How does the venturi principle work without a rate increase of the fluid? As a mechanic, I inherently know if there is a crack in the exhaust of a vehicle prior to the O2 (lambda) sensor, fresh air will get in and cause the system to read a false lean state (lean meaning, oxygen content greater than stoic). Typical logic would dictate since the exhaust is under greater pressure than the outside air, exhaust would be pushed out of the crack and no outside air could get in. In practice, however, I know the outcome is quite different.
My understanding is, the venturi principle comes into effect here. There is something about how, when air passes over a hole (or the crack in this case) it will draw the outside air along with it. Something to do with the speed of the gasses as it flows over the hole pulling from the hole as it goes over it. 
My questions are:


*

*Am I right in this being a venturi effect?

*Can someone explain the exact phenomenon?

*Is there a mathematical formula which explains any of the relationships? (ie: size of the hole v. speed of the exhaust produces this much intake of air)
I understand the Bernoulli principle may have something to do with this, as well. The part about it is in all the cases which I've seen explained, they talk about there being a need for the fluid (exhaust in this case) to be sped up as it passes the hole, thus causing a low pressure area at the hole (layman's terms, sorry) which will create a draw.  Reading this Q/A explains it through this diagram:

The diagram and attached question has to do with a boat hull and it allowing it to drain water. In my example of an exhaust, there's no lump/bulge/area which extends into the exhaust flow causing the fluid flow rate change ... in fact, due to turbulence, it probably slows it down. 
Wikipedia does nothing to help with my understanding in this situation.
 A: I hope you get a better answer than this from an experimentalist. This was always my understanding, but as I self study, there's never a professor around when you need one. (Not complaining,  just saying is all :)
The part I don't follow is that the picture below shows an obvious constriction, whereas a crack in say, the constant diameter rear exhaust box/muffler, is just a crack, not a narrowing.
Anyway, the  venturi effect makes sense to me in terms of air molecules movement.

As they enter the narrow part, the air molecules must speed up to maintain continuity of flow. So instead of exerting pressure randomly in all directions, now a lot of them are forced in the direction along the long axis of the exhaust, so less are available to "point" upwards, so static pressure drops and the outside air flows in.
The theoretical pressure drop at the constriction is given by this formula below, which is based on Bernoulli's equation:
$${\displaystyle p_{1}-p_{2}={\frac {\rho }{2}}\left(v_{2}^{2}-v_{1}^{2}\right)}$$
where ${\displaystyle \scriptstyle \rho \,}$ is the density of the fluid, ${\displaystyle \scriptstyle v_{1}}$ is the (slower) fluid velocity where the pipe is wider, ${\displaystyle \scriptstyle v_{2}}$ is the (faster) fluid velocity where the pipe is narrowed.
A: Following answer is speculative.
I don't know what exactly there is inside the exhaust pipe that may offer resistance to flow of gases, so I am going to assume that exhaust pipe is just a hollow pipe. If this is the case then (static) pressure of exhaust gases inside the pipe will be very close to atmospheric pressure, only slightly higher (enough to overcome viscous resistance within the flow). Where the pipe is broken, an eddying region may form in the wake of the broken piece, and flow being turbulent, is able to scoop in atmospheric air, while also simultaneously exhaust is leaking out from the broken region into the ambient. In other words I think, the effect you have observed owes more to turbulent entrainment rather than venturi effect.
