Why all the torque from the diesel engine?

[RustyJC]

If anyone is interested in working through the derivation to get to HP = (Torque x RPM)/5252, here's the root equation:



BHP = (P x L x A x N)/33000



Where:



P = mean effective pressure acting on the piston(s) during the power stroke (this applies to a 2-cycle or 4-cycle engine) expressed in PSI. In this case, since we're measuring brake horsepower, this factor would be the brake mean effective pressure (BMEP). If we were measuring IHP (indicated horsepower), the factor would be IMEP. This is a purely calculated value but can be used to compare relative output of engines of varying sizes. If you want a number to use for this exercise, try 250 PSI for a 4-cycle engine.



L = length of stroke, expressed in feet



A = area of cylinder, expressed in square inches



N = number of power strokes per minute (example - for a 6-cylinder 4-cycle engine running at 3000 RPM, this would be 3000/2 x 6, or 9000 since there is a power stroke in each cylinder only every other revolution.



33,000 = 1 BHP (33,000 ft-lb/min)



It's very interesting to work through this derivation paying special attention to the units of measure and how they eventually resolve themselves. For instance, the P x A reduces to average pounds force acting on the top of the connecting rod during the power stroke. Just working through the formula really helps understand how horsepower is developed - think in terms of Watt's original frame of reference - how many pounds of water could a horse lift out of a coal mine shaft that's a given vertical depth in one minute?



Well, I'll shut up now and watch the pencils fly!



Rusty

 

[RustyJC]

Further to the equation I posted earlier, if anyone wants a "plug" number to check it out, the 2002 Cummins HO rated 245 BHP @ 2700 RPM and 505 ft-lb torque @ 1600 RPM has a BMEP ("P" in the equation) of 200 PSI at peak BHP and 211. 85 PSI at peak torque.



Not mentioned in our discussion to date is an inherent challenge in a spark-ignited engine. As the BMEP goes up, the energy required to get the spark to jump the plug gap reliably goes up as well. For many years, production spark ignited engines were limited to BMEP's around 180-190 max by the energy the ignition system could deliver to the spark plug gap - above that BMEP, coils would start breaking down as would secondary lead insulation, and arcing and sparkover on the plug insulators became real problems. That's one of the reasons for the "coil on plug" ignition systems today.



If you play around with the equations {hint - since both (Torque x RPM)/5252 and (P x L x A x N)/33000 equal BHP, let (Torque x RPM)/5252 = (P x L x A x N)/33000 and solve for Torque (Q) and BMEP (P)}, you'll find that torque is ultimately a function of BMEP, bore diameter and stroke length. Since bore diameter and stroke length are constants for the 5. 9L Cummins, the relationship between torque and BMEP is BMEP = 0. 4195 x Q. So the moral of this story is:



Without changing bore or stroke, the only way to increase torque is to increase BMEP!



Oh, and the "5252" is not some mumbo-jumbo fudge factor - it is just 33,000 ft-lb divided by 2 x PI radians, necessary to convert linear work into radial (angular or rotating) work.



Rusty

 

[rbattelle]

I stand corrected!

 

[Hohn]

Originally posted by rbattelle

Hohn,



We see how the diesel has higher efficiency because of it's higher compression ratio and it's higher cutoff ratio (due to larger stroke). This allows it to extract more energy from a fuel that otherwise has less energy content than gasoline.



It would be interesting to see someone build a gasser with a compression ratio of 16 and a cutoff ratio closer to that of a diesel, just to compare it's characteristics to that of a similar diesel.

Hey, no problem. Like I said, I am often wrong, but I am trying to reduce the frequency of that occurring.



It's worth mentioning that a high cutoff ratio is not strictly a factor of a long stroke. If you built a VERY oversquare (huge bore, short stroke) engine, you could still have the high cutoff ratio you need/want/desire. You can build a given compression ratio (and cutoff ratio) into just about anything between WAY undersquare and way oversquare. Since cutoff ratio is just comparing volumes (as explained earlier to me), then HOW you get the volumes doesn't matter.



