Fundamental explanation for different torque curve SI-CI 3

[Xplode]

Diesel engines reach their maximum torque at lower engine speeds than SI engines.
What is the fundamental explanation for this difference?
Is it just because a diesel is built heavier than a SI engine? Or is it caused by the different combustion?
What happened to the "SI vs CI" thread that is mentioned in other questions?

 

[strokersix]

I didn't see mention of valve timing specifically above. The OP asked about SI engines having lower torque than CI engines at low speeds. For similar application, say automotive/light truck, an SI engine likely will have later intake closing than a CI engine. The SI will lose torque at low speed because it pumps the fresh charge back into the inlet manifold. A CI engine has to trap as much air as it can to build pressure for combustion, therefore it must have early intake closing.

I do not have cam timing specifications in front of me so please correct if I'm wrong here.

 

[JSteve2]

Torque is determined in SI engine by the amount of air in the cylinder at the time of combustion. In a CI engine, torque is only limited by the air, but is determined by the fuel.

SI - low engine speed throttle is typically closed so very low air flow, and further the low engine speed reduces air draw by the engine. High engine speed the throttle is typically open and the engine has a fast stream of air coming in (ram air as well if in a moving vehicle).

CI - torque is whatever you want as long as you have air. At high engine speeds, you are limited by the rate of combustion due to the slower flame mechanism as mentioned previously. However, the real torque peak is typically programmed a little lower than it could be to encourage drivers to drive at particular speeds.

 

[Lou Scannon]

"If we had identical SI and CI engines (bore, stroke, Vs) running at the same CR - say 14:1, would the diesel still produce more torque at WOT at low rpm? Yes."
This too is an oversimplification. Other posts above have alluded to this already.
At the very least the fuel rate also needs to be constrained for this comparison. If you do that (i.e. make the chemical LHV delivery rate the same for both engines, then I tend to agree. But it's hard to reconcile that constraint with WOT operation, since if we're talking about a normally aspirated engine, the spark ignition engine may well have the advantage (assuming this identical engine design is a compromise between SI and CI optimum design), since it will be running at or near stoich, which is difficult to do with a diesel as we all know.

I also think the 14:1 CR proposed for this comparison is unfair to the SI engine, assuming pump fuel. Perhaps a better way to pose the comparison would be to allow each engine to optimize CR (as I assume is the case for ignition/injection timing) while not materially affecting swirl/turbulence etc.

 

[Warmington]

Somptingguy:

The integral is just the area under the curve as already mentioned a few times in the statement that the pressure is held higher for longer over the whole cycle than in an SI engine... Nothing new there! If you are considering the PV diagram, how do you assume you translate that pressure into a torque? Through geometry of the engine! Would you care to give YOUR different explanation to the OP?

 

[SomptingGuy]

Uh? Do the maths. You can't increase the thermodynamic efficiency of a diesel/otto cycle by fiddling with the crank/rod ratio.

- Steve

 

[SomptingGuy]

... fiddling with the bore/stroke ratio and/or crank length may have secondary effects on torque and efficiency (ok, I admit a that). However, any discussion that uses the word "leverage" is bunk. Any job applicant using that word to compare cI vs SI would be shown my door.

- Steve

 

[Lou Scannon]

hear hear!

 

[al1]

SomptingGuy
Apparently there is some kind of semantics going on here regarding the word leverage. Whether you like it or not, the crankshaft is in fact a lever arm --- but you can pick the word.

Regarding rod/crank kinematics & BMEP being one size fits all, why do you think that instant torque is typically around 42 degrees ATDC, even though the typical best crank angle is around 72 degrees (18 degrees of angle for the rod makes 90). And why do you think that no matter how high the pressure is no torque is produced at TDC? And yes I know that this all averages out through out the power cycle.

