In a hydroelectric dam, water falls 24.0 m and then spins a turbine to generate electricity.
a.What is Delta U of 1.0 kg of water?
b.Suppose the dam is 80% efficient at converting the water’s potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 45.0 MW of electricity? This is a typical value for a small hydroelectric dam.
you mean ∆E ?
PE = mgh = 1 x 9/8 x 24 = 235 J
235 / 0.8 = 294 J input.
in one second that is 294 watts, for 1 kg
45e6 / 294 = 153000 kg of water.
Typical Hydroelectric Dam
a) The potential energy for 1.0 kg of water can be calculated using your equation for gravitational potential energy. GPE = mgh. M is mass (1.0 kg), g is 9.8 m/s^2, and h is the height (24.0 m). Plug your variables in and solve. Your answer will be in Joules.
b) 45.0 MW = 4.5 * 10^7 Watts, a unit of power. If the dam is only 80% efficient, you need to divide that number by 0.8 in order to figure out the total power that needs to be generated in order to create that much usable power.
Once you have that value, power is equal to work (which uses energy, measured in Joules) divided by time. As your work here is all due to gravitational potential energy, so P = (mgh)/t. P is your power you just calculated. t is 1.0 seconds, h is 24.0 m, and g is 9.8 m/s^2. Plus all those in and solve for m, the mass of water needed.
a) ΔU=-mgh=-(1*9.8*24)= 235.2