Here's a little quiz to all (Rusty please don't ruin it):



You have two identical displacement engines, both 5. 9 liters. Both are gasoline v-8s. One is very long stroke, narrow bore, the other is very short stroke, large bore. Cam timing, static compression, and rod/stroke ratio and all OTHER design parameters are identical. The ports in the cylinder heads are identical, and relatively restrictive.



Question: which engine will have more power? Or would they have the same power output? WHY?



This should stir things up a wee bit...



HOHN

 

[fbaurley]

If the turned at the same speed the longer stroke engine should have more power due to more torque. The increase in torque would be due to a larger offset on the crank. Now if you you kept increasing the rpm the shorter stroke would perform better at a higher rpm, where the longer stroke would start to fall off. In a short stroke engine the linear piston speed is less than in a longer stroke engine,(less distance to travel) therefore the crank can turn at more rpm, since crank speed is related to linear speed. The linear speed is a trade off, because the faster the linear speed the sooner the piston and cylinder wear out. so if you had a high rpm long stroke engine it would wear quicker than a high rpm short stroke engine. In a V8 configuration the geometry(60 or 90 degree) of the design has some play in stroke lengths. This is why v8 are usually short stroke.

 

[Hohn]

Not quite true. Torque is composed of force and distance. The longer stroke gives you more distance (true), BUT the much smaller bore (since we are keeping displacement the same) means there is much less force acting on the piston.



Remember, these engines have identical rod ratios and other internal geometry. Thus, the long stroke engine has much higher deck height.



EDIT: You are correct in what you say about linear speed. That it PART of the correct answer, BUT the engines will have identical airflow because displacement is identical. So even though the long stroke has a faster piston velocity, it doesn't necessarily mean it has more torque.



RPM in this example is fixed at 2500RPM.





Anyone else want to take a stab at it?



HOHN

 

[fbaurley]

Hohn so youre saying the compression ratio is the same. If so then what you are saying is that the power would be the same? If rpm is held constant then we are only concerned with torque. Which engine would be more efficient, though? An engine that is as close to square as possible should be the most efficient.

I think to answer this question correctly you would need some more info such cam timming etc. There would be a cam that could make either one out perform that other, (that is the optimum cam for the stroke length. ) There are way to many variables in engine building.

 

[rbattelle]

Seems to me they'd have roughly the same performance, since each is an equally poor design. fbaurley is right, that nearly-square is a good rule of thumb. Both of these are very far from this rule.



Yeah, my guess is that they'd both have the same power. The larger bore takes better advantage of available MEP, but the longer stroke takes better advantage of leverage on the crank. It balances out.

 

[Hohn]

OK, it's not what you'd think.

The long stroke engine DOES have more torque, but NOT necessarily because of how you would think of it.



Let me explain: though they have very different bore and stroke measurements, the air requirements of two engines are identical. If you graph the cylinder volume vs crankshaft position (which I did), you will find that it is identical to within a couple hundreths of a cubic inch. This is mostly because the rod:stroke ratios were identical at 1. 75 (in my example).



I ran numbers for two identical displacement cylinders of 60 cubic inches.

Engine Bore Stroke

Engine A 3. 56" 6"

Engine B 5. 528" 2. 5"



As you can see, they are VERY different in bore and stroke, yet the cylinder volume in every crankshaft position is the same (within ~. 01 cu inches). This is because the rod ratios were the same@1. 75. One rod was only 4. 375 long, the other was 10. 5"!!



So what about torque? Well, torque (or work) is force times distance. Force in this case is BMEP, the average pressure acting on the piston over the 180° travel from top to bottom. Distance is simply the stroke. If we figure a BMEP of 100 PSI for BOTH engines, then which will produce more torque (done more work)?



