# Homework Help: Marine power

1. May 25, 2013

### thebest99

A tidal barrage contains an area of of sea of 200km2 and the tidal range 6m at spring tides and 2m at neap tides.

What would be the average power output over the year?

Any help would be great,

2. May 25, 2013

### lewando

You can start by coming up with the tidal activity during a lunar cycle based on the definitions of spring and neap tides.

3. May 27, 2013

### thebest99

What I have worked out: ep=mgh

P2hgh/6x60x60

1025x2x2x9.81x2/ 6x60x60

=745gw

4. May 27, 2013

### lewando

I get something significantly different. If you provide more detail in your reasoning and calculations (units), I could better understand how we have such different results. I can tell you that your equation for potential energy is suspect.

5. May 27, 2013

### thebest99

Firstly thanks for your help. I have 1025 as my mass. 9.81gravity. 4m difference in height. PE=mgh.

This is what we had I'm class:

1025x2x2x9.81/6x60x60

200x10 to the power of 6.

Slightly confused how they got this.

Please could you help work through

6. May 27, 2013

### thebest99

4x1025x200x10to the power6x 4x9.81x2
=64353.6x10 to the power 9 J

7. May 27, 2013

### lewando

You need to start (and never stop) using units in your calculations. PE = mgh only applies if all the mass is at h. With a volume of water, mass is distributed along h [edit: from 0 to h]. "1025", by itself, is meaningless-- certainly not mass. 1025 kg/m3 is the density of seawater.

Last edited: May 27, 2013
8. May 29, 2013

### thebest99

Could you show me how you work this out then please

9. May 29, 2013

### lewando

Can't do that due to the forum's rules. You need to do the work. But anyway, let's work with what you have posted:
Potential energy for a column of water is not mgh. It is less than this. Water at the bottom of the container does not contribute as much as water at the top. But you are correct to look for the potential energy. Try to research this or at least think about it. After you have done that, you can look here to confirm:
PEwater_column = mgh/2
I was hoping you would say something like "to find average power, I need to find the energy collected over a specific interval and divide that energy by that time interval". Would you agree with that? If so, what specific interval are you considering (and why)? Anyway, it looks like you are trying to divide energy by time to get power (average power). But lack of units make that very unclear. I get that P is density of saltwater, but how do you get "P2hgh"? And what about "6x60x60"--why would you use 6? Again: use units.

Well I get 3.72 with no units, so please re-work this.

One final note: rather than banging out any more unitless expressions, why not take some time and elaborate on your method of extracting energy from the ocean. That way we can tell if your on the right track or not.