Modelling a ram air intake on a car

[2457]

Sorry for such an easy question LOL..

So.. I need a bit of help,
I was thinking of modelling a ram air intake on a car.

But it seems like I'm not educated enough..

What I know is:
the car goes with 100 km/h speed,
the engine consumes 1800 liters of air / minute,
and the intake at the engine side is 60 mm diameter.

Question is, if I make an intake (a ram air intake) how could I calculate the pressure that should build up?

I understand pressure is froce divided by area,
and I do have the density of air, but how could I calculate force out of this?
I just don't get it..

Car mod forums say raim air effect is allmost zero until high speeds,
and that is something I don't quite belive. Thats why I had in mind to do some calculations,
but I had to face that I'm not a genius, LOL...

 

[Topher925]

This is not a simple problem it is going to be very complex to do accurately. Start with considering the throttle body as a simple nozzle or venturi. Then do a energy balance across the nozzle. On the incoming side, you need to model the air and how it is entering the throttle body, like through a nozzle, diffuser, kinetic energy will be significant. On the exiting side you need to model the piston on the intake stroke and consider it an expanding volume. The affect of ram air is not necessarily putting more air into the engine, its reducing the negative work placed on the engine during the intake stroke thereby making the process more efficient. Hope you like differential equations!

 

[2457]

christ...
Thats sounds a real complex thing.
Well, efficiency is what I'm after.
I done some work on the exhaust and some other things,
so this 1991 make old ford consumes 5.2-5.5 liter diesel every 100 km.

I travel about 150 km each day to my work and back (its a total there and back)
so every tiny bit of fuel is importent LOL.. (it ain't cheap here, hungary has the highest prices in europe)

So.. my.. naive/noob approac would be to calculate the pressure air makes becouse of the movment, that should be the kinetic energy divided by the are of the intake "flare".
And consider air as an uncomppressable materia, and calculate the pressure as:

A1 / A2 = p1 /p2

where A1 is the intake of the flare, A2 is the small pipe going into the engine,
and p1 is the pressure at the A1 surface, p2 would be the pressure at the engines intake pipe.

This may be a terrible approach, but that's all I could came up with..

Question is how em I supposed to calculate the kinetic energy of the air at the intake side of the flare? I have a surface and speed, and specific gravity, but the surface is.. 2 dimensional, and there for is not enough to calculate the weight of the air, how can I solve this??

 

[RonL]

christ...
Thats sounds a real complex thing.
Well, efficiency is what I'm after.
I done some work on the exhaust and some other things,
so this 1991 make old ford consumes 5.2-5.5 liter diesel every 100 km.

I travel about 150 km each day to my work and back (its a total there and back)
so every tiny bit of fuel is importent LOL.. (it ain't cheap here, hungary has the highest prices in europe)

So.. my.. naive/noob approac would be to calculate the pressure air makes becouse of the movment, that should be the kinetic energy divided by the are of the intake "flare".
And consider air as an uncomppressable materia, and calculate the pressure as:

A1 / A2 = p1 /p2

where A1 is the intake of the flare, A2 is the small pipe going into the engine,
and p1 is the pressure at the A1 surface, p2 would be the pressure at the engines intake pipe.

This may be a terrible approach, but that's all I could came up with..

Question is how em I supposed to calculate the kinetic energy of the air at the intake side of the flare? I have a surface and speed, and specific gravity, but the surface is.. 2 dimensional, and there for is not enough to calculate the weight of the air, how can I solve this??

I would think you can find some good information, if you look for a few websites that deal with drag racing, they have worked with this type problem from day one of the sport.

As Topher925 stated, any forward motion decreases the work that the engine has to do and only at higher speeds, will you start to see much increase in efficiencies.

The small things can make a difference, so keep at it.

Ron

 

[Q_Goest]

Hi 2457,

Question is, if I make an intake (a ram air intake) how could I calculate the pressure that should build up?

You can make a good estimate of the increase in pressure according to Bernoulli's equation just as one would for example, calculating a pitot tube.

http://www.grc.nasa.gov/WWW/K-12/airplane/pitot.html

 

[2457]

That pitot pipe is really helpfull, other question is what happens when some accelerates.

about small differences.. this car is old. When I bought it, it consumed 7.5-9 liter deiesel every 100 km. I'm down to 5.2-5.5 with a big handfull of small changes.
for example, the cooling water tubes had been replaced with a much thicker one, and the old pump was removed, the one I use was made for 2.5 liter diesel truck.
The speed of the pump was decreased, and the bearings of thisone are much better too.
That alone made a change, almost 0.5 liter every 100 km.
bearings are verry imortent, my son gived me this advice. So i think if removeing such a small resistence from the system gave such a nice difference, this one should do it too..

I allready ordered an other stock air filter, so I can duole the surface, which means less work for the pistons to bread. Allso the axels which drive the wheels have been balanced, so they are not shakeing, that's an important thing too.

I can make a pretty decent surface to try ram air, at the sides of the engines cooling radiator I have 2 huge gaps. About 30 cm height, and a good 10 cm wide, on both sides.
Should be a nice one, comaperd to the 60mm diameter pipe. (surface is = 28,3 sqcm, and the 2 intakes I plan to make are 600 sqcm).

The other major tweak I done is reduceing drag. The exhaust pipe is short, it is under the car. So the smog fills he low pressure area. Is simple thinking, but gives mesureable amount of fuel efficiency increase, my next week projekt is to try ad tune the exhaust pipes for y travel speed.

 

[RonL]

One thing I forgot to mention, Look into electric driven superchargers, you can get air flow of considerable velocity, even if your moving just a few miles per hour.

Lots of people out there are trying to promote cheap stuff, so be careful.

