Steady Flow of an Incompressible Navier-Stokes Fluid down an Inclined Plane

In order to study this problem, it is more convenient to choose coordinate axes with $x_1$-axis along the inclined plane and $x_2$-axis perpendicular to it. We assume that the pressure and velocity fields are independent of $x_3$, and

$$v_i(x_1,x_2) = v(x_2)\delta_{i1}\, .$$ (6.3.12)

That is, the fluid is flowing parallel to the inclined plane. Thus, the top surface of the fluid will be parallel to the inclined plane. Such a flow is called laminar.

The balance of mass or the continuity equation, $v_{i,i}=0$, is identically satisfied. Also, the convective part of acceleration, $v_jv_{i,j}$, vanishes. Thus equations expressing the balance of linear momentum are

$$0 =\ -\frac{\partial p}{\partial x_1} + \mu \frac{d^2v}{dx^2_2} + \rho g\sin\theta\, ,$$ (6.3.13)

$$0 =\ -\frac{\partial p}{\partial x_2} - \rho g\cos\theta\, .$$ (6.3.14)

The associated boundary conditions are

$$\begin{split}&v_i(x_1,0) = 0,\ T_{ij}n_j\bigg\vert _{x_2=h} =... ...(x_2),\ T_{ij}n_j\bigg\vert _{x_1 = L} = q_i (x_2). \end{split}$$ (6.3.15)

That is, the fluid adheres to the stationary inclined plane, and the surface tractions are prescribed on the top surface and two arbitrarily chosen surfaces $x_1 = 0$ and $x_1 = L$.

A solution of equation (6.3.14) is

$$p = -\rho g\cos\theta x_2 + f(x_1),$$ (6.3.16)

where $f$ is an arbitrary smooth function of $x_1$. Substitution of (6.3.16) into (6.3.13) gives

$$\frac{df}{dx_1} = \mu \frac{d^2v}{dx^2_2} +\rho g\sin\theta.$$ (6.3.17)

Since the left-hand side of (6.3.17) is a function of $x_1$ and the right-hand side a function of $x_2$, therefore each must equal a constant $b$. Thus

$$\begin{split}f=\ & bx_1 + c,\\ v=\ & \frac{b -\rho g\sin\theta}{\mu} \frac{x^2_2}{2} + e_1x_2 + e_2,\end{split}$$ (6.3.18)

where $c,\ e_1$ and $e_2$ are constants of integration. Boundary condition (6.3.15)$_1$ requires that $e_2 = 0$. Substitution from (6.3.12), (6.3.18) and (6.3.16) into the constitutive relation (6.1.9) gives

$$T_{ij} =\ -(-\rho gx_2\cos\theta + bx_1 + c)\delta_{ij}$$

$$+ (\delta_{i1}\delta_{2j} + \delta_{j1}\delta_{2i})\left(\frac{b-\rho g\sin\theta} {\mu}x_2 + e_1\right)\, .$$ (6.3.19)

Equations (6.3.15)$_2$ and (6.3.19) yield

$$-(-\rho gh\cos\theta + bx_1 + c)\delta_{i2} + \delta_{i1}\left(\frac{b-\rho g\sin \theta}{\mu} h+e_1\right) = - p_a\delta_{i2}\, .$$ (6.3.20)

Hence

$$\begin{split}&-(-\rho gh\cos\theta + bx_1 + c) = -p_a\, ,\\ &\frac{b -\rho g\sin\theta}{\mu} h + e_1 =0\end{split}$$ (6.3.21)

Because (6.3.21)$_1$ must hold for all values of $x_1$, therefore

$$b=0,\ c=\rho gh\cos\theta + p_a,\ e_1 = \frac{\rho g\sin\theta}{\mu} h\, .$$ (6.3.22)

With these values of $b,\ c$ and $e_1$, equation (6.3.19) gives

$$T_{ij} = -(\rho g\cos\theta (-x_2 + h) + p_a)\delta_{ij} + (\delt... ...elta_{2j} + \delta_{j1}\delta_{2i})\frac{\rho g\sin \theta}{\mu} (-x_2 + h)\, .$$ (6.3.23)

Noting that on the surface $x_1 = 0,\ n_j = -\delta_{1j}$, equations (6.3.23) and (6.3.15) give

$$(\rho g\cos\theta (-x_2 + h) + p_a)\delta_{i1} + \delta_{2i}\frac{\rho g\sin\theta}{\mu}(x_2 - h) = t_i(x_2).$$ (6.3.24)

One can similarly find that $q_i(x_2) = - t_i(x_2)$. Unless normal and tangential tractions given by (6.3.24) are supplied on the planes $x_1 =$   const., the assumed flow $v_i = v(x_2)\delta_{i1}$ can not be maintained on the inclined plane. The inclined plane is usually assumed to be infinitely long and tractions necessary to maintain the flow are presumed to act on the planes $x_1 =$   const.