johnc Posted March 11, 2009 Share Posted March 11, 2009 Given the same displacement and the same levels of technology, why would a V12 engine make more power then a V10 or a V8 engine? Quote Link to comment Share on other sites More sharing options...
Administrators RTz Posted March 11, 2009 Administrators Share Posted March 11, 2009 Rpm Quote Link to comment Share on other sites More sharing options...
johnc Posted March 11, 2009 Author Share Posted March 11, 2009 And... ? Quote Link to comment Share on other sites More sharing options...
calpoly-z Posted March 11, 2009 Share Posted March 11, 2009 Do they make more power? I thought the typical reason to increase number of cylinders, especially for relatively small displacements i.e. car engines, was to increase the smoothness of power delivery by shortening the duration between power strokes in the engine. If they do make more power, then maybe the smaller cylinders allow higher compression ratios. With larger bores knock become more of an issue because the flame front has more distance to cover. By decreasing piston bore diameter this become less of an issue (to an extent). We happened to be talking about this concept in my combustion engine design class this week as it pertains to aircraft piston engines and why they needed 100LL with relatively low CR (8:1). Apparently the large bores of these engines (~5-6") means they are very detonation prone. They also have 2 spark plugs to help with this issue as well. Just an idea. Quote Link to comment Share on other sites More sharing options...
johnc Posted March 11, 2009 Author Share Posted March 11, 2009 (edited) With larger bores knock become more of an issue because the flame front has more distance to cover. Current F1 engine technology has hit a bore size limit of about 98mm (3.89") where anything more decreases thermodynamic efficiency. The combustion process is very time based and a larger bore requires more crank angle to achieve a complete and efficient burn. Power and fuel economy falls off beyond 98mm with the current ignition and fuel delivery technology available in F1. But this isn't the "And...?" I'm looking for. Edited March 11, 2009 by johnc Quote Link to comment Share on other sites More sharing options...
atomflatz Posted March 11, 2009 Share Posted March 11, 2009 More valves? Quote Link to comment Share on other sites More sharing options...
JMortensen Posted March 11, 2009 Share Posted March 11, 2009 Shorter stroke. Quote Link to comment Share on other sites More sharing options...
S130Z Posted March 11, 2009 Share Posted March 11, 2009 Might I ask why you are interested in this question? Is there something that you are basing this concern off of? I don't think that we have enough information to stake a claim because the same level of technology would wield the same amount of horse power in any motor. I.E. if you have the level of technology to make a 500hp V12 then you must also have a level of technology to produce a 500hp V8. Although I want to contradict myself from that statement because I understand what you are trying to say. I guess I just need an example. But I think it all depends on the combination of stroke, compression,friction, and centrifugal resistance. Quote Link to comment Share on other sites More sharing options...
Administrators BRAAP Posted March 11, 2009 Administrators Share Posted March 11, 2009 (edited) 1) More available RPM due to shorter stroke, for given displacement, all else being equal, = more power, (covered!) 2) Depending on available real-estate in the chamber roof, more overall valve area = more air flow, (sort of covered, falls under the "more valves" post) 3) Induction and/or exhaust pulse tuning/scavenging attributes of the V-12 allowing more "natural supercharging" effect, =more power Edited March 11, 2009 by BRAAP Quote Link to comment Share on other sites More sharing options...
johnc Posted March 11, 2009 Author Share Posted March 11, 2009 (edited) Its real simple guys: two (or four) more cylinders providing power during the 720 degrees of crank revolution needed to complete one Otto cycle. Friction losses from rings and cylinder walls are basically the same (a 4" bore 8 cylinder has 100.66" of cylinder wall/ring circumference and a 2.67" bore 12 cylinder has 100.53" of cylinder wall/ring circumference assuming same stroke - if we assume same bore then stroke is reduced which still gives equal friction due to stroke distance differences). Edited March 12, 2009 by johnc Quote Link to comment Share on other sites More sharing options...
jt1 Posted March 12, 2009 Share Posted March 12, 2009 So why is 10 small pulses better than 8 larger ones if the displacement is the same? Are we assuming equal rpm? Edit: 10, not 12. But what about a V12? More power yet? jt Quote Link to comment Share on other sites More sharing options...
at-jefft Posted March 12, 2009 Share Posted March 12, 2009 1) More available RPM due to shorter stroke, for given displacement, all else being equal, = more power, (covered!) 2) Depending on available real-estate in the chamber roof, more overall valve area = more air flow, (sort of covered, falls under the "more valves" post) 3) Induction and/or exhaust pulse tuning/scavenging attributes of the V-12 allowing more "natural supercharging" effect, =more power Thanks for clearing that up, I have wondered this for a while. Quote Link to comment Share on other sites More sharing options...
johnc Posted March 12, 2009 Author Share Posted March 12, 2009 (edited) Given equal technology, there's not a significant difference in IMP (indicated mean pressure) based on bore size. IMP is strictly related to volumetric and thermodynamic efficiency. Larger bores lead to reduction in thermodynamic efficiencies as mentioned above. Larger bores do increase volumetric efficiency (larger valves and ports) but the thermodynamic difficulties tend to trade off the VE gains. FYI... I'm pulling my discussion points from a great article in Racecar Engineering covering the design of Formula 1 engines. These engines have a bore/stroke ratio of .4 - typically a 98mm bore and a 40mm stroke for a V12. A V8 using the same bore/stroke ratio would need even larger bores. FYI2... I'm assuming equal rpm, but if we don't assume that, then more cylinders have an even greater advantage. Edited March 12, 2009 by johnc Quote Link to comment Share on other sites More sharing options...
