Exhaust gas volume
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Joined: Sep 2002
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From: In the middle of Weather Dry Creek Farm in Avilla, Arkansas
Exhaust gas volume
Has anyone seen any info on the volume (cubic feet per minute) of exhaust gasses produced by the 5.9 at various throttle and load settings?
EXAMPLE: If the engine is rotating at 2000 rpm and not getting any fuel (for all purposes, acting like an air pump) it will expel around 208 cu ft of air per minute. Adding fuel, the mixture burns and expands a significant amount, thus increasing the volume. But by what amount? Looking at what the maximum exhaust flow rate is going to be.
If you're wondering, the info is for an exhaust system project.
maybei'vegotjusttoomuchtimeonmyhandsShortround out
EXAMPLE: If the engine is rotating at 2000 rpm and not getting any fuel (for all purposes, acting like an air pump) it will expel around 208 cu ft of air per minute. Adding fuel, the mixture burns and expands a significant amount, thus increasing the volume. But by what amount? Looking at what the maximum exhaust flow rate is going to be.
If you're wondering, the info is for an exhaust system project.
maybei'vegotjusttoomuchtimeonmyhandsShortround out
I'm sure the turbo manufacturer can provide the data you're looking for (or close enuf for govt work). I don't think the engine will consume any more air than what the turbo can provide. What comes out can't be any more than the sum of what goes in.
What law is that? Matter cannot be created or destroyed? The only one I can think of right now is the one that says brown stuff flows down hill.
What law is that? Matter cannot be created or destroyed? The only one I can think of right now is the one that says brown stuff flows down hill.
That's true, it can't consume any more air...but more gas will come out than went in because of the coversion of diesel fuel from a hydrocarbon liquid to carbon dioxide, carbon monoxide, etc. Mass flow is constant, the sum of the mass of fuel and air in equals the mass of exhaust out. I think it's called a mass balance equation.
A chemist could tell you. 30 years ago, when chemistry was somewhat fresh, I would have tried. I probably wouldn't have gotten it right, but I would have tried.
A chemist could tell you. 30 years ago, when chemistry was somewhat fresh, I would have tried. I probably wouldn't have gotten it right, but I would have tried.
First you would have to find out how much fuel is being injected into each cylinder under a certain amount of load. More load = more fuel needed to run @ 2000 rpm's. Maybe you can find out how many gallons per minute of fuel you're burning. Then you would have to find out exactly what hydrocarbons and other chemicals are in diesel fuel. Then you would have to do all the calculations for how much H2O and CO2 would be produced under perfect conditions. I'd say those numbers would get you close.
Then again, there might be some kinda CFM machine/instrument you could hook up to your tailpipe that would tell you. In fact, I'd check w/ some of the heating/air-conditioning contractors in your area to see if they have such an instrument...and if so, they might let you use it. Seems to me I remember seeing one used on a job somewhere...
Then again, there might be some kinda CFM machine/instrument you could hook up to your tailpipe that would tell you. In fact, I'd check w/ some of the heating/air-conditioning contractors in your area to see if they have such an instrument...and if so, they might let you use it. Seems to me I remember seeing one used on a job somewhere...
Might be able to use a bass-ackwards approach for a given horsepower. If you're truck develops 250 hp on the dyno you're developing roughly 12500 btu/min (if my conversion factors are ok). I understand diesel fuel contains about 130,000 btu's gallon, but small engines are maybe 33% efficient? so triple the fuel btu's to 37500 btu's/min convert that to fuel used 37500/130,000~.29gal/min, diesel weighs 7.4lbs/gallon so you're burning 2.15lb/min. If air fuel ratio is 25:1( at full throttle) you have roughly 55lbs/min of stuff going into you're engine . to keep it simple assume it's all air (at near room temp thats roughly 750cfm at the air filter). Now heat it to 1200F? the density of air at atm pressure at 1200 degrees is .024 lb/ft3 so 55lbs/.024lb/ft3 ~2300 CFM at the exhaust manifold. Pardon the SWAGs guesses and assumptions.
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From: In the middle of Weather Dry Creek Farm in Avilla, Arkansas
Some where from the depths of my mind, a 1:7 ratio pops up for turbo/supercharged diesel engines. 1:5 for N/A and somewhere around a 1:4 for gassers. Remember. I'm pulling out "old" hot rod stuff from 35 years ago and there is probably an error factor of + or - 25%. We used those formulas to figure the flow rate of mufflers.
With that in mind, a 5.9 @ 3200 rpm would have a WOT volume of ~2300 cfm. That matches close to the rated flow of a number of aftermarket mufflers. Dynomax has a good listing of muffler flow rates.
Like I said earlier, I've got too much time on my hands.
Shortround out
With that in mind, a 5.9 @ 3200 rpm would have a WOT volume of ~2300 cfm. That matches close to the rated flow of a number of aftermarket mufflers. Dynomax has a good listing of muffler flow rates.
Like I said earlier, I've got too much time on my hands.
Shortround out
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Actually now that I think of it we use a formula to measure water cfm though an open ended pipe by measuring the distance the water comes out before dropping times the pipe diameter.
Bet if you hung a weight on a string and measured how far the string moved from vertical when placed in the exhaust steam that it could be mathematically converted to cfms.
There you go, that should keep you busy for awhile.
Bet if you hung a weight on a string and measured how far the string moved from vertical when placed in the exhaust steam that it could be mathematically converted to cfms.
There you go, that should keep you busy for awhile.
Shortround, see if a local A/C contractor in your area has a rotating vane anemometer. You take a timed reading (say 10 seconds worth) multiply times the discharge size in square feet (.0872 sq ft for a 4" exhaust) multiply that by 6 (to get a 1 minute reading variable) and you will get a cubic feet per minute reading. You could do this for varying rpms to assimilate volumes at different operating speeds but the only drawback is that you are calculating at a no load condition
Another thing you could do is let me know how you want tests conducted and I can use my hot tip anemometer on my stock 02 and I can give you readings.
Gary
Another thing you could do is let me know how you want tests conducted and I can use my hot tip anemometer on my stock 02 and I can give you readings.
Gary
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From: In the middle of Weather Dry Creek Farm in Avilla, Arkansas
Gonna do it the easy way.
Tomorrow, I should have a couple of big ole pipes running out the back to get rid of the stuff. Muffler flows 2200+ cfm and should be enough.
All I had to do was use the old formula for explosives; P=E (Plenty = Enough).
Shortorund out
Tomorrow, I should have a couple of big ole pipes running out the back to get rid of the stuff. Muffler flows 2200+ cfm and should be enough.
All I had to do was use the old formula for explosives; P=E (Plenty = Enough).
Shortorund out
Here's something I found on the web...scroll down to the bottom of the page.
http://www.merequipment.com/Exhaust%...ncersizing.htm
MikeyB
http://www.merequipment.com/Exhaust%...ncersizing.htm
MikeyB
Wow! Did any one cruise that website? The Yanmar 440 horse 6 cylinder is pretty impressive at around 850 lb. torque
and the 4 cylinder, 240 horse doesn't sound bad either for a conversion. Now I need to find an old Landcruiser! We all can dream once in a while!! Jim
and the 4 cylinder, 240 horse doesn't sound bad either for a conversion. Now I need to find an old Landcruiser! We all can dream once in a while!! Jim



