Tidal Power Generation

Discussions on classical and modern physics, quantum mechanics, particle physics, thermodynamics, general and special relativity, etc.

Tidal Power Generation

Postby elliottr1 on May 7th, 2007, 2:51 pm 

Hi, I am new to this forum but have had a quick look around and am impressed at the level of knowledge out there. My initial question may not be much of a challenge to some of you, but I expect to research this subject deeply and hope to push the bounds of knowledge in this area in doing so.

I am looking for a formula to calculate the electrical power that can be generated from a tidal source when a lagoon type system is built off-shore to trap water for release on a falling tide and to hold water out ready to flow in on a rising tide. Flow is generated by creating a latency at each side of high and low tide so as to create a head of water. I have gathered a simple formula from Internet research, but am looking for a more accurate method of measurement.

The formula I have currently is: flow (metres cubed/s) * head (metres) * efficiency of turbine (59%).
elliottr1
Forum Neophyte
 
Posts: 6
Joined: 07 May 2007
Location: South Wales


tidal energy

Postby megachristo on May 7th, 2007, 7:49 pm 

A system was developed and patented a year or so ago. The system was designed to work with a buoys travel up and down the wave. If you can find that their might be some info their. Also look here maybe you can find something. http://www.oceanrenewable.com/.
megachristo
Member
 
Posts: 131
Joined: 29 Apr 2007
Location: Usa, South florida


Postby Nick on May 7th, 2007, 7:58 pm 

megachristo, I think that what elliottr1 has in mind is different from the wave power generation you are thinking of. The method that elliottr1 has in mind is to place a barrier in tidal waters. This stops the tide either coming on or going out, building up a head of water that can be used for energy generation akin to a normal dam.

As for an equation for the power available I will have to have a little think....
User avatar
Nick
Resident Expert
 
Posts: 1954
Joined: 18 Jun 2006
Location: Oxford
Blog: View Blog (4)


Postby elliottr1 on May 7th, 2007, 8:40 pm 

Antisocial Boris - you have gathered the idea correctly - I look forward to your further thoughts.

Megachristo - thanks for your response but you have misunderstood the idea I am pursuing, but please grasp the idea and see if you can help further. If you have any questions let me know.
elliottr1
Forum Neophyte
 
Posts: 6
Joined: 07 May 2007
Location: South Wales


Postby Heron on May 8th, 2007, 8:30 am 

head*flow*eff? doesnt give J/s?

This should do it with out looking in my ref.
Power = npghQ
Where n = turbine efficiency
Q= flow rate per sec m^3/s
pgh = potential energy per unit volume.
Note h = head in meters.
p = 10^3 kgm^-3 for water.
g= 9.81 m/s^2

And pay close attention to the units.
This will only predict max power. You'll want a design that allows for maximum rate of flow. Since power will be dependent on the flow rate.
Last edited by Heron on May 8th, 2007, 8:47 am, edited 1 time in total.
User avatar
Heron
Member
 
Posts: 400
Joined: 20 Nov 2006
Location: Near Oakridge but not close enough to glow.


Postby Heron on May 8th, 2007, 8:35 am 

Hmmm which leads to the best way to calculate q ..let me respond at dinner.
User avatar
Heron
Member
 
Posts: 400
Joined: 20 Nov 2006
Location: Near Oakridge but not close enough to glow.


Postby elliottr1 on May 8th, 2007, 11:58 am 

Heron - Thanks for the formula - looks good. No doubt when looking at electrical power supply, this formula could be extended to include loss of power through transfer distance and transformation etc.

For an example I am thinking of a 10 square Km lagoon with an average tidal range of 5 metres, and a head of 2.5 metres running for a 12 hour period per day.

How would you see the flow rate being calculated for this example. Hope this gives a pointer for you?

I am looking for the units of measurement of Max Power to be in MW in order to perform a comparison with other systems.

Have a good dinner.
elliottr1
Forum Neophyte
 
Posts: 6
Joined: 07 May 2007
Location: South Wales


Postby Nick on May 8th, 2007, 12:05 pm 

Hmm, flow rates can be very tricky to calculate, fluid dynamics isnt the easiest thing in the world. You need to know more than just the head, you need to know about what exactly the water will be flowing through. Say you know it is going to be flowing through a circular pipe of known radius then you can work out whether you expect planar or terminal flow for a given pressure. You cross your fingers, hope it is planar and you might be able to get the mean flow rate through the tube fairly simply. You find out it is turbulent and life is a lot trickier I think (I havent actually done any of this in a few of years).

This situation is obviously a little more complicated than just a pipe, because you have to generate power somehow so it can get pretty hard. Basically the flow rate is gonna depend on the design on your generator.
User avatar
Nick
Resident Expert
 
Posts: 1954
Joined: 18 Jun 2006
Location: Oxford
Blog: View Blog (4)


Postby Heron on May 8th, 2007, 12:58 pm 

elliottr1 wrote:Heron - Thanks for the formula - looks good. No doubt when looking at electrical power supply, this formula could be extended to include loss of power through transfer distance and transformation etc.

