Startup of laminar pipe flow

Posted in math on Wednesday, July 23 2014

I've spent a fair bit of time examining various kinds of pipe flow, but so far only at steady state. This time I take a kick at the transient flow cat, looking at start-up of laminar pipe flow (Transport Phenomena 4D.2)

Suppose the same old cylindrical coordinates, assuming incompressible Newtonian flow, and that the only velocity component that matters is in the z-direction. I'm going to further suppose that the pressure gradient is constant.

pipe flow

Taking these assumptions, the equation of motion is: $$ \rho \frac{D}{Dt} \mathbf{v} = - \nabla p + \mu \nabla^2 \mathbf{v} $$ $$ \rho \frac{\partial}{\partial t} v_z = - \frac{\Delta P}{L} + \frac{\mu}{r} \frac{\partial}{\partial r} v_z + \mu \frac{\partial^2}{\partial r^2} v_z $$

I already know what the velocity distribution looks like at steady state: $$ v_{z} = - \frac{\Delta P}{L} \cdot \frac{R^2}{4\mu} \left( 1 - \left( \frac{r}{R} \right)^2 \right) $$

Now for some boundary conditions:

  1. Initially, the velocity is zero everywhere $ v_{z}(r,t=0) = 0$
  2. At large time the velocity goes to the steady state solution $$ v_{z}(r,t \to \infty) \to v_{z,ss}{r} = - \frac{\Delta P}{L} \cdot \frac{R^2}{4\mu} \left( 1 - \left( \frac{r}{R} \right)^2 \right) $$
  3. No slip boundary condition $ v_{z}(r=R,t) = 0 $
  4. The derivative of the velocity distribution is zero at r=0 $$ \left[ \frac{\partial v_{z}}{\partial r} \right]_{r=0} = 0 $$

The steady state distribution gives me some ideas on how to write the equation of motion in dimensionless form. I can write the steady-state version in dimensionless form by making the substitutions: $$ \phi = -\frac{L}{\Delta P} \cdot \frac{\mu}{R^2} v_z $$ $$ \xi = \frac{r}{R} $$

At steady state: $$ \phi = \frac{1}{4} \left( 1 - \xi^2 \right) $$

To complete the transformation I need some dimensionless "time". As a way to get there, I'm going to take some partial derivatives: $$ \frac{\partial}{\partial \xi} \phi = \frac{\partial r}{\partial \xi} \cdot \frac{\partial}{\partial r} \phi = R \cdot \frac{\partial}{\partial r} \left( -\frac{L}{\Delta P} \cdot \frac{\mu}{R^2} \cdot v_z\right) = -\frac{L}{\Delta P} \cdot \frac{\mu}{R} \cdot \frac{\partial}{\partial r} v_z $$ $$ \frac{\partial}{\partial \tau} \phi = -\frac{L}{\Delta P} \cdot \frac{\mu}{R^2} \cdot \frac{\partial t}{\partial \tau} \cdot \frac{\partial}{\partial t} v_z $$

$$ \frac{\partial}{\partial t} v_z = -\frac{\Delta P}{L} \cdot \frac{R^2}{\mu} \cdot \frac{\partial \tau}{\partial t} \cdot \frac{\partial}{\partial \tau} \phi $$ $$ \frac{\partial}{\partial r} v_z = -\frac{\Delta P}{L} \cdot \frac{R}{\mu} \cdot \frac{\partial}{\partial \xi} \phi $$ $$ \frac{\partial^2}{\partial r^2} v_z = -\frac{\Delta P}{L} \cdot \frac{1}{\mu} \cdot \frac{\partial^2}{\partial \xi^2} \phi $$

Plugging these into the equation of motion: $$ \rho \left( -\frac{\Delta P}{L} \cdot \frac{R^2}{\mu} \cdot \frac{\partial \tau}{\partial t} \cdot \frac{\partial}{\partial \tau} \phi \right) = - \frac{\Delta p}{L} + \frac{\mu}{r} \left( -\frac{\Delta P}{L} \cdot \frac{R}{\mu} \cdot \frac{\partial}{\partial \xi} \phi \right) + \mu \left( -\frac{\Delta P}{L} \cdot \frac{1}{\mu} \cdot \frac{\partial^2}{\partial \xi^2} \phi \right) $$

$$ \frac{\rho R^2}{\mu} \frac{\partial \tau}{\partial t} \frac{\partial}{\partial \tau} \phi = 1 + \frac{1}{\xi} \cdot \frac{\partial}{\partial \xi} \phi + \frac{\partial^2}{\partial \xi^2} \phi $$

Which then implies $$ \frac{\rho R^2}{\mu} \frac{\partial \tau}{\partial t} = 1 $$ $$ \frac{\partial \tau}{\partial t} = \frac{\mu}{\rho R^2} $$ $$ \tau = \frac{\mu t}{\rho R^2} $$

So, with the substitutions: $$ \tau = \frac{\mu t}{\rho R^2} $$ $$ \xi = \frac{r}{R} $$ $$ \phi = -\frac{L}{\Delta P} \cdot \frac{\mu}{R^2} v_z $$

