Work, force, power turbine pump problem

Click For Summary

Homework Help Overview

The problem involves calculating the volume of water that needs to be pumped from a lower reservoir to an upper reservoir to store energy produced by a power plant. The context is centered around concepts of work, energy, and gravitational potential energy in relation to a turbine pump system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between power, work, and energy, with some attempting to calculate the work done based on the power output over time. Questions arise about the conversion of energy forms and the assumptions regarding kinetic and potential energy.

Discussion Status

Participants are actively engaging with the problem, exploring various aspects of energy conversion and calculations. Some have provided calculations for work done, while others are questioning the relationships between kinetic energy, potential energy, and the parameters involved in the problem.

Contextual Notes

There is an assumption that the flow of water has a constant velocity, and participants are considering the implications of this on the calculations. The density of water is noted as a factor in converting mass to volume.

joseg707
Messages
32
Reaction score
0

Homework Statement


Some electric-power companies use water to store energy. Water is pumped by reversible turbine pumps from a low to high reservoir. To store the energy produced in 1.0 hour by a 120*106 electric-power plant, how many cubic meters of water will have to be pumped from the lower to the upper reservoir? Assume the upper reservoir is 520 m above the lower and we can neglect the small change in depths within each. Water has a mass of 1000 kg for every 1 m3


Homework Equations


P=W/t
W=Fd
F=ma


The Attempt at a Solution


I multiplied power by the total number of seconds in an hour to get work. I then divided the work by the distance of 520 m and I get force. This is where I'm stuck at. I have that force is equal to (120000000 kg*m2/s3)*(3600s)/(520m). I don't know how to solve for mass because it seems that the flow has a constant velocity and there is no acceleration. Can someone please point me in the right direction?
 
Physics news on Phys.org
One has a 120 MW plant. One has to find how much water must be stored at 520 m, to provide the energy produced in one hour.

Energy = Power * time.

The mechanical energy in the turbine comes from the kinetic energy in the water flow, which is transformed from the gravitational potential energy of the water stored 520 m above the turbine. What is the GPE of the water at 520 m.

Once one finds the mass, the volume is easy to find given the density.
 
Are you saying that the kinetic energy is equal to the output energy of the turbine, 120MW, and the the potential energy is equal to the kinetic energy?
KE=120 MW

PE(2)=KE(2)
mgh=120W
m(9.8)(520)=120MW?

I don't feel like I understood that right.
 
Power = 120 MW

Energy = Power * time. How many J of energy is produced by 120 MW in one hour?
 
Oh! Haha yeah, I completely jumped right over that didn't I.
(120e6W/s)(3600s)=work
work=4.32e11J
 
So does this hold true?

m(9.8)(520)=4.32e11
m=84772370.49 kg*1m3/1000kg=84772 m3
 

Similar threads

Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
4K
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
5K
Replies
1
Views
2K
Replies
3
Views
8K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K