I expect that the greatest power loss to a convenctional valvetrain is friction at very low speeds, but inertia at very high speeds, with the obvious extreme at valve float, were all of the power used to compress the valve spring is lost.
It's a long time since I did any text book physics, but I guess any effect that gasflow past the valves has on valvetrain drag is also inertia.
I expect that if the range of adjustment is sufficient, then controlling valve lift and duration will give suitable speed control, with less overall valve train losses, particularly at low speed, however I am not aware of the details for the valvetronic system Regards
pat
I'm still not sure how you guys expect valvetrain inertia to have any direct effect on valvetrain power consumption (excluding the friction increase due to larger contact forces). By the same reasoning, would you say that a larger flywheel consumes more power than a small one?
with the obvious extreme at valve float, were all of the power used to compress the valve spring is lost
I assume that the assumption you're making here is that when the follower comes back into contact with the cam, the cam has turned so far that none of the impact force is converted into torque on the cam, and all of the energy is lost as heat? That would be a pretty extreme case of "valve float," and I doubt that an engine could really power itself up to such a speed. Usually separation happens "over the nose," with an impact that occurs on the closing flank. Depending on the cam shape, a great deal of the energy can be recovered. With some cams, even when "floating," there are no significant impacts to speak of (think of a skier landing on a steep downhill after a jump), and in these cases I would expect nearly all of the energy to be recovered.
Of course, if your valves are not closing when you want them to, you're probably going to get less-than-optimal power output...
No, not a flywheel a steady revs, of course. But is it wrong to consider that a heavier vehicle (or flywheel) consumes more power to be accelerated than a lighter one? In the same way heavier followers and valves would consume more power than lighter ones to be accelerated at high revs. As Patprimmer says, at valve float the power used to compress the spring is totally lost. Can we consider this loss as being friction? I would rather think it's lost in kinetic energy to accelerate the valvetrain components, the same way as accelerating the vehicle consumes some power that you don't get back during deceleration. Am I wrong?
Oops, sorry Ivymike, my post was gone before I could read your last. I don't think all the energy used to accelerate the valve and follower can be recovered on the backside of the cam because of the inertia of these components, even below valvefloat revs - unlike in a crank gear system.
But is it wrong to consider that a heavier vehicle (or flywheel) consumes more power to be accelerated than a lighter one?
Sure, it takes more energy to accelerate. That's why you get more energy back from it when it decelerates. The energy is stored, not consumed.
When you decelerate in a vehicle, the energy that is dissipated by the (friction) brakes is lost to the atmosphere. You can recover the energy if you slow the vehicle down in another manner. Driving an electric generator is one way. If you come to a stop by coasting to the top of a big hill, then you haven't lost the energy (except the standard losses of a moving vehicle). If you come to a stop by lifting big weights to the top of a tower, or by accelerating a flywheel, then you haven't lost the energy.
In a valvetrain, when a cam lobe forces the valve line to move, torque applied to the cam lobe is converted into kinetic energy of the valvetrain components, and stored energy in the spring(s). Some of the energy is lost to friction, of course. As the valvetrain components are slowed by the spring, their kinetic energy is transferred into it. On the closing flank, when the valve gear begins to accelerate back towards the closed position, the stored energy in the spring is transferred back to both the moving components and the camshaft (negative drive torque). As the components slow to a stop (with their velocity controlled by the cam lobe), their kinetic energy is converted back into cam torque. In a frictionless valvetrain with no separation, you would have zero net drive torque regardless of the component inertia. If you have separation, you'll have some amount of energy lost in the impact when the components come back together, but not all of it.
Try to imagine the drive torque required if the energy stored in the valve spring and the moving components was not recovered. I can do a detailed calc when I get back to work on monday, but here's a quicky for the time being:
Peak lift: 8mm (intake and exhaust)
valves per cylinder: 4
4 cylinders
spring fitted force: 150N
max working force: 500N
crank rpm: 3000
valve spring stiffness (constant rate spring): (500N-150N)/8mm = 43.75 N/mm
work to lift a single valve = f_0*x + 0.5*k*x^2
w = 150N*8mm + 0.5*(43.75*N/mm)*(8*mm)^2 = 2.6J
work to lift all 16 valves = 16*2.6J = 41.6J
frequency at which all 16 are lifted = 0.5*3000*cyc/min
freq = 25/sec
power consumption (frictionless) = 1.04kW
so a frictionless valvetrain as described above (all spring energy is lost) would consume 1.04kW to run.
