Ian C Posted February 28, 2003 Share Posted February 28, 2003 What's the conversion rate from CFM to lbs/min when considering airflow? I can't find a conversion factor anywhere, and I'm working out a 3l engines' boosted CFM's to around 750 at 7,000rpm, and that doesn't fit in with turbo compressor maps that use lbs/min, ranging from 10 to 80ish on that scale... I suppose cfm/10 is too simplistic...? -Ian Quote Link to comment Share on other sites More sharing options...
Nathan Posted February 28, 2003 Share Posted February 28, 2003 Now that is one seriously bloody good question..... I'm not even sure that the conversion is possible? CFM will not take into account air density but lbs/min will, which is why I guess turbo manufacturers would use it as I think it would be more accurate. I must admit I'm pretty unsure on this. I have 16 pages of A4 in front of me with conversion factors for just about everything on the planet and yet there is no mention of this conversion. The nearest is CFM>LBs of WATER (62.4242). I would think that using this equation could mean some funny figures... Maybe find the conversion for water to air and use the above formula? Thing is, we are back to density again, which will affect the weight... AARGGHHH!!! Nathan TDI PLC Quote Link to comment Share on other sites More sharing options...
Chris Wilson Posted February 28, 2003 Share Posted February 28, 2003 Originally posted by Ian C What's the conversion rate from CFM to lbs/min when considering airflow? I can't find a conversion factor anywhere, and I'm working out a 3l engines' boosted CFM's to around 750 at 7,000rpm, and that doesn't fit in with turbo compressor maps that use lbs/min, ranging from 10 to 80ish on that scale... I suppose cfm/10 is too simplistic...? -Ian I make it 62.28, so agree with Nathan, but air pressure, humidity and stuff will have some effect, but the above should give you an idea. Quote Link to comment Share on other sites More sharing options...
Digsy Posted February 28, 2003 Share Posted February 28, 2003 The density of air will depend on the temperature and atmospheric pressure (or the pressure of wherever the CFMs of air that you want to weigh are). A book or someone who knows gas laws should be able to help out. Quote Link to comment Share on other sites More sharing options...
Nathan Posted February 28, 2003 Share Posted February 28, 2003 Hmm....I'm not sure Chris. The 62.42 figure assumes you are dealing with water as your CFM side of the equation, which is obviously quite far removed from the air that Ian is trying to convert. On top of that, theres no mention of 'per min' in the 62.42 conversion either. In short, I don't think the 62.42 figure is of any use whatsover. Ian, don't go anywhere yet..... Cheers, Nathan TDI PLC Quote Link to comment Share on other sites More sharing options...
Digsy Posted February 28, 2003 Share Posted February 28, 2003 These might be worth a look: http://wright.nasa.gov/airplane/airprop.html (WTF is a slug?) http://www.chem.sunysb.edu/hanson-foc/lesson12.htm Quote Link to comment Share on other sites More sharing options...
Adam W Posted February 28, 2003 Share Posted February 28, 2003 I've got some fluid dynamics textbooks at home, I'll have a look this evening. Quote Link to comment Share on other sites More sharing options...
Chris Wilson Posted February 28, 2003 Share Posted February 28, 2003 The "slug" is the unit for mass using pounds per square foot. At the Earth's surface, a slug is about 32.2 pounds. SEE HERE Quote Link to comment Share on other sites More sharing options...
Chris Wilson Posted February 28, 2003 Share Posted February 28, 2003 Originally posted by Nathan Hmm....I'm not sure Chris. The 62.42 figure assumes you are dealing with water as your CFM side of the equation, which is obviously quite far removed from the air that Ian is trying to convert. On top of that, theres no mention of 'per min' in the 62.42 conversion either. In short, I don't think the 62.42 figure is of any use whatsover. Ian, don't go anywhere yet..... Cheers, Nathan TDI PLC I'm not sure now, either. Hmm, got a good pass in physics A levels too... Much alcohol and carbon monoxide has taken its toll though, since then. Quote Link to comment Share on other sites More sharing options...
