# 8.7.1: Interfacial Instability

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Fig. 8.21 Flow of liquid in partially filled duct.

In Example 8.9 no requirement was made as for the velocity at the interface (the upper boundary). The vanishing shear stress at the interface was the only requirement was applied. If the air is considered two governing equations must be solved one for the air (gas) phase and one for water (liquid) phase. Two boundary conditions must be satisfied at the interface. For the liquid, the boundary condition of "no slip''  at the bottom surface of liquid must be satisfied. Thus, there is total of three boundary conditions
to be satisfied. The solution to the differential governing equations provides only two constants. The second domain (the gas phase) provides another equation with two constants but again three boundary conditions need to satisfied. However, two of the  boundary conditions for these equations are the identical and thus the six boundary conditions are really only 4 boundary conditions. The governing equation solution for the gas phase ($$h \geq y \geq a\,h$$) is
$\label{bmI:eq:gas} {U_x}_g = \dfrac{g\sin\theta}{2\,\nu_g} \,y^2 + c_1\,y + c_2 \tag{118}$
Note, the constants $$c_1$$ and $$c_2$$  are dimensional which mean that they have physical units ($$c_1 \longrightarrow \left[1/sec\right]$$ The governing equation in the liquid phase ($$0 \geq y \geq h$$) is
$\label{bmI:eq:liquid} {U_x}_{\ell} = \dfrac{g\sin\theta}{2\,\nu_{\ell}} \,y^2 + c_3\,y + c_4 \tag{119}$
The gas velocity at the upper interface is vanished thus
$\label{bmI:eq:gbc1} {U_x}_g\left[ (1+a)\,h\right] = 0 \tag{120}$
At the interface the "no slip'' condition is regularly applied and thus
$\label{bmI:eq:gbc2} {U_x}_g(h) = {U_x}_{\ell} (h) \tag{121}$
Also at the interface (a straight surface), the shear stress must be continuous
$\label{bmI:eq:gbc3} \mu_{g} \dfrac{\partial {U_x}_g}{\partial y} = \mu_{\ell} \dfrac{\partial {U_x}_{\ell}}{\partial y} \tag{122}$
Assuming "no slip'' for the liquid at the bottom boundary as
$\label{bmI:eq:lbc1} {U_x}_{\ell} (0) = 0 \tag{123}$
The boundary condition (120) results in
$\label{bmI:eq:gbcs1} 0 = \dfrac{g\,\sin\theta}{2\,\nu_g}\,h^2\,(1+a)^2 + c_1 \,h\,(1+a) + c_2 \tag{124}$
The same can be said for boundary condition (123) which leads
$\label{bmI:eq:gbcs1c} c_4 = 0 \tag{125}$
Applying equation (122) yields
$\label{bmI:eq:gbcs3} \overbrace{\dfrac{\mu_g}{\nu_g}}^{\rho_g} {g\,\sin\theta}\, h + c_1\,\mu_g= \overbrace{\dfrac{\mu_{\ell}}{\nu_{\ell}}}^{\rho_{\ell}} {g\,\sin\theta}\, h + c_3\,\mu_{\ell} \tag{126}$
Combining boundary conditions equation (??) with (124) results in
$\label{bmI:eq:gbcs2} \dfrac{g\sin\theta}{2\,\nu_{g}} \,h^2 + c_1\,h + c_2 = \dfrac{g\sin\theta}{2\,\nu_{\ell}} \,h^2 + c_3\,h \tag{127}$

The solution of equation (124), (126) and (127) is obtained by computer algebra (see in the code) to be
$\label{bmI:eq:gSol} \displaystyle c_1= -\dfrac{\sin\theta\,\left( g\,h\,\rho_{g}\,\left( 2\,\rho_{g}\,\nu_{\ell}\,\rho_{\ell}+1\right) +a\,g\,h\,\nu_{\ell}\right) } {\rho_{g}\,\left( 2\,a\,\nu_{\ell}+2\,\nu_{\ell}\right) } \\ \displaystyle c_2= \dfrac{\sin\theta\,\left( g\,{h}^{2}\,\rho_{g}\,\left( 2\,\rho_{g}\,\nu_{\ell}\,\rho_{\ell}+1\right) -g\,{h}^{2}\,\nu_{\ell}\right) }{2\,\rho_{g}\,\nu_{\ell}} \\ \displaystyle c_3= \dfrac{\sin\theta\,\left( g\,h\,\rho_{g}\,\left( 2\,a\,\rho_{g}\,\nu_{\ell}\,\rho_{\ell}-1\right) -a\,g\,h\,\nu_{\ell}\right) }{\rho_{g}\,\left( 2\,a\,\nu_{\ell}+2\,\nu_{\ell}\right) } \tag{128}$

