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My Bonneville Sunbeam's Fuel Economy

To: tigers@Autox.Team.Net
Subject: My Bonneville Sunbeam's Fuel Economy
From: DrMayf@aol.com
Date: Thu, 8 Oct 1998 16:36:17 EDT
Well, who was it that wondered about my Bonneville Sunbeam's fuel economy?
Here is what I have estimated. I performed two set of calculations, one at
14.7 psig and the other at 22 psig. I had previously determined the engine air
flow in lbm/minute at 6500 rpm, my target engine speed (or is that motor?). At
14.7 psig, the engine is fed 45.29 lbm/min; at 22 psig it gets 51.76
lbm/minute. All flows have been computed using the appropriate fluid dynamic
and thermodynamic equations. And these were verified through two differen
turbocharger publications. I'll be happy to cite these should anyone believe
otherwise. I may even be wrong.

So given that a good torque air to fuel ratios is 12.5 to about 15.0 ( best
torque, best lean cruise) I can easily calculate  the amount of fuel required
and hence the fuel expenditure. I used 12.5 because I want to err on the side
of a rich burn, not lean piston frying economy.

At 14.7 psig:  (45.29 lbm air / min) / 12.5  = lbm fuel / min = 3.623 lbm fuel
/ min
                     (3.623 lbm fuel / min) x 60 min / hr = 217.39 lbm fuel /
hour

                         since fuel is approximately 7.5 lbm / gallon. I get
                         (217.39 lbm fuel / hour) / (7.5 lbm / gallon) = 28.986 
gal / hour

                        if the car actually has enough get up and go to reach 
max speed,
                        then, the fuel economy would be
                        (240 miles / hour) / 28.986 gal / hour = 8.28 miles / 
gallon

At 22.0 psig: (51.76 lbm air / min) / 12.5  = lbm fuel / min = 4.141 lbm fuel
/ min
                    (4.141 lbm fuel / min) x 60 min / hr = 248.45 lbm fuel /
hour

                        since fuel is approximately 7.5 lbm / gallon. I get
                        (248.45 lbm fuel / hour) / (7.5 lbm / gallon) = 33.13 
gal / hour

                        if the car actually has enough get up and go to reach 
max speed,
                        then, the fuel economy would be
                        (260 miles / hour) / 33.13 gal / hour = 7.85 miles / 
gallon

Assuming that the full runup distance and shutdown is covered at full gallop,
then I can
determine the amount of fuel required and thus minimum tank size.

At 260 mph I get 7.85 miles per gallon. If the track length was 7.85 miles it
would take one gallon: a pretty small tank. I plan on using a 5 or 8 gallon
RCI tank with foam internals. I do this because to make a valid run, you have
to turn around and run the course backwards within an hour. Since I am a crew
of one, I will just turn the car around and go, after repacking the chute (if
I deployed it). I suspect that I will make many one way runs before I get to
the two way record runs. I may not be fast enough for a record, but if I can
better the old Sunbeam LSR record of 203.792 mph, then I will be a happy
camper.

Class dismissed.

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