Question

In a hydroelectric power station, the water is flowing at \(2 \, {ms}^{-1}\) in the river which is \(100 \, {m}\) wide and \(5 \, {m}\) deep. The maximum power output from the river is:

• A. \(1.5 \, {MW}\)
• B. \(2 \, {MW}\)
• C. \(2.5 \, {MW}\)
• D. \(3 \, {MW}\)

Hint:

The actual power output in hydroelectric plants depends significantly on the efficiency and height drop, which should be factored into realistic calculations.

Answer 1

The Correct Option is B

The maximum power output from the river can be calculated using the formula for the kinetic energy of flowing water. The power \( P \) is given by:

\[ P = \frac{1}{2} \rho A v^3 \]

where:

• \( \rho = 1000 \, \text{kg/m}^3 \) is the density of water,
• \( A \) is the cross-sectional area of the river,
• \( v = 2 \, \text{m/s} \) is the velocity of water.

The cross-sectional area \( A \) of the river is:

\[ A = \text{width} \times \text{depth} = 100 \times 5 = 500 \, \text{m}^2 \]

Now, substituting the values into the power formula:

\[ P = \frac{1}{2} \times 1000 \times 500 \times (2)^3 \] \[ P = \frac{1}{2} \times 1000 \times 500 \times 8 \] \[ P = 2000000 \, \text{W} = 2 \, \text{MW} \]

Thus, the maximum power output from the river is 2 MW.

Answer 2

To solve this problem, we need to calculate the maximum power output from a hydroelectric power station, given the velocity, width, and depth of the river.

1. Understanding the Given Data:
We are given the following information:
The water flow velocity, \(v = 2 \, \text{m/s}\),
The width of the river, \(w = 100 \, \text{m}\),
The depth of the river, \(d = 5 \, \text{m}\),
The gravitational acceleration, \(g = 10 \, \text{m/s}^2\).
We need to calculate the maximum power output using the formula for the power of flowing water:

2. Formula for Maximum Power Output:
The maximum power that can be extracted from flowing water is given by the formula:

\[ P = \frac{1}{2} \rho A v^3 \] where:
\(P\) is the power in watts,
\(\rho\) is the density of water (\(\approx 1000 \, \text{kg/m}^3\)),
\(A\) is the cross-sectional area of the river, and
\(v\) is the velocity of the water.

The cross-sectional area \(A\) is given by the product of the width and depth of the river: \[ A = w \times d = 100 \times 5 = 500 \, \text{m}^2. \] Now, substituting the values: \[ P = \frac{1}{2} \times 1000 \times 500 \times (2)^3 = \frac{1}{2} \times 1000 \times 500 \times 8 = 2000000 \, \text{W}. \] So, the power in watts is \(P = 2,000,000 \, \text{W}\), or 2 MW.

Final Answer:
The correct answer is Option B: 2 MW.