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Joined: Mon Jul 10, 2006 6:57 am Posts: 204 Location: Sydney, Australia
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I can't say I have any knowledge of electrochemistry, but presumably you are talking energy input to produce a quantity of product. If 0.215 Joule per ml of product is 100% efficiency, then 0.43 Joule per ml would be 50% efficient. Your efficiency is thus 0.215 divided by your actual energy requirements. 0.215 / 8 = 0.026875, so that's 2.6875% efficiency.
_________________ The Famous Brett Watson -- brett.watson@gmail.com
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