Engine A: area of top of piston=27. 96 in3

Engine B: area of top of piston=43. 42 in3



Multiplying by the 100psi BMEP AND each engine's stroke gives this result:



Engine A: 1677. 61 in-lbs

Engine B: 1085. 42 in-lbs



Staggering! The long stroke engine has WAY more torque! The long stroke more than offsets the lack of force because of the smaller diameter piston.



So the long stroke engine has more torque in this ideal example, but here's where things start to fall apart for the long stroke engine in the REAL world:



First, the swept area of the two engines is WAY different. The fact the the long stroke cylinder has about 50% more swept area (67. 1 vs 43. 3) means that is loses a LOT more thermal energy as it bleeds off heat to the coolant.



Second, the long stroke engine would NOT have the same BMEP as the short stroke engine because the combustion gases would have to travel further (distance) in the long stroke, even though the volume is the same. The short stroke engine would have marginally higher BMEP. The thermal losses of the long stroke engine might reduce the BMEP to the point where the short stroke had a considerable advantage.



Furthermore, to keep the compression ratios the same, the long stroke engine would have to have a MUCH larger combustion chamber, FURTHER increasing heat loss and decreasing efficiency.



It's also worth mentioning that the 6" crankshaft and the 10. 5" connecting rod are MANYtimes heavier than their counterparts in the short stroke engine.



Finally, even though they have IDENTICAL cylinder heads (in terms of flow) feeding IDENTICAL cylinder volumes, the long stroke engine will breathe much less efficiently. This is because the air in the long stroke cylinder has to travel much further in a given amount of time (which takes energy to accelerate the air). The higher piston speed causes inefficiency. To breathe as well, the long stroke cylinder would need larger ports and bigger valves even though it has the same volume displacement. Unfortunately, it has VERY little room for valves because of its narrow bore.



Another advantage of the long stroke is that the cylinder bores are much more stable. Ever wonder why anything that has to contain high pressure is a long skinny cylinder? (welding tanks, fire extinguishers, etc. . ). A tall thin cylinder can contain a given pressure with less stress on the tank's material than another shape. That's why a long skinny bullet casing can handle much higher pressure (think . 223 vs . 45ACP). The same pressure that is routine in a . 223 case might be +P or higher in a . 45.



Anyway, you can now see a little more into the nature of all the compromises that engine designers make.



BTW- The most thermally efficient design is the sphere, which has the lowest ratio of surface area to volume. The closer to a sphere (the closer to being "square" as well) that an engine cylinder is, the more that the respective compromises are averaged.



Sorry this isn't better-- my quiz was poorly worded.



HOHN

 

[rbattelle]

a couple comments

Hohn,



I have one question about your analysis of the problem. Why did you multiply the piston force (BMEP*area) by the stroke length? It seems more appropriate to me to multiply by the sine of the angle the crank arm makes with vertical? That quantity would vary over one complete stroke, and by integrating we could obtain the true torque.

 

[Hohn]

That's why I kept the rod ratios the same. If you were to compare engines with different rod ratios, then I think your method MIGHT be more accurate.



The BMEP method is simple and accurate. Since work=forcexdistance, it gives us the work performed on the piston.



IF you ran the integral, you would find that your answer would not be significantly different. Why? The torque delivery characteristics of the engines will be the same if rod ratio is the same. It's a simple equation if you want to run the integral, since the piston position graph is a sine wave.



I didn't do so well in calculus, so maybe you will tackle this for me? I would like to see if there is a substantial difference, if any at all.



It's like gears: you can multiply torque, but not HP because HP is torque over time in any combination. You can trade one for another at a given HP level.



So it is with cylinder pressure and "leverage" of the stroke. the in-between stuff doesn't really matter. The pressure and the stroke are all that really count.



Hohn

 

[Hohn]

I could accomplish the same thing as integrating in Excel if you want me to.



All I would have to do is graph the torque at each crank position from 0-180 and insert a trendline.



