01102013, 08:37 PM  #1 
R3VLimited
Join Date: Aug 2009
Location: Colorado Springs, CO
Posts: 2,931

Volumetric Efficiency
I am trying to look at turbo compressor maps for turbo selection. The formula is straightforward for airflow of a piston engine.
CFM = L x RPM x VE x Pr The problem I am having is trying to get realistic numbers from this formula for a boosted engine. Since forced induction can increase the volumetric efficiency of an engine the amount of air an engine can flow (CFM) should not only increase based on RPM, but also Volumetric Efficiency (VE). Displacement and Pressure Ratio stay the same. A soild value for 2 valves VE is 85%, and 9095% for 4 valve heads. 5660 What I am missing is where the engine begins to increase over 100% volumetric efficiency. For Example an M20B25 at 10 PSI at Sea Level: PR = 14.7 psi + 10 psi/ 14.7 psi = 1.68 (2.5 L x 2500 RPM x 85 VE x 1.68 )/5660 = 157.7 CFM Now at the RPM we want to be spooled by, say 3500 RPM (2.5 L x 3500 RPM x 85 VE x 1.68 )/5660 = 220.7 CFM And Now the same engine at 6000 RPM (2.5 L x 6000 RPM x 85 VE x 1.68 )/5660 = 378.5 CFM But shouldn't the engine begin building boost, say at 3500 RPM, and somewhere cross 100% VE and begin nearing 110% VE or higher? Let assume we get to 100% VE while building boost at 3500 RPM. (2.5 L x 3500 RPM x 100 VE x 1.68 )/5660 = 259.7 CFM And now the same engine at 6000 RPM assuming we are now at 110% VE. (2.5 L x 6000 RPM x 110 VE x 1.68 )/5660 = 489.8 CFM While spooling this is a 17.7% increase in flow and near redline this is 29.4% more flow. I belive this is an under approximation of the VE, but the point is the same, typical engine flow calculations should be higher due to the turbo's effect on the engines VE. Take a look at the map below at a 1.68 PR and the difference in flow on the compressor map. This turbo is either correct or may be too small for the m20, but with more accurate calculations shouldn't the effect of the turbo on the VE incorporated into the calculations make turbo selection more accurate? The PR is incorporated but it doesn't take into account how it affects the VE, just the AMOUNT of air in the engine, not the EFFICIENCY. I know im spewing, but what do you guys think? NINJA EDIT: Volumetric Flow rates for some BMW V12 Engines. Interesting read, considered looking at the M70 engine specifically as it is basically two M20B25s siamesed together. It has very good info on some turbo charged V12s. http://forum.roadfly.com/threads/112...iency...(Long)
__________________
Last edited by downforce22; 01102013 at 10:00 PM. 
01102013, 11:28 PM  #2 
E30 Modder
Join Date: Apr 2012
Location: Monster Island
Posts: 937

one day im going to know exactly what your talking about.
Is there a difference between VE and efficiency? I thought the amount of air filling the cylinder is VE. Wouldnt that be the same as the amount of air in the engine? School me on this. 
01102013, 11:36 PM  #3 
Moderator
Join Date: Nov 2003
Location: bellingham, wa
Posts: 32,771

VE is the difference between an engine's displacement and how much volume of air it's intaking relative to that displacement. efficiency usually refers to thermal efficiency but it can mean a lot of things.

01112013, 04:48 PM  #4 
R3VLimited
Join Date: Aug 2009
Location: Colorado Springs, CO
Posts: 2,931