Graham Rudd Posted February 28, 2003 Share Posted February 28, 2003 Not sure if this is what you want or not but: CFM * 0.076 = lbs/min Graham Quote Link to comment Share on other sites More sharing options...
Nathan Posted February 28, 2003 Share Posted February 28, 2003 Blimey Gra, where did that come from? Chris, I know what you mean about alcohol & monoxide....and I'm only 31. Lord knows what I'll be like at your age.....probably be a recluse in a big house in the middle of nowhere with a bloody great dog to keep me company.... Cheers, Nathan. TDI PLC Quote Link to comment Share on other sites More sharing options...
Digsy Posted February 28, 2003 Share Posted February 28, 2003 Check it out (I really should be working, but this is much more interesting). To calculate the mass of air at any temperature or pressure you need: T: The temperature in K (=degrees C + 273). P: The pressure in N/m^2 V: The volume of air in m^3 And - this is the key to it all: R: Universal gas constant. But it needs to be expressed in J/kg/K and not any other unit. This value for air is 286. Then use the rearrangement of the Ideal Gas Law: n=(PxV)/(RxT) Where n=mass in kg. The reason why the units for R are so important is that normally this law is applied to a single gas, whereas air is a mixture of gases. Also, it is more usually expressed in "J/mol/K" which would give you the answer in Moles, which again is only applicable to a single pure gas. Using the value 286 rigs the equation to spit out its answer in kg. Thankfully all the information is on the Nasa page I posted earlier - except in the metric column they have @r5ed up all the units and I have just had to spend an hour working it all out. So, you can either assume fixed standard temperature and pressure and work out a fixed conversion factor of (P/T) which would look like: n= factor x (V/R) Or you can plug in whatever you think the real temperature and pressure figures will be. Sorry but I can't work this out in imperial - it was har enough in metric (thanks, NASA!) Quote Link to comment Share on other sites More sharing options...
Graham Rudd Posted February 28, 2003 Share Posted February 28, 2003 Nice info Darren. http://www.grapeaperacing.com/GrapeApeRacing/tech/turbosize.cfm They have a conversion factor of 0.069 which is different to the one I posted earlier, although I suspect there is quite a bit of tolerance on the range of values which will 'do'. Quote Link to comment Share on other sites More sharing options...
Nathan Posted February 28, 2003 Share Posted February 28, 2003 Ahh..that'll be the answer then Graham!! Great info, and I think will give Ian everything he needs. I would offer you a job but you're probably better off where you are.. Like the breakdown too Darren, interesting. Cheers, Nathan TDI PLC Quote Link to comment Share on other sites More sharing options...
Digsy Posted February 28, 2003 Share Posted February 28, 2003 I've just taken my calcs a step further and confirmed your earlier post, Graham. If you work out the factor (P/T) for the standard conditions on the NASA page you get 351.736 Since R is a constant, you can incorporate it into the factor also: n= 351.736 x (V/286) becomes n = 1.229 x V Note that the 1.229 is actaully the density figure from the NASA site, so its all working so far. If we make V=0.0283m^3 (1 cubic foot) then n = 1.229 x 0.0283 = 0.0348kg and the kg to lb conversion is to multiply by 2.2, so 0.0348 x 2.2 = 0.076lb So at this temperature and pressure 1 cubic foot or air weighs 0.076lb. Quote Link to comment Share on other sites More sharing options...
Adam W Posted February 28, 2003 Share Posted February 28, 2003 I just followed the link of Grahams and worked through my engine (a 3l engine revving to 6500 at 1.5 bar gives 38.9lb/min at 2.5:1 pressure ratio by my calcs). However, theres no allowance for intercooling in that formula. I know my IC gets my charge temps down to around 20-25C at low air temperatures, surely that means my density ratio isn't as high as that formula suggests? (It's giving compressor outlet temps of 330 fahrenheit.) Quote Link to comment Share on other sites More sharing options...
