I have a 2" pipe running over a distance of 600 feet with a 100ft drop?
set with no motor attached, i would use off peak power to run the water into the top holding tank, using off peak power,it would be 1 third of the cost of normal power what presure would the water have to be at the lowest point could you
please answer.
Answer:
First off you will want a professional engineer to design the impeller specifically for you. This is the difference between 20% and 95% efficiency.
Losses in your pipeline will be minuscule compared to the efficiency of your impeller.
Fluid Potential Energy of water with 100foot of head=
1Kilogram/L * 9.81 m/s^2 * 100 ft
100 feet = 30.48 meters
= 299 Joules/ Liter
In order for your to power your generator at max power, your flow would have to equal
15kva=15000Watts = 15000J/s
15000 / 299 = 50.17L/s
Thats a lot of water through a 2"pipe. You may want to go to 3 inch if that is an option.
Fortunately though, if your AC generator is linked to the grid, you will generate power even if your are only moving 1L/s, just the current output will be much smaller.
Your idea to use off peak power is a good idea and is done on large scales, but on a smaller scale the efficiency and investment will not be a money saver. Certainly a good conversation starter though!
I heard of an idea you might want to link with yours. In some places they use windmills (that generate energy at random times) to pump water up to a holding pond where it is stored till its energy is needed.
Best of luck to you!
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