5KWhr pumped storage 'kit' detail?

 How about a pumped storage kit that has 1KW of PV panels, a pump, two
big water tanks, and a pelton wheel water turbine. A 'solar battery'.
It pumps water uphill for 5 or 6 hrs using the 1KW PV, and at night
you could have 1KW for 5 hrs, or 5KW for 1hr. (Run the fridge all
night or something). So my question is: how big do the tanks need to
be? I guess the top tank has mgh joules of energy? Whats that in
gallons or tank depth and diameter? And about how much would all that
equipment cost, not counting the site prep (digging a big hole for one
tank, putting the other one on the big pile of dirt)  and all this for
75 cents worth of electricity every night? Well, it might be useful in
some very remote location. All the equipment seems simple and rugged.

 
Just use two L16H batteries at 2.2 kWh per battery. About 80 percent
charge eff using MPP controller.
One kW of panels will be 10 square meters, that will net about 4 kWh
electrical energy per day at a good solar site in December/January.
Capital cost about 500 dollars using proven techmology. For an average
solar site you would need more batteries in December/January.

Your system efficiency will be terrible, and would be enourmously
larger than two batteries
Therefore the capital cost is not even worth calculating.

 What would be inefficient? The pump impeller? Pump motor could be very
efficient. The pelton wheel turbine is pretty efficient. The turbine
could spin a pma alternator that is pretty efficient. The water and
the tanks wont wear out like a couple of batteries would. Can you pin
down 'enormously larger than two batteries' a little more
quantitatively? Like tank depth and diameter for 5KWhrs? Volume of
equipment doesnt count... its a fixed installation. Help me with the
numbers folks.

 BobG wrote:

Well, I imagine that there will be some friction loss in the pipes,
pumps and other plumbing in addition to the losses in the motors
and generators. If the tanks have open tops then there are losses
due to evaporation.

How big for 5KWHrs? I seem to recall doing this calculation for
concrete just recently over at alt.solar.photovoltaic and sometime
in the past using railroad locomotives but we can do it for water
just as easily.

If I might quote from myself...
: The formula we are looking for here is energy = force x distance.
: http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html
: In this case we could use kg x m x 9.8 = joules
: 1 watt-hour is 3600 joules so we could go further and use...
: kg x m x 9.8 / 3600 = watt-hours (wh)

So, for 5000 watt-hours we would need... (showing my work...)
kg x m x 9.8 / 3600 = 5000 watt-hours
kg x m = 5000 / 9.8 x 3600
kg x m = 1,836,734

Any combination of kg x m that comes out to 1,836,734 (or so).
The density of pure water is 1000 kg/m3 and the volume of a
cylinder is pi x r^2 x h (r=radius, h=height).

Let's make the math simple and say that the one tank is 10m
higher than the other. This means we'll need a volume of water
of (1,836,734 / 1000 / 10) 184 m^3. Let's further say that
the water tanks are only 3m (9.9 feet) tall. That makes them
some 4.42m in radius (8.84m diameter, 29 feet).

So, your tanks need to be roughly 30 feet in diameter, 10 feet
tall and the higher one is roughly 3 stories up in the air.
This is assuming that there are no losses.

Two L16H style batteries store almost 5kWhs and each battery
measures 295 x 178 x 424 mm (11 5/8 x 7 x 16 3/4 inches).

Anthony

 On Aug 11, 3:43?am, Anthony Matonak

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Yep, those are some big tanks. I guess if you put those big ol boogers
in your back yard, the zoning inspector would write you a ticket for
having an unauthorized tank farm. You bring the neighbors over to see
your alternative energy system and brag that "this produced 75 cents
of electricity per day!" they'd stop coming around and bothering you.
I guess a city with some hills might be able to use something like
this for a peaking device scaled up appropriately. Thanks for the help
with the numbers.