Administrators BRAAP Posted March 12, 2009 Administrators Share Posted March 12, 2009 (edited) Its real simple guys: two (or four) more cylinders providing power during the 720 degrees of crank revolution needed to complete one Otto cycle. Friction losses from rings and cylinder walls are basically the same (a 4" bore 8 cylinder has 100.66" of cylinder wall/ring circumference and a 2.67" bore 12 cylinder has 100.53" of cylinder wall/ring circumference assuming same stroke - if we assume same bore then stroke is reduced which still gives equal friction due to stroke distance differences). I’m waving the BS flag. Number of firing pulses “alone” per full otto cycle, does not equate to more power! I wont get into the friction aspect, we'll just assume that is equal. A 12 cylinder has more pulses per cycle vs a 10 cylinder, yes, but, and this is a huge butt, each pulse has less energy due to each of those pulses coming from less displacement. For a “given BMEP”, regardless of the number cylinders 2, 4, 6, 8, 10, or 12 cylinders, all having the same overall displacement, all at the same RPM, all will generate the same power! The more cylinders the engine has, the smoother that power is being transferred to the crank/wheels, none the less, overall power at the end of each otto cycle will still be the same. Now if that same BMEP is being made at higher RPMs, more horse power will be made, which the 12 cylinder variant may very well be capable of due to its shorter stroke, if piston speed is the limiting factor of RPM. Also, if that 12 cylinder has some inherent natural supercharging effect that the 8 and 10 cylinder variants don’t have, then the 12 cylinder will have more BMEP, hence more power. Edited March 12, 2009 by BRAAP Typos. Quote Link to comment Share on other sites More sharing options...
jt1 Posted March 12, 2009 Share Posted March 12, 2009 Assuming "equal technology" combustion chambers, at some cylinder displacement the compression ratio is going to force detonation. More displacement to a given chamber means more comp ratio. So it's a matter of fuel quality to chamber design. This is irrefutable evidence my V8 is superior to the sixers. jt Quote Link to comment Share on other sites More sharing options...
Administrators RTz Posted March 12, 2009 Administrators Share Posted March 12, 2009 (edited) Assuming "equal technology" combustion chambers, at some cylinder displacement the compression ratio is going to force detonation. More displacement to a given chamber means more comp ratio. So it's a matter of fuel quality to chamber design. That's not whats being said. Smaller displacement (per cyl.) could mean smaller chambers (if shooting for a specific CR target). Smaller chambers, of the same 'type', will normally be less detonation prone. This is irrefutable evidence my V8 is superior to the sixers. Good one Edited March 12, 2009 by RTz Quote Link to comment Share on other sites More sharing options...
Gritz Posted March 12, 2009 Share Posted March 12, 2009 Easy... 6 beats 4, 8 beast 6, 10 beats 8, 12 beats 10...it's really that simple...If you had all these engines on a engine dyno, all built to the hilt, v12 would make the most power...Just like a v16 would beat a v12...But, you do have to realize that v8's are most known...so a v8 would be the cheapest and more bang for the buck alternative.. Quote Link to comment Share on other sites More sharing options...
lazycat5 Posted March 12, 2009 Share Posted March 12, 2009 Its real simple guys: two (or four) more cylinders providing power during the 720 degrees of crank revolution needed to complete one Otto cycle. Friction losses from rings and cylinder walls are basically the same (a 4" bore 8 cylinder has 100.66" of cylinder wall/ring circumference and a 2.67" bore 12 cylinder has 100.53" of cylinder wall/ring circumference assuming same stroke - if we assume same bore then stroke is reduced which still gives equal friction due to stroke distance differences). What about the number of rings in a v12 vrs. a v8, how much friction loss can that have. I understand that up to twenty or more hp can be found in a v8 from one set of rings to another. Quote Link to comment Share on other sites More sharing options...
PalmettoZ Posted March 12, 2009 Share Posted March 12, 2009 Do they make more power? I thought the typical reason to increase number of cylinders, especially for relatively small displacements i.e. car engines, was to increase the smoothness of power delivery by shortening the duration between power strokes in the engine. If they do make more power, then maybe the smaller cylinders allow higher compression ratios. With larger bores knock become more of an issue because the flame front has more distance to cover. By decreasing piston bore diameter this become less of an issue (to an extent). We happened to be talking about this concept in my combustion engine design class this week as it pertains to aircraft piston engines and why they needed 100LL with relatively low CR (8:1). Apparently the large bores of these engines (~5-6") means they are very detonation prone. They also have 2 spark plugs to help with this issue as well. Just an idea. Hey CalpolyZ- Not quite correct. Recip aircraft engines have 2 spark plugs for another reason, and do not fire simultaneously. Normally there is 2 ignition systems independent of each other in case one fails the other system is there as a back-up. System redundancy. Quote Link to comment Share on other sites More sharing options...
EMWHYR0HEN Posted March 12, 2009 Share Posted March 12, 2009 Given the same levels of technology: The V12 will have more bearing surface area than a V8 or V10 (longer crank, more connecting rods) meaning more friction. There's also an increased amount of friction from the cylinder heads (longer cams, more lifters,valves, etc. Clearly, we can't assume friction levels are equal even with the same ring/cylinder circumference. Quote Link to comment Share on other sites More sharing options...
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