Gents, you got me thinking about the whole flow rate issue. This is the physical explanation I can come up with:
Watch out guys, switching on smart idiot mode now!
Bernoulli tells us: p + rho/2 * C^2 = const., where p is pressure, rho is the fluid's viscosity (which we assume being constant) and C is the fluid's velocity.
As both big ends are exposed to the same surrounding pressure p inside the crankcase, you can derive from Bernoulli's principle, that the fluid's velocity must be the same at both big ends, regardless of their cross-sectional area. We are neglecting line loss here, as the sludge trap is not very long this should be insignificant enough.
The volumetric flow Q is defined as
Q = A * C.
C is the same for both big ends, like stated before.
So with an increase in the cross-sectional area A, the flow rate MUST increase as well.
So YES, the with the drilling oil flow to the lhs big end WILL increase.
In case the oil pump is not able to cope with the higher flow rate (at low rpm for instance), this will lead to a reduction of volumetric flow at the rhs big end due to the principle of continuity accordingly:
Q_oil_pump = Q_big_end1 + Q_big_end2
***smart idiot mode off***
Jeez, now I really need a cold Goesser to clear my mind...
Cheers!