Date:         Sun, 25 Nov 2007 20:55:34 -0800
Reply-To:     Kim Springer <kimspringer@RCN.COM>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Kim Springer <kimspringer@RCN.COM>
Subject:      Calculate fuel consumption when idling?
Content-Type: text/plain; charset="iso-8859-1"

Volks,

I'm with Scott, measure out a quart and run it 'till it's gone and watch the
clock.  It would be interesting to compare the calculted with the real
world.

Kim

----- Original Message -----
From: "Scott Daniel - Shazam" <scottdaniel@TURBOVANS.COM>
To: <vanagon@GERRY.VANAGON.COM>
Sent: Sunday, November 25, 2007 4:13 PM
Subject: Re: Calculate fuel consumption when idling?


Just want to say two things -
One, this method, figuring the air amount, the mixture ratio, and so forth -
I'm very impressed, and that you came up with a figure that seems in the
ball park - right on ! I am inclined to think this could be about right.
1/6 gallons per hour.
   # 2 - well.............it would be child's play to run the engine off a
one gallon can of gas.  You just undo the supply line to the fuel pump,
stick that hose in the can.  Then undo the return hose from the engine and
stick that in the can. Then fire 'er up, and time how long it idles on one
gallon of gas.   Then use that figure to get the fuel consumed in one hour.
Do everything possible to ensure accurate measurement of the fuel of course.


 You might appreciate this story. Thomas Edison gave a new assistant a
project, to see what the new assistant could do. He assigned him the task of
determining the volume of a light bulb. The new assistant worked for several
hours with formulas and calculations to get the volume of the odd shape.
When done, Edison looked at the result, and said ....that's close but not
quite right.       He then drilled a hole in the side of the light bulb,
filled it with a liquid, then poured the liquid out into a calibrated
measuring beaker, where the volume was easy to determine.   It took just
minutes, and was considerably more accurate.   Think simple first.
Scott
www.turbovans.com

-----Original Message-----
From: Vanagon Mailing List [mailto:vanagon@gerry.vanagon.com] On Behalf Of
Michael Elliott
Sent: Sunday, November 25, 2007 3:28 PM
To: vanagon@GERRY.VANAGON.COM
Subject: Re: Calculate fuel consumption when idling?

855 liters per minute? Let's see ... 1.3g per liter = 1,110 grams of air
per minutes / 15 = 74 grams of gasoline per minutes = 4.4kg of gasoline
per hour, which is about 6 liters per hour = roughly 1.6 gallons U.S. per
hour. Same number I came up with before. I might have munged my units, but
the result appears to be the same-o.

--
Mike "Rocket J Squirrel" Elliott
71 Type 2: the Wonderbus
84 Westfalia: Mellow Yellow ("The Electrical Banana")
74 Utility Trailer. Ladybug Trailer, Inc., San Juan Capistrano
KG6RCR



On 11/25/2007 9:39 AM Zoltan Kuthy wrote:

> That would be 855 000 cc, if I multiply them all together.  Which is 855
> liter.  Is that more likely?
> Z
> ----- Original Message -----
> From: "Michael Elliott" <camping.elliott@gmail.com>
> To: "Zoltan Kuthy" <zolo@foxinternet.net>
> Cc: "Vanagon Mailing List" <vanagon@GERRY.VANAGON.COM>
> Sent: Saturday, November 24, 2007 8:38 PM
> Subject: Re: Calculate fuel consumption when idling?
>
>
>> 900 (revolutions) x 2 (cylinder firings per revolution) x 475 (cc of air
>> per cylinder) = 855 cc (or air sucked in per minute). That's what I had
in
>> mind. Yup.
>> --
>> Mike "Rocket J Squirrel" Elliott
>> 71 Type 2: the Wonderbus
>> 84 Westfalia: Mellow Yellow ("The Electrical Banana")
>> 74 Utility Trailer. Ladybug Trailer, Inc., San Juan Capistrano
>> KG6RCR
>>
>>
>>
>> On 11/24/2007 4:55 PM Zoltan Kuthy wrote:
>>
>>> 900x2x475=855 ?
>>> Z
>>> Unless you were thinking of something else.  I liked the way you
>>> calculated though.
>>> ----- Original Message -----
>>> From: "mike elliott" <camping.elliott@GMAIL.COM>
>>> To: <vanagon@GERRY.VANAGON.COM>
>>> Sent: Saturday, November 24, 2007 4:26 PM
>>> Subject: Re: Calculate fuel consumption when idling?
>>>
>>>
>>>> Do you win? Heck, I don't know. I'll take a shot at it. Watch me get
>>>> it spectacularly wrong.
>>>>
>>>> The 1.9L engine has 475cc displacement per cylinder. I think two
>>>> cylinders fire per revolution, so at 900 rpm we got 900 x 2 x 475 =
>>>> 855cc of air per minute being sucked into the engine. If the mixture
>>>> is a good stoichiometric one, we'd have about a 1:15 fuel:air mass
>>>> ratio. Air has a mass of 1.3g per liter, so that's .855 x 1.3 = 1.1kg
>>>> of air per minute, with being one-fifteenth of that, or 74g per
>>>> minute. At 60 minutes that's 4.4kg of fuel. Gasoline masses roughly
>>>> 740 grams per /liter, so that's 6 liters, roughly 1.6 gallons U.S. per
>>>> hour.
>>>>
>>>> Those are the numbers I come up with and I bet they're within an order
>>>> of magnitude of being right.
>>>>
>>>> Mike "Rocket J Squirrel" Elliott
>>>>
>>>> On Nov 24, 2007 3:52 PM, Matthew Snook <matt@snooksband.com> wrote:
>>>>> 0.9375?  Is that right?  Do I win?
>>>>>
>>>>> :)
>>>>>
>>>>> At 60mph, mine's turning @ 3200rpm.  It will burn 3 gallons doing
that.
>>>>> 3
>>>>> gallons per hour at 3200rpm.  But it idles at 1000rpm.  So that comes
>>>>> to
>>>>> 0.9375 gallons per hour at 1000rpm.  Of course there's no wind
>>>>> resistance at
>>>>> that speed, so maybe less...
>>>>>
>>>>> Matthew Snook (@ ~3000 ft)
>>>>>
>>>>>
>>>>>
>>>>> -----Original Message-----
>>>>> From: Michael Elliott
>>>>> Subject: Calculate fuel consumption when idling?
>>>>>
>>>>> A properly-tuned 1.9L WBX engine would consume how many gallons of
>>>>> gasoline per hour when idling at sea level? Would this be
significantly
>>>>> different at 6,000 feet?
>>>>>
>>>>>
>>>> --
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>>>>
>>>>
>>
>> --
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>>
>