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Transport Phenomena - Fluid mechanics Problem : Newtonian fluid flow in a falling film

Problem.

Consider a Newtonian liquid (of viscosity μ and density ρ) in laminar flow down an inclined flat plate of length L and width W. The liquid flows as a falling film with negligible rippling under the influence of gravity. End effects may be neglected because L and W are large compared to the film thickness δ.

      Figure. Newtonian liquid flow in a falling film.

a) Determine the steady-state velocity distribution.

b) Obtain the mass rate of flow and average velocity in the falling film.

c) What is the force exerted by the liquid on the plate in the flow direction?

d) Derive the velocity distribution and average velocity for the case where x is replaced by a coordinate x' measured away from the plate (i.e., x' = 0 at the plate and x' = δ at the liquid-gas interface).

Solution.

Click here for stepwise solution

a)

Step. Shear stress distribution

For axial flow in rectangular Cartesian coordinates, the differential equation for the momentum flux is (click here for derivation)

dτxz

dx  =  

ΔP

L(1)

where P is a modified pressure, which is the sum of both the pressure and gravity terms, that is, ΔP = Δp + ρ g L cos β. Here, β is the angle of inclination of the z-axis with the vertical. Since the flow is solely under the influence of gravity, Δp = 0 and therefore ΔP/L = ρ g cos β. On integration,

τxz  =  ρ g x cos β + C1 (2)

On using the boundary condition at the liquid-gas interface (τxz = 0 at x = 0), the constant of integration C1 is found to be zero. This gives the expression for the shear stress τxz for laminar flow in a falling film as

τxz  =  ρ g x cos β (3)

It must be noted that the above momentum flux expression holds for both Newtonian and non-Newtonian fluids (and does not depend on the type of fluid).

The shear stress for a Newtonian fluid (as per Newton's law of viscosity) is given by

τxz  =  − μdvz

dx(4)

Step. Velocity distribution

To obtain the velocity distribution, there are two possible approaches.

In the first approach, equations (3) and (4) are combined to eliminate τxz and get a first-order differential equation for the velocity as given below.

dvz

dx =  − 

ρ g cos β

μ x (5)

On integrating, vz = − ρ g x2 cos β/(2μ) + C2. On using the no-slip boundary condition at the solid surface (vz = 0 at x = δ ), C2 = ρ g δ2 cos β/(2μ). Thus, the final expression for the velocity distribution is parabolic as given below.

vz   =  ρ g δ2 cos β

2μ

1 −

x

δ

2

 

(6)

In the second approach, equations (1) and (4) are directly combined to eliminate τxz and get a second-order differential equation for the velocity as given below.

− μ d2vz

dx2 = ρ g cos β (7)

Integration gives dvz

dx =  − 

ρ g cos β

μ x + C1 (8)

On integrating again,

vz =  − ρ g x2 cos β

2μ + C1 x + C2 (9)

On using the boundary condition at the liquid-gas interface (at x = 0, τxz = 0 ⇒ dvz/dx = 0), the constant of integration C1 is found to be zero from equation (8). On using the other boundary condition at the solid surface (at x = δ, vz = 0), equation (9) gives C2 = ρ g δ2 cos β/(2μ). On substituting C1 and C2 in equation (9), the parabolic velocity profile in equation (6) is again obtained.

b)

Step. Mass flow rate

The mass flow rate may be obtained by integrating the velocity profile given by equation (6) over the film thickness as shown below.

w   =   δ   ρ vz W dx

  =   ρ2 g W δ3 cos β 1

∫

1 −

x

2

d

x

(10)

∫ 0

2μ0

δ 

δ

w   =  ρ2 g W δ3 cos β

3μ(11)

When the mass flow rate w in equation (11) is divided by the density ρ and the film cross-sectional area (W δ), an expression for the average velocity <vz> is obtained. Also, the maximum velocity vz, max occurs at the interface x = 0 and is readily obtained from equation (6). Thus, equations (6) and (11) give

<vz>   =  ρ g δ2 cos β

3μ  =  

2

3 vz, max (12)

c)

Step. Force on the plate

The z-component of the force of the fluid on the solid surface is given by the shear stress integrated over the wetted surface area. Using equation (3) gives

Fz  =   L W τxz|x = δ   =   ρ g δ L W cos β (13)

As expected, the force exerted by the fluid on the plate is simply the z-component of the weight of the liquid film.

d)

Step. Choice of coordinates

Consider the coordinate x' starting from the plate (i.e., x' = 0 is the plate surface and x' = δ is the liquid-gas interface). The form of the differential equation for the momentum flux given by equation (1) will remain unchanged. Thus, equation (2) for the new choice of coordinates will be

τx'z  =  ρ g x' cos β + C1 (14)

On using the boundary condition at the liquid-gas interface (τx'z = 0 at x' = δ), the constant of integration is now found to be given by C1 = − ρ g δ cos β. This gives the shear stress as

τx'z  =  ρ g (x' − δ) cos β (15)

Note that the momentum flux is in the negative x'-direction. On substituting Newton's law of viscosity,

dvz

dx'  =  

ρ g cos β

μ (δ − x') (16)

On integrating, vz = (ρg/μ) (δ x' − ½ x'2) cos β + C2. On using the no-slip boundary condition at the solid surface (vz = 0 at x' = 0), it is found that C2 = 0. Thus, the final expression for the velocity distribution is

vz   =  ρ g δ2 cos β

μ

x'

δ − 

1

2

x'

δ

2

 

(17)

The average velocity may be obtained by integrating the velocity profile given in equation (17) over the film thickness as shown below.

w   =  1

δ

δ

∫ 0

  vz dx'

  =  ρ g δ2 cos β

μ

1

∫

0

x'

δ − 

1

2

x'

δ

2

 

d

x

δ

(18)

Integration of equation (18) yields the same expression for <vz> obtained earlier in equation (12).

It must be noted that the two coordinates x and x' are related as follows: x + x' = δ. By substituting (x/δ)2 = [1 − (x'/δ)]2 = 1 − 2(x'/δ) + (x'/δ)2, equation (17) is directly obtained from equation (6).

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