Vertical ocean tidal movement power generation

  • Thread starter Paul Seaton
  • Start date
  • #1
Paul Seaton
2
0
TL;DR Summary
Alaska’s Cook Inlet with 2nd highest tide change in world at max about 30ft. 200 x 80 ft dry dock weight of 2000 tons. install series of double acting hydraulic cylinders to generators. dry dock held up/down 1 foot. How much energy could we get?
Designing harbor expansion in Homer Alaska’s Cook Inlet with Second highest tide change in world at max about 30 feet twice daily (minimum daily about 15feet) to include a 200 ft by 80 ft floating dry dock with gross weight of 2000 tons. Can we install serries of double acting hydraulic cylinders and pressure relief valves to hydraulic drive generators by holding down dry dock 1 foot on rise and holding up 1 foot on dropping tide. How much energy could we harvest.
 
Engineering news on Phys.org
  • #2
We'll need to know the buoyancy of the dry dock. i.e. is the 2Ktons displacement or gross weight? We also need to know the profile, like buoyancy vs. submerged height. Maybe post a drawing with materials identified?

For example, if your drydock is made out of water balloons, it takes no force to submerge it, even though it may weigh a lot. OTOH, it takes a lot of force to lift it out of the water. The opposite if it's made of air filled balloons.
 
  • #3
Rather than energy per unspecified, let's have a go at a crude power approximation that can be compared with other forms of generator.
200' x 80' delayed by 1'.
Area of piston = 200' * 80' = 16,000 sq ft = 1486.4 m2
Hydrostatic pressure of 1' of (fresh) water = 3.1 kPa
Say it sweeps a height of 20' up and down twice daily. Less 1' hysteresis.
( 20 - 2 ) x 4 = 72' per day = 22 metre per day.
22 m / 8640 sec = 2.55 mm/sec.
Volume per second = 0.00255 * 1486.4 = 3.79 m3/s.
Power in watts = volume * pressure = 3.79 * 3.1 kPa = 11.75 kW

Did I go wrong somewhere ?
Why delay 1' when you could delay 2', and so get more power.
 
Last edited:
  • #4
I don't see how this device harvests any energy, but we can set some limits. You have a mass of 2000 tons, you say, and it moves up and down 30 feet twice a day. I am going to call this 2 million kilograms, and 10 meters. So mgh is 200 MJ. You get this twice every 86400 s, for 4.6 kW.

That's what there is to extract. What you actually get will be no more.
 
  • #5
DaveE said:
We'll need to know the buoyancy of the dry dock. i.e. is the 2Ktons displacement or gross weight? We also need to know the profile, like buoyancy vs. submerged height. Maybe post a drawing with materials identified?

For example, if your drydock is made out of water balloons, it takes no force to submerge it, even though it may weigh a lot. OTOH, it takes a lot of force to lift it out of the water. The opposite if it's made of air filled balloons.
Dry dock is generally concrete with void spaces to flood or pump out. I am going with specs for smaller drydock in Hawaii which is 1972 gross tons. Of course if base is submerged say 10 foot then ship loaded and pumped out to make ship dry for work, it would have greater weight depending on vessel but could be double. I will attempt to get design specs for that existing unit but federal government may not comply. I am starting to understand that the energy is essentially related to the displaced water rather than the gross weight moved up and down with tide. The idea was to get force by delaying movement of gross weight throughout the tide cycle except a period at slack high or low tide change. The question of delaying by 2 ft instead of 1 ft means the period of slack about doubles before getting hydraulic pressure up to a setting for that movement delay.
A calculation of 4 to 5 kw total power available does not seem right. Although this would be near constant throughout the cycle it is just about the same during daylight as my personal 22 solar panel array.
Thanks to everyone for all your help!
 
  • #6
Paul Seaton said:
The question of delaying by 2 ft instead of 1 ft means the period of slack about doubles before getting hydraulic pressure up to a setting for that movement delay.
The duty cycle falls slightly, but the pressure is doubled, so more energy.
For 20' of movement, with 0' of hysteresis, pressure 0 * (20 - 0) = 00 units.
For 20' of movement, with 1' of hysteresis, pressure 1 * (20 - 2) = 18 units.
For 20' of movement, with 2' of hysteresis, pressure 2 * (20 - 4) = 32 units.
For 20' of movement, with 3' of hysteresis, pressure 3 * (20 - 6) = 42 units.
For 20' of movement, with 4' of hysteresis, pressure 4 * (20 - 8) = 48 units.
For 20' of movement, with 5' of hysteresis, pressure 5 * (20 - 10) = 50 units. ***
For 20' of movement, with 6' of hysteresis, pressure 6 * (20 - 12) = 48 units.
For 20' of movement, with 7' of hysteresis, pressure 7 * (20 - 14) = 42 units.
For 20' of movement, with 8' of hysteresis, pressure 8 * (20 - 16) = 32 units.
 
  • #7
Paul Seaton said:
TL;DR Summary: Alaska’s Cook Inlet with 2nd highest tide change in world at max about 30ft. 200 x 80 ft dry dock weight of 2000 tons. install series of double acting hydraulic cylinders to generators. dry dock held up/down 1 foot. How much energy could we get?

How much energy could we harvest.
I imagine that PF has not been your only port of call. Have you found any examples of projects similar to your idea? If it were a worthwhile idea then wouldn't you expect vertical tidal motion to be used all over the world for energy harvesting? Every ship that's moored to a quay in tidal waters could use this 'free energy' as it rises and falls relative to the quay. Your example seems to present a very attractive scheme, on the face of it.
There have been many successful new inventions so I cant say your idea wouldn't work but it's a massive project that would need a lot of justification for the sort of investment needed. A small scale test with a floating mass (private pleasure boat) moored against a quay would be a good start. The calculations would be easy.
 

Similar threads

  • Electrical Engineering
Replies
9
Views
3K
Back
Top