Question

The South Holston Dam uses hydroelectric energy to produce 38,500 kilowatts of power. If the density of water is 997 kg/m^3 and the dam allows 3,290 m^3/s of water through. How high does the waterfall from to produce that much power a second

Answer 1

The waterfall must 1.197 meters high to produce the abovementioned power in a second.

Step-by-step explanation:

Let suppose that South Holston Dam is a conservative system, that is, there is no non-conservative forces that dissipates energy. According to the Principle of Energy Conservation, final kinetic energy equals to initial gravitational potential energy, whose form per unit time is:

`\dot W = \dot K = \dot U_(g)`

Where:

`\dot W` - Power, measured in watts.

`\dot K` - Instantaneous change in kinetic energy in time, measured in watts.

`\dot U_(g)` - Instantaneous change in gravitational potential energy in time, measured in watts.

Now, let suppose that final speed is constant, so that:

`\dot W = \dot m \cdot g \cdot \Delta h`

`\dot W = \rho \cdot \dot V \cdot g \cdot \Delta h`

Where:

`\dot m` - Mass flow, measured in kilograms per second.

`\dot V` - Volume flow, measured in cubic meters per second.

`g` - Gravitational constant, measured in meters per square second.

`\Delta h` - Height change, measured in meters.

The change is finally cleared in the latter equation:

`\Delta h = (\dot W)/(\rho \cdot \dot V \cdot g)`

Given that
`\dot W = 38.5* 10^(6)\,W`,
`\rho = 997\,(kg)/(m^(3))`,
`\dot V = 3290\,(m^(3))/(s)` and
`g = 9.807\,(m)/(s^(2))`, the height change is:

`\Delta h = (38.5* 10^(6)\,W)/(\left(997\,(kg)/(m^(3)) \right)\cdot \left(3290\,(m^(3))/(s) \right) \cdot \left(9.807\,(m)/(s^(2)) \right))`

`\Delta h = 1.197\,m`

The waterfall must 1.197 meters high to produce the abovementioned power in a second.