Pumped hydoelectric storage equation problem

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The discussion focuses on calculating the efficiency of a reversible hydraulic generator/pump station, with specific power and flow rates provided for both the pump and turbine. The user is struggling to set up an equation to determine the percentage of energy that can be delivered back to the grid, noting that their calculations yield unrealistically high efficiencies around 99%. Participants suggest that typical efficiencies range from 70-80%, and highlight that the pump's capacity is significantly lower than the turbine's due to operational dynamics and the nature of energy consumption during peak and low-cost periods. The conversation also touches on the importance of considering the design of the penstock and the relationship between pumping and generating power. Overall, the efficiency calculations and system design considerations are critical for optimizing performance in pumped hydroelectric storage systems.
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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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Typical efficiencies are 70-80% overall. Why is pumped capacity only 10% of turbine capacity?
 
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
 
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?
 
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...
 
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.
 
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.
 
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.
 
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.)
 
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  • #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.
 
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