Potential energy lost to non conservative forces?

Question

A pumped-storage reservoir sits 155 m above its generating station and holds 8.5 109 kg of water. The power plant generates 350 MW of electric power while draining the reservoir over an 8.0 hour period. What fraction of the initial potential energy is lost to nonconservative forces (i.e., does not emerge as electricity)?

Answer 1

potential energy of water reservoir is mgh where m=mass, g=acceleration due to gravity and h is the height. Power is energy divided by time. So the total power available is

mgh/t = (8.5e9kg)(10 m/s^2)(155m)/(8*60*60) =457 MW

so 107 MW is lost to non conservative forces.

Answer 2

The potential energy
PE = 8.5*10^9 *9.8*155 Joule =
The energy generated by the power plant is
350*10^6*8 *3600 joule
The difference is 2.8315*10^12 joule energy lost to nonconservative forces

Answer 3

Hi. I am not a physics major. I hope you get a better answer than mine. 1. General Relativity (Einstein's) says: Mass-Energy-Stress-Momentum is conserved. But this topic is for advanced students, and should be ignored until you understand classical physics. 2. The two and only two categories of energy are P.E. and K.E.; E = P.E. + K.E. (k.e. = kinetic energy, p.e. = potential energy). In Classical Physics, the (total) Energy of an isolated system is conserved. 3. The energy of a spring or an elastic substance is conserved only in the ideal case. In the real world, energy is lost (non-conserved) to heat (friction or chemical changes). I suspect this is what your lecturer is talking about. That a vibrating spring (with a weight attached) has components of energy that change second to second but that the total energy is conserved E = P.E. + K.E. = constant but P.E. = f(t) and of course K.E. = E - f(t) = g(t). Also note that there is a dependence on displacement x {or on the strain, if you are dealing with more complex cases (volumes) ← (this is usually not a 'first year' topic in school because it needs integral calculus)} So if E = 1, for a displacement x at time t=0 (the time the spring is released K.E. = 0, E = P.E. = f(x,k) =1 (where k is the spring constant) ((and the displacement is smaller than the elastic limit)). At the point of maximum speed E = ½mv² = K.E., P.E. = 0. See Wikipedia on Hooke's Law See Wikipedia on Harmonic Oscillator