How to use the hydroelectric power equation

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In summary, the conversation discusses the power equation for hydroelectric dams and how to calculate the necessary flow rate for a certain amount of power generation. It also mentions the size and location of the reservoir in relation to the turbine. The conversation concludes with a discussion about the feasibility and costs of building a hydroelectric dam.
  • #1
gloo
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I would like to get some help with the power equation that derives the power generated in Watts

P= density (water) * head (meters) * flow rate (cubic meter per second) * g * efficiency

I understand how to plug in everything except for the flow rate variable. I would like to get help in deriving 2 things for the following set up: If I have a reservoir that is 20m by 20m by 20m (8000 cubic meters of water) and it is 70 meters high.

1. How big would my pen stock tunnel diameter half to be achieve a 50 cubic meter per second drop? Or is that a function of the head size and the diameter?

2. If I achieved the 50 cubic meter per second...would the reservoir empty out in about 160 seconds? (8000 cubic meter / 50 seconds??)
 
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  • #3
Is this something you are building yourself? If so, 50 cubic meters per second is a very high flow rate. To get a reasonable flow velocity (say 5 ft/sec or 1.5 m/sec) you'd want a pipe 6.5 meters diameter. Yes, it would drain an 8000 m3 reservoir in 160 seconds.

when you say 20 x 20 x 20 meters, 70 meters high, do you mean the 8000 m3 tank is on a 70 meter hill, above your turbine?
 
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  • #4
gmax137 said:
Is this something you are building yourself? If so, 50 cubic meters per second is a very high flow rate. To get a reasonable flow velocity (say 5 ft/sec or 1.5 m/sec) you'd want a pipe 6.5 meters diameter. Yes, it would drain an 8000 m3 reservoir in 160 seconds.

when you say 20 x 20 x 20 meters, 70 meters high, do you mean the 8000 m3 tank is on a 70 meter hill, above your turbine?

Yes those assumptions are correct. I was just working out some theoretical costs of building a hydro electric dam and calculating payback periods etc.
 
  • #5
gloo said:
2. If I achieved the 50 cubic meter per second...would the reservoir empty out in about 160 seconds? (8000 cubic meter / 50 seconds??)

Yes. If there was no inflow.

50 cubic meters per second is pretty fast. What's the flow rate into the reservoir?
 

FAQ: How to use the hydroelectric power equation

1. How do I calculate the power output of a hydroelectric power plant?

The power output of a hydroelectric power plant can be calculated using the equation P = ρghQ, where P is power in watts, ρ is the density of water in kg/m^3, g is the acceleration due to gravity in m/s^2, h is the height of the water in meters, and Q is the flow rate of water in cubic meters per second.

2. What is the significance of the variables in the hydroelectric power equation?

The variables in the hydroelectric power equation represent different factors that affect the power output of a hydroelectric power plant. ρ represents the density of the water, which can vary depending on factors such as temperature and salinity. g represents the gravitational pull, which is a constant. h represents the height of the water, also known as the head, which is the key factor in determining the potential energy of the water. Q represents the flow rate of the water, which is the volume of water passing through the plant per unit time.

3. Can the hydroelectric power equation be used for all types of hydroelectric power plants?

Yes, the hydroelectric power equation can be used for all types of hydroelectric power plants, including traditional dams, run-of-river systems, and tidal power plants. However, the specific values for the variables may vary depending on the design and location of the plant.

4. How can the hydroelectric power equation be applied to real-world situations?

The hydroelectric power equation can be applied to real-world situations to calculate the potential power output of a hydroelectric power plant based on the specific conditions at that plant. It can also be used to compare different potential sites for a hydroelectric power plant and determine which location would be most suitable for maximum power output.

5. What are the limitations of the hydroelectric power equation?

The hydroelectric power equation does not take into account factors such as the efficiency of the power plant, which can vary depending on the design and condition of the equipment. It also does not account for external factors that may affect the flow rate of water, such as droughts or floods. Additionally, it assumes a constant flow rate, which may not always be the case in a real-world setting.

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