Question on volumes of water, excess mass, potential energy and power

AI Thread Summary
The discussion revolves around calculating the excess volume, mass, potential energy, and power related to water in a harbour during high and low tides. The user initially assumed a low tide of 1.5m, leading to calculations that resulted in large numbers, which raised concerns about accuracy. Key errors identified include incorrect unit conversions, particularly in calculating volume, as well as the need to clarify assumptions made during calculations. It is suggested to consider low tide as 0m for simplicity and to ensure proper unit handling throughout the calculations. Overall, the user is encouraged to focus on unit consistency and to clearly state assumptions for better accuracy in their calculations.
shiftthatbogi
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Ive done this question but from the answers I am getting I can see that I am going wrong, but I don't know where. There are lots of different parts to this question, I will show them all, followed by my attempt at working them out.

A Harbour has an average water covered area of approx. 10km^2. On a particular day there is a 4m tide. Making clear any assumptions that you make calculate:

a) The excess volume of water in the harbour at high tide compared with low tide:

Here I didnt know whether I had to assume my own low tide? So I did and assumed that
low tide was 1.5m...if I am being stupid and that's wrong please let me know, so I worked
out

Volume of water at high tide = 10km x 4m = 10000m x 4m = 40,000m^3

Volume of water at low tide = 10km x 1.5m = 15,000m^3

Excess volume of water = 40,000 - 15,000 = 25,000m^3

b) The excess mass of water in the harbour at high tide compared with low tide:

Mass=density x volume = 1027 x 25,000 = 25,675,000 kg

c) The increase in potential energy of the water in the harbour

This is where my numbers start getting rediculously big, making me think I must have
gone wrong somewhere

Potential energy at low tide = mass x g x height = 15,405,000 x 9.8 x 1.5 = 226,453,500

Potential energy at high tide = 41,080,000 x 9.8 x 4 = 1,610,336,000

Increase = 1,610,336,000-226,453,500= 1,383,882,500 J

d) If all of this energy could be extracted by a generator, how much power could be
produced:

Because all my numbers before this step are so big, I don't even know where to start with
this question.

Please help me by telling me where and why I am going wrong with this, I really can't work it out. Thanks alot
 
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The numbers are even larger than you think. 10km^2=10*(1000m)^2=10^7m^2. 10km^2 of water piled to meter high depths is a LOT of water. Notice you are doing stuff like "10km x 4m = 10000m x 4m = 40,000m^3". The units don't work out there. m*m=m^2, not m^3. You could probably just assume low tide=0m. But this whole thing is just an order of magnitude estimate anyway. You are basically doing ok. Just watch the units. And as they said, state the assumptions you are making, which you already doing pretty well.
 
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