Inclined Plate

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SOLUTION: Gravity-driven viscous flow over an inclinedplate

1 Configuration

We consider the situation represented in Fig.1. Due to gravity, a thin liquid film(assumed of constant thickness h) flows steadily and at low speed (a laminarflow can be assumed) along an inclined plate (inclination angle ). The fluidproperties (density , dynamic viscosity ) are known and constant.

Figure 1: Gravity-driven viscous flow over an inclined plate.

2 Questions and answers

2.1 Velocity field

Determine the velocity profileu(y).We are in a typical situation where we may employ the simplest formulation ofthe Navier-Stokes equations, since many assumptions have been made:

the flow is steady, so that all time derivatives will disappear;

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the flow is incompressible (liquid);

the liquid is Newtonian (constant dynamic viscosity );

the flow is indeed two-dimensional, since no specific information is givenconcerning the third spatial direction, perpendicular to the figure. Onlythe directions x and y will be considered, neglecting all possible changesalong z.

In order to fully determine this flow, we just need to determine the x and y-components of the velocity, written respectively u and v, and the pressurep (thedensity is constant and therefore not a variable). In principle, the three variablesof interest u, v and p all depend onx and y .

What are our starting conservation equations? The generic conservation equa-

tions for mass and momentum, simplified using the above assumptions.For mass, one gets v= 0, here:

u

x+

v

y = 0 (1)

Now, what are the boundary conditions for the y-component v? The no-slipboundary condition at the wall reads v(y= 0) = 0. Since the film thickness h isconstant, the flow velocity at the contact point with the surrounding atmospheremust necessarily be parallel to the wall, leading again to v(y = h) = 0. Startingfrom 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, majorresult:

v= 0 (2)

Plugging back this result in the mass conservation equation, Eq.(1), one getsautomatically:

u

x= 0 (3)

Hence, the x-component of the velocity is constant along the x-direction andchanges only with y :

u= u(y) (4)

What are the boundary conditions for u? Again, along the plate, the no-slipboundary conditions reads

u(y= 0) = 0 (5)

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 andthe air molecules near the surface lead to the same no-slip condition: the velocityin the liquid must be the same as the velocity in the gas near the contact surface.

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What is the consequence? The air near the surface of the liquid film will beentrained 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 isthe result of a velocity difference. Therefore, if the velocity in liquid and gas areidentical, there are no friction forces along the film surface. The only componentof the friction force along the film surface, obtained by simplifying the friction

tensor for the conditions considered here, is proportional to u

y. Finally, the

boundary condition we have to enforce at the film surface reads:

u

y(y= h) = 0 (6)

Let us now write the standard Navier-Stokes equations taking into account thespecific assumptions of this problem. It comes:

uu

x+ v

u

y = gx

1

p

x+

2u

x2+

2u

y2

(7)

uv

x+ v

v

y = gy

1

p

y+

2v

x2+

2v

y2

(8)

Sincev = 0, the second equation just leads to

p

y =gy =

gcos (9)

The boundary condition for the pressure reads classically at the contact surfacebetween liquid and gas

p(y= h) =pa (10)

This is the same at everyx-position. Considering this condition and the previousequation (9), one obtains directly:

p= p(y) =pa+g(hy)cos (11)

This is indeed the same as hydrostatics (the pressure increases linearly with thedepth (hy) within the fluid), which is not a surprise, since the flow is completelyperpendicular to the pressure gradient. Note that the change in pressure will beindeed extremely small! The overpressure at the wall is gh, which for =1000 kg/m3 and h = 1 cm would mean about 100 Pa, or about 1/1000 of theatmospheric pressurepa. Therefore, you understand why we would rightly assumep= pa= constant when analyzing globally this flow in the light of the Bernoulliequation!

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Now, taking into account this (small) pressure variation, the only flow velocitycomponentu can be easily obtained by considering what remains of Eq.(7):

2u

y2 =

gsin

. (12)

Since the right-hand side is a constant value, this can be easily integrated twiceas

u(y) = gsin

y2

2 +C1y+C2 (13)

The two integration constantsC1 and C2 are easily determined by enforcing thetwo boundary conditions, Eq.(5) and (6), leading to:

C1 =

gh sin

(14)C2 = 0 (15)

Finally, the x-component of the velocity is now perfectly known:

u= u(y) =gsin

y

h

y

2

(16)

Of course, for = 0 (a horizontal plate), there is no gravity-driven flow (u= 0),as expected!

2.2 Volumetric flow rate

Determine the resulting volumetric flow-rateq.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:

q:= y=hy=0

bu(y)dy (17)

Using the known solution for u, Eq.(16), one becomes successively:

q = bgsin

h y=hy=0

ydy 12

y=hy=0

y2dy

(18)

= bgsin

hh2

2

1

2

h3

3

(19)

Finally, the answer reads:

q= bgsin

3 h3 (20)

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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 = pa 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

y =

gsin

(y h) (21)

is not zero). By expressing that the flow velocity (assumed, rightly, to be only inthe 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 energyis exactly what is needed to compensate the friction loss in the flow. In thismanner, it is indirectly possible to get back the mean flow velocity in the film.

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