I should have mentioned that those lift and force values are in the approximate range of what you'd see on a 2.0L gasoline engine. Note that the calculated power consumption for the frictionless valvetrain was larger than the measured power consumption of a real valvetrain from a similar-size engine.
If we want to calculate the drive power given the assumption that the kinetic energy of the valvetrain components is lost instead of recovered, let's try the following approach:
same engine
average valve velocity = freq*2*8mm = 0.4m/s
let's guess that the peak valve velocity is 2.4m/s
valve mass = 60gm
peak valve KE = 0.173J
valve "power loss" = 69.1W
So the energy lost (if this really happened) would not be of great interest anyway...
The results of the second calc above, KE loss, are unfortunate for my effort to demonstrate my point, since they make your expected effect (if it really occured) almost negligible anyway. Thus it doesn't help my argument to say that in my experience, changes to the valvetrain component inertias do not have a significant effect on the valvetrain drive torque (analytically or measured), since I wouldn't likely notice the effect anyway.
Yes, of course my example was flawed and stupid; the kinetic energy spent for accelerating a vehicle or a flywheel is stored!
Of course, in case of valve float, the landing of the follower occurs on the closing side of the cam and not by an impact on the came base circle.
But it's difficult to understand how at high revs the torque recovered on the closing side can be as great as on the torque spend to open the valve, even if there's no separation. So the difference would only be due to friction?
There's a direction reversal, I mean the valve is projected down and some of the spring force is used to counterbalance the kinetic energy and to send back the valve. OK, at low rev and if we neglect frictional losses, the torque recovered is the same. But before separation occurs, before the valve floats wildly, there must be a gradual reduction of the force applied on the closing side at high lift. So, as revs increase, more and more of the energy stored in the spring is used to accelerate back the valve and follower and is recovered later by the camshaft, but nearer to the shaft axis and so with less leverage.
Do you mean that the produce of this force by this leverage would produce anyway the same recovered torque as long as the force isn't applied to the cam base circle, so that only an extreme valve floating wouldn't allow the recovery of the full torque – still neglecting the friction losses, of course?
I think it might help you to understand me if you were to draw lift, velocity, and acceleration diagrams for a simplified cam profile. On both of the flanks, when valve acceleration is positive (in the opening direction), there will be a greater force applied to the cam than would be applied if the valvetrain were massless. Over the nose of the profile, when valve acceleration is negative (in the closing direction), there can be force between the follower and cam, but there doesn't necessarily have to be. For a given spring, geometry, and lift profile, the larger the inertia of the valvetrain, the less force will be applied between the cam and follower over the nose. Nonetheless, the closing flank will see a large follower force. With a typical cam profile, the middle of the negative acceleration part of the profile (nose) is where the drive torque will switch from positive to negative. Inertia will reduce the force a little bit on both sides of the nose. If there is no separation at all, then the force between the cam and follower will start increasing immediately when the flank starts. Even with a moderate-to-large amount of separation, the follower should be back on the cam before you move very far along the flank.
A good way to picture this is to hold a ball in your hand, and make a motion as though you are going to toss it. If you move your hand slowly up and down, gravity holds the ball in your hand. If you gradually increase the speed of this motion, you will find that there is less and less force between your hand and the ball, not when your hand is moving down, but when it's acceleration is in the downward direction. Eventually you will find that the ball leaves your hand completely for a while, and you must catch it. If it weren't for wind resistance, the energy transferred from your hand to the ball when you accelerate it will be the same as the energy transferred from the ball to your hand when you decelerate (catch) it. There may be a period of time when the ball leaves your hand completely, but the end result will be the same.
I am sure we all realise that valve train losses are quite small in the overall scheme of things, and as the discussion got very theoretical, I deliberately chose an extreme case to make a point, but it is a point if we want to get into extremes.
