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Calculating exhaust gas CFM.............Helpful Member!(10) 

JadedSilver (Automotive) (OP)
2 Oct 04 17:21
I'm working on a little project here.  Nothing serious or new like that, but I do need to figure out how to calculate the exhaust flow volume of a particular sized engine.  the project is based around extracting power through efficiency, and as such we are gathering every last bit of information from the motor now before any other work commences.

I'd imagine there has to be some sort of formula or series of formulas that can solve this, similar to that of finding the charge temp of a force inducted motor, or the CFM intake of an engine(which I have already completed).

I'm also assuming there are multiple variables including but not limited to a/f, fuel octane, displacement, RPM, exhaust temp, OAT,ATM, humidity, etc.  Even if I can locate some formulas or charts that would get me a rough idea of the exhaust CFM then that would probably suffice for what I need to do.

These would be based around a standard DOHC gasoline engine that is turbocharged.  I have located a chart about this topic for turbo desiels, but they are based around different operating parameters, and fuels of course.

If you have any information I'd greatly appreciate the effort on your part in lending me a hand.

Sincerely,

-Adam
R247 Motorsports  
Helpful Member!  franzh (Automotive)
2 Oct 04 22:03
The exhaust volume is the same as the intake volume, just hotter.  If the temperature were normalized to intake temperature, the volume would be the same, plus the minute amount of fuel.  You cannot create more mass just by heating the medium, it simply expands.
Franz

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JadedSilver (Automotive) (OP)
3 Oct 04 0:29
Oops, I worded the post wrong.  I need to know how to calculate the exhaust CFM, not the volume.  Air combining with fuel and burning has to create a larger CFM flow going out than coming in so that's where I need to know the math to figure it out.

franzh (Automotive)
3 Oct 04 0:31
I stand by my original reply.  You should review your Thermo 101 notes.  First and Second laws apply.
Franz

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wr180561 (Electrical)
3 Oct 04 2:28
You should start by determining the displacement of exhaust gas at each exhaust stroke by determining  the cylinder volume when the exhaust valve starts to open in one particular cylinder (this should equal to the piston displacement minus that volume from BDC to the opening of exhaust valve) and let’s call it L, this volume should represent the volume of exhaust gas displaced by the piston in each exhaust stoke.  Then you can count the number of exhaust stokes per minute by dividing the engine’s rpm  over 2 (since one exhaust stoke takes place every two rpms) and let us call it n/min. Now by multiplying L X n/min you will get the exhaust flow in L/min for each cylinder. Finally multiply this figure by the number of cylinders you will get the exhaust flow in L/min for that engine.

Note,  this calculation is relevant to a 4-stroke engine without considering the loses or the leaks that may take place during operation. WR

GregLocock (Automotive)
3 Oct 04 2:43
Ooer Franz. Ingoing mixture: 4N2+1.5O2+CH2

That is 5.5 moles of gas (+ liquid hydrocarbon)

Outgoing gas 4N2 + CO2 + H2O

That is 6 moles

Or have I tripped over myself?

Cheers

Greg Locock

franzh (Automotive)
4 Oct 04 21:37
Sorry for the delay, had trouble logging on last night and had to teach today.

Let's try this out:
The exhaust CFM is mainly a function of temperature.  The mass of the air and fuel is the same going in and out.

Using the Ideal Gas Law:
PV = RT

then;
V2 = V1 * T2/T1 where T = absolute temp.

There is a little conversion of O2 and HC to CO2 and H2O, but with the constant laws, matter cannot be created or destroyed, thus the mass remains constant.  I guess I should rephrase my earlier statement:

"The exhaust mass is the same as the intake mass, just hotter.  If the temperature were normalized to intake temperature, the mass would be the same, plus the minute amount of fuel.  You cannot create more mass just by heating the medium, it simply expands."

If you take the standard equation of calcuating the CFM of the engine, do the math for thermal expansion, you should have the exhaust flow in CFM.  If the mass were heated it would expand in volume but not in mass.  It would be less dense (hence the smaller exhaust valves and exhaust ports as opposed to the intake valves and ports).

I guess this really makes the water cloudy, there is more to it than simple equations.  Time to get out my chemical engineering handbook, but I hope this gets you in the right direction.

Thanks to my director for a reality check.

