Tigers List Archive

This read-only message was archived from a public mail list.
Mail From: (email redacted)

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.

This read-only message was archived from a public mail list.
Mail From: "Bob Hokanson" <(email redacted)>



----------
> From: (email redacted)
> To: (email redacted)
> Subject: My Bonneville Sunbeam's Fuel Economy
> Date: Thursday, October 08, 1998 1:36 PM
> 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.

Son, If you believe you can get by on 7mpg with the setup you're building
including full runup, then you've probably bought a lot of cow magnets in
your time too.

Skeptical

This read-only message was archived from a public mail list.
Mail From: (email redacted)

Rather than a cryptic comment, how about sharing your knowledge with us all
and show me where I went wrong? I at least put my numbers where all can see.