Nando nailed it. VE is amount the engine can flow, so by definition the VE of a FI Spark Ignition engine is going to at some point be higher than 100% because the compressed air is moving more air than the engine can theoretically move at ambient pressure (14.7 psi). So for the N/A m20b25 say it can flow 250 cfm at sea level at 85% VE. (Doing the math that is at 6660 RPM) With a VE increase to 110% we get 323.5 CF. So we are flowing more because the engine is matched well with exhaust, intake, valves, cam etc. You probably wouldn't see these big of gains on any engines without forced induction, nonetheless a 2 valve per cylinder m20. The Mclaren F1 has 105% VE as a natural aspirated V12 (Built by BMW ), so it is possible to achieve over 100% VE on an NA engine.
CFM = L x RPM x VE x Pr This m20 example is just to get an idea of how much the volumetric efficiency affects the output of the engine. But remember, this is just making the engine more efficient. A byproduct of boost is increasing the pressure ratio, so on a FI engine the numbers change a bit. Since Displacement, RPM, and VE are multiplied by the pressure ratio the increase in air flow is also multiplied meaning there is more flow due to the higher the pressure ratio. This is why most people think cranking up the boost will help, because it will increase flow if the turbo can handle flowing more. But when PR goes up, I believe so does VE to an extent. The purpose of this thread is to shed some light onto where a forced induction engine increases in VE, at what point it crosses over 100% or ideal volumetric efficiency and beyond, as well as how far it goes, (150% VE, or 200% VE?).5660 The main reason I brought this up is for turbo selection. If you calculate a certain flow from the engine, but it is actually flowing more because an increase in VE has not been taken into account in the calculations, then the turbo you choose will be too small. Refer to the attached pictures, but I converted the CFM rating into lb/min numbers by dividing by 14.5. 85% VE at 3500 and at 6000 RPM, PR= 1.68 220.7 CFM = 15.2 lb/min 378.5 CFM = 26.1 lb/min 100% VE at 3500 and 110% VE at 6000 RPM, PR= 1.68 259.7 CFM = 17.9 lb/min 489.8 CFM = 33.8 lb/min Comparing the graphs below for the same exact turbo at the same exact pressure ratio; At the usually suggested 85% VE for 2 valve engines you can see not only does the engine flow less, but it is in a low efficiency at 3500 RPM, meaning the turbo is adding more than optimum heat. It is also just getting into the most efficient zone near redline around 6000 RPM. The second map shows the engine being more efficient at 3500 RPM and just starting to run out of steam at 6000 RPM. It is important to note the efficiency of the turbo is right in the powerband, about 40005500 RPM, Using a higher VE for the same engine at the same conditions produces the dilemma of what value to use. Because one formula shows the turbo being just right for the turbo, right in the rev range, pulling to redline and spooling quickly while the other shows some lag and being a bit big. Probably not much fun unless in the higher revs, like on a race track. So what VE are we at during FI and during spooling? You could see how this could make somebody believe their turbo will surge, when it really is incresing VE and not going to be a problem.
__________________

01122013, 11:52 AM  #5 
E30 Modder
Join Date: Apr 2012
Location: Monster Island
Posts: 937

hmm..
so i just plugged in my numbers for a 2.0l m42. Also, i have 7mm valve, some porting, etc. Could i assume 95% VE? Im not seeing where your getting your VE value. Are we just assuming? Im having a hard time grasping this... PR = 14.7 psi + 9 psi/ 14.7 psi = 1.61 2500 RPM (2.0 L x 2500 RPM x 90 VE x 1.61 )/5660 = 128.00 CFM = 8.8lb/min 3500 RPM (2.0 L x 3500 RPM x 90 VE x 1.61 )/5660 = 179.20 CFM = 12.3lb/min 6000 RPM (2.0 L x 6000 RPM x 90 VE x 1.61 )/5660 = 307.20 CFM = 21.1 lb/min So im almost positive im reading it wrong but in that first map it seems that it would hold well through the powerband? fuck... im going to try this again and ill be back to edit this post. 
01122013, 12:00 PM  #6  
Moderator
Join Date: Nov 2003
Location: bellingham, wa
Posts: 32,771

Quote:
it's really not at all the same. the M70 has completely different cams, intake, exhaust, and heads. I'd put money on the M20 being more efficient. the M70 is meant to have a lot of low end, lazy grunt. it's a cruising motor meant to be quiet and refined, but have lots of torque everywhere. the only thing the M20 and M70 really share are the 91mm bore spacing, 75mm stroke and 84mm bore. but most BMW engines share some of these characteristics.. 

