Question

In a hydroelectric power station, the height of the dam is $10\text{ m}$. How many kilograms of water must fall per second on the blades of a turbine in order to generate $1\text{ MW}$ of electrical power? $[g = 10\text{ m/s}^2]$

• A. $10^3\text{ kg/s}$
• B. $10^4\text{ kg/s}$
• C. $10^5\text{ kg/s}$
• D. $10^6\text{ kg/s}$

Hint:

Always keep track of prefixes! $1\text{ kW} = 10^3\text{ W}$, $1\text{ MW} = 10^6\text{ W}$, and $1\text{ GW} = 10^9\text{ W}$. Hydroelectric problems are simply an application of the Work-Energy theorem over time.

Answer 1

The Correct Option is B

Generating power in a hydroelectric plant involves converting the gravitational potential energy of falling water into electrical energy. Step 1: Identify Given Information

• Power required ($P$) = $1\text{ MW} = 10^6\text{ W}$ (Joules per second)

• Height ($h$) = $10\text{ m}$

• Acceleration due to gravity ($g$) = $10\text{ m/s}^2$

Step 2: Establish the Power Formula Power is the rate of doing work or the rate of energy change. Potential energy is $mgh$. Therefore, Power = $\frac{mgh}{t}$. We are looking for the mass per second ($\frac{m}{t}$), which we can call $R$. $$P = R \cdot g \cdot h$$

Step 3: Solve for $R$ Rearranging the formula: $$R = \frac{P}{g \cdot h}$$ Substituting the values: $$R = \frac{10^6}{10 \times 10}$$ $$R = \frac{10^6}{100} = \frac{10^6}{10^2}$$ $$R = 10^4\text{ kg/s}$$ Therefore, $10,000\text{ kg}$ (or $10^4\text{ kg}$) of water must fall every second to maintain a $1\text{ MW}$ output, assuming $100\%$ efficiency.