Question

Hoover Dam on the Colorado River is the highest dam in the United States at 221 m, with an output of 680 MW. The dam generates electricity with water taken from a depth of 150 m and an average flow rate of 650m3/s. (a) Calculate the power in this flow. (b) What is the ratio of this power to the facility’s average of 680 MW?

Answer 1

the avalable power \(P_A\), is given by the subsequent formula:
\[\Large{P_A} = \frac{{\delta gQH}}{{{{10}^6}}}\left( {MWatt} \right)\]
where \(\delta=10^3\;Kg/m^3\) is the water density, \(g=9.8\;m/sec^2\) is gravity, \(Q\) is the flow rate, and \(H\) is the depth.
So, substituting numeric data, we get:
\[\Large {P_A} = \frac{{\delta gQH}}{{{{10}^6}}} = \frac{{1000 \cdot 9.8 \cdot 650 \cdot 150}}{{{{10}^6}}} = ...?MW\]
the requested ration \(\eta\), is:
\[\Large \eta = \frac{{{P_A}}}{{680}} = ...?\]

Answer 2

I tried this and I got the wrong answer. The number I came out with was 955500000 watts.

Answer 3

please you have to divide by \(10^6\) as I wrote in my first formula, so you get this:
\(P_a=955.5\) \(MW\)

Answer 4

so, we have:
\[\Large \eta = \frac{{{P_A}}}{{680}} = \frac{{955.5}}{{680}} = ...?\]

Answer 5

no it wants the answer in watts not MW

Answer 6

I asked my friend and all we had to do was round. So final answer was 956000000

Answer 7

it didn't say to round which was the annoying part.

Answer 8

thanks though!

Answer 9

ok! :)