Date:         Sun, 25 Nov 2007 15:28:08 -0800
Reply-To:     Michael Elliott <camping.elliott@GMAIL.COM>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Michael Elliott <camping.elliott@GMAIL.COM>
Subject:      Re: Calculate fuel consumption when idling?
Comments: To: Zoltan Kuthy <zolo@foxinternet.net>
In-Reply-To:  <06F02EFEB7D043A3987824CF0EA2D63A@ZoltanPC>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed

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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>>
>