- $9.56 \times 10^8 \textrm{ W}$
- $1.41$

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This is College Physics Answers with Shaun Dychko. Since the Bernoulli's Equation has units of Joules per cubic meter, that means if we multiply by the flow rate, which has units of cubic meters per second, we'll end up with Joules per second. Let me show that a bit better. So we have Joules per cubic meter from Bernoulli's Equation, so which of these terms has those units. And then, we're going to be multiplying by <i>Q</i> which is cubic meters per second. And the cubic meters will cancel leaving us with Joules per second which is abbreviated as Watts. And so we're getting power by multiplying Bernoulli's Equation by the flow rate. And now, each of this terms in here represents differences between two different points. So, in the Hoover Dam we're looking for the difference in pressure between the surface of the water and where it exits the dam at a depth of 150 meters. Now, both of those locations will have the same atmospheric pressure. And so the pressure difference is zero. And we'll assume that where the water is entering, the hole in the dam, it's going to have some speed. But at the very entrance its speed will be zero. And so <i>V</i> is zero. And it's just this difference in height then that's going to result in the power. So the power is going to be the density of water times <i>g</i> times the height difference between the surface of the water and the hole in the dam times the volume flow rate. So that's one times ten to three kilograms per cubic meters times 9.81 meters per second squared times 150 meters depth times 650 cubic meters per second of volume flow rate giving us 9.56 times ten to the eight Watts. And then we take that total possible power. This is the theoretical maximum power you could get from this water and divided by the actual power output in electricity of the dam. We got a ratio of 1.41 which actually shows fairly good efficiency. 1.0 would be perfect efficiency.