Work, force, power turbine pump problem

In summary, the conversation discusses the use of water to store energy in electric-power companies. The question is how many cubic meters of water would need to be pumped from a lower to higher reservoir, 520 m apart, to store the energy produced by a 120*106 electric-power plant in 1.0 hour. The solution involves finding the gravitational potential energy of the water at 520 m, which is then used to find the mass of water needed to produce the desired energy output. The final answer is approximately 84772 m3 of water.
  • #1
joseg707
32
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?
 
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  • #2
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.
 
  • #3
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.
 
  • #4
Power = 120 MW

Energy = Power * time. How many J of energy is produced by 120 MW in one hour?
 
  • #5
Oh! Haha yeah, I completely jumped right over that didn't I.
(120e6W/s)(3600s)=work
work=4.32e11J
 
  • #6
So does this hold true?

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

1. What is the difference between work, force, and power?

Work refers to the energy expended or transferred by a force acting on an object. Force is a push or pull that can cause an object to accelerate or change direction. Power is the rate at which work is done or energy is transferred.

2. How do you calculate work, force, and power?

Work is calculated by multiplying the force applied by the distance moved in the direction of the force. Force is calculated using the formula F=ma (mass times acceleration). Power is calculated by dividing work by time or multiplying force by velocity.

3. What is a turbine pump and how does it work?

A turbine pump is a type of centrifugal pump that uses a rotating impeller to create suction and move fluid through the pump. As the impeller rotates, it creates a low pressure area that draws in fluid from the inlet and then pushes it out through the outlet at a higher pressure.

4. What factors affect the performance of a turbine pump?

The performance of a turbine pump is affected by factors such as the pump's speed, size and shape of the impeller, fluid viscosity and density, and the efficiency of the motor driving the pump.

5. How do you solve a problem involving work, force, power, and a turbine pump?

To solve a problem involving work, force, power, and a turbine pump, you first need to determine the values of each variable given in the problem. Then, you can use the appropriate equations to calculate the unknown variables. It is important to pay attention to the units of measurement and use the correct conversion factors if needed.

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