Ian C Posted February 28, 2003 Author Share Posted February 28, 2003 Christ, I was expecting one reply with a ratio figure! OK, before my head explodes, can I say that at sea level (hey, I'm in Norfolk, we don't 'do' hills) at about 15degC I can convert 1cfm to 0.076lbs/min? My "divide by ten" rule doesn't really work, then, no wonder all those turbos looked too small Thanks for the highly technical input, folks. That's my £10 justified -Ian Quote Link to comment Share on other sites More sharing options...
Graham Rudd Posted February 28, 2003 Share Posted February 28, 2003 Originally posted by Ian C Thanks for the highly technical input, folks. That's my £10 justified -Ian £10? Its paid subscription based now? Should I have sent a cheque off to someone? Sorry if I missed something, I haven't been around for a while. Quote Link to comment Share on other sites More sharing options...
Nathan Posted February 28, 2003 Share Posted February 28, 2003 Originally posted by Graham Rudd £10? Its paid subscription based now? Should I have sent a cheque off to someone? Sorry if I missed something, I haven't been around for a while. FREELOADER!!! FETCH THE GALLOWS!! Anyone would think you have actually put some input into the BBS, like answering one of the more techincal questions for a long time with a one-line bit of text.... Nathan TDI PLC Quote Link to comment Share on other sites More sharing options...
Graham Rudd Posted February 28, 2003 Share Posted February 28, 2003 Bah! Where do I send my money? I leeched off this BBS for long enough when I owned a soop. I'd be more than happy to contribute to its upkeep. If I get another I can then continue asking dumb questions without feeling guilty. Actually I feel ripped off, I used to have to write to you seperately Nathan. I get rid of the car and suddenly you pop up here; are you trying to tell me something? Quote Link to comment Share on other sites More sharing options...
Nathan Posted February 28, 2003 Share Posted February 28, 2003 Chaps, Darren, I sent your mail to someone far smarter than I (my bro and his cronies he met at Uni) who I must say haven't the foggiest clue about cars or engines, but are generally pretty clued up when it comes to maths. He posted the below to me: expressed as J/K/kg is foolish, as it's only true at one pressure and temperature. And as you want to work out pressure/volume changes, you're up $hit street without a paddle this way. You need to stay in moles. Not a problem that it's a mixture of gasses. Air deviates from ideal gas laws at about 230 bar. As regards to converting the result back to imperial, thats all you need to do, not the whole equation system.>> Then, I sent him a compressor map from a T04 and asked for his comments. They were: temperature and pressure. You know pressure, from the left hand axis. You know your mass flow from the horizontal axis. You know your engine RPM. Right? So, look at the pressure and the mass flow. For example, at 84000 rpm (surely not, or is this rpm of the turbo?), mass flow of 25lb/min, you have a pressure ratio of 2.0. I'm guessing this is ratio compared to atmospheric. So your gas is at 2 bar absolute. In free air, the mass flow would be 25x2 bar= 50 lb/min. (Assuming no temperature change.) Now, At 1 bar, dry air weighs 1.173 kg/m^3 Or 2.6 lb/m^3 1 ft = 0.3m 1 cuft = 0.027m^3 Air weighs 2.6 x 0.027 = 0.07 lb/cuft So you can work out the flow by 50 / 0.07 = 714 CFM Does that make sense? The numbers don't look too insane... Rich>> I haven't pulled apart his formula yet but it might help. Really don't need this on a Friday afternoon. A little story; both my brother and 'Rich' studied applied maths for years, side by side. My brother decided to take a different path when he left Uni and is now working as an I.T. solutions director for Merril Lynch (after being at Deutsche Bank and Warburg). Rich is using his credentials in the hospital field. My bro has several properties, gets bonuses that would buy my house, runs 3 cars (plus a tractor), an autogyro and generally has an extremely comfortable life. All of which comes from a leeching merchant bank. Rich lives in a 2 up/2 down terraced house in Sheffield, drives an old diesel Peugeot 406 and can just about afford his love of diving. He works on cures for blocked arteries and other old-people related diseses. He has personally extended the lives of more than 200 people. Sorry about that. Totally off topic I know but goes to show that the world is crazy. Have a good weekend all. Nathan. TDI PLC Quote Link to comment Share on other sites More sharing options...