Franz

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MaxRaceSoftware (Automotive)
4 Oct 04 22:35
The mass of the air and fuel is the same going in and out.
The exhaust mass is the same as the intake mass, just hotter
======================================================

yes..thats what i also thought ?
i was Porting a pair of BBC heads for a College Student
and he was going for a mechanical engineering degree ..and he was asking me all kind of questions and that subject came up..i told him the same thing ...but he came back the next day and said his Professor told him different ??

its been 10+ years ago..so i can't remember what the Professor told him  :)





Larry Meaux (maxracesoftware@yahoo.com)
Meaux Racing Heads - MaxRace Software
ET_Analyst for DragRacers
Support Israel - Genesis 12:3
http://www.maxracesoftware.com/

GregLocock (Automotive)
4 Oct 04 22:39
Sure, the mass is equal, but the number of molecules of gas increases slightly, because each O2 becomes 2/3 of a molecule of CO2 and 2/3 of a molecule of H2o, with a hydrocarbon that has roughly twice as many hydrogen atoms as carbon atoms.

Cheers

Greg Locock

Helpful Member!(9)  SBBlue (Automotive)
4 Oct 04 23:46
Okay, class. Excellent question, Jaded. Pay attention now, because this will be covered on the test.

Several assumptions: 1) This is a gas engine we are talking about, 2) it is a four-cycle gas engine, 3) Combustion will be stochiometric and complete, 4) the compression ratio is about 10:1, 5) the engine is throttled (no variable valve timing), 6) normal aspirated engine (no turbocharger), and 7) volumetric efficiency (the amount of air that makes it into the cylinder during the induction stroke) is 1.00 (actually it depends upon the RPM and intake manifold pressure, but work with me here.)

First, it should be intuitively obvious to the most casual observer that the amount of air that passes through the engine in will be equal to the engine displacement times the RPM divided by 2. For an engine of 3 liter displacement going at 3000 RPM, the amount of air pumped for minute will be 4500 liters.

That will approximately be the intake volume flow for an engine with the throttle wide open. If we assume that the throttle is only open 33%, the intake volume flow will still be 4500 liters, but the pressure will be one-third of an atmosphere. The equivalent mass of air will be the same as 1500 liters at one atm of pressure.

Neglecting the addition of the fuel mass, the mass of the exhaust gas will be the same as the mass of the intake gas. From the ideal gas law we know that the increase in volume of the exhaust gas will be proportional to the increase in absolute temperature. If we assume an intake temperature of 80 deg F, and an exhaust temperature of 1800 deg F (reasonable assumption, depends upon compression ratio), the absolute temperature will be 540 and 2260 deg Rankine, respectively. The volume increase will therefore be 2260/540, or 4.185.

For the hypothetical 3 liter engine running at 3000 RPM and full throttle, the exhaust gas volume will be about 4500*4.185, or 18,833 liters/min. At one third throttle the corresponding flow is 6277 liters/min. Since one cubic foot is equal to 28.3 liters, the respective CFM flows will be 665.4 and 221.8, respectively.

How about the contribution from combustion products? Assuming stoichometric combustion, there will be one pound of fuel burned for each 14.55 lbs of air. Air is 21% oxygen, so there is 3.05 lbs of oxygen available to burn each pound of gas.

A reasonable chemical approximation for gasoline is octane, which has a chemical of C8H18. The molecular weight is (12*8+18*1)= 114.

The combustion formula is C8H18 + 12.5 O2 ==> 8 CO2 + 9 H20. For each 114 grams of C8H18, there will be 12.5 moles of oxygen consumed, producing 8 moles of CO2 and 9 moles of H2O. For gas volume purposes, since equal moles of gas produce equal volume, the volume of exhaust gas replacing oxygen will be equal to 17/12.5 = 1.36.

The volume percentage of oxygen in air is about 21% (not exact, but work with me here). This volume will be removed, and replaced by exhaust gas with a "volume" of (21*1.36) = 28.56%. The resulting post combustion volume is (79% + 28.56% = 107.56%) of the pre- combustion volume -- assuming no temperature increase.

So what do we have? Combining the increase in volume from combustion reactions and thermal expansion, an engine with a 3 liter displacement running at 3000 rpm with the throttle wide open will have an exhaust volume (at 1800 deg F) of 665.4*1.0756 ~~ 715 cubic feet per minute. For the throttle one-third open, the exhaust flow will be 238.6 cfm.

I'm tired, so I may have made some calculation errors, but this should get you in the ball park for exhaust flow.
Majik (Automotive)
5 Oct 04 3:17
Perfect :)  hadn't factored in for that post combuston volume but it makes perfect sense.  definitly something to consider when redesigning exhaust ports for flow...  I love this forum.. learn something new everyday
patprimmer (Publican)
5 Oct 04 3:40
You especially learn something when SBBlue gets out the calculator.

