# Water is flowing in a river at 2.5 m/s. The river is 40 m wide with an average depth of 6.0 m....

## Question:

Water is flowing in a river at 2.5 m/s. The river is 40 m wide with an average depth of 6.0 m. Compute the power available from the water current.

## Energy Conservation:

In a closed system, the total energy is conserved. That means the sum of the kinetic energy and the potential energy of the system (mechanical energy) doesn't change during the course of events. When heat, sound or friction are involved, the energy conservation should consider those effects.

Parameters given;

• {eq}v = 2.5 \ \textrm{m/s} {/eq} is the velocity of water in the river,
• {eq}d = 40 \ \textrm m {/eq} is the width of the river,
• {eq}h = 6.0 \ \textrm m {/eq} is the average depth of the river,
• {eq}\rho = 1000 \ \textrm{kg/m}^3 {/eq} is the density of water.

The total mass of water in 1 sec of flow is;

{eq}\displaystyle{ \begin{align} M &= \rho V\\ &= \rho dh v\\ &= 1000 \ \textrm{kg/m}^3 \times 40 \ \textrm m \times 6.0 \ \textrm m \times 2.5 \ \textrm{m/s}\\ &= 6.0 \times 10^5 \ \textrm{kg/s} \end{align} } {/eq}

Thus, the power of the water flow P is the kinetic energy of water per second;

{eq}\displaystyle{ \begin{align} P &= \frac{1}{2}Mv^2\\ &= \frac{1}{2} \times 6.0 \times 10^5 \ \textrm{kg/s} \times (2.5 \ \textrm{m/s})^2\\ &= \boxed{1.875 \times 10^6 \ \textrm W} \end{align} } {/eq}