But to best understand why some engine combinations are better than others, I believe you have look at the details on say a 1 degree basis. The peak torque at the 42 degree area is because it’s the best combination of available pressure at the most advantageous (like that word better) crank position at this point in the stroke. Obviously at 72 degrees the crank is in the best position, but the pressure is much lower at this point. And if the highest peak pressure takes place at exactly TDC there is no leverage (there is that word again) and no work is produce as only rotational inertia makes it all happen. As a matter of fact, if the peak pressure takes place to much before TDC the engine will turn backwards. As a result, most engines are calibrated to have the peak pressure around 15 degrees ATDC.

Yes I understand the laws of energy conservation, but you can change the power profile characteristics of an engine by changing the mechanical configurations. And when you skew the pressure later in the power cycle you get a combination of higher pressure with a more advantageous crank angle and thus more torque.

The best example I can think of at this time is how do you think a supper charged engine can double the power over a normally aspirated engine without doubling the peak pressure? Yes the peak pressure is higher, but not twice as high. It’s the charge density being higher in a boosted engine and therefore taking longer to complete the combustion process is one of the real reasons for the additional power. This increases combustion time can permit twice the pressure at 72 degree crank angle over a normally aspirated engine, but still have lower pressure at this point then the peak pressure of a normally aspirated engine. The results are twice the power potential without twice the structural mass and/or integrity of the engine all because of the combination of best combustion pressure with the most advantageous crank angle.
al1

 

[Lou Scannon]

Show us your reasoning on a P-V diagram.

 

[al1]

Per your request

 

[SomptingGuy]

I didn't see a P-V diagram in that link.

- Steve

 

[JSteve2]

Al1,
You are laboring under a common mis-conception about engine torque.

Engine power is the amount of work performed by the engine in a given time. Engine torque is the amount of work performed by the engine in a given crank angle (typically, the average work performed over one radian).

Therefore, the "leverage" and actual TORQUE exhibited by the engine on the crankshaft is irrelevant for calculating torque of the engine as that term is used in industry. The supercharged engine has double the area under the curve and/or a higher engine speed to produce double the power.

Area under the P-V curve, minus efficiency losses, is proportional to the engine "torque". That's it.

 

[SomptingGuy]

... except as I mentioned earlier if you are concerned with stress/durability calcs where instantaneous forces and torques can be important.

- Steve

 

[JSteve2]

SomptingGuy / Steve:
Absolutely.

 

[al1]

Look the original post was about why CI engine have more torque then SI engines. I used the constant volume verses constant pressure PV charts as why this can happen. I am not selling instant torque; I used it in an attempt to depict the dynamic pressure changes laid over the dynamically changing mechanical advantages of the crankshaft position. Efficiency was not included as that was not the question.

Speaking of efficiency, not only does the real world diesels have more torque at low RPM, they also have more efficiency as well. A spark engine can be calibrated to have the same peak pressure later in the cycle, but the efficiency becomes even poorer all things being equal.
al1

 

[Kier14]

JSteve2

Could you explain your definition of Torque and Power in greater detail please?

I thought Torque was a force multiplied by a distance hence its units being 'Nm', therefore implying that for a given force an increase in the length away from centre of rotation will increase torque.In this case the force is used to turn a shaft. As the engine is of the reciprocating type, the length away from centre of rotation changes every degree.

Thus the torque changes every degree(regardless of cylinder pressure). The mean of the calculated torque for the cycle will be the same as the measured torque.

Is this wrong??

 

[JSteve2]

Kier14,
If you were a physicist, you would be right on the money. However, as torque has been used in the engine business (I'll call that e-torque), it's a measure of how hard the engine tugs on the transmission. It happens to have units of force*distance, which is either torque or work.

If you could actually calculate the instantaneous net torque (let's call that p-torque for physics-torque) on the crankshaft, you could possibly determine the "mean of the calculated torque for the cycle" as you suggest. However, you have to remember that there are typically multiple cylinders acting on the crankshaft at any moment, some putting work onto the crankshaft, and some taking work off of the crankshaft. Therefore as a practical matter e-torque cannot be determined by your method - at least not without exotic modeling that no one would bother to perform. Realize that "peak torque" is a steady-state concept referring to the net action of all cylinders acting on the crankshaft at the maximum fueling per stroke, and not the maximum p-torque exerted by any cylinder on the crankshaft.

