can someone explain the water power equation in the IB physics study guide?
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project 4 for physics watch
- Thread Starter
- 29-01-2010 16:32
- 30-01-2010 04:41
Max power available = (1/2)A^(2)Lpgv
A is area. L is length of the wavefront. Rho (funny looking p) is density. g is gravitational field strength. v is the wave's velocity.
It's derived from PE=mgh
Mass is calculated by density (p) times volume [(λ/2) x A x L]. NB: this is the volume of the shaded bit across L. h is just A which explains the A^2.
This gives PE=(λ/2)A^(2)Lg
Power is energy per unit time, so we divide the above by time to get the final term. This is the same as multiplying by the waves frequency (f). Now f=v/λ, so the lambdas cancel and we are left with the power equation.
It's a bit of a rubbish model in my opinion. Everyone knows waves aren't square. Also I can't see how they can justify just using half of one trough of a wave (the shaded bit). Surely the potential energy is spread across the whole wave? I guess if it's experimentally consistent then fair enough, but I've seen no experiment that verifies it nor can I think of a way to test it to a suitable degree of uncertainty.