Partial Differential Equations/Parallel Plate Flow: Easy IC

Parallel Plate Flow: Easy IC

Formulation

As with ODEs, separation of variables is easy to understand and works well whenever it works. For ODEs, we use the substitution rule to allow antidifferentiation, but for PDEs it's a very different process involving letting dependencies pass through the partial derivatives.

A fluid mechanics example will be used.

Consider two plates parallel to each other of huge extent, separated by a distance of 1. Fluid is smoothly flowing between these two plates in only one direction (call it x). This may be seen in the picture below.

Visualization of the parallel plate flow problem.

After some assumptions, the following PDE may be obtained to describe the fluid flow:

$\frac{\partial u}{\partial t}=\nu \frac{{\partial }^{2}u}{\partial y^2}-\frac{P_x}{\rho }$

This linear PDE is the result of simplifying the Navier-Stokes equations, a large nonlinear PDE system which describe fluid flow. u is the velocity of the fluid in the x direction, ρ is the density of the fluid, ν is the kinematic viscosity, and P_x is the pressure gradient. Note that u = u(y, t), there is no dependence on x. In other words, the state of the fluid upstream is no different from the state downstream.

u(y, t) is a velocity profile. Fluid mechanics typically is concerned with velocity fields, contrary to rigid body mechanics in which the position of an object is what is important.

The ratio P_x/ρ describes the driving force, it's a pressure change (gradient) along the x direction. If P_x is negative, then the pressure downstream (positive x) is smaller then the pressure upstream (negative x) and the fluid will flow left to right, ie u(y, t) will generally be positive.

Now on to create a specific problem: let's say that a constant negative pressure gradient was applied for a long time, until the velocity profile was steady (steady means "not changing with time"). Then, the pressure gradient is suddenly removed, and without this driving force the fluid will slow down and stop.

Initial flow profile.

Let's say that before the pressure was removed, the velocity profile was u(y, t) = sin(π y). This would make sense: the friction dictates less motion near the plates (see next paragraph), so we could expect a maximum velocity near the centerline (y = 1/2). This assumed profile isn't really correct, but will serve a better example for now. It's graphed at right in the domain of interest.

Before getting into the math, one more thing is needed: boundary conditions. In this case, the BC is called the no slip condition, which states that the velocity of a fluid at a wall (boundary) is equal to the velocity of the wall. Since the velocities of the walls (or plates) in this problem are both zero, the velocity of the fluid must be zero at these two boundaries. The BCs are then u(0, t) = 0 (bottom plate) and u(1, t) = 0 (top plate).

The IBVP is:

$\frac{\partial u}{\partial t}=\nu \frac{{\partial }^{2}u}{\partial y^2}\text{(PDE)}$

$u(y,0)=\sin (\pi y)\text{(IC)}$

$u(0,t)=0$

$u(1,t)=0\text{(BCs)}$

Separation

Variables are separated the following way: we assume that u(y, t) = Y(y)T(t), where Y and T are (unknown) functions respectively of y and t. This form is substituted into the PDE:

$\frac{\partial }{\partial t}(u(y,t))=\nu \frac{{\partial }^2}{\partial y^2}(u(y,t))$

$\frac{\partial }{\partial t}(Y(y)T(t))=\nu \frac{{\partial }^2}{\partial y^2}(Y(y)T(t))$

$Y(y)\frac{\partial }{\partial t}(T(t))=\nu T(t)\frac{{\partial }^2}{\partial y^2}(Y(y))$

$Y(y)T^{\prime }(t)=\nu T(t)Y^{″}(y)$

$\frac{T^{\prime }(t)}{\nu T(t)}=\frac{Y^{″}(y)}{Y(y)}$

Look carefully at the last equation: the left side of the equation depends strictly on t, and the right side strictly on y, and they are equal. t may be varied independent of y and they'd still be equal, and y may be varied independent of t and they'd still be equal. This can only happen if both sides are constant. This may be shown as follows:

$\frac{d}{dt}\left(\frac{T^{\prime }(t)}{\nu T(t)}\right)=\frac{d}{dt}\left(\frac{Y^{″}(y)}{Y(y)}\right)$

$\frac{d}{dt}\left(\frac{T^{\prime }(t)}{\nu T(t)}\right)=0$

$\frac{T^{\prime }(t)}{\nu T(t)}=\text{a constant}$

Integration of the ordinary derivative recovers the left side but leaves the right side a constant. It follows by similarity that Y''/Y is a constant as well.

