In the absence of surface tension, two incompressible viscous fluids are conned between a pair of parallel flat plates $y = -h$ and $y = h$. The region $-h<y<0$ is occupied with a fluid of density $\rho_1$ and viscosity $\mu_1$. The region $0<y<h$ is occupied with a fluid of density $\rho_2$ and viscosity $\mu_2$. A steady fully developed flow is set up with the lower plate $(y = -h)$ fixed and the upper plate $(y = h)$ moving with a constant velocity $U_0$ parallel to the x-axis. Prove that the velocity distribution in the two fluids is: $$\boldsymbol{u} = u(y)\underline{\hat\imath}\quad\text{where}\quad u(y) =\left\{ \begin{array}{11} \frac{\mu_2}{(\mu_1 + \mu_2)}U_0(1+\frac{y}{h})\quad-h\le y\le0\quad(1)\\ \frac{\mu_1}{(\mu_1 + \mu_2)}U_0(\frac{\mu_2}{\mu_1}+\frac{y}{h})\quad\,\,\,\,0\le y\le h\quad(2)\\ \end{array} \right.$$

I write the problem out: $$\boldsymbol{u} = u(y)\underline{\hat\imath} =\left\{ \begin{array}{11} u_2(y)\underline{\hat\imath}\quad\,\,\,\,\,\,\,0\le y\le h\quad(1)\\ u_2(y)\underline{\hat\imath}\quad-h\le y\le 0\quad(1)\\ \end{array} \right.$$ I have the boundary conditions: $$y=-h,\quad u=0\\ y=h,\quad u=U_0$$ I am not given a pressure gradient so I assume $p=p_0$. I check continuity holds: $$\nabla\cdot\boldsymbol{u} = \frac{\partial u}{\partial x} = 0.$$ I then use N-S equations: $$\frac{\partial u}{\partial t} + (\boldsymbol{u}\cdot\nabla)\boldsymbol{u} = -\frac{1}{\rho}\frac{\partial p}{\partial x} + \boldsymbol{X} + \nu\nabla^2u$$ The body force $\boldsymbol{X}=0$ and $\frac{\partial u}{\partial t}=0$ because steady flow. This then comes down to: $$u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = -\frac{1}{\rho}\frac{\partial p}{\partial x} + \nu\left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right)$$ Here, both $\frac{\partial u}{\partial x}$ and $v$ are $0$ so the LHS $=0$. And I think that the pressure is $p=p_0$ so that makes $\frac{\partial p}{\partial x} = 0$ leaving: $$\frac{\mu_2}{\rho_2}\frac{\partial^2 u_2}{\partial y^2} = 0,\quad\frac{\mu_1}{\rho_1}\frac{\partial^2 u_1}{\partial y^2} = 0$$


1 Answer 1


For fully developed, steady, unidirectional flow, the velocity component $u$ does not depend on $x$ and $v=0$. The flow is driven by the motion of the upper plate, so you can assume the pressure is constant. In this case, the Navier-Stokes equations reduce to

$$\frac{\partial^2u_1}{\partial y^2} = 0, \,\,\,\frac{\partial^2u_2}{\partial y^2} = 0$$

where $u = u_1$ for $-h \leqslant y \leqslant 0$ and $u = u_2$ for $0 \leqslant y \leqslant h$. Integrating twice we obtain,

$$u_1 = Ay +B, \,\,\, u_2 = Cy +D.$$

The natural boundary conditions are used to determine the constants $A,\, B,\, C,\, D$.

We require no-slip at solid surfaces

$$\tag{1}u_1(-h) = 0,$$ $$\tag{2}u_2(h) = U_0,$$

and continuity of velocity and stress at the interface:

$$\tag{3}u_1(0) = u_2(0), $$ $$\tag{4}\mu_1 \left.\frac{\partial u_1}{\partial y}\right|_{y=0} = \mu_2 \left.\frac{\partial u_2}{\partial y}\right|_{y=0}$$

Applying these conditions we get

$$\qquad \,\,\,\,\,\,(1) \implies -Ah + B = 0\\ \qquad \,\,\,\,\,\,(2) \implies Ch + D = U_0 \\ (3) \implies B = D \\ \quad \,\,\,\,\,(4) \implies \mu_1A = \mu_2 C$$

From (1) and (3) we get $D = B = Ah$, and from (4) we get $C = \mu_1A/\mu_2$ leading to

$$u_1 = Ay + Ah =Ah\left(1 + \frac{y}{h}\right), \\ u_2 = \frac{A\mu_1}{\mu_2}y + Ah =\frac{\mu_1}{\mu_2}Ah\left( \frac{\mu_2}{\mu_1} + \frac{y}{h}\right) $$

From (4) we get $U_0 = Ch +D = (\mu_1/\mu_2)Ah +Ah = Ah(1 + \mu_1/\mu_2)$ leading to

$$Ah = \frac{\mu_2 U_0}{\mu_1 + \mu_2},$$


$$u_1 = \frac{\mu_2 U_0}{\mu_1 + \mu_2}\left(1 + \frac{y}{h}\right), \\ u_2 = \frac{\mu_1 U_0}{\mu_1 + \mu_2}\left( \frac{\mu_2}{\mu_1} + \frac{y}{h}\right)$$

(At the start, a clue that inertial terms and the pressure gradient are negligible is the absence of the densities $\rho_1$ and $\rho_2$ in the answer given to you.)

  • $\begingroup$ in you're first line you have what i have but without $\mu_2/\rho_2$ etc. Why is this? $\endgroup$
    – MRT
    Apr 15, 2018 at 17:59
  • $\begingroup$ using the conditions I have: $$0 = -Ah + B,\quad U_0 = Ch + D$$ But how do I determine these? $\endgroup$
    – MRT
    Apr 15, 2018 at 18:04
  • $\begingroup$ Divide both sides by the nonzero constant $\mu_2/\rho_2$ and you get what I have. There are four boundary conditions including the two at $y=0$ to solve for the four constants. Does that make it clear or should I add some more. $\endgroup$
    – RRL
    Apr 15, 2018 at 18:18
  • $\begingroup$ $\frac{\mu}{\rho} \frac{\partial^2u}{\partial y^2} = 0 \implies \frac{\partial^2u}{\partial y^2} = 0$. I finished for you. Were you really unable to do the algebra for the constants or was there some other confusion? $\endgroup$
    – RRL
    Apr 15, 2018 at 19:03
  • $\begingroup$ I do just get lost in the algebra and can't see the subtle connections. Like after you have those 4 equations, seeing that to get $Ay$ I just need to do $Ah\cdot \frac {y}{h}$ can just go right over my head. And then going back to $U_0$ to find the fraction equal to $Ah$ I just can't see it sometimes. But you've helped clarify me that what I was thinking was good. I think and try for so long and end up questioning myself with how can this not be the right way? And it is but I'm not reading inbetween the lines $\endgroup$
    – MRT
    Apr 15, 2018 at 19:46

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