# Question help

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#1

What have i done wrong here?
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6 years ago
#2
in part b you've multiplied the electrical output of the station by the efficiency to get the thermal output of the reactor

you should have multiplied the electrical output by 1/efficiency
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#3
(Original post by Joinedup)
in part b you've multiplied the electrical output of the station by the efficiency to get the thermal output of the reactor

you should have multiplied the electrical output by 1/efficiency
So e=output x 100/input

And we are finding the input?
What is the overall efficiency?
It produceds 600MW but most of this is wasted?
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6 years ago
#4
(Original post by Zenarthra)
So e=output x 100/input

And we are finding the input?
What is the overall efficiency?
It produceds 600MW but most of this is wasted?
The useful output is 600MW electrical and this is 35% of the input. You need to find the input to solve the rest of the question.

fwliw it's a realistic level of efficiency, often you'll see power stations with both the electrical and thermal output quoted. thermal is the rate at which the reactor produces heat, electrical is the electrical power coming out of the station.

e.g. http://econtent.unm.edu/cdm/singleit...n/nuceng/id/41

Wylfa:
1180MW electrical
3751MW thermal

efficiency is 1180/3751
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#5
(Original post by Joinedup)
The useful output is 600MW electrical and this is 35% of the input. You need to find the input to solve the rest of the question.

fwliw it's a realistic level of efficiency, often you'll see power stations with both the electrical and thermal output quoted. thermal is the rate at which the reactor produces heat, electrical is the electrical power coming out of the station.

e.g. http://econtent.unm.edu/cdm/singleit...n/nuceng/id/41

Wylfa:
1180MW electrical
3751MW thermal

efficiency is 1180/3751
Ahh that clears it up, thanks!
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