Question 3.2: Film Coating Consider “film coating”, i.e., a liquid fed fro......

Concerning the Solution Steps (i)–(iii) just discussed, an inclined x– y coordinate system is attached to the plate, i.e., the x-momentum equation is of interest (Note, the y- and z- momentum equations are both zero). Based on the sketch and assumptions, \vec {\bf v} = (u, 0, 0), ∇p ≈ 0 (thin-film condition, or no air-pressure variations), and ∂/∂t = 0 (steady-state). With v = w = 0, continuity \nabla\cdot\vec {\bf v}=0 indicates that:

\frac{\partial\mathrm{u}}{\partial\mathrm{x}}+0+0=0                         (E.3.2.1)

which implies fully-developed (or parallel) flow; thus, u = u(y) only. Consulting the equation sheet (App. A) and invoking the postulates u = u(y), v = w = 0, and ∂p / ∂x = 0 , we have with the body force component ρg sin \phi ,

0=\mu{\frac{\mathrm{d}^{2}\mathrm{u}}{\mathrm{d}y^{2}}}+\mathrm{\rho g}\,\sin\phi                      (E.3.2.2)

subject to u(y = 0) = 0 <no-slip> and \tau_{\mathrm{interface}}={\mu}\,{\frac{\mathrm{du}}{\mathrm{dy}}}\bigg|_{\mathrm{y=h}}\approx0 which implies that at y = h → du/dy = 0 <zero velocity gradient>. Double integration of Eq. (E.3.2.2) in the form u′′ = K (see App. A), yields after invoking the two B.C.s:

\mathrm{u}({\rm y})=-\frac{\mathrm{\rho gh}^{2}\;\mathrm{sin}\;\phi}{2\mathrm{\mu}}[\,\left\lgroup\frac{{\rm y}}{\rm h}\right\rgroup ^{2}-2{\rm y}/\mathrm{h}]                       (E.3.2.3a)

Clearly,

\rm{u}({y}={h})={u}_{max}={\frac{\rho \rm{gh}^{2}}{2\mu}}\sin\phi                    (E.3.2.3b)

and the average velocity

\rm{u}_{av}=\frac{1}{A}\int_A{\vec{\bf v}}\cdot{d}{\vec{\bf A}}:={\frac{1}{h}}\int_{0}^{h}{u\,d}{y}={\frac{\rho  gh^{2}\sin\phi}{3\mu}}                        (E.3.2.4a)

Hence,

\mathrm{u}_{\mathrm{av}}={\frac{2}{3}}\mathrm{u}_{\mathrm{max}}                      (E.3.2.4b)

In order to determine the film thickness, we first compute the volumetric flow rate Q, which is usually known.

\mathrm{Q}=\int{{\vec {\bf v}}\cdot\mathrm{d}\vec{\bf A}}:=\mathrm{b}\operatorname*{\int}_{0}^h\mathrm{u\,d}{\bf y}:={\rm u}_{\rm a v}(\mathrm{hb})                          (E.3.2.5a)

where b is the plate width. Thus,

\mathrm{Q}={\frac{\mathsf{\rho }\,\mathrm{ g\,b}\,\textrm{sin}\,\phi }{3\mu}}\mathrm{h}^{3}                     (E.3.2.5b)

from which

\mathrm{h}=\left\lgroup{\frac{3\ \mathbf{\mu}\ \mathrm{Q}}{\rho \,\mathrm{{g}\,\mathrm{b}\ \sin\ {\mathbf{\phi}}}}}\right\rgroup^{1/3}                       (E.3.2.5c)

The shear force (or drag) exerted by the liquid film onto the plate is

{\rm F}_{s}=\int_{\rm A}\tau_{\rm {yx}\, }\mathrm{d}{\rm A}={\rm b}\int_{0}^{\rm l}\left\lgroup{-\mu }\left.{\frac{\mathrm{d}{\rm u}}{\mathrm{d}y}}\right|_{\rm y=0}\right\rgroup \rm dy                             (E.3.2.6a)

Thus,

\mathrm{{F}_{s}}=\mathrm{\rho ~\rm g\,b~h~\rm l~sin~}\phi                                  (E.3.2.6b)

which is the x-component of the weight of the whole liquid film along 0 ≤ x ≤ l .

Graph:

Comments:

• \rm{u_{average}} can be used to compute the film Reynolds number, \mathrm{Re}_{\mathrm{h}}={\frac{\mathrm{u}_{a\rm v}\,\mathrm{h}}{\mathrm{\nu}}} , and check if \mathrm{Re}_{\mathrm{h}}\lt \rm{Re_{critical}}\approx 20

• Given Q (or \dot m=\rho Q), the coating thickness can be estimated

Note: The case of wire coating, i.e., falling film on a vertical cylinder is assigned as a homework problem (see Sect. 3.8).