Calculating Ram Effect

[HaveBlue]

Here is some info from some engineering sites.

The calculation is quite trivial if you have had exposure to gas dynamics and understand the Isentropic Flow tables. Take the speed of the car and divide by the speed of sound. This gives you the Mach number of the air relative to the car and presumably the air intake. Go to the Isentropic Flow table and find the value of P/Pt for the given Mach number. Take the inverse of P/Pt this gives the ratio of the stagnation pressure to the atmospheric pressure. For choked flow the maximum flow rate is directly proportional to the stagnation pressure Pt. And it I calculate the change in pressure as 1.2% at 100 MPH which makes ram air meaningless for these cars.

Josh Gillett
Oregon State FSAE

RAM AIR

Operating Conditions
Temperature = Tair = 20C = 293K
Atm. pressure = Pair = 14.7PSIA
Cpair = 1005J/KgK
K = 1.4
The Cpair and K are constants for air.

Case 1 @ 100Kmh (62Mph) = 27.78m/s

Calculating temperature of the ram air

Tramair = ((Vcar^2/2gc)/cp) + Tair
Tramair = (((27.78m/s)^2/2(1kgm/Ns^2)/1005J/kg) + 293K
Tramair = 293.4K

The temperature increased by 0.4K or 0.4C.

Pram = Pair (Tram/Tair)^(k/(k-1))
Pram = 14.7PSIA (293.4K/293K)^(1.4/(1.4-1))
Pram = 14.75PSIA - 14.7PSIA
Pram = 0.05PSIG (gauge pressure)

So as you can see driving 100kmh will only have a gain of 0.05 psi! now lets try for 200kmh.

Case 2 @ 200Kmh (124Mph) = 55.5m/s

Calculating temperature of the ram air

Tramair = ((Vcar^2/2gc)/cp) + Tair
Tramair = (((55.5m/s)^2/2(1kgm/Ns^2)/1005J/kg) + 293K
Tramair = 294.5K

The temperature increased by 1.5K or 1.5C.

Pram = Pair (Tram/Tair)^(k/(k-1))
Pram = 14.7PSIA (294.5K/293K)^(1.4/(1.4-1))
Pram = 14.97PSIA - 14.7PSIA
Pram = 0.27PSIG (gauge pressure)

By seeing how the velocity of the car increases the ram air effect...it is barely anything! I'll conclude by saying that in racing circles where a 1/100th of a second counts, it's worth it, but not on the street.

[DGthe3]

I look at it from an airflow perspective (instead of pressure). Its a little less precise, but a lot easier to follow.

Find the flow rate of air the inlet would be collecting by multiplying its area by the speed of the car. Find the air consumption rate of the engine at WOT (displacement*RPM/2). If the car needs more air than is being forced in through the scoop there is no significant effect.

For my Grand Am (which says Ram Air right on the fender) the twin openings are roughly 7.5"x3" which gives an area of 290 cm^2. At 60 mph that would give 46,700 L/min. Now, how much air does my engine need? My 3.4L makes its peak power @ 4900 rpm so it would require 8,200 L/min. There should be a rather significant ram air effect in my car. I have no clue how much extra power I would get from it but clearly more air could be forced into the engine than it would ever need.

Contrast that with a 9"x1/2" mailslot opening feeding a 6.2L engine. That would give 9300 L/min at 60 mph but the engine needs over 18,000L/min of air at its power peak of 5900 rpm.

[live2well]

Originally Posted by DGthe3

I look at it from an airflow perspective (instead of pressure). Its a little less precise, but a lot easier to follow.

Find the flow rate of air the inlet would be collecting by multiplying its area by the speed of the car. Find the air consumption rate of the engine at WOT (displacement*RPM/2). If the car needs more air than is being forced in through the scoop there is no significant effect.

For my Grand Am (which says Ram Air right on the fender) the twin openings are roughly 7.5"x3" which gives an area of 290 cm^2. At 60 mph that would give 46,700 L/min. Now, how much air does my engine need? My 3.4L makes its peak power @ 4900 rpm so it would require 8,200 L/min. There should be a rather significant ram air effect in my car. I have no clue how much extra power I would get from it but clearly more air could be forced into the engine than it would ever need.

Contrast that with a 9"x1/2" mailslot opening feeding a 6.2L engine. That would give 9300 L/min at 60 mph but the engine needs over 18,000L/min of air at its power peak of 5900 rpm.

No disrespect intended, but check out this link and then rethink your calculation using The fluid mechanics theory. The results may surprise you. http://www.camaro5.com/forums/showthread.php?t=48033

[DGthe3]

Originally Posted by live2well

No disrespect intended, but check out this link and then rethink your calculation using The fluid mechanics theory. The results may surprise you. http://www.camaro5.com/forums/showthread.php?t=48033

None taken. It was just my common sense figurin' about what was involved. Fluid mechanics was a while ago and I haven't used it since, and I haven't taken any courses related to automotive design. I know enough to realize that simply having more air available isn't enough on its own to do much beyond maybe counteracting some restrictions in the intake, assuming it could even do that.

[live2well]

Originally Posted by DGthe3

None taken. It was just my common sense figurin' about what was involved. Fluid mechanics was a while ago and I haven't used it since, and I haven't taken any courses related to automotive design. I know enough to realize that simply having more air available isn't enough on its own to do much beyond maybe counteracting some restrictions in the intake, assuming it could even do that.

WoW I didnt know Our northern brothers were so smart.


(please hear the sarcasm in my voice)