Awesome work Bluey

Regards
pat   pprimmer@acay.com.au
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notnats (Mechanical)
5 Oct 04 6:40
Yeah, spot on, apart from those old fashioned units. I had forgotten about °R

Well said

Jeff
franzh (Automotive)
5 Oct 04 7:10
Bluey:
Sir, I doff my hat!
Good job.

Franz

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DesignOutsourcing (Automotive)
7 Oct 04 4:01
Excellent work! I actually remembered some of this from high school chemistry but was really surprised to see how little of the increase was due to the change in composition-- just assumed it was higher than 7.5%.  thanks for the clarity.
crysta1c1ear (Automotive)
7 Oct 04 21:51
I think its the hydrogen that does it.
Let's say 3 oxygens molecules get used up in combustion.
A couple will end up as CO²s and the third will end up doubling into two H²Os.

If air is say 21% oxygen, then 1/3 of that is 7%.
So its 1/3 of the 21% oxygen is changing from
O² to
H²O + H²O.

SBBlue's maths is far more accurate than mine, so I hope this posting isn't a step backwards!
Rob45 (Automotive)
8 Oct 04 10:33
SBBBlue gives the best explanation,  but I have found over many years of flow-bench testing exhaust systems,  then measuring on-car (or -truck) exhaust backpressures,  that a very good simple approximation of the exhaust gas flowrate is to use the engine volume x rpm x .5 (for a four-stroke) then correct for gas temperature.  The error in assuming 100% volumetric efficiency about equals the error in neglecting the products of combustion.
For example, a ZR-1 Corvette engine running 6000 rpm with a gas temperature of 1400 F showed an exhaust flow rate around 2000 cfm.
This worked well for sizing exhaust systems,  and as a "reality check" when comparing bench to vehicle backpressure data.
franzh (Automotive)
12 Oct 04 9:17
This thread keeps bugging me.  Although I agree with SBBlue's comments, I felt there was something left out.  Therefore, I contacted a professor friend of mine (UT-Austin) who sent me his interpretation, but unfortunatly I was unable to cut and paste the math figures here.  I have posted it on a seperate website for any to review.

http://franzh.home.texas.net/rdmattcalc.doc

Credit for the formula's on the document.

Franz

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SBBlue (Automotive)
12 Oct 04 12:46
I agree with the good professors ruiminations.

The only thing I would add, is that he is addressing things that rapidly become very complicated.

For example:

He mentioned the use of a turbocharger and it's effects on volumetric efficency, which is all well and good. However, the exhaust turbine will greatly effect the temperature and hence the volume of the exhaust gas. Upstream of the turbo the exhaust gas will be hotter but the pressure will be higher; downstream of the turbo the exhaust gas temperature, and hence volume will be lower. But this also depends on the load is on the exhaust turbine, which of course will vary according to engine load and throttle opening.

Another thing to consider is how much engine cooling affects the exhaust temperature. As an extreme, a perfect adiabatic engine (no heat loss through cylinder walls) will have a higher exhaust gas temperature than a more traditional water cooled engine operating at a low load and low rpm.

The ignition timing also affects exhaust gas temperature. The longer it takes for complete fuel burn after top dead center on the combustion stroke, the more energy will be converted to heat and the less will be converted to mechanical energy.

I would suggest it would be much more reasonable to measure the exhaust gas temperature for different conditions rather than to try to calculate it. His comments about the "slow exhaust stroke" are true, but are less a factor the more cylinders an engine has, since the exhaust flow from the various cylinders tends to mix and average out.

Warpspeed (Automotive)
12 Oct 04 17:25
I suppose it depends on how accurate you want to be, and at what point you are measuring things.

A quick and dirty rule of thumb figure, might be 1.5 CFM of induction air per horsepower, and 2.2 CFM of exhaust flow per horsepower. That is a ballpark figure, usually quite sufficient to size a common exhaust pipe or muffler.

On the other hand, the temperatures close to the exhaust valve can cycle quite a bit along with port pressure during the actual exhaust cycle. The flow will obviously be pulsing fairly violently as well.

