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Goat
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#1
Report Thread starter 15 years ago
#1
Just did the June 2003 Physics unit 4 paper (AQA b) and I made a mistake on what appeared to be a simple question, and I dont know why.

"A turbine is linked to a generator that produces 800MW of electrical power. Calculate the efficiency of conversion of the internal energy of the steam into electrical enrgy."

I thought efficiency = output/input

From part (a) you know that 3.53GW is used to heat the water, so

efficiency = 800E6/3.53E9 = 22.7%

but apparently the answer is 18.5% (?). It says eff = 0.8/(0.8+3.53)

I dont understand why!
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queen-of-pain
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#2
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#2
Its the useful energy divided by the total energy thats been produced.
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Goat
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#3
Report Thread starter 15 years ago
#3
(Original post by queen-of-pain)
Its the useful energy divided by the total energy thats been produced.
OK that would make sense but in the text book it says

efficiency = useful power output / power input

Isn't power input is different from total produced?
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john !!
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Your textbooks defines it more generally, "total energy thats been produced" is a bit ambiguous.

I don't understand why the useful power output would be used in the part of the equation that should represent the total power input.
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Golden Maverick
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#5
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#5
(Original post by Goat)
"A turbine is linked to a generator that produces 800MW of electrical power. Calculate the efficiency of conversion of the internal energy of the steam into electrical enrgy."

I thought efficiency = output/input

From part (a) you know that 3.53GW is used to heat the water, so

efficiency = 800E6/3.53E9 = 22.7%

but apparently the answer is 18.5% (?). It says eff = 0.8/(0.8+3.53)

I dont understand why!
From part a are you getting the energy lost when the steam re-cools ar the actual energy used to heat it?
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Goat
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#6
Report Thread starter 15 years ago
#6
(Original post by Golden Maverick)
From part a are you getting the energy lost when the steam re-cools ar the actual energy used to heat it?
Its the energy lost by the steam when is condensed by cold water.
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