| Gasoline & Fuel Economy Energy and strategy |
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Thu, May 21st, 2009, 12:08 pm
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| Making E30 in an E85/E10 world Hi, I threw together a quick spreadsheet and near as I can tell: 3 - part E10 1 - part E85 Gives 4 parts, E29%, which should be close enough to E30. Yes, given a nap, I could probably wake up and solve the simultaneous equations but I'm a little tired right now. Anyone have a more precise ratio? I've got a 5 gallon, spare can so I could mix at the station. Then drive to fuel exhaustion and refill from the 5 gallon spare. It won't be perfect but it should be 'good enough.' Thereafter, I'll just do 3/1 or 6/2 when I fill up. Thoughts? Comments? I know there is about a 5% difference between straight Shell/Chevron/Exxon and other straight gas. Yes, I'll survey them again this summer. But I wanted to baseline E30, now that E10 is the universal standard. Thanks, Bob Wilson |
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Thu, May 21st, 2009, 02:48 pm
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| Bob, Your numbers don't add up for me. By Volume (US) 87 Octane Gas == 125,000 BTU/gal Ehanol == 84,600 BTU/gal SANITY CHECK: E10 Gasohol Mix == 125,000* .9 + 84,600*.1 ==120,960 BTU/gal (agrees with published numbers) E85 Gasohol Mix == 125,000*.15 + 84,600*.85 == 90,660 BTU/gal (also agrees) CALC E30 Gasohol Mix == 125,000*.7 + 84,600*.3 == 112,880 BTU/Gal Your ratio == 120,960 (E10)* .75 (3 Parts in 4) + 90,660 (E85)*.25 (last Part) = 113,385 BTU/Gal, which is a little too high in energy. The actual ratio is .7333... to .2666... My ratio 120,960 * ..73333 + 90,660 * ..26666 == 112.878 BTU/gal For five gallons a reasonable mix at the pump is 3.6 gallons of E10 to 1.3 gallons of E85 for a BTU total of 112.921 BTU/gal and 4.9 gallons. (E-almost-30?) If they still have 87 Octane gas, you can mix 4.5 gallons of straignt gas and one-half gallon of E85 for a BTU content of 112,500 BTU/gal and five gallons (E30-point-1) That's why Bill Gates invented Excel Solver <grin>. Last edited by FastMover; Thu, May 21st, 2009 at 03:17 pm. |
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Fri, May 22nd, 2009, 12:59 am
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| Ok, I've had a nap and a cuppa coffee so my question is: Given 1 gallon of E85, how many gallons of E10 are needed to make E30? Once we know this ratio, it will hold for any quantity. (1*(.85e + .15g)) + (n*(.10e + .90g)) = (1+n)*(.30e + .70g) ## adding 1 gallon of E85, "n" gallons of E10, to make E30 .85e + .15g + .10en + .90gn = (1+n)*.30e + (1+n)*.70g .85e + .15g + .10en + .90gn = .30e + .30en + .70g + .70gn .85e + .15g - .30e -.70g = -.10en + .30en -.90gn + .70gn (.85 - .30)e + (.15 - .70)g = (-.10en + .30en) + (.70gn - .90gn) (.55)e + (-.55)g = (.20)en + (-.20)gn (.55)e + (-.55)g = ((.20e) + (-.20)g)) *n ((.55)e + (-.55)g) / ((.20e) + (-.20)g) = n .55 * (e-g) / .20* (e-g) = n ## the "(e-g)" term confirms nothing was lost in the earlier algerbra .55 / .20 = n 2.75 = n Ok, so with this approach:
Now if the local Jet station will let me pre-pay for exact amounts, it should be fairly straight forward. Estimate the amount to fuel to add, right column, and pre-pay the amounts of E85 and E10 needed. Bob Wilson |
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Fri, May 22nd, 2009, 09:30 am
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| ...better ! |
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Fri, May 22nd, 2009, 10:04 pm
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| So today I mixed:
Bob Wilson |
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Sat, Jun 6th, 2009, 06:44 pm
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| Preliminary data:
I can not rule out that E30 is as effective as straight gas in my NHW11 but the association is weak and within experimental error. I can confirm it is a 'pain in the *ss' as mixing E30 in the car is not a trivial effort. I'm not happy with my methodology and won't be happy until I have a good model to identify what is going on. It is not enough to suggest the engine systems have 'tuned' to the new fuel without having some empirical data, something from the Graham scanner or other sources that shows what is going on. Is the spark greatly advanced? Is something else going on? Most of all, I need automatic data collection and analysis to factor out temperature effects. Bob Wilson Last edited by bwilson4web; Sun, Jun 7th, 2009 at 03:15 am. |
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Sun, Jun 21st, 2009, 12:39 pm
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| Bob- I applaud your efforts, and want to encourage you to continue. Long-term averages will show what's "reality". The reality is, the btu's in a gallon of gas are never the same from station to station, or even from delivery to delivery. Then there's the whole temperature thing. ( fewer btu in "hot" gas, etc... ) Then, of course, the E85 is not ever perfectly 85% either. And E10 can be as low as 6.5% and as high as 12%. So if you are in this for the long-haul... then I look forward to your data. If your data are "close" then I agree, that "close" is "good enough" to draw some conclusions, due to all the uncertainty involved. Of course, you could always rent a University testing lab for a weekend. ![]() Best of luck, and I appreciate your efforts. |
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Sun, Jun 21st, 2009, 12:49 pm
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| I drive many "corn belt" states. In some states, they allow the consumer to choose between gas, E10 or "other". (Some parts of IA, KS, MN, SD sell E20 and E30 at the pump, but labeled as "for Flex Fuel Vehicles Only". ) I have a ScanGauge, a very useful tool. In those states selling both, gas is always more costly than E10. ( as it should be ) But, by using a ScanGauge to monitor fuel trim, I've found more than a few stations selling E10 ( or close ) in the pure gas pump, and charging me $0.10 a gallon more for the privledge. I'd like to report this, but don't know if it's exactly illegal and/or if there is actually any agency out there policing this. I just learned that Iowa has 7 fuel station inspectors for the entire state, one is going to retire soon, and they won't replace him due to budget cuts. This equals 1 inspector visit to a gas station every 2 years, on average. I believe they inspect the pumps and test that 1 gallon on the pump = 1 gallon metered into your tank. I doubt they even check the % ethanol. So much uncertainty... it is frustrating.... Last edited by HybridDriver; Sun, Jun 21st, 2009 at 12:52 pm. |
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Sun, Jun 21st, 2009, 07:53 pm
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| I ran into a hiccup. Friday evening we had a four hour, power outage. My NHW11 is on straight gas but had to run to provide power for lights, fans and TV. I'll have to restart the tank, dang it! Right now, I can't call E30 equivalent to straight gas. There may be an optimum ratio but I'm going to have to come up with a better testing methodology. Bob Wilson |
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