water, g is the acceleration due to gravity (9.8 m/s^2), and h is the height difference between the two reservoirs.

In this case, we are looking for the increase in specific potential energy, which is the potential energy per unit mass. So we can rewrite the equation as ΔP/Δm = gh, where ΔP is the change in potential energy and Δm is the change in mass.

To find the change in mass, we can use the density of water (1000 kg/m^3) and the volume of water that was pumped from the first reservoir to the second reservoir. We know that the volume of water is the same in both reservoirs, so we can use the formula for volume of a cylinder (V = πr^2h) to find the volume of water in the second reservoir. The radius (r) is not given, so we can assume it to be 1m for simplicity.

Therefore, the volume of water in the second reservoir is V = π(1m)^2(20m) = 20π m^3.

Now, we can calculate the change in mass:
Δm = density * change in volume = (1000 kg/m^3) * (20π m^3) = 20,000π kg.

Finally, we can plug in the values in the equation ΔP/Δm = gh to find the increase in specific potential energy:
ΔP/Δm = (9.8 m/s^2) * (100m) = 980 J/kg.

Therefore, the increase in specific potential energy of the water is 980 J/kg.

I hope this helps you understand the problem better. If you have any further questions, please don't hesitate to ask.

Best,