Now, let's look into how much energy could be stored.
The potential energy stored is Mgh:
where g is the acceleration due to gravity.
h is the height of the high lake above the low lake.
M is the mass of water, which is evaluated by multiplying the volume of one lake times the density of fresh water (1,000 kg per cubic meter).
We'll also introduce an efficiency factor because the system won't be perfectly efficient.
Inputs | ||
Population | people | |
Average Power Per Person | Watts | |
Lake Radius (r) | meters | |
Lake Depth (d) | meters | |
Altitude Difference (h) | meters | |
Efficiency | % | |
Number of Lake Pairs | ||
Results | ||
Annual Energy Consumption | TW hours / year | |
Energy Stored | TW hours | |
Days of Energy Stored | days. |
If we were to use tidal power, which goes through almost 2 complete cycles per day, then we may need about half a day's worth of demand to be stored, which may be achievable with a series of pumped storage power stations. On the other hand, if we were to try to rely on wind as the sole energy source and we wanted to store enough energy to see us through say six weeks of calm weather, then the number of pumped storage power stations required is likely to be unacceptably large.
Now suppose we choose to harness tidal power. One approach would be to put a sea-wall across the entrance to some bays or inlets. Let the water flow in as the tide rises. The sea water is then blocked as the tide starts to ebb. To extract a maximum amount of energy from this, we would wait until the tide at its lowest point and then instantaneously release all the sea water, passing it through a generator. But it is likely to be much more practical to release the water before low-tide. Also if we want to use much of the electricity as it is being used, then we would want the generation of electricity to continue through much of the tidal cycle.
We can set a target of what proportion of our energy needs we want to come from tidal and then we can work out what area of the sea needs to be put behind sea walls with tidal generators.
In a 27 day lunar cycle we have 2 tides per day due to the spin of the earth, less 2 tides due to orbit of the moon. For now we won't go into the details of why that's true. But our statement here is that we have 2 * ( 27 - 1 ) tides in 27 days, i.e. 1.93 tides per day. So a tidal cycle is (24 / 1.93) hours = 12.44 hours. But we can generate power both as the tide flows in and as the tide flows out. So we'll use the half tidal period which is a little over 6 hours.
The target amount of energy to generate in a half tidal cycle is:
population * powerPerPerson * durationOfHalfTidalCycle * targetTidalPowerProportion.
On the other hand we see how much energy is available from the tidal movement using the equation:
Potential Energy: E= Mgh
In this case M is the mass of the sea water which is the area times the tidal height times the density of sea water (approximately 1030 kg per cubic meter).
So we can evaluate the area of sea that is required:
Inputs | ||
Tidal Height | meters | |
Tidal Power Efficiency | % | |
Proportion of Total Energy from Tidal | % | |
Result | ||
Sea Area Required | square KM |
Tidal power is something that can provide predictable renewable energy. However, a significan area of sea needs to be harnessed.