Question

In a hydroelectric power station, the water is flowing at 3 ms$^{-1}$ in the river which is 150 m wide and 4 m deep. The maximum power output from the river is :

• A. 1.74 kg
• B. 2.26 kg

Answer 1

The maximum power output from the river is 8.1 MW.

Answer 2

Solution:

1. Given:
     - Flow velocity, $v = 3\,m/s$
     - Width of river, $w = 150\,m$
     - Depth of river, $d = 4\,m$
     - Density of water, $\rho = 1000\,kg/m^3$

2. Calculation:
     - Cross-sectional area, $A = w \times d = 150 \times 4 = 600\,m^2$
     - Maximum power from moving water is given by:
       $P = \frac{1}{2} \rho A v^3$
     - Substitute values:
       $P = \frac{1}{2} \times 1000 \times 600 \times (3)^3 = 0.5 \times 1000 \times 600 \times 27$
     - Compute step-by-step:
       $0.5 \times 1000 = 500$
       $500 \times 600 = 300000$
       $300000 \times 27 = 8,100,000\,W$
       Thus, $P = 8.1 \times 10^6\,W = 8.1\,MW$.

Answer:
The maximum power output from the river is 8.1 MW.