So if you have the pressure ratio of each turbo and the PSI discharge of each turbo, I feel like there should be an equation out there that can convert that to lb/min. I'm really interested in this as well.
Do these numbers look right...
- Thread starter chevyburnout1
- Start date
well flow is different from pressure. i thought the same but searching around and reading, thats when i saw that.
Flow and pressure are both relevant, as long as you know the restriction (aka: engine displacement, volumetric efficiency, rpm, etc). I though I had found an equation that somewhat factored all of those variables in. It said at those ratios the stock charger was running about 57lb/min, however it only said the large charger was at 50ishlb/min. So I'm not sure if I did that right haha.
Here is what I found that I was basing it off of. I may have not factored in the correct variables in regards to our engine's however.
CALCULATING AIRFLOW RATE AT REDLINE
cid = Cubic Inches Disp.
VE = Volumetric efficiency in percent
.5 = (given) 4-stroke engine fills cylinder only on one-half the revolutions
1728 = converts cubic inches to cubic feet
Airflow in CFM = (cid x rpm x .5 x VE%) / 1728 = CFM no boost
CFM x PR = CFM under boost
now since most compressor maps have their flow rates in LB/MIN we need to convert CFM to LB/MIN. A cubic feet of air (length+width+height) weighs different at diff. Altitudes and different temperatures. to simplify it well just assume we are at sea level and the air temp is 112 *. the conversion number is 0.070318.
so for CFM TO LB/MIN = CFM x 0.070318 = LB/MIN
CALCULATING AIRFLOW RATE AT REDLINE
cid = Cubic Inches Disp.
VE = Volumetric efficiency in percent
.5 = (given) 4-stroke engine fills cylinder only on one-half the revolutions
1728 = converts cubic inches to cubic feet
Airflow in CFM = (cid x rpm x .5 x VE%) / 1728 = CFM no boost
CFM x PR = CFM under boost
now since most compressor maps have their flow rates in LB/MIN we need to convert CFM to LB/MIN. A cubic feet of air (length+width+height) weighs different at diff. Altitudes and different temperatures. to simplify it well just assume we are at sea level and the air temp is 112 *. the conversion number is 0.070318.
so for CFM TO LB/MIN = CFM x 0.070318 = LB/MIN
but the stock turbo will be sucking down the pressure the 480 would normally push without a turbo infront while still moving the same amount of air so to me, the pressure i see in the 480 with a stocker in front may in fact flow the same amount as the 480 by its self pushing more psi.
so if i see 15psi right now from the 480 with the stocker in front, it could be that the 480 at 26psi without another turbo would be pushing the same amount of CFM. obviously not that far apart in psi and this only an assumption.
i was thinking, if we know where a stock turbo really looses efficiency at around 27+gpsi boost, then as long as i keep the calculated boost pressure of the stock turbo at or below that, i would not be pushing the small turbo too far.
but then how do you know how to split the cfm up for one turbo or the other? that what im trying to figure out
This is definitely true! Which may be why my lb/min readings didn't seem correct for the S480.
That equation seemed to work for the small charger, which from what I have been told is the most important to get in the 'efficiency island' of the turbo map. However I'm still at a loss for figuring out lb/min for the large charger
Since the airflow from the large charger is what is making the hp could you base lb/min off of what the truck actually dynos? So if you do end up making that 600hp you can roughly assume you're around the 80-90lb/min range? All of this assuming there are no charged air/exhaust leaks
Thats what i was thinking but i was really looking for a way to find that split. from what Mark is saying, we need to know turbine speed to figure it out which sucks cause unless we got an EFR turbo i cant see how you would know the speed.
i figure the 480 is flowing quite a bit of CFM but i was looking to see where the stock turbo is going and if it may be going to far one way or the other in its map. ive noticed that if i back the regulator down to say 45 psi total boost, the 480 hits 20-22psi. if i bump it to 50, it drops to 15psi. if i bump it to 55 psi, its still 15psi out the 480. thats where i was thinking if i keep the stock one on its calculated pressure below the max of 32 psi on a stock single truck, i would be good though i know its outside its map and i should see EGT rise when its starts running outside its map. would it be fair to assume that or does it not work that way?
so flow should essensially be 85 lbs/min out of the 480 into the stocker and then 85 lbs/min out of the stocker into the motor? i was thinking the CFM from the 480 would be less than the full 85 lbs/min and the stock charger then increased it some to the full 85 lbs/min
if thats the case, a map of the stock charger does me no good as its way off plot and instead must create a whole new map for its self while under a different atmosphere
Turbo maps are based on a single setup only. I don't believe they make any 'compound' maps. So this is why we have to do math to figure out the calculated lb/min the stock charger would be making if it was being ran as a single.