When solving this kinds of mathematical problem the engineers reduce it to minimum amount of parameters to reduce the labor involve. So equation (124) transformed by some simple rearrangement to be
$\label{bmI:eq:gbcsd1} \left(1+a\right)^{2} = \overbrace{\dfrac{2\,\nu_g\,c_1}{g\,h\,\sin\theta}}^{C_1} + \overbrace{\dfrac{2\,c_2 \,\nu_g}{g\,h^2\,\sin\theta}}^{C_2} \tag{129}$
And equation (126)
$\label{bmI:eq:gbcsd3} 1 + \overbrace{\dfrac{\nu_g\, c_1}{g\,h\,\sin\theta}}^{\dfrac{1}{2}\,C_1} = \dfrac{\rho_{\ell}} {\rho_g} + \overbrace{\dfrac{\mu_{\ell}\,\nu_g\, c_3}{\mu_{g}\,g\,h\,\sin\theta}} ^{\dfrac{1}{2}\dfrac{\mu_{\ell}}{\mu_g}\,C_3 } \tag{130}$
and equation (127)
$\label{bmI:eq:gbcsd2} 1 +\dfrac{2\,\nu_g\,\cancel{h}\,c_1}{h^{\cancel{2}}\,g\,\sin\theta} + \dfrac{2\,\nu_{g}\,c_2} {h^2\,g\sin\theta} = \dfrac{\nu_g}{\nu_{\ell}} + \dfrac{2\,\nu_{g}\,\cancel{h}\,c_3} {g\,h^{\cancel{2}}\,\sin\theta} \tag{131}$
Or rearranging equation (131)
$\label{bmI:eq:gbcsd2a} \dfrac{\nu_g}{\nu_{\ell}} - 1 = \overbrace{\dfrac{2\,\nu_g\,c_1}{h\,g\,\sin\theta}}^{C_1} + \overbrace{\dfrac{2\,\nu_{g}\,c_2} {h^2\,g\sin\theta}}^{C_2} - \overbrace{\dfrac{2\,\nu_{g}\,\,c_3} {g\,h\,\sin\theta} }^{C_3} \tag{132}$
This presentation provide similarity and it will be shown in the Dimensional analysis chapter better physical understanding of the situation. Equation (129)  can be written as
$\label{bmI:eq:gbcsdd1} \left(1+a\right)^2 = C_1 + C_2 \tag{133}$
Further rearranging equation (130)
$\label{bmI:eq:gbcsdf3} \dfrac{\rho_{\ell}} {\rho_g} -1 = \dfrac{ C_1}{2} - \dfrac{\mu_{\ell}}{\mu_g}\, \dfrac{C_3}{2} \tag{134}$
and equation (132)
$\label{bmI:eq:gbcsdd2} \dfrac{\nu_g}{\nu_{\ell}} - 1 = C_1 + C_2 -C_3 \tag{135}$
This process that was shown here is referred as non&ndash; dimensionalization. The ratio of the dynamics viscosity can be eliminated from equation (135) to be
$\label{bmI:eq:gbcsdd2m} \dfrac{\mu_g}{\mu_{\ell}} \,\dfrac{\rho_{\ell}} {\rho_g} - 1 = C_1 + C_2 -C_3 \tag{136}$
The set of equation can be solved for the any ratio of the density and dynamic viscosity. The solution for the constant is
$\label{bmI:eq:solC1} C_1 = \dfrac{\rho_{g}}{\rho_{\ell}}-2 +{a}^{2}+2\,a\dfrac{\mu_{g}}{\mu_{\ell}}+2\dfrac{\mu_{g}}{\mu_{\ell}} \tag{137}$
$\label{bmI:eq:solC2} C_2=\dfrac{-\dfrac{\mu_{g}}{\mu_{\ell}}\,\dfrac{\rho_{\ell}}{\rho_g}+ a\,\left( 2\,\dfrac{\mu_{g}}{\mu_{\ell}}-2\right) + 3\,\dfrac{\mu_{g}}{\mu_{\ell}}+{a}^{2}\,\left( \dfrac{\mu_{g}}{\mu_{\ell}}-1\right) -2}{\dfrac{\mu_{g}}{\mu_{\ell}}} \tag{138}$
$\label{bmI:eq:solC3} C_3=-\dfrac{\mu_{g}}{\mu_{\ell}}\,\dfrac{\rho_{\ell}}{\rho_g}+{a}^{2}+2\,a+2 \tag{139}$
The two different fluids have flow have a solution as long as the distance is finite reasonable similar. What happen when the lighter fluid, mostly the gas, is infinite long. This is one of the source of the instability at the interface. The boundary conditions of flow with infinite depth is that flow at the interface is zero, flow at infinite is zero. The requirement of the shear stress in the infinite is zero as well. There is no way obtain one dimensional solution for such case and there is a component in the $$y$$ direction. Combining infinite size domain of one fluid with finite size on the other one side results in unstable interface.

### Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.