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# Mailing List Logs for From: Amonchare economy Equation Tue, 23 Feb 93 23:55:58 EST
Well here's what i got for the economy question:

I started with the amount of chemical energy available in a liter of gasoline.
3.17 x 10^7 J/l according to my undergrad physics book. In order to account
for better refinement techniques and other improvements( additive, etc.), I
increased this by 20% so in 2054 it would be 3.8 x 10^7 J/l.

1 J = 1 Kg x m x m/s^2 therefore J/l = Kg x m/s^2 x m/l

by using the value for gas we get

3.8 x 10^7 J/l = kg x m/s^2 x m/l

if we divide both sides of the Eq. by mass of the vehicle we get

3.8 x 10^7 J/l kg x m/s^2 x m/l
-------------- = ---------------- = m/s^2 x m/l
mass(kg) mass(kg)

the m/s^2 on the right-hand side of the Eq. is an acceleration. I'll discuss
this in more detail later. But for now I'm going to divide both side of the
Eq. by this acceleration.

3.8 x10^7 J/l m/s^2 x m/l
---------------------- = ----------- = m/l
mass(kg) x acc.(m/s^2) acc.(m/s^2)

now we have an expression of economy (meters/l). We want to convert this to
km/l. In order to do this we must divide both sides by 1000.

3.8 x 10^7 J/l m/l
----------------------------- = --------- = km/l
mass(kg) x acc.(m/s^2) x 1000 1000 m/km

now I'm going to the two constants (3.8 x 10^7 and 1000) to simplify the Eq.
also I'm going to remove the unit from the left side of the Eq. because if I
did the math right them cancel to km/l.

3.8 x 10^4
----------- = km/l
mass x acc.

The acc. term is the average acceleration of the engine. In the discussion of
max. velocity I used an acc. of 1 m/s^2 to maintain a speed. The acceleration
in this case must be greater than the 1 m/s^2 necessary to maintain a velocity
because there is acceleration necessary to reach these velocities. After
fiddling with the above Eq. awhile I believe 1.4 m/s^2 would give value on the
order of those given in RBB ( that's if we want to follow those.) This (acc.)
value may seem low at first glance but remember that this is an average acc.
Most of time driving is spent maintaining a velocity (acc. = 1 m/s^2) while
only a relatively small amount of time is spent at much higher acceleration.
(For 0-60 MPH in 6 sec the acc. would be approximately 17 m/s^2.) Of course in
stop and go driving more time is spent accelerating and the acc. would be
higher.
Anyway by holding the acc. constant at 1.4 m/s^2 we get the formula for
economy:

3.8 x 10^4 28000
---------- = ----- = economy (km/l)
mass x 1.4 mass

According to this formula economy is only depends on the mass of the vehicle.
Realistic? Let me know what you think.

Also, the economies for multifuel engines in RBB are about 1.5 times those of
gas engine. This would make the economy Eq. for multiF:

28000 x 1.5 42000
----------- = ------ = economy multiF (km/l)
mass mass

The following is a table which uses my formulas and my ideas of the mass of
vehicles and compares my calculated economies to those of RBB. The table is in
the metric system. Values in () are the english system equivalents.

my Eq. gas RBB gas my Eq. MultF RBB MultiF
Size mass kg(lb) eco kpl(mpg) eco kpl(mpg) eco kpl(mpg) eco kpl(mpg)
-------------------------------------------------------------------------------
Compact 525 (1050) 53 (126) 60 (142) 80 (190) 90 (213)
Small 750 (1650) 37 (88) 50 (119) 56 (133) 75 (178)
Standard 1000 (2200) 28 (66) 30 (71) 42 (100) 45 (106)
Large 1300 (2860) 22 (51) 20 (47) 32 (77) 30 (71
Luxury 1700 (3740) 16 (39) 15 (36) 25 (59) 20 (47)

Again do these seem reasonable? What do I have to change?
I believe the eqs. qualify as simply and are reasonable. If the values aren't
what we want we can vary the numerator (e.i. change acc.)? What about the
car masses? Are these the masses we should be shooting for in cars without
accessories.
]-----------------------\--------------------------/---------------------------[
]Guak \ Lehigh University / 670 Atlantic Str. [
]A.K.A. Amonchare \ Chemistry Department / Bethlehem, PA 18015-3538[
](Michael A. Kauffman) \ 6 E. Packer Ave. / (215) 758-4706 (w) [
]mak9@******.edu \ Bethlehem, PA 18015 / (215) 868-5043 (h) [
]-----------------------\--------------------------/---------------------------[