01132013, 01:04 AM  #7 
No R3VLimiter
Join Date: Nov 2005
Location: Sydney, Oz
Posts: 3,834

you can't get VE from a dyno graph alone so the roadfly link is not correct and the numbers should be taken with a grain of salt, also his very definition is wrong. An engine always breathes 100% of the volume of the cylinder it is impossible to not being a gas. What ever amount whether it be a large or small amount that goes in fills the cylinder exactly. The easiest way to imagine VE is to not use volume at all, it is the density in the cylinder versus what the density would be at ambient conditions (STP).
mass flow through the engine is more relevant than volume flow
__________________
89 E30 325is Lachs Silber  currently M20B31, M20B33 in the works... new build thread http://www.r3vlimited.com/board/showthread.php?t=317505 
01132013, 11:33 AM  #8  
R3VLimited
Join Date: Aug 2009
Location: Colorado Springs, CO
Posts: 2,931

This is going to be long, so get some popcorn.
Quote:
Quote:
Quote:
But perhaps you can chime in on your thoughts on the subject of volumetric efficiency in turbocharged application if/when/as the VE increases. Actually I did some digging myself and found this website with the following equation, http://hpwizard.com/volumetricefficiency.html VEboost=VEunboost*((boost gauge pressure / atmospheric pressure) + 1)^(1/1.7) It is just as I suspected that boost raises VE multiplicatively (or in reality exponentially) with boost pressures. So what does this mean? For my example above of the m20b25: It would be accurate to assume VEboost at 2500 RPM is 85% because the we have selected a turbo already that will be big enough to not spool yet. (2.5 L x 2500 RPM x 85 VE x 1.68 )/5660 = 157.7 CFM From the last calculation, say 3500 RPM and we are boosting say we just spooled up to full pressure. (2.5 L x 3500 RPM x XY VE x 1.68 )/5660 = ?? XY = 85 ((10psi/14.7) +1)^ 1/1.7 = 115.33 VE (2.5 L x 3500 RPM x 115.33 VE x 1.68 )/5660 = 299.53 CFM Remember my last post showed a guess of VE at 100% at 3500 RPM with a flow of 259.7 CFM. It was lower than what we will see if we are actually spooled by 3500 RPM. At this point, our flow will actually be 300 CFM when the initial formula showed 220 CFM at the same RPM. Using the initial formula (in the first post) we actually underestimated the flow by 36%!! Simply because we didn't factor in the VE changing. Lets see how we are up top of te rev range. (2.5 L x 6000 RPM x XY VE x 1.68 )/5660 = ?? We already found the boosted VE: XY = 115.33 VE (2.5 L x 6000 RPM x 115.33 VE x 1.68 )/5660 = 513.5 CFM The inital formula showed a CFM of 378.5, and my guess of VE at 110 in the last post was pretty close to the 115 resulting in 489.8 cfm. Again, this is a 36% gain in flow from the original equation. So what does this mean? Lets plot it again with actual values for this m20b25 example at seal level and 10 psi. 3500 rpm = 299.53 CFM = 20.69 lb/min 6000 rpm = 513.5 CFM = 35.41 lb/min PR = 1.68 This map looks great for the car, at full boost and 3500 RPM we are in the sweet spot of the turbo. We are far from surge and starting to run out of steam at higher RPM. This turbo would be great on the street, responsive and perfect for daily driving. It would not like the track much where hitting 6000 rpm continuously may happen, but would feel great on the highway. I also plotted the flow at 2500 RPM and 0 boost and it shows that line from 20% redline and PR = 0 is a pretty good estimate of how the turbo will spool. Thanks for reading, I answered my own question! I may make a spreadsheet or another thread on how to PROPERLY select a turbo based on VE and how VE changes with different boost settings at different altitudes.
__________________
Last edited by downforce22; 01132013 at 11:42 AM. 

