Laminar closed-channel flows [Fluid Mechanics #8]

[@alecfeller5687]

When I solved for viscosity driven flow, I ended with u=(y/2μ)(dP/dx)(y-a) + (V(not)*y/a), but in a later video I see that this would be correct without that first term. Is there a quick explination you can give me as to how that first term cancels out? (also your videos are saving my life thank you!)

[@prof.vanburen]

Hey Alec, nice to see you working through it! In this case, the flow is purely viscosity driven so there is no pressure gradient in the streamwise direction (dP/dx = 0). This was probably not stated as clearly as it should have. You can have flow with a moving plate and a pressure gradient, as in your solution, which just adds complexity.

[@akarshshetty9262]

In the gravity driven flow scenario, I am confused as to why the dP/dX term does not go to zero as the flow is being primarily driven by the action of gravity. What is the reason for this assumption ?

[@haiminghuang1542]

I am confused : In the gravity driven case, shouldn't the flow go faster and faster? That means flow is NOT fully developed

[@prof.vanburen]

Gravity driven flow can definitely become fully developed. E.g., gravity provides a force and viscosity provides resistive forces. When those two balance you have fully developed flow. Think of it as similar to terminal velocity, something falling through the air eventually reaches a steady velocity because the drag force balances the force of gravity.

[@haiminghuang1542]

@@prof.vanburen Got it. Thank you.