For an example I am thinking of a 10 square Km lagoon with an average tidal range of 5 metres, and a head of 2.5 metres running for a 12 hour period per day.

How would you see the flow rate being calculated for this example. Hope this gives a pointer for you?

I am looking for the units of measurement of Max Power to be in MW in order to perform a comparison with other systems.

Have a good dinner.

Boris listed some of the problems with calculating flow rates. A hypothetical calculation could be made given the dimensions of the penstock (inlet.) But it will give only a max value.
There is also the matter of the generator which will vary the flow output as its electrical load changes.
User avatar
Heron
Member
 
Posts: 400
Joined: 20 Nov 2006
Location: Near Oakridge but not close enough to glow.


Postby Heron on May 8th, 2007, 4:31 pm 

Heron wrote:head*flow*eff? doesnt give J/s?

This should do it with out looking in my ref.
Power = npghQ
Where n = turbine efficiency
Q= flow rate per sec m^3/s
pgh = potential energy per unit volume.
Note h = head in meters.
p = 10^3 kgm^-3 for water.
g= 9.81 m/s^2

And pay close attention to the units.
This will only predict max power. You'll want a design that allows for maximum rate of flow. Since power will be dependent on the flow rate.


I'm afraid I'm wrong here. I was thinking about how a tidal lagoon would work and the head isn't constant. Therefore P = np(delta)hg(delta)Q
You will still need to supply more dimensions to do any ballpark calculations.
User avatar
Heron
Member
 
Posts: 400
Joined: 20 Nov 2006
Location: Near Oakridge but not close enough to glow.


Postby elliottr1 on May 11th, 2007, 7:46 pm 

OK got that and it works well. Now to extend the problem slightly.

What would the impact on max power potential if I was to divide the lagoon into 2 sections and allow the water to flow through the first into the second and then in reverse on the ebb tide? Would it be possible to generate more than the max power that could be generated if the lagoon was not divided. If not then what are the limiting factors?
elliottr1
Forum Neophyte
 
Posts: 6
Joined: 07 May 2007
Location: South Wales


Postby elliottr1 on May 14th, 2007, 12:39 pm 

The head of the lagoon could be constant if it was only running for 50% of the time and was turned on and off at the half tide mark, when rising and falling.
elliottr1
Forum Neophyte
 
Posts: 6
Joined: 07 May 2007
Location: South Wales


Postby Heron on May 14th, 2007, 12:57 pm 

elliottr1 wrote:OK got that and it works well. Now to extend the problem slightly.

What would the impact on max power potential if I was to divide the lagoon into 2 sections and allow the water to flow through the first into the second and then in reverse on the ebb tide? Would it be possible to generate more than the max power that could be generated if the lagoon was not divided. If not then what are the limiting factors?


Potential energy is still the same in the above scenario with a divided chamber.
The advantage would be continious power generation.
User avatar
Heron
Member
 
Posts: 400
Joined: 20 Nov 2006
Location: Near Oakridge but not close enough to glow.


Postby elliottr1 on May 23rd, 2007, 10:01 am 

If each section had a turbine that generated power as the water flowed into and out of each section, then I think you will find that the dynamics change completely and that a greater energy potential is available. What is the view on this?
elliottr1
Forum Neophyte
 
Posts: 6
Joined: 07 May 2007
Location: South Wales


Postby Rettaw on May 23rd, 2007, 10:40 am 

well as Heron said the potential energy, ie the maximum allowed by theory, is still the same, any gain in production would depend on the exact build of turbine and transformators and such...
User avatar
Rettaw
Chatroom Operator
 
Posts: 1213
Joined: 31 Mar 2005
Location: Sweden


Re: Tidal Power Generation

Postby rossweinberg21 on April 9th, 2018, 1:41 am 

Thanks for the formula, Heron! It looks great. I am a newbie in this forum but there was a great discussion here. Lots of information I gained.
rossweinberg21
Forum Neophyte
 
Posts: 3
Joined: 05 Mar 2018


Re: Tidal Power Generation

Postby Braininvat on April 9th, 2018, 9:12 am 

The OP's last login was six years ago, so he/she may not receive your reply. You could try responding via email, at their profile page.

Be interesting if a thread that's dormant for 11 years will revive. It has happened here.
User avatar
Braininvat
Forum Administrator
 
Posts: 6483
Joined: 21 Jan 2014
Location: Black Hills


Re: Tidal Power Generation

Postby Watson on April 9th, 2018, 9:27 am 

I think it depends on the length, width and dept of the connecting channel. I see a synergy of the greater whole? The whole is greater than the sum of the two parts?

Yes I saw that.
User avatar
Watson
Resident Member
 
Posts: 4605
Joined: 19 Apr 2009
Location: Earth, middle of the top half, but only briefly each 24 hours.



Return to Physics

Who is online

Users browsing this forum: No registered users and 5 guests