Averaged CFM right at the valve seat will not tell you much because temperatures, pressures, and velocities are constantly changing. A bit further down the pipe, flow conditions will still not be steady, but an averaged CFM figure might be more relevant.
kmb1949 (Automotive)
28 Nov 04 16:52
I am developing a new diesel engine design and we are begining to machine and assemble the prototypes. During our dyno test research we found that we could make 340hp with 575cfm of ambient air. This was accomplished in a 305ci four cylinder at 2800rpm running 37psi boost. The new designs should be capable of much greater air flows. I do not have an engineering degree but am very interested in fuel system applications and cylinder combustion. The last post seemed a little short of our dyno results and I was wondering if you guys might help me understand more fully the relationship between air and fuel for optimum combustion. Our indicated BSFC was .32.We do not yet have the equipment to test emmisions but I was also wondering what a good BSFC might be for diesel engines.
franzh (Automotive)
29 Nov 04 9:13
kmb1949
This might be better addressed as a new thread.

Franz

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sacem1 (Mechanical)
29 Nov 04 15:42
I have just read this thread and I find the discussion quite interesting and the analisys of the variables not known are really the difference between values obtained by the different methods but I don't get to understand this reasoning from SBBLUE (Oct 4):

"That will approximately be the intake volume flow for an engine with the throttle wide open. If we assume that the throttle is only open 33%, the intake volume flow will still be 4500 liters, but the pressure will be one-third of an atmosphere. The equivalent mass of air will be the same as 1500 liters at one atm of pressure"

To my understanding if we are at 1/3 throttle the engine will be at 1/3 rpm's and still be filled at each stroke by the same amount of air at the same atmospheric pressure than at full throtle because the slight additional resistance in air flow at the throttle will automatically be compensated by opening the butterfly a little bit more, we are really setting the 1/3 throttle position by the 1/3 rpm's shown on a tach because the position of the carburetor butterfly is not proportional to rpm's so I do not see where is the 1/3 of an atmosphere pressure coming from.

Can anyone clear up this?

SACEM1

ivymike (Mechanical)
29 Nov 04 16:39
I think that when he said 1/3 throttle he meant butterfly opening = 1/3 of max (on an effective flow area basis).  That's really the only answer that would seem to make the statement approximately correct, and is fairly consistent with my understanding of how the terms "wide open throttle" and "part throttle" are used (independent of RPM).

sacem1 (Mechanical)
29 Nov 04 18:13
I agree with you on that, what I don't get is the drop in inlet presure, to me the atmospheric pressure is maintained what cuts the gases output is that the engine is going at one third the rpm's working at the atmospheric pressure.

SACEM1
GregLocock (Automotive)
29 Nov 04 18:36
I understand your confusion. At low rpms you are at full throttle (zero MAP) even though the throttle angle is only 30 or 45 degrees.

That's why we plot and calibrate everything against MAP, not throttle position. The throttle position sensor is mostly used for some predictive transient stuff, and fault detection.

Cheers

Greg Locock

ivymike (Mechanical)
29 Nov 04 20:00
I agree with you on that, what I don't get is the drop in inlet presure, to me the atmospheric pressure is maintained what cuts the gases output is that the engine is going at one third the rpm's working at the atmospheric pressure.

While the actual throttle angle to achieve the desired pressure drop may be rpm-sensitive, I think that's beside the point that SB was trying to make.  At any given rpm, it's still intake flow restriction via the throttle valve that sets the power output (up to the WOT power @ that rpm).  If you wanted to estimate the CFM exhaust gas, you could use power/(WOT power) at the given rpm to estimate it.

sacem1 (Mechanical)
30 Nov 04 13:08
I'm not worried about the throttle angle, the point I don't get is if there should be a drop in inlet pressure to 1/3 atmospheric when you are at 1/3 lets call equivalent throttle opening, maybe just in the moment the driver "floors it" (pushes the pedal all the way down opening the butterfly completly) and the rev's haven't gone up yet but once stabilized at lets say 1000rpm's the inlet pressure should be atmospheric, am I wrong or not?

Cheers

SACEM1
ivymike (Mechanical)
30 Nov 04 13:31
I'm not sure I understand what you're asking.  
  * If the throttle is wide open and the rpms are low, then intake pressure should be close to atmospheric.  How quickly the throttle was opened and how long it has been open should have insignificant effects (other intake restrictions may be more important).
  * If the engine is being "throttled" to reduce power output to 1/3 of max power at a given rpm, then the intake pressure should be about 1/3 of the WOT pressure at that rpm.
  * If the engine is throttled to reduce power output to 1/3 of max power at 1000rpm, and you subsequently snap the throttle wide open, intake pressure will increase very rapidly and engine power output will increase almost immediately to the WOT power output @1000rpm.  If you simultaneously increase load on the engine to keep rpm constant, then the intake manifold pressure will remain at atmospheric and the power output will remain constant at (wot@1000) from then on out.  

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