Dynamic Cylinder Deactivation

[christofoo]

Originally Posted by t vago

...
Now, how much work does each cylinder consume? Not looking for numbers, but a general answer.

Not a fixed number and not sure what answer you're looking for.

[t vago]

Originally Posted by christofoo

Not a fixed number and not sure what answer you're looking for.

The piston in that cylinder has to work to push out gases into the exhaust manifold, on the exhaust stroke, right?

The piston in that same cylinder then has to work to suck in gases from the intake manifold, on the subsequent intake stroke, right?

[christofoo]

Originally Posted by t vago

The piston in that cylinder has to work to push out gases into the exhaust manifold, on the exhaust stroke, right?

The piston in that same cylinder then has to work to suck in gases from the intake manifold, on the subsequent intake stroke, right?

Affirmative.

[t vago]

So, the piston has to push against an absolute exhaust gas pressure, and then has to suck in against an absolute intake gas pressure.

Using thermodynamic work equation for constant pressure processes, it can be shown that the work consumed is proportional to the difference between the exhaust gas pressure and the intake manifold pressure.

This is true in all cases - rich, lean, stoich, or no fuel at all.

[christofoo]

Originally Posted by t vago

So, the piston has to push against an absolute exhaust gas pressure, and then has to suck in against an absolute intake gas pressure.

Using thermodynamic work equation for constant pressure processes, it can be shown that the work consumed is proportional to the difference between the exhaust gas pressure and the intake manifold pressure.

This is true in all cases - rich, lean, stoich, or no fuel at all.

Agreed. But if you want to evaluate DCD or lean-burn relative effects have to be considered now.

[christofoo]

Originally Posted by christofoo

Agreed. But if you want to evaluate DCD or lean-burn relative affects have to be considered now.

And, don't neglect the power required to generate the vacuum.

[christofoo]

Originally Posted by t vago

...
Using thermodynamic work equation for constant pressure processes, it can be shown that the work consumed is proportional to the difference between the exhaust gas pressure and the intake manifold pressure.
...

(I don't think you mean constant pressure process - doesn't seem to matter for the moment I know what you mean (adiabatic, I'd think).)

[t vago]

Originally Posted by christofoo

Agreed. But if you want to evaluate DCD or lean-burn relative affects have to be considered now.

Nope.

The work consumed is proportional to the difference between the exhaust gas absolute pressure and the intake manifold absolute pressure. This is true in all cases - rich, lean, stoich, or no fuel at all. We call this pumping work, or pumping loss, if you will.

You're still not quite at the step where you're evaluating DCD or lean-burn, but you're close.

[christofoo]

Getting ahead of you for a moment, don't forget there is a reason I'm asking you to compare with lean-burn. You need to be able to explain why a lean-burn engine, having a higher air flow mass rate, gets better FE than than a stoichiometric engine.

[t vago]

Originally Posted by christofoo

(I don't think you mean constant pressure process - doesn't seem to matter for the moment I know what you mean (adiabatic, I'd think).)

In reality, no, it isn't a constant pressure process. However, for the sake of this argument, humor me and assume that in an ideal world, the exhaust absolute pressure and the intake absolute pressure are held constant for the intake and exhaust strokes of an ideal Otto cycle engine.