Digsy Posted February 28, 2003 Share Posted February 28, 2003 Originally posted by Nathan OK, the 286 may be right (haven't worked it out). Using R expressed as J/K/kg is foolish, as it's only true at one pressure and temperature. And as you want to work out pressure/volume changes, you're up $hit street without a paddle this way. You need to stay in moles. Not a problem that it's a mixture of gasses. Air deviates from ideal gas laws at about 230 bar. As regards to converting the result back to imperial, thats all you need to do, not the whole equation system. Okay, I'm keeping an open mind about this. I was cr@p at thermodynamics at college, but something I do remember is that the Mole is an atomic or molecular parameter - i.e. it applicable only to atoms of pure gases, so as far as I know you cannot relate it directly to a mixture of gases like air. It can be defined as the number of atoms needed such that the number of grams of a substance equals the atomic mass of the substance. As far as I can see, for Ian to work in Moles he would need to know how much a "mole of air" weighted to convert N back to kg. I had a look about and reckon its 28.964 grams per mole. I should be able to use R in J/mol/K and then use this to convert at the end, but I'm going for a curry in a minute. But R is the Gas Constant. It is the measure of energy in a unit amount of a gas per degree, therefore it shouldn't change with temperature or pressure, rather energy changes are determined by it. While writing this I have come across another similar value for R for air of 287.05 Joule/Kg/K. If it does change with temperature and pressure then it is foolish for the sites I pulled this off to call it "R" - it should be called something else. Anyway it all seemed to work when I layed it out, but as your freind correclty points out I only used one temperature / pressure point so it might all go to hell if either of these figures change. I'm not dissing what your mate says. If he works in chemistry then I'm sure he's forgotten more about moles than I will ever know, but I'm not quite ready to say I got it wrong yet. Quote Link to comment Share on other sites More sharing options...
Adam W Posted February 28, 2003 Share Posted February 28, 2003 I thought I understood it all until Nathan posted that My slightly inebriated take on it . . . R is a CONSTANT, it does not change regardless of temp, pressure or whatever. You can do dimensional analysis and some algebra to change the units you write it in, but it won't change your result (if you do it properly). All the constant really does is allow you to use sensible units for your answer. The part that will change dependant on air temp and pressure etc is the simple equation, x CFM = y lb/min If you take a cubic foot of air and heat it, if it's enclosed in a fixed volume it's pressure will go up as each molecule has more energy and wants to vibrate about more. If it's not, its density will go down as it expands and the molecules spread apart. (All basic stuff.) So one CFM of hot air could be 5 lb/min, or it could be 50 lb/min, depending on how hot and how pressurised it is. You should be able to come up with a workable conversion ratio for the air being sucked into the inlet as the pressure is atmospheric, and the temperature will be within a fairly narrow range. At the compressor outlet though, the three variables of temp, pressure and density are constantly changing over a wide range in a sort of equilibrium thingy. Quote Link to comment Share on other sites More sharing options...
Ian C Posted February 28, 2003 Author Share Posted February 28, 2003 My head just blue-screened. I grasp a straw: "So you can work out the flow by 50 / 0.07 = 714 CFM" That 0.07 looks similar to the 0.076 conversion factor I was clutching at further up the thread. So I can use that? "for Ian to work in Moles" - well, that's not going to happen I draw the line somewhere! Oh, and yes, I have a paramedic friend who's saved more lives than I care to think about, and seen others fade away in various circumstances that I don't *want* to think about, and I get paid much more for sitting on my arse rebooting server. Being aware of this injustice, I give blood in an attempt to give something back to society! -Ian Quote Link to comment Share on other sites More sharing options...
Adam W Posted February 28, 2003 Share Posted February 28, 2003 Yes, but as I say this is a pretty good approximation of the airflow into the turbo (while it's still at atmospheric pressure and ambient temp). After it's been compressed by the turbo that conversion factor will be totally different. Quote Link to comment Share on other sites More sharing options...
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