The constant in question is called the separation constant. We could simply give it a letter, such as A, but a good choice of the constant will make work easier later. In this case the best choice is -k². This will be justified later (but it should be reemphasized that it may be notated any way you want, assuming it can span the domain).

$\frac{T^{\prime }(t)}{\nu T(t)}=\frac{Y^{″}(y)}{Y(y)}=-k^2$

$\Downarrow$

$\frac{T^{\prime }(t)}{\nu T(t)}=-k^2$

$\frac{Y^{″}(y)}{Y(y)}=-k^2$

The variables are now separated. The last two equations are two ODEs which may be solved independently (in fact, the Y equation is an eigenvalue problems), though they both contain an unknown constant. Note that ν was kept for the T equation. This choice makes the solution slightly easier, but is again completely arbitrary.

$\frac{T^{\prime }(t)}{\nu T(t)}=-k^2\Rightarrow T^{\prime }(t)=-k^2\nu T(t)\Rightarrow T(t)=C_1{e}^{-k^2\nu t}$

$\frac{Y^{″}(y)}{Y(y)}=-k^2\Rightarrow Y^{″}(y)=-k^{2}Y(y)\Rightarrow Y(y)=C_2\cos (ky)+C_3\sin (ky)$

The solution, still with unknown constants, is the product of Y and T:

$u(y,t)=Y(y)T(t)=C_1{e}^{-k^2\nu t}(C_2\cos (ky)+C_3\sin (ky))$

$u(y,t)=e^{-k^2\nu t}(A\cos (ky)+B\sin (ky))$

Note that C₁ has been multiplied into C₂ and C₃, reducing the number of arbitrary constants.

The IC or BCs should now be applied. If the IC was applied first, coefficients would be equated and all of the constants would be determined. However, the BCs may or may not have been fulfilled (in this case they would, but you're not generally so lucky). So to be safe, the BCs will be applied first:

$u(0,t)=e^{-k^2\nu t}(A\cos (k\cdot 0)+B\sin (k\cdot 0))=0$

$A\cdot 1+B\cdot 0=0\Rightarrow A=0\text{(no slip along lower boundary)}$

$u(1,t)=e^{-k^2\nu t}B\sin (k\cdot 1)=0$

$B\sin (k\cdot 1)=0\Rightarrow B=0\text{or}k=n\pi \text{(no slip along upper boundary)}$

If we took B = 0, the solution would have just been u(y, t) = 0 (often called the trivial solution), which would satisfy the BCs and the PDE but couldn't possibly satisfy the IC. So, we take k = nπ instead, where n is any integer. So far:

Decaying flow.

$u(y,t)=e^{-(n\pi {)}^2\nu t}B\sin (n\pi y)$

The IC may finally be applied.

$\sin (\pi y)=u(y,0)$

$\sin (\pi y)=e^{0}B\sin (n\pi y)$

Which can only hold if B = 1 and n = 1. That's all, the complete solution is:

$u(y,t)=e^{-{\pi }^2\nu t}\sin (\pi y)$

It's worth verifying that the IC, BCs, and PDE are all satisfied by this. Also notice that the solution is a product of a function of t and a function of y. A graph is illustrated at right. Observe that the profile is plotted for different values of νt, rather than specifying some ν and graphing different values for t. Looking at the solution, t and ν appear only once and they're multiplying, so it's natural to do this. A dimensionless time could've been introduced from the beginning.

So what happens? The fluid starts with its initial profile and slows down exponentially. Note that with x replaced with y and t replaced with νt, this is exactly the same as the result of heat flow in a rod as shown in the introduction. This is not a coincidence: the PDE for the rod describes diffusion of heat, the PDE for the parallel plates describes diffusion of momentum.

Take a second look at the separation constant. The square is convenient, without it the solution for Y(y) would've involved square roots. Without the negative sign, the solution would've involved exponentials instead of sinusoids, so the constant would've come out imaginary.

The assumption that u(y, t) = Y(y)T(t) is justified by the physics of the problem: it would make sense that the profile would keep its general shape (thank Y(y)), only it'd get flattened over time as the fluid slows down (thank T(t)).