Question

Each second, $\mathbf{1200}{\mathbf{m}}^{\mathbf{3}}$of water passes over a waterfall 100 mhigh. Three-fourths of the kinetic energy gained by the water in falling is transferred to electrical energy by a hydroelectric generator. At what rate does the generator produce electrical energy? (The mass of$\mathbf{1}\mathbf{}{\mathbf{m}}^{\mathbf{3}}$of water is 1000 kg.)

Short answer

$8.8\times {10}^8\mathrm{W}$is the rate of the generator to produce electrical energy.

Step-by-step solution

Given

The volume of the water is$\mathrm{V}=1200{\mathrm{m}}^3$

The height of the waterfall is h = 100 m

Mass of of water is 1000 kg

To understand the concept

Calculating density from mass and volume, find the Potential Energy of the water at that height usingthecorresponding formula. Using the law of conservation of energy, find the kinetic energy of the water. From this, we can get the energy transferred to the generator. Using this, find the rate of generation of electricity by the generator.

Formula:

$$\begin{gathered} PE=mgh \\ p=\frac{m}{V} \\ P=\frac{Energy}{Time} \end{gathered}$$

Calculate the rate at which the generator produces electrical energy

The density of the matter with mass m and volume V can be expressed as,

$p=\frac{m}{V}$

Therefore, substituting the given values in the above expression, we get,

$$\begin{gathered} \mathrm{p}=\frac{1000\mathrm{kg}}{1{\mathrm{m}}^3} \\ ={10}^3\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

The potential energy of the water at h = 100 m can be expressed as,

$$\begin{gathered} PE=\mathrm{mgh} \\ =\mathrm{pVgh} \\ ={10}^3\mathrm{kg}/{\mathrm{m}}^3\times 1200{\mathrm{m}}^3\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 100\mathrm{m} \\ =1.17\times {10}^9\mathrm{J}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ =1.17\times {10}^9\mathrm{J} \end{gathered}$$

From the principle of conservation of energy, we can write that the,

PE = KE

And therefore from the above expression, we have,

$\mathrm{KE}=1.17\times {10}^9\mathrm{J}$

Only three fourth amount is transferred to electrical energy, therefore, the power generation rate will be,

$$\begin{gathered} \mathrm{P}=\frac{{\mathrm{KE}}_{\mathrm{transferred}}}{\mathrm{time}} \\ =\frac{\left({\displaystyle \frac{3}{4}}\right)\left(\mathrm{KE}\right)}{\mathrm{t}} \\ =\frac{\left({\displaystyle \frac{3}{4}}\right)(1.17\times {10}^9\mathrm{J})}{1\mathrm{s}} \\ =8.8\times {10}^8\mathrm{J}/\mathrm{s}\left(\frac{1\mathrm{W}}{1\mathrm{J}/\mathrm{s}}\right) \\ =8.8\times {10}^8\mathrm{W} \end{gathered}$$