Couette Flow - Laminar Flow Past Two Plates

Last edited: 2021-05-04 08:06:57

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The Navier Stokes equation is famously known for being unsolvable if there is any turbulence in a fluid but there are certain scenarios where laminar flow exist and a solution for the velocity profile can be derived. One of these scenarios is flow past two plates either caused by one plate moving, Couette flow, or a pressure gradient.

Before solving these two problems we need to make some assumptions. First we assume that our fluid is incompressible and Newtonian so we can use the Navier Stokes equation and the mass continuity equation. Secondly, the flow has to be stationary (so all time dependent terms in the Navier Stokes equation is zero) and in one direction, the xx-direction, so the velocity vector only has one component (v=w=0v=w=0). The plates also has to be considered infinite. Lastly, the flow has to be laminar which has already been mentioned and the impact from gravity has to be considered negligible (g=0g = 0).

Couette Flow - One Plate Moves

So lets start with the case where one of the plates is moving. From our assumptions that fluid is Newtonian, stationary and goes in the xx-direction we get the following mass continuity equation:

∂u∂x+∂v∂y+∂w∂z=∂u∂x=0,\frac{\partial u}{\partial x} + \cancel{\frac{\partial v}{\partial y}} + \cancel{\frac{\partial w}{\partial z}} = \frac{\partial u}{\partial x} = 0,

where uu, vv and ww is the velocity in the xx-, yy- and zz-direction. It is then safe to assume that the velocity profile uu is only dependent on yy because the plates' areas are infinite. Now lets look at the Navier Stoke equation in all directions and see which terms we can cancel out because of our assumptions and the mass contuinity equation:

{ρ(∂u∂t+u∂u∂x+v∂u∂y+w∂u∂z)=−∂p∂x+μ(∂2u∂x2+∂2u∂y2+∂2u∂z2)+ρgxρ(∂v∂t+u∂v∂x+v∂v∂y+w∂v∂z)=−∂p∂y+μ(∂2v∂x2+∂2v∂y2+∂2v∂z2)+ρgyρ(∂w∂t+u∂w∂x+v∂w∂y+w∂w∂z)=−∂p∂z+μ(∂2w∂x2+∂2w∂y2+∂2w∂z2)+ρgz.\left\{ \begin{aligned} & \rho \left( \cancel{\frac{\partial u}{\partial t}} + \cancel{u \frac{\partial u}{\partial x}} + \cancel{v \frac{\partial u}{\partial y}} + \cancel{w \frac{\partial u}{\partial z}} \right) = \cancel{- \frac{\partial p}{\partial x}} + \mu \left(\cancel{\frac{\partial^2 u}{\partial x^2}} + \frac{\partial^2 u}{\partial y^2} + \cancel{\frac{\partial^2 u}{\partial z^2}} \right) + \cancel{\rho g_x} \\ & \rho \left( \cancel{\frac{\partial v}{\partial t}} + \cancel{u \frac{\partial v}{ \partial x}} + \cancel{v \frac{\partial v}{\partial y}} + \cancel{w \frac{\partial v}{\partial z}} \right) = \cancel{- \frac{\partial p}{\partial y}} + \mu \left(\cancel{\frac{\partial^2 v}{\partial x^2}} + \cancel{\frac{\partial^2 v}{\partial y^2}} + \cancel{\frac{\partial^2 v}{\partial z^2}} \right) + \cancel{\rho g_y} \\ & \rho \left( \cancel{\frac{\partial w}{\partial t}} + \cancel{u \frac{\partial w}{ \partial x}} + \cancel{v \frac{\partial w}{\partial y}} + \cancel{w \frac{\partial w}{\partial z}} \right) = \cancel{- \frac{\partial p}{\partial z}} + \mu \left(\cancel{\frac{\partial^2 w}{\partial x^2}} + \cancel{\frac{\partial^2 w}{\partial y^2}} + \cancel{\frac{\partial^2 w}{\partial z^2}} \right) + \cancel{\rho g_z} \end{aligned} \right. .

As you can see, a bunch of terms cancel out and all that remains is just one single expression:

μ∂2u∂y2=0⇒∂2u∂y2=0.\mu \frac{\partial^2 u}{\partial y^2} = 0 \Rightarrow \frac{\partial^2 u}{\partial y^2} = 0.