If you like to do your own designing you can find lots of components from surplus dealers (a lot of this is new equipment) and things can be assembled for a very attractive price.

Ron

 

[2457]

elektronic chargers--- no way.

Consider that my old 1.8 ford consumes at 3000 RPM
about 2700 liter of air every single minute.

Just moving this volume of air if hard.
Hard to achieve the airflow.

Besides that I need no power gain.
I just want to ride as cheap as possible.

On my sons audi s4 we reduced the turbo lag by adding a nice dc motor to the turbo,
120 watt power rateing. It reduces the turbo lag. Butz it eats the batterys...
120 watt is not much power, but still needs a bit more than 10 amps from the battery..

 

[rcgldr]

Doing the math, assuming no losses:

pressure increase = 1/2 density speed ^2

pressure of air at sea level at 70 F = 14.696 lb / in^2

density of air at sea level at 70 F = .074887 lbmass / ft ^3
1 slug = 32.174 lb sec^2 / ft
1 foot = 12 inches
density of air at sea level at 70 F = .000001346969 slug / in ^3

if speed is feet / sec

pressure increase = 1/2 .000001346969 (slug / in ^3) speed ^2 (ft^2/sec^2)
1 slug = 32.174 lb sec^2 / ft
pressure increase = 1/2 .000001346969 (lb sec^2 / (ft in ^3)) speed ^2 (ft^2/sec^2)
pressure increase = 1/2 .000001346969 (lb / in ^3) speed ^2 (ft)
1 foot = 12 inches
pressure increase = 1/2 .000001346969 12 (lb / in ^2) speed ^2
pressure increase = .00000808181 (lb / in ^2) speed ^2

1 kph = 0.621371192 mph = .9113444 ft / sec

at 100kph, pressure increase = 0.067123 lb / in^2, less than 1/2% increase

To check the math, I compared with an article on ram air:

150mph = 220 ft / sec, and pressure increase = 0.391160 (2.66% increase)
1 psi = 68.948 mb (millibar)
150mph, pressure increase = 26.97 mb

For a speed of 150 mph, the resulting maximum theoretical pressure would be about 27mb (approximately .4 psi).

http://www.sportrider.com/tech/146_9910_ram/index.html

The gain in power is more than the pressure increase over ambient, becasue normally there's a reduction of pressure below 1 atm at the intake of an engine. Some tests have shown a 6.5% increase in power at 150mph. At 62mph (about 100kph) this would drop to about 1.1%. These are power increases though, I don't know how efficiency is affected.

Part 2 of the above article includes some graphs but not % increase in power.
http://www.sportrider.com/tech/146_9912_ram/index.html

 

[2457]

WOW!
You sure know something, that's for sure.

From the article it seems that the size of the intake does not mather, as it will fill up with air, so gigant size will make no difference.

Other thing I would be interested is how.. do these ram air intakes behave while accelarateing?

the low pressure is not really a problem, I do not want more power from the cars engine.
If I would need that kind of stuff an intercooler and any small supercharger would be my choice :) This old diesel was never designed to handle extra induction, but if negative pressure gets eliminated at travel speed I may gain an other 0,1-0,2 liter less fuel consumption, in my situation it is a big value.
(dayli about 150 km I drive at travel speed of 100 km/h, 300 days a year. If It consumes just 0,1 less, its 45 liters a year. Thats about 800 km free ride, LOL.. )

Well, my question was answered, I dun't know how to thank You for it.

I would not like to spam this forum,
but I allso have an other gadget in mind that could help to lover fuel consumption a bit.
If I could cool down the air 1 degree, that would increase the oxigen by 1/273 for every unit of air the engine sucks.
now.. for the peak power the enigne runs at about 2000 RPM, it takes 1800L air every minute, 108000 L every hour, and in exchange it produces 44 Kw power.
soo... are my calculations are o.k.?
44000 * (1/273) = 161 watt hour
1 Joule is 2.7778 ×10−4 watt hour.
To cool down 108000L air, I would need (air cp ~ 0.001297 J cm-3 K-1)
108000L air is 108000000 ccm,
108000000*0,001297 = 140046 joule
and 140046 joule * 2.7778 ×10−4 = 38 watt hour

Allso diesel engines provide a good amount of heat, I have a few dozens of 200 watt/hour TECs laying around, a simple metal maze is easy task to do for air intake (so the surface of the air touching the "cold-plate" is big) is no big deal, and the ram air intake should be enough to eliminate the air resistence it would make.
If my calculation is right, I could gater about 1000-1200 watt with 10 TECs without any problems from the cooling water, and power other TECs to cool down the air.
1000 watt cooling power each hour would decrease the air temperature (if I did not screw my calculations) by 26 degree, that should give 26*160 watt = ~4 Kw of power to the engine, a nice free (4kw *100 / 44Kw) 9% power. That would be some nice thing, instead of 60 Hp I would have 65 Hp.
Keeping the TECs cold is no problem, the hood is made out of metal, and is a nice heat spreader, when doing 100 Km/h constantly. well not the best, but it could work.
Question is, did I estimate things well? efficiency of the electronic stuff can be omitted, if I like I can add 12 of those TECs to the engines cooling water,
it has loads of energy in it. The cooant holds 7 liter water, that's 7000 ccm, and the temperature under normal conditions running on the highway with 100Kph is 85 celsius degree, even at winter time. Thats about (water cp = 4.184 J cm-3 K-1) 7000*4.184 joule for each degree, and it gets this hot in a few minutes.

 

[rcgldr]

Oops, definition of slug messed up, but ok as used in equations:

1 slug = 32.174 lbmass

1 lbforce = 1 slug ft / sec^2 (force = mass x acceleration)

1 slug = 1 lbforce sec^2 / ft