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

Answer:

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

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:

$\Delta h = \frac{\dot W}{\rho \cdot \dot V \cdot g}$

Given that $\dot W = 38.5\times 10^{6}\,W$, $\rho = 997\,\frac{kg}{m^{3}}$, $\dot V = 3290\,\frac{m^{3}}{s}$ and $g = 9.807\,\frac{m}{s^{2}}$, the height change is:

$\Delta h = \frac{38.5\times 10^{6}\,W}{\left(997\,\frac{kg}{m^{3}} \right)\cdot \left(3290\,\frac{m^{3}}{s} \right) \cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta h = 1.197\,m$

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