What I’ve called nu is your eta over rho, unless you meant kinematic viscosity, in which case my nu is your eta.

]]>Using big D derivative, with del dot v = 0, assuming x is perpendicular to the inclined plane and nu=constant –

0 = g sin(theta) + nu d^2v/dx^2

v = – (g sin(theta) / 2nu) x^2 + c_1 x + c_0

Assuming the flow is stuck on the bottom (v = 0 at x = 0) and there is no shear at the air boundary (dv/dx = 0 at x = d)

v = (g sin(theta) / 2nu) ( 2xd – x^2 )

(fastest at the top)

The flow rate is integral of this over x, times width w

= (g sin(theta) / 3 nu) d^3 w

This looks right dimensionally (m^3 s^-1)

]]>Only up to dimensionless factors (like, e.g., 1/3)…

]]>I must have completely missed the last bit “as well as the acceleration due to gravity, g”, sorry! Isn’t this a case where a dimensional argument gives you the answer?

]]>I doubt if oil would flow on Jupiter…

]]>