So if we integrate this expression twice we get:

u(y)=C1y+C2,u(y) = C_1 y + C_2,

where C1C_1 and C2C_2 are constants. To determine these constants we have to use the no-slip boundary conditions of the plates. If we assume that the origin is located exactly in between the two plates we know that:

{u(+h)=upu(−h)=0,\left\{ \begin{aligned} & u(+h) = u_p \\ & u(-h) = 0 \end{aligned} \right. ,

where hh is the shortest distance from the origin to a plate and upu_p is the speed of the moving plate. This gives the constants:

{C1=up2hC2=up2.\left\{ \begin{aligned} & C_1 = \frac{u_p}{2 h} \\ & C_2 = \frac{u_p}{2} \end{aligned} \right. .

So our velocity profile becomes:

u(y)=up2(1+yh).u(y) = \frac{u_p}{2} \left(1 + \frac{y}{h} \right).

This solution is called the Couette flow which is named after the french physicist Maurice Couette.

Laminar Flow Past Two Plates Caused by Pressure Gradient

So what solution do we get if the two plates are stationary and we instead have a constant pressure gradient? The assumptions mentioned earlier are still the same so we get the same mass continuity equation. So lets see what cancels out in our Navier Stokes equations:

{ρ(∂u∂t+u∂u∂x+v∂u∂y+w∂u∂z)=−∂p∂x+μ(∂2u∂x2+∂2u∂y2+∂2u∂z2)+ρgxρ(∂v∂t+u∂v∂x+v∂v∂y+w∂v∂z)=−∂p∂y+μ(∂2v∂x2+∂2v∂y2+∂2v∂z2)+ρgyρ(∂w∂t+u∂w∂x+v∂w∂y+w∂w∂z)=−∂p∂z+μ(∂2w∂x2+∂2w∂y2+∂2w∂z2)+ρgz.\left\{ \begin{aligned} & \rho \left( \cancel{\frac{\partial u}{\partial t}} + \cancel{u \frac{\partial u}{\partial x}} + \cancel{v \frac{\partial u}{\partial y}} + \cancel{w \frac{\partial u}{\partial z}} \right) = - \frac{\partial p}{\partial x} + \mu \left(\cancel{\frac{\partial^2 u}{\partial x^2}} + \frac{\partial^2 u}{\partial y^2} + \cancel{\frac{\partial^2 u}{\partial z^2}} \right) + \cancel{\rho g_x} \\ & \rho \left( \cancel{\frac{\partial v}{\partial t}} + \cancel{u \frac{\partial v}{ \partial x}} + \cancel{v \frac{\partial v}{\partial y}} + \cancel{w \frac{\partial v}{\partial z}} \right) = - \frac{\partial p}{\partial y} + \mu \left(\cancel{\frac{\partial^2 v}{\partial x^2}} + \cancel{\frac{\partial^2 v}{\partial y^2}} + \cancel{\frac{\partial^2 v}{\partial z^2}} \right) + \cancel{\rho g_y} \\ & \rho \left( \cancel{\frac{\partial w}{\partial t}} + \cancel{u \frac{\partial w}{ \partial x}} + \cancel{v \frac{\partial w}{\partial y}} + \cancel{w \frac{\partial w}{\partial z}} \right) = - \frac{\partial p}{\partial z} + \mu \left(\cancel{\frac{\partial^2 w}{\partial x^2}} + \cancel{\frac{\partial^2 w}{\partial y^2}} + \cancel{\frac{\partial^2 w}{\partial z^2}} \right) + \cancel{\rho g_z} \end{aligned} \right. .

We can see that the pressure gradient is zero in the yy- and zz-direction. So we get the following differential equation:

μ∂2u∂y2=∂p∂x.\mu \frac{\partial^2 u}{\partial y^2} = \frac{\partial p}{\partial x}.

With some integration we get:

u(y)=1μ∂p∂xy22+C1y+C2,u(y) = \frac{1}{\mu} \frac{\partial p}{\partial x} \frac{y^2}{2} + C_1 y + C_2,

where C1C_1 and C2C_2 are constants. The velocity at both plates are now zero because of the no-slip condition, so:

{u(+h)=0u(−h)=0.\left\{ \begin{aligned} & u(+h) = 0 \\ & u(-h) = 0 \end{aligned} \right. .

This gives the constants:

{C1=0C2=1μ∂p∂xh22.\left\{ \begin{aligned} & C_1 = 0 \\ & C_2 = \frac{1}{\mu} \frac{\partial p}{\partial x} \frac{h^2}{2} \end{aligned} \right. .

So our velocity profile becomes:

u(y)=1μ∂p∂xh22(1+(yh)2).u(y) = \frac{1}{\mu} \frac{\partial p}{\partial x} \frac{h^2}{2} \left(1 + \left(\frac{y}{h} \right)^2 \right).

The velocity profile is a parabola with its maximum at the middle of the plates.

Image of pressure gradient flow

So that is it, that is how you calculate fluid flow past two plates!

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