Turbo maps are based on a single setup only. I don't believe they make any 'compound' maps. So this is why we have to do math to figure out the calculated lb/min the stock charger would be making if it was being ran as a single.
josh is staying its 85 lbs/min from the outlet of that 480 all the way to the motor though. i dont follow on that or maybe i didnt quite understand his answer. if i understood him correctly, you cant find what lbs/min the stocker is moving as its not doing any work in the CFM department and its all on the 480 which sounds strange in my head thinking about it. i have that same map up in one of my tabs currently and going back and forth looking at it as we have been discussing it
:roflmao:
Josh is correct. Airflow is based on the large charger. The lb/min readings of a turbo map will not be correct when there is more than atmosphere pressure being introduced into the inlet of the turbo. That's why you have to do some math and figure out what it would be if there wasn't a second charger pressurizing the inlet.
Yeah, all the air through the system is being pulled in by the big charger. That's why the big charger is the limiting factor on how much power a set of compounds can make.
*Please don't take anything from here on out as fact because I'm merely hypothesizing on how this works*
If I was to guess, and I am, I would say you could get a rough idea of where the small charger is running in it's map by dividing the lbs/min by the pressure ratio the big charger is running at.
example: Say the big charger is putting 30 psi of pressure into the cold pipe. That would be a pressure ratio of 3.1:1 ((30+14.5)/14.5=3.06). Now, say the big turbo is flowing 90 lbs/min. Divide 90 by 3.1 and you get 29.0 lbs/min. Now we just need the pressure ratio of the small turbo. Say the output is 60 psi, that would give you a pressure ratio of about 1.7:1 ((60+14.5)/(30+14.5)=1.67).
Here's a map I found for an S366:
Plugging in the numbers, you can see it would be running in the middle of it's efficiency range, but right at the bottom of the map. You could push it up further into the middle by driving the small charger harder. That would drop the pressure ratio of the big charger, moving you further to the right, and it would raise the pressure ratio of the small charger, moving you higher up.
Take all this with a grain of salt. I very well could be completely wrong on how it works.
Yeah, all the air through the system is being pulled in by the big charger. That's why the big charger is the limiting factor on how much power a set of compounds can make.
*Please don't take anything from here on out as fact because I'm merely hypothesizing on how this works*
If I was to guess, and I am, I would say you could get a rough idea of where the small charger is running in it's map by dividing the lbs/min by the pressure ratio the big charger is running at.
example: Say the big charger is putting 30 psi of pressure into the cold pipe. That would be a pressure ratio of 3.1:1 ((30+14.5)/14.5=3.06). Now, say the big turbo is flowing 90 lbs/min. Divide 90 by 3.1 and you get 29.0 lbs/min. Now we just need the pressure ratio of the small turbo. Say the output is 60 psi, that would give you a pressure ratio of about 1.7:1 ((60+14.5)/(30+14.5)=1.67).
Here's a map I found for an S366:
Plugging in the numbers, you can see it would be running in the middle of it's efficiency range, but right at the bottom of the map. You could push it up further into the middle by driving the small charger harder. That would drop the pressure ratio of the big charger, moving you further to the right, and it would raise the pressure ratio of the small charger, moving you higher up.
Take all this with a grain of salt. I very well could be completely wrong on how it works.
if your right, thats exactly what i was looking for. it sure seems feasable but obviously i dont know enough on the subject to confirm or deny
x2
running the math the way you state josh really seems to fall in line with what i see egt wise and how the truck acts. dropping the psi to 45 psi would put the stock charger near the choke edge and at the low end of its graph which the truck has higher egts and smokes more. bumping it up to 50psi brings it closer to the middle of the graph and better all around where i saw a drop in egts and less smoke. 55psi shows it should work even better.
all as a hypothesis though.
I'd like to point out that I accomplished nothing at work today after I started crunching numbers haha.