View Single Post
Old 05-23-2014, 11:11 AM   #1 (permalink)
mateospeed
10MPG w/ 8.5'x16' Trailer
 
Join Date: Dec 2010
Location: Morehead City, NC
Posts: 11
Thanks: 1
Thanked 1 Time in 1 Post
What determines an engine's stall speed?

Folks, since there's more technical discussion on this forum than any other I've found, I'll pose this question here.

WHAT DETERMINES AN ENGINE'S STALL SPEED??

I have a good feeling that this discussion could, and will be, closely related to one involving engine lugging; where and why it happens. I'm defining stall speed as the engine speed at which the engine, under no load and no throttle, will no longer run, shutting itself off.

A few observations: different engines appear to have different stall speeds, and I'm not sure what determines that. Most automobile engines, regardless of size, seem to have a stall speed of around 400 to 450 RPM. Other applications have much different stall speeds. The old hit-and-miss engines, like you'll see at the state fair sometimes, actually operate around 200 RPM, IIRC. Large marine diesels (like those that power ships) operate at very slow speeds as well, on the order of 150-200 RPM.

So, I have a theory about what determines the speed at which an engine can run, and it's simple: cam design. I'm not completely convinced, though, and would love if someone would attempt to verify, or counter my idea, with good technical backing.

A little background on my theory, to get you thinking on the same lines as I am. Consider a single-cylinder engine, running normally. A normal camshaft's events are timed relative to crankshaft position, in degrees of rotation. For instance, it might have the following specs:
-Intake valve(s) open @ 4 degrees Before Top Dead Center (*BTDC),
-Intake valve(s) close @ 20 degrees After Bottom Dead Center (*ABDC)
-Exhaust valve(s) open @ Bottom Dead Center (BDC)
-Exhuast valve(s) close @ 15 degrees After Top Dead Center (*ATDC)

organized in order of crankshaft rotation, these go:
-Exhaust valve(s) open @ Bottom Dead Center (BDC)
-Intake valve(s) open @ 4 degrees Before Top Dead Center (*BTDC),
-Exhuast valve(s) close @ 15 degrees After Top Dead Center (*ATDC)
-Intake valve(s) close @ 20 degrees After Bottom Dead Center (*ABDC)

so, notice a few things off the bat:
-at the top of the piston stroke, both the intake and exhaust valves are open at the same time, for a short period of time, from 4* BTDC to 15*ATDC. This is "valve overlap."
-at the bottom of the piston stroke, the piston starts moving back UP 20* of crank rotation BEFORE the intake valve closes.

Both of these situations seem counter-intuitive:
-"Why would you open both intake and exhaust valves at the same time? Wouldn't that blow fresh intake charge out the exhaust port, or exhaust gas into the intake?"
-"Why would you close the intake valve AFTER the piston reaches BTC? Wouldn't that push air back OUT the intake, since the piston is moving back up?"

The answer to these questions is the same: it has to do with the momentum of the air entering or exiting the cylinder. Consider the first case: Exhaust valves are open, piston is moving upwards, forcing the air out the exhaust valves. The air moving out the exhaust ports moves with some velocity, 'v'. Air has mass, so it cannot be stopped immediately. It takes some time to slow down and stop, even though it happens very quickly (air doesn't have much mass). When you're talking about engine events, you're working in 0.01 second intervals, so that small time it takes the air to slow down and stop is significant. SO, the intake valves are open, but the exhaust is already moving OUT of the cylinder. This moving exhaust gas creates a venturi effect, drawing intake charge into the combustion chamber, and yes, some out the exhaust valves. The piston reaches TDC and starts back down, but the air hasn't stopped moving out of the exhaust valve(s) yet (it has momentum, and takes time to stop!). It slows down and stops, and only THEN do we shut the exhaust valve(s) - in this case, @ 15*ATDC.

-Now, the piston travels down to the bottom of its stroke, drawing fresh intake charge with it. It reaches BDC and starts heading back upward, but the intake valves are still open! Why? The incoming air is still moving; still has momentum. We wait until it stops moving to shut the intake valve, in this case @ 20*ABDC.

-Consider this: the faster the engine is turning, the faster the piston is moving in its cylinder. The faster the piston moves, the faster it forces air out and in of the cylinder (exhaust and intake, respectively). The faster the air is being forced out of and into the cylinder, the more momentum is has, and thus, the longer it takes to slow down and stop.

-So, your cam designer faced a compromise: "The faster the engine spins, the longer I can keep the intake and exhaust valves open, allowing momentum to be used to its full potential. "So, they pick an engine speed when it will run the most efficiently, but everywhere else must be a compromise.

-Consider this scenario: Assume your car's engine was designed for maximum efficiency at 3500RPM, which you can tell from your dyno graph because that's where the TORQUE peak is. Ignore power for this discussion; it doesn't come into play. Therefore, the intake and exhaust valve closing events will happen at some point AFTER BDC and TDC, respectively. Now, take this engine and slow it down. A lot. to 400 RPM. Think through the 4-stroke cycle again.

-Momentum has very little effect at this speed, because the piston isn't moving very quickly, relative to 3500RPM. So, around TDC, you suck in just as much exhaust as you do intake air from TDC to 15*ATDC. Then, around BDC, you actually force air OUT of the intake from BDC to 20*ABDC. This greatly reduces your compression, since you just blew a significant part of your intake charge right back out of the cylinder! You effectively make your engine less and less efficient the slower it goes.

-At some point, then engine will move slowly enough that it's no longer making much cylinder pressure. If the cylinder pressure gets low enough, the fuel/air charge will simply burn, instead of explode, and won't do much to push the cylinder downwards. So, you essentially have a misfire. Add a few of those together, and the engine quits running.

How's that for a theory? Can anyone shed some light on whether that's correct or incorrect?


Last edited by mateospeed; 05-23-2014 at 11:22 AM..
  Reply With Quote