Pump head.

[LoganPrice]

I have a submersible pump that I won't be using submerged. It has 90" (7.5') as maximum head. I am wanting to pump water from my lake to a 300 gallon water storage container. Input to output is about 50 yards at 5 degrees up. Would the pump be able to handle that?

The water will almost be going right back into the lake. I am using the water to turn a generator for power and a small wind turbine to power the pump 30W.

 

[Valveman]

I don't know how far up it would be in 50' at a 5 degree rise. But if it anywhere close to 7.5', a pump with a max pressure of 7.5' won't move a drop.

You would probably be better off to use the wind generator for power. You will always lose some in the transfer. It will tank more energy to pump the water than the water can make when running back through a turbine generator.

 

[Sponsor]

 

[LoganPrice]

I don't know if it's even 5 degrees. but a slight incline. I was planning on using the wind turbine for energy as well, but I have no place to store expensive batteries. I can however store water and later turn it into energy when needed.

 

[Reach4]

Your submersible pump must be under water -- at least the intake. And you would want the whole thing under water if you ran it long. It does not take extra pumping power to get from 2 or 3 ft down up to the surface. You start measuring from the surface.

 

[LoganPrice]

Your submersible pump must be under water -- at least the intake. And you would want the whole thing under water if you ran it long. It does not take extra pumping power to get from 2 or 3 ft down up to the surface. You start measuring from the surface.

I was just going to attach a hose/tube to the intake to the water. But submerging the pump could work. The power from the wind turbine would power the pump until the container is full to the float switch and then it wouldn't power the pump until the container needed more water.

 

[Boycedrilling]

If I remember my trigonometry correctly, you multiple the angle distance, by the sine of the angle in degrees. The answer is the rise in elevation. So if you take a distance of 150 ft and multiply it by the sine of 5 degrees, your answer is a rise in elevation of 13.07 feet.

 

[Valveman]

If I remember my trigonometry correctly, you multiple the angle distance, by the sine of the angle in degrees. The answer is the rise in elevation. So if you take a distance of 150 ft and multiply it by the sine of 5 degrees, your answer is a rise in elevation of 13.07 feet.

I'm impressed! I can't even remember how to spell trignomtry anymore.

 

[Boycedrilling]

I didn't take trig in school either. I mean who needs math, right? I had to teach myself when I learned how to lay out center pivot irrigation systems with a Topcon GTS2B total station.

 

[Valveman]

My tough one was figuring wire to water efficiencies on line shaft turbines for bid jobs when I was about 13. Just had to figure it out the hard way. Still don't know what kind of math that is called.

 

[Craigpump]

The angle of the dangle is directly proportionate to ............

 

[DonL]

The angle of the dangle is directly proportionate to ............

... The mass of the ass.

I think ?

 

[Craigpump]

Exactly