01132013, 04:57 PM  #9 
E30 Mastermind
Join Date: Dec 2009
Location: Bay Area, CA, USA
Posts: 1,623

http://www.squirrelpf.com/turbocalc/ I'll throw this in here for those who want to mess around with visualizing some of this data on normalized compressor maps.
__________________

01192013, 10:33 AM  #10  
E30 Modder
Join Date: Apr 2012
Location: Monster Island
Posts: 937

Quote:
I make it by without any surge, it doesnt fall off near redline, stays in the efficient islands, correct? 

01192013, 10:39 AM  #11 
Moderator
Join Date: Nov 2003
Location: bellingham, wa
Posts: 32,771

what does boost increasing VE mean? it means, that the more VE you make N/A, the less boost you require to make the same amount of HP.
so, spend your money on two things: the head, and an efficient turbo. 
01192013, 11:05 AM  #12 
E30 Modder
Join Date: Apr 2012
Location: Monster Island
Posts: 937

head is 7mm titanium valve retainers, stainless steel valves, bigger valve guides,
VAC valve springs. efficient turbo and cams soon. trying to move to SD so i have no spending money. being broke is always a good time to learn though and downforce, im glad you posted this thread. good info here. 
01292013, 02:51 PM  #13 
E30 Addict
Join Date: Sep 2011
Location: Socal
Posts: 503

i love guys smarter than me. thanks for the education..

02022013, 02:30 PM  #14 
Banned
Join Date: May 2010
Location: 98310
Posts: 9,837

Excellent post! Just learned alot! thanks
EDIT: Just re read the whole thread and have a question Why is there a VE and a Pressure calculation? If the pressure s higher than 1.0 (like 1.6) then doesnt that in turn increase the VE? Why would you need to calculate the VE of a boosted engine if the pressure value is higher than 1.0? Somebody please explain Last edited by 5Toes; 02032013 at 12:33 PM. 
02122013, 12:18 PM  #15 
R3VLimited
Join Date: Aug 2009
Location: Colorado Springs, CO
Posts: 2,931

By pressure I assume you mean pressure ratio. I think what you are asking is why the VE would change when boosted. Or why the VE wouldn't changed directly proportionally to the PR. To say it simpler, if your VE was 85%, why wouldn't the boosted VE at PR 1.6 be .85 * 1.6?
The answer to this solution is not simple. Looking at the below equation, that is already factored into the boosted air flow. By definition, a turbo is an air pump, and that will be a restriction of the exhaust however, it will also compress the air into the engine. There will be losses. One problem I have with the below equation is that air density is not constant at different temperatures and pressures. Hence the term density ratio. Not only is atmospheric pressure lower at altitude, but also the density is slightly lower. So multiplying by a pressure ratio does not take the difference in density into account. BSFC also changes which is related to how much power the engine can make. And when you change one parameter, it slightly alters the output of the engine. CFM = L x RPM x VE x Pr/ 5660 At the start of this thread I was investigating the affects of turbo applications on VE and ultimately power output. The only true way to find VE is to work backwards from a dyno graph and use the actual power output at a specific rpm to find the actual air flow and divide that by the theoretical air flow of the engine displacement. For all of these situations there are assumptions that must be made and these assumptions make a solution tough to obtain, we could be off by 1015% or more on the numbers above. I did find this calculator that factors in a lot of these parameters depending on how detailed you want to be. The corrected air flow should be able to make it useful on any turbo map you may find. Look at the videos for a tutorial, though they are a bit dry. Again you have to make some assumptions for VE putting us back to where we started but you can estimate those from a dyno of a stock m20b25 with this equation. 100% VE AIRFLOW (scfm) = DISPLACEMENT (ci) x RPM / 3456 http://www.turbodriven.com/performan...6_wrsin=92044& For further reading I would recommend this website: http://www.epieng.com/piston_engine...efficiency.htm
__________________

Thread Tools  

