Calculating exhaust gas CFM............. 10

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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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.

I stand by my original reply. You should review your Thermo 101 notes. First and Second laws apply.
Franz

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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

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

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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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

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

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.

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

You especially learn something when SBBlue gets out the calculator.

Awesome work Bluey

Regards
pat pprimmer@acay.com.au
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Yeah, spot on, apart from those old fashioned units. I had forgotten about °R

Well said

Jeff

Bluey:
Sir, I doff my hat!
Good job.

Franz

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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.

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!

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.

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.


Credit for the formula's on the document.

Franz

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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.

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.