But back to the original point - e-torque is defined as the amount of work the engine puts onto the crankshaft over a given rotational angle of the crankshaft, and not the physics based concept of rotational force exerted on the crankshaft at any particular moment.

In my opinion, the term torque for engines as it is used is a bit misleading. However, it is what it is, and it's been that way for so long I doubt anyone is going to change it. It is useful to think of it as p-torque when designing transmissions, because the transmission sees the entire net effect of all cylinders acting - which is probably why we are where we are. However, it's not helpful to imagine the p-torque provided by a particular cylinder at a particular moment in time, because that's just not the same concept.

 

[Warmington]

JSteve,

Just because the automotive world confuses the engine torque with the phisical torque exerted on the shaft does not give an answer to the OP in my opinion.

Imagine the case of a one cylinder diesel engine, how would you explain the higher torque there compared to SI?

If we had a one cylider engine connected to a brake with a line plotter showing torque throughout the cycle at a low rpm, the CI engine would hold a higher value for longer than the SI engine on the plotted line through the combustion stroke, this is agreed.

If you then averaged the torque value for the whole trace, the diesel would still come out on top. Your concepts of e-torque and p-torque are not mutually exclusive.

Although the averaged value (e-torque) includes the time frame for the average, the axis units of the dyno plot will cancel this out, so effectively, at a certain rpm the theory states that you could stop the engine dead with a bar and a weight long enough to match the torque being produced.

But on this one cylinder engine, reading the averaged torque value how would you know when the engine was producing exactly this value? It may be more or alot less at any one time.

The torque is always changing due to the geometry of the engine.

Diesels benefit from this by being able to hold the pressure for longer.

The overall area under the curve (minus pumping loop and friction) on a diesel will be higher, yes. But if you could find a diesel engine and a gas engine that had the same overall areas, the diesel WOULD STILL PRODUCE MORE LOW RPM TORQUE!

Its not the magnitude of the force on its own, but rather the effectiveness of the timing of the push on the piston to generate maximum torque at the crankshaft, there are two separate components that combine to give THE SAME TORQUE READING. Torque is torque, no matter the application.

 

[JSteve2]

Warmington,
If you build them to have the same area under the curve (and all that other stuff you said), they will in fact produce the same torque (e-torque).

I think what you're getting at is that the gasoline engine would have more crank angle degrees where it didn't "push", which is true. That does mean the gasoline engine would be more prone to stall, but not because it was producing less torque. That's why they have flywheels.

I agree that e-torque and p-torque are not exclusive. It's just that getting e-torque from p-torque is difficult, because you have to sum the p-torque for all cylinders through an entire engine rotation (probably two to get a combustion event on each cylinder since the cylinders won't be identical), then divide the total by 2-pi to get your e-torque. That's not exactly what people are thinking of when they start talking p-torque.

 

[Warmington]

I dont beleive they will. The work done on the piston by the gas is converted into the plotted torque line by multiplication of the distance of the acting force to the crank centreline. The PV diagram shows the amount of work done on the piston, but not the timing of this work essentially (the diesel will be fatter towards the V2 end). The later this pressure occurs in the cylider, the more instantaneous torque is produced at the crankshaft due to the beneficial effects of the rod/crankshaft geometry.

The gas engine will still use the same work on the piston earlier on to produce a lower value of torque, but can obviously maintain this at higher rpm. How COULD it have the same value of torque if the force produced is the same yet it has less distance?

If you say that this is a function of time and that the force is applied over a shorter period of time though at a smaller angle which gives an average torque for the whole combustion period similar to the diesel (which is what I assume you are getting at) then a simple integration of instantaneous torque by time within the combustion window will show that is not true.

I imgaine it like spinning a bike wheel; spinning by hand either the tyre or the hub of the spokes. You can obviously start it spinning easier at the tyre, but can only maintain acceleration to a point, then you are better off spinning at the inside with the same work input you can acheive a much higher speed. The work you are doing to spin the wheel is exactly the same. Diesel at the tyre, gas at the hub.

Greg