GRA said:
They're a lot more expensive, especially as the storage and/or peak load requirement rises, than including a genset. PV and wind costs have dropped dramatically since I was doing this, but storage costs haven't.
For pumped storage, the size of the battery is controlled by the size of the reservoirs. Let's see how big of a battery they have.
In this case, both of the reservoir sizes have been reduced, with the lower reservoir ending at 150,000 m^3. Assuming you have access to all of that water at 700 meters of height, you have a storage potential of:
Energy stored = (150,000 m^3 * 1000 liters/m^3 * 1 kg/liter) * 9.8 m/sec^2 * 700 m = 1 TJ * 1 MWh/3.6GJ = ~290 MWh
Assuming about 50% efficiency can be converted into electricity, that provides a battery of about 150 MWh.
So, how long can that much energy run the island?
According to the article, the peak power is about 7.5 MW and minimum power is about 3 MW. Let's assume the average power level is 5 MW. So the storage on the island is enough for about 30 hours of storage with NO WIND.
If you can drain the lower pool, then perhaps you could use most of the water in the top pool, which would double this amount to 60 hours. Perhaps higher efficiencies are possible, maybe getting you to a maximum of four days. That's probably the outside edge of the total storage if the upper reservoir is full.
So, I'll have to agree this is a fairly small amount of storage to try to achieve 100% renewable generation.
TimLee said:
RegGuheert said:
... (I'm guessing that six months is needed mainly to pump enough water into the reservoir to have a large amount of energy stored for future use.)...
That seems a bit unlikely.
US pumped storage volume at TVA and Virginia Power is at most a few days.
If reservoir is also for irrigation, might be weeks but six months seems unlikely.
Let's remember that this water is being pumped uphill by wind power with a MAXIMUM output of 11.5 MW and the ability to pump up to 6 MW when the wind blows. After supplying half of the island's needs, lets assume that there is an average amount of power left over for pumping of 3 MW. So, how long would it take to pump 380,000 m^3 of water at 66% efficiency?
Energy to fill lake = (380,000 m^3 * 1000 l/m^3 * 1 kg/l) * 9.8 m/sec^2 * 700 / 0.66 = ~4 TJ * 1 MWh/3.6GJ = 1100 MWh
At 3 MW, that would take 365 hours or about 15 days.
So, you are correct, it only takes weeks. The six months must be for the purposes of testing and learning to operate the system (or some other purpose).
Anyone see any math or physics errors in the above?
BTW, at Smith Mountain Lake, they can generate up to 600 MW and they can lower the lake level a couple of feet in just a few minutes! I've only ever seen it used for a few minutes at a time.