## The purpose of Air-intake/Scoop?

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As this web-site is called F1Technical, the purpose of this thread was to comment on my draft calculations, see top. No fun that?
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

xpensive wrote:As this web-site is called F1Technical, the purpose of this thread was to comment on my draft calculations, see top. No fun that?

Ouch)) Sorry if I sounded harsh. Tried to add couple of things to consider.
Anyway I have a question - imagine a car with engine at idle revs. It still generates some thrust on exhausts, may it generate some thrust also on engine intake by consuming air and lowering the pressure?
timbo

Joined: 22 Oct 2007

That's ok timbo, no offense taken.

Obviously, every pressure differential, multiplied with the exposed area will generate a force, but in the case of the scoop I think extremely small.

The exhausts on the other hand, can generate a considerable pressure, consider the turbo days when 6 or 7 atmospheres on the inlet side was nothing uncommon.
Not exactly the same as the exhaust pressure obviously, depending on difference in turbine diameter, but not far away.
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

Just extrapolating the points made a bit...yes the dynamic pressure at the face of the inlet would be governed by the speed of the car, but this isn't a straight tube. From what I recall about ram-pipes, quite a trumpet shaped profile had quite an effect and a straight pipe was quite detrimental and turbulent to airflow, also as the pipe cross section changed there would be a change in pressure and airspeed - remember a flow in a pipe on reaching a smaller pipe the flow speed increases and the pressure drops. So the next question comes - what about a big face to the inlet to 'gather' a greater air volume in and increase air density to the engine. Inevitable trade-off with aero, depending how big you went - I'm guessing 4ft would be unhelpful to say the least. Any takers on the variable inlets of days gone by?
I am an engineer, not a conceptualist
alexbarwell

Joined: 20 Mar 2008
Location: London

I guess no matter the size or shape of the engine air-intake, you can never get a higher additional pressure on the trumpets than the generated dynamic such, Rho * speed^2/2? Or can you?

At 288 km/h and 1.2 kg/m^3, 3840 N/m^2 or 0.038 Bar.
Last edited by xpensive on Fri May 08, 2009 10:17 am, edited 1 time in total.
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

When I've a bit of time to run through the calcs I'll get you lot an answer (rather than post a ham fisted effort).
kilcoo316

Joined: 9 Mar 2005
Location: Kilcoo, Ireland

xpensive wrote:I guess no matter the size or shape of the engine air-intake, you can never get a higher additional pressure on the trumpets than the generated dynamic such, Rho * speed^2/2? Or can you?

At 288 km/h and 1.2 kg/m^3, 3840 kPa or 0.038 Bar.

You may want to re-check that conversion

'10-'11 Head of Powertrain - Glasgow University Formula Student
Scotracer

Joined: 22 Apr 2008
Location: Edinburgh, Scotland, UK

Obviously, that "k" was left there on purpose, just to see if there were any engineers in Scotland.
No typing-error whatsoever.
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

xpensive wrote:Obviously, that "k" was left there on purpose, just to see if there were any engineers in Scotland.
No typing-error whatsoever.

Such excuses could only ever come from an engineer

'10-'11 Head of Powertrain - Glasgow University Formula Student
Scotracer

Joined: 22 Apr 2008
Location: Edinburgh, Scotland, UK

Don't mention it, while I have to admit being mighty impressed it was spotted by an "imperial" engineer?
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

Finally, I'm back online.

Pt/P = (1+ [{(gamma-1)/gamma}*Mach^2]) ^[gamma/(gamma-1)]

Pt = total pressure
P = static pressure
gamma ~ 1.4 (gas const)
Mach = obvious

Simply put, that is the total pressure at the entry of the airbox.

Envisaging the IC engine as a turbojet - the airbox will equate to the inlet RAM pressure, while the 'suck & squeeze' phase of the cylinder stroke will equate to the compressor of the turbojet.

For a turbojet, the overall pressure ratio (OPR) is calculated as

Pr(tot) = Pr(inlet)*Pr(compressor)

The pressure ratio of the cylinder stroke would be measured using conservation of mass and the perfect gas law, i.e.

m = rho*V
P = rho*R*T

-or rearranging-

(P1)(V1)/(R)(T1) = (P2)(V2)/(R)(T2)

V1 = volume of cylinder at downward stroke limit
V2 = volume of cylinder immediately prior to ignition

P = pressure
R = gas const (287 J/kg/K)
T = Temperature

Someone talked about overfill earlier. Apply it as a factor to the numerator on the LHS.

Also, if the airbox were removed and the air entry to the engine placed parallel to the flow, then you have pressure working against you (approximate by Bernoulli), reducing your effective fill rates further.
kilcoo316

Joined: 9 Mar 2005
Location: Kilcoo, Ireland

Impressive kilcoo, but you need to crack some numbers in order to prove you can do different than what I wrote above:

I guess no matter the size or shape of the engine air-intake, you can never get a higher additional pressure on the trumpets than the generated dynamic such, Rho * speed^2/2? Or can you?

At 288 km/h and 1.2 kg/m^3, 3840 kPa or 0.038 Bar.

"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

xpensive wrote:Impressive kilcoo, but you need to crack some numbers in order to prove you can do different than what I wrote above:

I guess no matter the size or shape of the engine air-intake, you can never get a higher additional pressure on the trumpets than the generated dynamic such, Rho * speed^2/2? Or can you?

At 288 km/h and 1.2 kg/m^3, 3840 kPa or 0.038 Bar.

@ 80 m/s (Mach 0.24), Pt/P ~ 1.06

1.06 * 100000 ~ 100000 + 6000

= 6kPa gain.
kilcoo316

Joined: 9 Mar 2005
Location: Kilcoo, Ireland

I'm a tool... it didn't sound right, so I wrote it down... looked wrong, and then I remembered...

Its Pt/P = [1+ {(gamma-1)/2}*M^2]^[gamma/(gamma-1)]

Which @ 80 changes it to Pt/P = 1.04

4kPa.
kilcoo316

Joined: 9 Mar 2005
Location: Kilcoo, Ireland

And my original estimate above was 3840 Pa, or 0.038 Bar, at Rho=1.2. Pretty close, eh?
"Bernoulli is a nine-letter name"
xpensive

Joined: 22 Nov 2008

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