The surface of the continental United States has an area of about$8\times {10}^{6}k{m}^2$and an average elevation of about 500 m(above sea level). The average yearly rainfall is 75 cm. The fraction of this rainwater that returns to the atmosphere by evaporation is$\frac{\mathbf{2}}{\mathbf{3}}$; the rest eventually flows into the ocean. If the decrease in gravitational potential energy of the water–Earth system associated with that flow could be fully converted to electrical energy, what would be the average power? (The mass of$\mathbf{1}{\mathit{m}}^{\mathbf{3}}$of water is 1000 kg.)

a) Area of the surface,$A=8\times {10}^{6}k{m}^2$

b) Average elevation of the surface, h = 500 m

c) Average yearly rainfall, d = 75 cm

d) Fraction of rainwater that returns to atmosphere is 2/3

e) Mass of $1m^3$water,$m_{water}=1000kg$

We use the concept of density to find the mass of water, which flows down to the sea. We can use gravitational potential energy, which will decrease and fully convert to electrical energy. Using the equation of power, we can find the average power per year from the water that flows down to the sea.

Formulae:

The volume of a body, V = Ad (i)

The density of a body in terms of volume, $\rho =\frac{m}{v}$ (ii)

The potential energy of a body at a height,$P.E=mgh$ (iii)

The power consumption by a body, $P=\frac{E}{t}$ (iv)

Using equation (i), the volume of the rainfall is given as:

$$\begin{gathered} V=\left(8\times {10}^6\times {10}^6{m}^2\right)\left(0.75m\right) \\ =6\times {10}^{12}{m}^3 \end{gathered}$$

We know density of water, $\rho =\frac{1000kg}{1m^3}$so we can find mass using equation (ii) as.

$$\begin{gathered} m=\rho v \\ =\frac{1000kg}{1m^3}\left(6\times {10}^{12}{m}^3\right) \\ =6\times {10}^{15}kg \end{gathered}$$

Mass of evaporated water using the fraction value is$\frac{2}{3}m$ ; that is given as:

$$\begin{gathered} m_1=\frac{2}{3}\times 6\times {10}^{15} \\ =4\times {10}^{15}kg \end{gathered}$$

So the mass of water, which flows down to ocean, can be given as:

$$\begin{gathered} m_2=m-m_1 \\ =\left(6\times {10}^{15}\right)-\left(-4\times {10}^{15}\right) \\ =2\times {10}^{15}kg \end{gathered}$$

Thus, this mass of water will have gravitational potential energy at 500m, which will fully convert into electrical energy.

Now, we can get the potential energy of the mass using equation (iii), as follows:

$$\begin{gathered} P.E=\left(2\times {10}^{15}\right)\left(9.8\right)\left(500\right) \\ =9.8\times {10}^{18}J \end{gathered}$$

This energy will convert into electrical energy per year. We get the average power using equation (iv) as:

$$\begin{gathered} P=\frac{\left(9.8\times {10}^{18}J\right)}{1year} \\ =\frac{\left(9.8\times {10}^{18}J\right)}{365\times 24\times 3600s} \\ =3.1\times {10}^{11}W \end{gathered}$$

This is the average power when gravitational potential energy of water flowing down to sea decreases.