The Grand Coulee Dam is 1270 m long and 170 m high. The electrical power output from generators at its base is approximately 2000 MW. How many cubic meters of water must flow from the top of the dam per second to produce this amount of power if 92 % of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of water has a mass of 1000 kg.)

The length of the dam is L = 1270 m.

The height of the dam is h = 170.

The power output is P = 2000 MW.

The relationship between the power and work is given by,

$\mathit{P}\mathbf{=}\frac{\mathbf{W}\mathbf{\hspace{0.33em}}}{\mathbf{t}\mathbf{\hspace{0.33em}}}\mathbf{\hspace{0.33em}}\mathbf{}\mathbf{}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, P, is power, W is work done, and t is time.

The work done in the form of potential energy is given by,

W = mgh ........(2)

Convert the power output in watt.

$$\begin{gathered} \mathrm{P}=\left(2000\mathrm{MW}\right)\left(\frac{{10}^6\mathrm{W}}{1\mathrm{MW}}\right) \\ =2\times {10}^9\mathrm{W} \end{gathered}$$

It is given that 92% of the work done on water by gravity is converted to electrical energy.

The work done on the water is calculated as follows.

$$\begin{gathered} W=Pt \\ 0.92W=\left(2\times {10}^{9}W\right)\left(1s\right) \\ W=\frac{\left(2\times {10}^{9}W\right)}{0.92} \\ =2.174\times {10}^{9}J \end{gathered}$$

The mass of water flowing from the top of the dam per second is calculated as follows.

$$\begin{gathered} \mathrm{W}=\mathrm{mgh} \\ 2.174\times {10}^9\mathrm{J}=\mathrm{m}\left(9,8\mathrm{m}/{\mathrm{s}}^2\right)\left(170\mathrm{cm}\right) \\ \mathrm{m}=\frac{2.174\times {10}^9\mathrm{J}}{\left(9,8\mathrm{m}/{\mathrm{s}}^2\right)\left(170\mathrm{cm}\right)} \\ =1.3\times {10}^6\mathrm{kg} \end{gathered}$$

Find the volume of the water as follows.

$$\begin{gathered} V=\frac{m}{P_{water}} \\ =\frac{1.3\times {10}^{6}kg}{1000kg/m^3} \\ =1.3\times {10}^3{m}^3 \end{gathered}$$

Therefore, the required volume of water is $1.3\times {10}^3{m}^3$.