Pumped hydoelectric storage equation problem

  • Thread starter Erik S
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In summary: Archimedian?No, Archimedian is a mathematician who studied fluid flow and was the first to develop the theory of buoyancy.It also has to do with the cost of power transmission. Transmission capacity is unnecessarily expensive for high power flows (plus or minus direction) for short periods of time.I hope this helps.
  • #1
Erik S
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I am trying to figure out the efficiency of a reversible hydraulic generator/pump station that will generate electricity when the price is high, and pump water up into the reservouar with power from the grid when electricity is cheap.

I have calculated the following information for the pump and turbine:

Ppump= 10 MW
Qpump= 7.5 m^3/s

Pturbine= 96 MW
Qturbine= 72.5 m^3/s

How can I figure out what percentage of the bought energy can be delivered back to the net?
I am having trouble figuring out how to set up an equation I can use to solve this problem, No matter what I go I get really high effeciency like 99% which must be wrong.

Thanks in advance.
 
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  • #2
Typical efficiencies are 70-80% overall. Why is pumped capacity only 10% of turbine capacity?
 
  • #3
The pump is rated at 10MW, the produced flow is calculated using Mannings equation with given Mannings constant.

This is because I don't have any information on the volumetric displacement/ rpm of the pump
 
  • #4
Erik S said:
The pump is rated at 10MW, the produced flow is calculated using Mannings equation with given Mannings constant.

This is because I don't have any information on the volumetric displacement/ rpm of the pump
So this system has open-conduit flow? Can you post a sketch of your system?
 
  • #5
You can easily just ratio all of those numbers against each other to find the efficiency.

...however, just by looking at them I can see that your 99% efficiency is about right. So your problem is in how you came up with those performance points.

By the way, this is engineering, not math. Moving thread...
 
  • #6
insightful said:
Why is pumped capacity only 10% of turbine capacity?
Maybe because water is pumped for longer periods while some power is available at low cost, than the shorter periods of peak turbine flow that generate power when the demand and price are higher.

Erik S, You need to find the efficiency of the motor, pump, turbine and alternator.
The penstock will need to be designed to operate with the higher flow expected during turbine operation.
 
  • #7
Baluncore said:
Maybe because water is pumped for longer periods while some power is available at low cost, than the shorter periods of peak turbine flow that generate power when the demand and price are higher.
Pumping and generating power are usually close:

https://en.wikipedia.org/wiki/Pumpe...ty#/media/File:Pumpspeicherkraftwerk_engl.png

For a 10:1 difference, you'd need a separate pump and turbine for high efficiency.
 
  • #8
insightful said:
Pumping and generating power are usually close:
But not in this case.

insightful said:
For a 10:1 difference, you'd need a separate pump and turbine for high efficiency.
Erik S said:
I have calculated the following information for the pump and turbine:

Ppump= 10 MW
Qpump= 7.5 m^3/s

Pturbine= 96 MW
Qturbine= 72.5 m^3/s
As you can see, there is a separate pump and turbine for the different flow rates.
 
  • #9
insightful said:
Pumping and generating power are usually close:
Where does that fact come from ? Do you have a reference ?
 
  • #10
Erik S said:
I am trying to figure out the efficiency of a reversible hydraulic generator/pump station that will generate electricity when the price is high, and pump water up into the reservouar with power from the grid when electricity is cheap.

I have calculated the following information for the pump and turbine:

Ppump= 10 MW
Qpump= 7.5 m^3/s

Pturbine= 96 MW
Qturbine= 72.5 m^3/s

How can I figure out what percentage of the bought energy can be delivered back to the net?
I am having trouble figuring out how to set up an equation I can use to solve this problem, No matter what I go I get really high effeciency like 99% which must be wrong.

Thanks in advance.

Easy, Net Work rate =Net Powerhydroplant= Pturbine-Ppump
If you are interested with Hydro Plant Efficiency = Net Powerhydroplant/Pturbine

Efficiency means what you get from the total effort you exert or gain over capital.
I hope this helps.
 
  • #12
Baluncore said:
Where does that fact come from ? Do you have a reference ?

Re: Pump versus generate powers.

Insightful didn't say always, he said usually.

It is simply that the number of hours in the day for high-peak power and low-peak power are similar. That alone means you want pumping power and generating power to be of similar magnitude.

It also has to do with the cost of power transmission. Transmission capacity is unnecessarily expensive for high power flows (plus or minus direction) for short periods of time. Both plus and minus flows are typically spread over several hours.
 
  • #14
insightful said:
I'm wondering if this system might be using Archimedian screws:

Sure, but remember that the energy stored is proportional to the quantity of water pumped, and the vertical height difference. You might need 200 meters of height to make it worth the trouble. I question the efficiency of a Archemedes Screw at that height.

Consider an example: The Blenheim-Gilboa Pumped Hydro Plant, has 19 million cubic meters of water storage, at 348 meters vertical head, to make 1134 MW electric.
 
  • #15
anorlunda said:
Sure, but remember that the energy stored is proportional to the quantity of water pumped, and the vertical height difference.
Cannot an Archimedes screw be very efficient at low head and high volume for both pumping and generating?

(Edit: Admittedly, this system would require a huge surface area reservoir.)
 
Last edited:
  • #16
anorlunda said:
Sure, but remember that the energy stored is proportional to the quantity of water pumped, and the vertical height difference. You might need 200 meters of height to make it worth the trouble. I question the efficiency of a Archemedes Screw at that height.

Consider an example: The Blenheim-Gilboa Pumped Hydro Plant, has 19 million cubic meters of water storage, at 348 meters vertical head, to make 1134 MW electric.

Archimedes Screw is ideally applicable to Low Head, High flow reservoirs. For high heads and high flow reservoir, ideal is the impulse and mix flow type of turbine. 200 m is very high elevation indeed.
 

Related to Pumped hydoelectric storage equation problem

What is the pumped hydroelectric storage equation problem?

The pumped hydroelectric storage equation problem is a mathematical formula used to calculate the energy efficiency of a pumped hydroelectric storage system. It takes into account factors such as the amount of water used, the height of the storage reservoir, and the efficiency of the pumping and generation processes.

How does pumped hydroelectric storage work?

Pumped hydroelectric storage works by using excess energy from a power grid to pump water from a lower reservoir to a higher reservoir. When there is a demand for electricity, the water is released back to the lower reservoir, passing through turbines that generate electricity. This process can be repeated, making it a form of energy storage.

What are the advantages of pumped hydroelectric storage?

Pumped hydroelectric storage has several advantages, including its ability to store large amounts of energy, its high efficiency, and its long lifespan. It also has a low environmental impact and can help stabilize the grid by providing backup power during times of high demand.

What are the limitations of pumped hydroelectric storage?

One limitation of pumped hydroelectric storage is that it requires specific geographic conditions, such as a suitable site for the upper and lower reservoirs. It also has a high upfront cost and may face opposition from local communities. Additionally, the process of pumping and releasing water can result in some water loss.

How is the efficiency of pumped hydroelectric storage calculated?

The efficiency of pumped hydroelectric storage is calculated using the pumped hydroelectric storage equation problem, which takes into account the potential energy of the water, the efficiency of the pumping and generation processes, and any energy losses. The equation is Efficiency = (Energy Produced / Energy Consumed) x 100%.

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