Biopowered - vegetable oil and biodiesel forum
Biodiesel => Chemistry and process => Topic started by: bradburypizza on May 10, 2014, 12:26:44 PM
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Quick question ,how critical is the 90 ml meth in the 10/90 test as my understanding is converted oil (bio) dissolves into the meths leaving the unconverted oil to dropout so why 90 is it purely for the purpose of calculation as as far as I can see as long as you have 10 ml of converted /part converted bio from your batch surely the dropout will be the same whether you have 70,80,90 ml of meth Too much time on my hands sorry
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Very good question in my opinion and one about which I've often wondered.
I don't think the maximum is critical, but I believe the minimum is ... ie I think a 10/100 would work and give a meaningful result, but a 1/80 wouldn't.
Just my opinion based on an idiots logic and I stand to be corrected ... I guess it would e easy enough to prove.
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I too have wondered about this.
At present I do an accurate 10/90 on the first stage but am not to fussy on the second stage.
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ok well today I've done a 10/80 on my first test and got 2.8 dropout added another 10 ml of meths making it 10/90 and the dropout result is 2.7 but the first test sat overnight so my feelings are that the extra .1 % would dropout if left longer. That being the case what would happen if I now at 10ml from the second reaction making a note of the first dropout surely if there's no more dropout it would say its a complete conversion
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Maybe the way to go is to start off with a known, 100% conversion and try that ...
1) with reduced amounts of methanol.
2) with known additions of new cooking oil.
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update tried simply added 10 ml second stage to the test and showed 1.5 extra dropout so did a new 10/90 test and that came out a clear pass
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Given amount of meth will only hold a given amount of bio? Too little meth drop more cause all bio did not dissolve?
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Thankyou that explains it nicely