# Power problem with water over the dam - why not Fv?

1. Aug 14, 2010

### Ryker

1. The problem statement, all variables and given/known data
$$10m^{3}$$ of water flows over a 50m high dam every second. What's the maximum power of the falling water at the bottom of the dam?

2. Relevant equations
$$W = mgh$$

$$P = \frac{\Delta W}{\Delta t}$$

$$P = Fv$$ - apparently not

3. The attempt at a solution
I used the equation

$$P = \frac{mgh}{\Delta t} = \frac{10kg\times 9.8\frac{m}{s^{2}}\times 50m}{1s} = 4,9 MJ$$

But why isn't it possible to use $$P = Fv = mgv$$ with $$m = 10000kg$$ and $$v = \sqrt{2gh}$$. I mean, say you were to drop a block, holding 10000kgs of water from the same dam. It would hit the bottom with $$v = 31,3\frac{m}{s}$$. But the power calculated this way is smaller than power calculated above? Why is that? The difference can't be the speed, since the water is hitting the bottom with the speed mentioned and in one second 10000kgs of it do that.

What am I not seeing here and why the difference?

Last edited: Aug 14, 2010
2. Aug 14, 2010

### javierR

Good question. It's a bit more subtle than at first sight. The idea you used to solve the problem the first way is to say a mass m=(density)*(Volume flow rate) is delivered by the dam in one second. Ignoring any other water delivered after that, that mass m of water then gains kinetic energy (work is done on it) and you calculated that for that 1 second of delivery, (at most) mgh energy comes out as useful for geneating electricity. I.E., the power is the energy delivered divided by the time to deliver it.
As for the alternative approach you were wondering about, the power would be calculated from Fv = mgv, where v is the speed of *any* object falling down the 50 m dam. The problem would then be to determine how much mass to enter here...but in the setup we were essentially given a mass delivery RATE: kg/s...the problem is to figure out how much mass to put in there...that's the only thing that the power will depend on using mgv, and we don't know how much o fit to put in since there's a constant flow of it. The tradeoff that has occured is now the time units are stuck inside of the "v" factor. (In other words, m=(density water)*(Volume flow rate)*(some time interval). If you just try to say "t=1 sec" and use P=mgv, you've snuck in a time amount, when the time issues should really be handled by "v" since it has units of m/sec).
So the given data is not useful for this latter approach using Fv. By just using Work=mgh, you were able to sneak in time units by working with mass RATE kg/s (instead of just the mass m) immediately giving units of Power.
Hope that helps.

3. Aug 15, 2010

### hikaru1221

I think javierR's explanation for the 2nd approach doesn't go to the core of the problem. F in P=Fv is not simply mg (even if you drop a single block down onto the bottom, not to mention the complicated case of continuous water flow). F is greater than mg, and may even be way greater, if the height is so high. When the block falls down onto the bottom, the job of the bottom is to exert an impulse to stop the block (or to reduce the block's momentum to zero), i.e. F=dp/dt. After that, when the block's speed is already zero, if the block remains on the bottom, the force on the bottom = mg; otherwise, if the block goes away to somewhere else (which is the case of water), the block no longer exerts any force on the bottom, and the bottom waits for another block to come and stops it and so on.

But there is another problem: v in P=Fv is not the velocity of the block (or water). It is the velocity of the bottom, since P is the mechanical energy rate that the bottom (or turbine) receives. If the water falls onto the ground, v=0 and so, the ground receives zero mechanical energy, and it doesn't move. This is intuitive, right? So all of the energy of water goes to heat: the temperatures of both water and the ground rise. However, in the case of the dam, it's simply impossible to determine v with so few information.

Some more things about the power: We have 2 equations:
(1): $$P=\frac{\Delta E}{\Delta t}=\frac{\Delta (mgh)}{\Delta t}$$
(I write E instead of W to discriminate energy and work)
(2): $$P=Fv$$
There is a difference between P in (1) and P in (2).
1/ P in (1) is deduced from the condition "maximum power". In order to achieve max power, all of the energy of water must goes to the bottom. So in 1 second, the bottom receives an amount of energy equal to the energy of the water E=mgh. So P in (1) means the max energy rate that the bottom receives.
2/ P in (2) is actually the amount of mechanical energy that the bottom receives in 1 second. You know that the mechanical work = Fds. If we divide that by the time dt, we get: Fds/dt = Fv = P. So P in (2) arises from mechanical work, which corresponds to the mechanical energy. But energy can be thermal energy, not just mechanical energy.

4. Aug 17, 2010

### Ryker

Thanks to both for the thorough explanation!