Electricity Homework: Calculating Energy, Volume & Distance

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SUMMARY

The discussion focuses on calculating the energy stored in the top meter of water in a hydroelectric dam with a surface area of 3.00×106 m2 and a water height of 130 m. The potential energy is calculated using the formula mgh, where m is the mass of the water in the top meter, derived from the volume and density of water (1000 kg/m3). The efficiency of energy conversion is 93.0%, leading to the conclusion that to produce 1000 kilowatt-hours of electrical energy, the mass of water required can be determined, and subsequently, the change in water level can be calculated.

PREREQUISITES
  • Understanding of gravitational potential energy (mgh)
  • Knowledge of hydroelectric energy conversion efficiency
  • Familiarity with basic fluid mechanics concepts
  • Ability to perform unit conversions (e.g., kilowatt-hours to joules)
NEXT STEPS
  • Calculate gravitational potential energy using the formula mgh for different heights
  • Learn about energy conversion efficiency in hydroelectric systems
  • Explore the relationship between volume, mass, and density in fluid mechanics
  • Investigate the impact of water level changes on hydroelectric energy production
USEFUL FOR

Students studying physics, particularly those focusing on energy calculations in hydroelectric systems, as well as educators and tutors assisting with related homework problems.

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Homework Statement


A hydroelectric dam holds back a lake of surface area 3.00×106 m^2 that has vertical sides below the water level. The water level in the lake is a height 130 m above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is converted into electrical energy with 93.0% efficiency.

If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the water in the lake? The density of water is = 1000 kg/m^3.

What volume of water must pass through the dam to produce an amount of electrical energy totalling 1000 kilowatt-hours of electrical energy?

What distance does the level of water in the lake fall when this much water passes through the dam?

Homework Equations


This is what I'm thinking but I'm not sure: deltaK + deltaU + delta(other energy) = 0


The Attempt at a Solution


I actually don't understand where to go with this problem. So far, I've used the surface area x height to get the volume and then divided by the density to get the mass. I don't know why I did that, it's probably not even relevant.

Then I started with my book with problems that look a bit similar and then I tried to integrate it with the integral and the resulting eqn I got is:

((h_2)^2 - (h_1)^2)/2 * dAg

d = density
A = area
g = gravity

Plugging in everything, I got the wrong answer obviously.

I know I'm completely lost on this problem, like I don't even understand the whole picture or what is electrical energy. Can someone pls shed some light for me?
 
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Youre given the surface area of the lake and asked to find the energy in the top meter of the lake. Multiply to find the volume and multiply volume with density to find the mass of the water in the first meter. This much mass held at a height of 130 meters contains mgh potential energy.

Now, at 93% efficiency, 93% of the potential energy of the water is converted into electrical energy. Therefore, 0.93*mgh=1000KW*3600 (for kw-hour). This will give you the mass of the water required PER hour and from there you can find the volume required (you're given the density).

You have the volume of water required and you have the surface area, so you can find the change in depth.
 
The system keeps telling me it's wrong. I took the surface area x height = volume. Then I took volume x density = mass. I got 3.9x10^11. I used the mgh eqn but I have the wrong answer. Isn't there something else I'm missing...?

Edit: Crap. >< I totally forgot to read the first meter. *hits head* I got it now I got it now. Many thanks!
 
Last edited:

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