Question

(II) Water is stored in an artificial lake created by a dam (Fig. 15–27). The water depth is 48 m at the dam, and a steady flow rate of\({\bf{32}}\;{{\bf{m}}{\bf{3}}}{\bf{/s}}\)is maintained through hydroelectric turbines installed near the base of the dam. How much electrical power can be produced?

FIGURE 15-27 Problem 55

Short answer

The produced electric power is \(1.5 \times {107}\;{\rm{W}}\).

Step-by-step solution

Understanding the electric power

The electric energy transferred per unit time from the source is termed electric power. The SI unit of electric power is Watt. It is also defined as the rate of doing work.

The expression for electric power is given as:

\(P = \frac{W}{t}\)

Here, W is the work done and t is the time.

Given data

The depth of water is \(h = 48\;{\rm{m}}\).

The flow rate is \(\frac{V}{t} = 32\;{{\rm{m}}3}{\rm{/s}}\).

Evaluation of the electric power produced

The mass of the water is given by:

\(m = \rho V\)

Here, \(\rho \) is the density of water and V is the volume.

The relation to calculate the electric power is given by:

\(\begin{array}{l}P = \frac{W}{t}\\P = \frac{{mgh}}{t}\\P = \frac{{\left( {\rho V} \right)gh}}{t}\end{array}\)

Here, g is the gravitational acceleration.

Substitute the values in the above expression.

\(\begin{array}{l}P = \left( {1000\;{\rm{kg/}}{{\rm{m}}3}} \right)\left( {32\;{{\rm{m}}3}{\rm{/s}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}2}} \right)\left( {48\;{\rm{m}}} \right)\\P = 1.5 \times {107}\;{\rm{W}}\end{array}\)

Thus, the electric power produced is\(1.5 \times {107}\;{\rm{W}}\).