Question

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each second. The water is released 220 m below the top of the reservoir.If the generators that the dam employs are 90% efficient, what is the maximum possible electric power output?Express your answer with the appropriate units.

Answer 1

The maximum electric power output is
`P_(max) =1.339*10^(9) \ W`

Step-by-step explanation:

From the question we are told that

The capacity of the hydroelectric plant is
`(V)/(t) = 690 \ m^3 /s`

The level at which water is been released is
`h = 220 \ m`

The efficiency is
`\eta =`0.90

The electric power output is mathematically represented as

`P = (PE_l - PE _o)/(t)`

Where
`PE_l` is the potential energy at level h which is mathematically evaluated as

`PE_l = mgh`

and
`PE_o` is the potential energy at ground level which is mathematically evaluated as

`PE_o = mg(0)`

`PE_o = 0`

So

`P = (mgh)/(t)`

here
`m = V * \rho`

where V is volume and
`\rho` is density of water whose value is
`\rho = 1000 kg/m^3`

So

`P = ((\rho * V) * gh)/(t)`

`P = (V)/(t) * gh \rho`

substituting values

`P =690 * 9.8 * 220 * 1000`

`P =1.488*10^(9) \ W`

The maximum possible electric power output is

`P_(max) = P * \eta`

substituting values

`P_(max) =1.488*10^(9) * 0.90`

`P_(max) =1.339*10^(9) \ W`