add hydro to an existing solar

Re: add hydro to an existing solar

Hydro is all about the numbers... How much pressure differential and what kind of volume are you looking at?

You probably saw this post referenced a few days ago:

BB. wrote: »

You can use a site like this:

http://www.oregon.gov/ENERGY/RENEW/Hydro/Hydro_index.shtml

To figure out how much power you can generate for 5,000 gallons of water:

Gross Head x Flow x System Efficiency (in decimal equivalent) x C = Power (kW)

C is a constant (the value is different in English and metric units).

• 100 psi / 2.13 psi per foot * 1 cf/s * 0.55 turbine eff * 0.085 = 2.19 kWatts
• note fixed 0.85 should be 0.085
• 5,000 gallons * 1/7.48 gallons per cuft * 1/1cf per sec = 668 seconds = 11 minutes of power

Looks like a lot of "work" for little "work"... 2kWatts for 11 minutes.

A pair of 225 AH 6 volt Lead Acid Batteries:

• 225 AH * 12 volts = 2,700 WH for batteries
• 2,190 WH * 11min/60min per hour = 402 WH for above water system

In this example, the batteries store 6.7x more energy than 5,000 gallons at 47 feet elevation.

If you could use a few hundred watts continuously--Then a smaller hydro system may work OK.

If you have a lot of water and low head (5 feet or so?)... There are systems that work with that too... But you need a huge amount of water flow because of the low head.

-Bill