MAF

[Greg Hermann]

><snip>
>>170 Gm/sec works out to 300 cfm (and about 205 HP) (at sea level), which
>>works out to 84% or so VE--
>
>Could you elaborate on the math??
>
>Thanks

454 grams = 1 lb. (conversion constant)
.075 lbs./cubic foot = typical sea level density of air at a reasonable
temperature

(1) therefore 170 grams per second /454 grams/ lb. = .374 lbs./sec

(2) therefore, .374 lbs./sec. x 60 seconds/minute  = 22.47 lbs./min. (of
air, number used in (3) & (4) below)

(3) therefore 22.47 lb/min. / .075 lbs./cubic ft. = 299.6 cubic feet/min.
(cfm) of air. (actual intake air volume)

(4) therefore, 22.47 lbs./ min. x 60 min/hour = 1348 lbs/hour of air.

(5) therefore 1348 lbs./ hour / 13.5 (typical WOT a/f ratio for an NA
engine) =  99.87 lbs./hour of fuel

(6) therefore 99.87 lbs./ hour of fuel/ .48 lbs. fuel/ brake horsepower
hour ( a pretty typical brake specific fuel consumption figure for an NA
engine) = 208 Horsepower.

(NA = naturally aspirated)

Regards, Greg

Sorry--spaced out showing the VE calc--so here it is:

(7)  3.8 liters x 61 cubic inches/liter = 232 cubic inches

(8) 232 cubic inches/revolution / 2 revolutions/cycle x 5300
revolutions/minute = 614800 cubic inches/minute.

(9) 614800 cubic inches/minute / 1728 cubic inches/cubic foot = 356 cubic
feet/minute (theoretical swept intake volume)

(10) 299.6 cfm/356 cfm = 84.2% VE (actual intake air volume/theoretical
swept intake volume) ((3)/(9))

That oughtta do it!

Regards, Greg