Or you could just calculate the peak torque (rod at 90° relative to crank) and multiply by . 707 which is the Root Mean Square for sine waves.



Hohn

 

[RustyJC]

Hint - mathematically speaking, other than instantaneous torque values, linear or rotating work doesn't matter. To wit:



Abbreviating torque as Q,



(Q x RPM)/5252 = BHP deals with rotating work/power



(P x L x A x N)/33000 = BHP deals with the linear aspect of work/power.



Remember I said that, since both equations equal BHP, let:



(Q x RPM)/5252 = (P x L x A x N)/33000



Now, replace 5252 with 33000 / (2 x PI)



So the equation becomes



(Q x RPM x 2 x PI)/33000 = (P x L x A x N)/33000



or, multiplying both sides by 33000,



Q x RPM x 2 x PI = P x L x A x N



Now let's replace A with PI x R^2 (R = 1/2 of cylinder bore)



Q x RPM x 2 x PI = P x L x PI x R^2 x N



Dividing by PI



Q x RPM x 2 = P x L x R^2 x N



Notice that the differentiation between radians (rotating work) and linear work just left the room!



But N = RPM x CYL x F



Where



CYL = # of cylinders



F = 1 for 2 cycle engines, 0. 5 for 4 cycle engines



So for a 4-cycle engine



Q x RPM x 2 = P x L x R^2 x RPM x CYL x 0. 5



Multiply both sides of the equation by 2 and divide by RPM (the last reference to rotation just left the building as well)



Q x 4 = P x L x R^2 x CYL



Dividing both sides of the equation by 4, for a 4-cycle engine:



Q = (P x L x R^2 x CYL)/4



The equation above also demonstrates that it's no coincidence that both torque (rotating) and linear work are measured in the same units (e. g. , foot-pounds or pound-feet) since they are mathematically interchangeable as shown.



(Before anyone flames me, yes, I've ignored piston thrust loading, peak pressure angle (PPA), piston velocity during valve overlap, and a number of other important technical nuances that determine frictional losses, achievable BMEP, etc. I'm just demonstrating what, from a purely mathematical standpoint, determines torque output. )



Rusty

 

[Hohn]

What Rusty said^^^^^



rbatelle: that's why i didn't do the integral. :-laf:



Rusty, no one will flame you for ignoring the stuff that you did, because this is THREORETICAL. Frictional losses and all that other stuff you mentioned are what separates theory from practice.



With these kinds of posts, does anyone know if we can get a degree from the TDR? I think we should be awarded for our efforts and have proof of all we have learned.



Prof Rusty, have any connections?



This is great stuff. Only in the TDR do you have everything from what's your favorite beer to dissertations on entropy and enthalpy.



HOHN

 

[rbattelle]

This was a great old-timey thread. Thought I'd bring it back up for the new guys.



-Ryan

 

[4xquadrod]

I missed it last time around and I didn't bother looking at the dates until your post that brought it back.

Good thread.

 

[WBusa]

Wow, so this is where the in-tee-lektuells hang out at TDR! And I thought I was alone! Carnot cycle... man, takes me back to my grad-school P-Chem course. I loved that course... decided that PV=nRT explained pretty much everything. rbatelle, thanks for recycling this thread... wouldn't have missed it for the world!

 

[barbwire]

Yeah, thanks Ryan. It's clear as mud now!!!! :-laf :-laf

 

[snowracer69]

HA! I actually started a worthwhile thread on the TDR for once... amazing!

 

[rbattelle]

BTW- The most thermally efficient design is the sphere, which has the lowest ratio of surface area to volume. The closer to a sphere (the closer to being "square" as well) that an engine cylinder is, the more that the respective compromises are averaged.

HOHN

I've always thought that the whole reason Chrysler came up with the hemispherical combustion chamber was to more closely approximate a spherical bomb calorimeter, which is used extensively in idealized combustion experiments.



Not to get off topic...



-Ryan