Question by Chris: What is the average efficiency of a nuclear fission power plant?
I’m writing a paper comparing nuclear fusion and fission. I’d like to toss in the fact that nuclear fission plants are terribly inefficient (at least that’s what I remember learning in class). What is the average efficiency of a nuclear fission power plant? What percentage of the energy given off by a fission-reaction do humans actually end up using?
Best answer:
Answer by PeteR
Well, any steam cycle power system is going to seem “inefficient” – I believe that that theoretical maximum efficiency of any steam cycle power system is somewhere between 40% and 50%.
This will be the case for any fusion reactor that operates by heating water to operate a steam turbine.
All systems have some parasitic loads – such as powering pumps and such – that would reduce the efficiency further.
The problem with fusion reactions is that they require more energy to create the reaction than they release from the fusion reaction – in other words, any fusion reactor built currently has a negative efficiency.
Fusion bombs work because the energy release does not have to be controlled – so they create the energy to start the fusion reaction with a nuclear fission explosive.
For electrical energy purposes – that would be undesirable.
The benefits of fusion in the long run – if it is possible to create a reactor – is not so much in whatever efficiency benefits it would have – but in clean reaction without hazardous radioactive byproducts – and in the cheapness and plentifulness of the fuel (hydrogen or helium – instead of mining for uranium).
Give your answer to this question below!
One Response
When speaking of efficiency – i get the impression that you mean how much energy is released by the reaction relative to how much energy could be released by the entire mass. (given by E = mc^2)
According to wikipedia – 202.5 MeV is released as heat from a uranium fission (on average).
We can also see that with a mass of 235 u (1 u = 235 g/mol = 235 / 6.022 * 10^23 g)
E = (3.9*10^-25 kg) * (3.00*10^8 m/s)^2 = 219 GeV
So basically – we see that the percentage given off is less than 0.1% of the total potential
Fusion, as I recall, releases over 1% – I had to do some of these calculations for a class awhile back :)