On a more practical note, maybe you can explain why valve springs get hotter than their environment if they do not absorb kinetic energy and convert it into heat by shear at the molecular or crystal interface. Shear as I understand it is friction, so I guess the losses due to inertia really come back to frictional losses in the end, but still, the greater the valve lift and acceleration and deceleration rates, the greater the loss (though still quite low) to friction
Re the original question, I guess this system can control speed effectivly without a throttle butterfly or slide in the traditional sense, but if the valves are only opened slightly to control airflow so as to control speed, doesn't the valve then become an effective throttle. As I understand it, a throttle is a device to restrict or control air flow.
Re flywheels.
As it is in the real world that valve float is never so extreme, that the follower lands on the base circle, also virtually no one drives a car with a system that converts energy stored in the flywheel "normally lost in brakeing" into an energy form that can be reused for acceleration. Everyone I know, simply applies a friction brake and converts energy stored in the flywheel into heat which is then, deliberately lost to atmosphere, as quickly as possible Regards
pat
I deliberately chose an extreme case to make a point, but it is a point if we want to get into extremes.
So the point was that you can possibly make the valvetrain do something such that the kinetic energy of the components would be lost?
On a more practical note, maybe you can explain why valve springs get hotter than their environment if they do not absorb kinetic energy and convert it into heat by shear at the molecular or crystal interface.
Sure, there is some hysteresis that happens in a valve spring. I wouldn't say that it has a whole heck of a lot to do with the pushrod or valve inertia, though. Do you disagree?
As it is in the real world that valve float is never so extreme, that the follower lands on the base circle, also virtually no one drives a car with a system that converts energy stored in the flywheel "normally lost in brakeing" into an energy form that can be reused for acceleration. Everyone I know, simply applies a friction brake and converts energy stored in the flywheel into heat which is then, deliberately lost to atmosphere, as quickly as possible
I brought up the regenerative braking example to illustrate the difference between energy stored and energy lost, because it was starting to seem like the distinction needed to be made.
An important difference between the valvetrain example and the braking example is that people don't want to do the former, and do want to do the latter, if possible. I expect that more and more people will be driving vehicles with regenerative braking systems in the future. At present, I know (personally) four people with regenerative braking systems on their "daily drivers." There's also a person who owns a Honda Insight living on my block, but I don't know him.
To try to sum up all this, it appears that all valvetrain mechanical losses can ultimately be reduced to friction, as Ivymike demonstrated. So my phrase about a loss of kinetic energy was irrelevant and anyway those losses are kept quite insignificant if roller followers are used.
Now about the pumping losses reduction with infinitely variable valve lift and duration, BMW say that:
" Variable valve lift adjusts the “effective cam” and, accordingly, the cross-section and opening times of the valves. This, in turn, eliminates two factors partly responsible for the pumping losses: First, the process of filling and emptying the collector between the throttle butterfly and the valve. Second, the need to build up a vacuum. While the shorter opening times under part load also generate a certain vacuum in the cylinders during the intake stroke of an engine with VALVETRONIC, this vacuum acts like a spring in the subsequent compression stroke, setting off the balance of energy."
I suppose they mean the required amount of charge is allowed early during the intake stroke, so that the cylinder is under a reduced vacuum in the first part of that stroke, most of the vacuum being established late in the stroke. As the intake valve is closed before BTC instead of several degrees after, the cylinder remains under vacuum during the first part of the compression stroke.
BMW phrasing again:
"Under normal driving conditions, this concept keeps valve lift relatively small between 0.5 and 2.0 mm.
With this kind of small valve lift, fuel is only able to reach the combustion chambers through a narrow slot. This ensures ideal atomisation of the fuel/air mixture, while a conventional engine draws in relatively large drops of fuel. Such a fine mist is however the best prerequisite for rapid, constant and effective ignition and combustion.
A further advantage is the unusual spontaneity of the engine transient response. This is because load management is near the point of action, that is right next to the combustion chambers. So there is no more delay between the process of giving gas and the actual acceleration attributable so far to the time required for filling up the intake manifold between the throttle butterfly and the combustion chamber"
