SOLUTION: Gravity-driven viscous flow over an inclined plate
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Con onfig figu ura rati tion on
We consider the situation represented in Fig .1 .1.. Due to gravity, a thin liquid film (assumed of constant thickness h) flows steadily and at low speed (a laminar flow can be assumed) along an inclined plate (inclination angle α). Th Thee fluid fluid properties (density ρ (density ρ,, dynamic viscosity µ viscosity µ)) are known and constant.
Figure 1: Gravity-driven viscous flow over an inclined plate.
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Ques Qu esti tion onss an and d an answ swer erss
2.1 2. 1
Vel eloci ocitty fie field ld
Determine the velocity profile u(y ).
We are in a typical situation where we may employ the simplest formulation of the Navier-Stokes equations, since many assumptions have been made: • the
flow is steady, so that all time derivatives will disappear; 1
• the
flow is incompressible (liquid);
• the
liquid is Newtonian (constant dynamic viscosity µ);
• the
flow is indeed two-dimensional, since no specific information is given concerning the third spatial direction, perpendicular to the figure. Only the directions x and y will be considered, neglecting all possible changes along z .
In order to fully determine this flow, we just need to determine the x and ycomponents of the velocity, written respectively u and v, and the pressure p (the density is constant and therefore not a variable). In principle, the three variables of interest u, v and p all depend on x and y. What are our starting conservation equations? The generic conservation equations for mass and momentum, simplified using the above assumptions. For mass, one gets ∇ · v = 0, here: ∂u ∂ v + =0 ∂x ∂y
(1)
Now, what are the boundary conditions for the y-component v? The no-slip boundary condition at the wall reads v(y = 0) = 0. Since the film thickness h is constant, the flow velocity at the contact point with the surrounding atmosphere must necessarily be parallel to the wall, leading again to v(y = h) = 0. Starting from zero, going back to 0, and without any particular reason to appear, the y-component of the velocity is therefore zero everywhere! This is the first, major result: (2) v=0 Plugging back this result in the mass conservation equation, Eq.(1), one gets automatically: ∂u =0 (3) ∂x Hence, the x-component of the velocity is constant along the x-direction and changes only with y: (4) u = u(y) What are the boundary conditions for u? Again, along the plate, the no-slip boundary conditions reads (5) u(y = 0) = 0 like for v. Now, at the contact surface with the atmosphere, the exchange of momentum at the molecular level between the molecules of the liquid near the surface and the air molecules near the surface lead to the same no-slip condition: the velocity in the liquid must be the same as the velocity in the gas near the contact surface. 2
What is the consequence? The air near the surface of the liquid film will be entrained by the movement of the film and will flow locally at the same velocity (at present unknown). How can we translate mathematically this property? We know that friction is the result of a velocity difference. Therefore, if the velocity in liquid and gas are identical, there are no friction forces along the film surface. The only component of the friction force along the film surface, obtained by simplifying the friction ∂u tensor for the conditions considered here, is proportional to . Finally, the ∂y boundary condition we have to enforce at the film surface reads: ∂u (y = h) = 0 ∂y
(6)
Let us now write the standard Navier-Stokes equations taking into account the specific assumptions of this problem. It comes:
∂ u ∂ u + ∂x ∂ v ∂ ∂yv
1 ∂p ∂u ∂u = gx − + ν u + v ∂x ∂y ρ ∂x 1 ∂p ∂v ∂v = gy − + ν u + v ∂x ∂y ρ ∂y
2
2
2
2
∂x 2
(7)
2
2
+
(8)
∂y 2
Since v = 0, the second equation just leads to ∂p = ρg y = − ρg cos α ∂y
(9)
The boundary condition for the pressure reads classically at the contact surface between liquid and gas (10) p(y = h) = p a This is the same at every x-position. Considering this condition and the previous equation (9), one obtains directly: p = p(y) = p a + ρg (h − y)cos α
(11)
This is indeed the same as hydrostatics (the pressure increases linearly with the depth (h − y) within the fluid), which is not a surprise, since the flow is completely perpendicular to the pressure gradient. Note that the change in pressure will be indeed extremely small! The overpressure at the wall is ρgh, which for ρ = 1000 kg/m3 and h = 1 cm would mean about 100 Pa, or about 1 /1000 of the atmospheric pressure pa . Therefore, you understand why we would rightly assume p = p a = constant when analyzing globally this flow in the light of the Bernoulli equation!
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Now, taking into account this (small) pressure variation, the only flow velocity component u can be easily obtained by considering what remains of Eq.( 7): ∂ 2 u g sin α = − . ∂y 2 ν
(12)
Since the right-hand side is a constant value, this can be easily integrated twice as g sin α y 2 + C 1 y + C 2 (13) u(y) = − ν 2 The two integration constants C 1 and C 2 are easily determined by enforcing the two boundary conditions, Eq.(5) and (6), leading to: gh sin α ν = 0
C 1 = C 2
(14) (15)
Finally, the x-component of the velocity is now perfectly known: g sin α y u = u(y) = y h − 2 ν
(16)
Of course, for α = 0 (a horizontal plate), there is no gravity-driven flow ( u = 0), as expected!
2.2
Volumetric flow rate
Determine the resulting volumetric flow-rate q ˙.
The volumetric flow-rate is obtained by integrating through the depth of the film, taking into account the flow length b perpendicular to the figure: y =h
q ˙ :=
y =0
bu(y)dy
(17)
Using the known solution for u, Eq.(16), one becomes successively:
y =h 1 bg sin α q ˙ = h ydy − 2 ν y =0 bg sin α h2 1 h3 = h − 2 2 3 ν
Finally, the answer reads: q ˙ =
bg sin α 3 h 3ν
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y=h
y =0
y 2dy
(18) (19)
(20)
2.3
Connection to Bernoulli
If we would try to solve this problem in a “Bernoulli-like” manner, then we would also obtain similar informations, but of course not at the same level of details! We would just write as usual p = p a throughout the flow. As we have seen, this is almost right (typically 0.1% error) for the present flow. We would not get the detailed velocity field, but we could write the Bernoulli equation (only valid along streamlines, since the vorticity ∂v ∂x
−
∂ u g sin α = (y − h) ∂y ν
(21)
is not zero). By expressing that the flow velocity (assumed, rightly, to be only in the x-direction) cannot change with the x-coordinate (same as stating that the film thickness h is constant), one would get that the change in potential energy is exactly what is needed to compensate the friction loss in the flow. In this manner, it is indirectly possible to get back the mean flow velocity in the film.
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