The turbines in a hydroelectric plant are fed by water falling from a 50 m height. Assuming 91% efficiency for conversion of potential to electrical energy, and 8% loss of the resulting power in transmission, what is the mass flow rate of water required to power a 200 W light bulb?

Question

What will be an ideal response?

Answer 1

We can either work forward, computing the energy output per mass of water, or work backward, starting from the energy requirement of the lightbulb. If we choose to work forward, then we compute that the potential energy change, per kilogram of water that falls is We are told that

91% of this energy is converted to electrical energy, and 92% of that electrical energy is transmitted to the lightbulb (8% is lost). So, the energy transmitted to the lightbulb is 490 J/kg * 0.91 * 0.92 = 410 J/kg. The 200-Watt light bulb requires 200 J per second. Dividing this by the energy production per kilogram gives Required water flow rate per minute. To get more familiar volumetric units for water, we can divide by the density of water (about 8.3 lbm per gallon) to get 7.7 gallons per minute.