Im sure we have been here b4

nigtomdawnigtomdaw Solar Expert Posts: 705 ✭✭
Ive currently got 2.4 kw solar array of off grid power Im seeing a full system recharge on a sunny day in less than 4 hours in March 2008. To compound this problem I intend to increase my PV power to 4.2 KW very soon with a MPPT charge controller. I can see a 2 hour window of sun b4 Im spilling power.;)

I have a lucky feature in that my land is on a step hill so I have an easy drop of land head hieght and I can easily provide a 5 meter head for a water turbine. If I decide to put in a tank to tank water transfer system what size of tank do I require to provide a 300 amp hour refresh capacity at 24 v nominal to gain an extra days capacity.

You may ask just add more batteries I have a 2100 ah bank at 24 v which could go 2 days with a 70% discharge but I never go below 85% .1 days discharge no sun !

So if I could construct a water battery bank what size would I need to replenish 15% of a 2100 ah bb at 24 v:confused:

Comments

  • icarusicarus Solar Expert Posts: 5,436 ✭✭✭✭
    Re: Im sure we have been here b4

    A REALLY big one!!!

    Icarus
  • nigtomdawnigtomdaw Solar Expert Posts: 705 ✭✭
    Re: Im sure we have been here b4

    Icarus lets not fly to close to the sun I can easily build a 1000 or 2000 or 3000 gallon tank so go figure a really big one. chum !Need sensible replys
  • BB.BB. Super Moderators, Administrators Posts: 32,921 admin
    Re: Im sure we have been here b4

    Well, here is a nice website that gives some numbers and has a few different turbines for sale (don't know anything about the site)... It will give you some idea of how much drop and how much water you will need for your "wet" battery...
    [FONT=Times New Roman,Times New Roman PS]To get an idea about available power in watts, multiply the head in feet, times flow in GPM, times 0.18 times turbine efficiency. Turbine efficiency ranges from 25% to 50%, with higher efficiency at higher heads. To get a rough idea, use 0.30 (representing 30%) as a multiplier for efficiency.[/FONT]
    Watts = Head (ft) * GPM * 0.18 (magic number) * turbine eff

    or

    GPM= Watts / (Head (ft) * 0.18 (magic number) * turbine eff)

    Assume 2 kWhrs over 10 hours (200 watts), 15 foot head, 30% turb eff...

    GPM = 200 watts / (15ft*0.18*0.30) = 247 gallons per minute

    10 hours of storage:

    Storage = 247 GPM * 10 hours * 60m/h = 148,200 gallons of water per 2kWhrs of storage at 15 foot of head... (if I got all of my numbers right)...

    The last part is how much pumping loss to raise the water (can you hit 80% efficiency?)...

    -Bill

    PS: I am not sure you can push that much water through a small turbine (nozzle diameter limit)--may not be achievable or you may need something like 5 or more small turbines to get 200 watts of power...

    PPS: Here is a generator for low head situations and pretty high output (linked from above page)... Perhaps this is more efficient and will cut down on your GPM requirements...
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
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