The equation of motion is: $$ \frac{\partial}{\partial \tau} \phi = 1 + \frac{1}{\xi} \cdot \frac{\partial}{\partial \xi} \phi + \frac{\partial^2}{\partial \xi^2} \phi $$

(I could have also done some dimensional analysis to find the dimensionless groups, but this was more fun)

So, in dimensionless variables the boundary conditions are:

  1. $ \phi(\xi,0) = 0$
  2. $ \phi(\xi, \tau \to \infty) \to \frac{1}{4} \left( 1 - \xi^2 \right) $
  3. $ \phi(1,\tau) = 0 $
  4. $ \left[ \frac{\partial \phi}{\partial \xi} \right]_{\xi=0} = 0 $

I'll tackle the second boundary condition by supposing a separation of variables: $$ \phi(\xi,\tau) = \frac{1}{4} \left( 1 - \xi^2 \right) - \phi_{t}(\xi,\tau) $$ Where, $$ \lim_{\tau \to \infty} \phi_{t}(\xi,\tau) = 0 $$

And a further separation of variables: $$ \phi_{t}(\xi,\tau) = \Xi(\xi) T(\tau) $$ $$ \phi(\xi,\tau) = \frac{1}{4} \left( 1 - \xi^2 \right) - \Xi(\xi) T(\tau) $$

Dropping this into the equation of motion: $$ -\Xi T^{\prime} = 1 + \frac{1}{\xi} \left( - \frac{1}{2} \xi - T \Xi^{\prime} \right) + \left( -\frac{1}{2} - T \Xi^{\prime \prime} \right) $$ $$ \frac{T^{\prime}}{T} = \frac{1}{\xi} \frac{\Xi^{\prime}}{\Xi} + \frac{\Xi^{\prime \prime}}{\Xi} = A $$

Where A is some constant. With this separation of variables I can easily solve for T(τ): $$ T^{\prime} - A T = 0 $$ $$ T(\tau) = B \exp \left( A \tau \right) $$

Where B is another constant. At this point for the second boundary condition to hold, $ A < 0 $ Or to pick something that will be convenient later $ A = - \lambda^2 $ $$ T(\tau) = B \exp \left( - \lambda^2 \tau \right) $$

Turning to the spatial part of the pde: $$ \frac{1}{\xi} \Xi^{\prime} + \Xi^{\prime \prime} + \lambda^2 \Xi = 0 $$

Multiplying through by $ \xi^2$ we get Bessel's equation: $$ \xi^2 \Xi^{\prime \prime} + \xi \Xi^{\prime} + \lambda^2 \xi^2 \Xi = 0 $$

Which has a general solution: $$ \Xi(\xi) = C J_{0}(\lambda \xi) + D Y_{0}(\lambda \xi) $$

Now turning to the last boundary condition, $$ \Xi^{\prime}(0) = - C J_{1}(0) - D Y_{1}(0) = 0 $$ Since $ J_{1}(0) = 0 $ and $ Y_{1}(0) = \infty $, the constant D = 0. The same conclusion could have been arrived at by noting that $ Y_{0}(0) = \infty $ and since the velocity distribution has to be finite at the origin, D = 0.

So we get: $$ \Xi(\xi) = C J_{0}(\lambda \xi) $$

The third boundary condition gives us: $$ \Xi(1) = C J_{0}(\lambda) = 0$$

Unfortunately there is an infinite sequence of λ that solve the equation J0(λ)=0, so the solution to our pde is some linear combination of these. $$ \phi (\xi, \tau) = \frac{1}{4} \left( 1 - \xi^2 \right) - \sum \limits_{i=1}^{\infty} C_{i} J_{0}(\lambda_{i} \xi) \exp \left( - \lambda_{i} \tau \right) $$

At this point I've used the 2nd, 3rd, and 4th boundary conditions, so why not use the 1st? First things last: $$ \phi (\xi, 0) = \frac{1}{4} \left( 1 - \xi^2 \right) - \sum \limits_{i=1}^{\infty} C_{i} J_{0}(\lambda_{i} \xi) = 0 $$

$$ \sum \limits_{i=1}^{\infty} C_{i} J_{0}(\lambda_{i} \xi) = \frac{1}{4} \left( 1 - \xi^2 \right) $$

The coefficients Ci are the coefficients of the Fourier-Bessel series. $$ C_i = { {\int \limits_0^1 \xi \cdot \frac{1}{4} \left(1 - \xi^2 \right) J_{0} \left( \lambda_{i} \xi \right) d\xi} \over {\frac{1}{2} \left( J_{1}(\lambda_{i}) \right)^2 }} $$

$$ \int \limits_0^1 \xi \cdot \frac{1}{4} \left(1 - \xi^2 \right) J_{0} \left( \lambda_{i} \xi \right) d\xi = \frac{1}{4 \lambda_{i}} \left[ \lambda_{i} J_{1} \left( \lambda_{i} \right) - 2 J_{2} \left( \lambda_{i} \right) + \lambda_{i} J_{3} \left( \lambda_{i} \right) \right] = \frac{J_{1} \left( \lambda_{i} \right)}{\lambda_{i}^{3}} $$

Where I've taken advantage of the recurrence relation: $$ J_{n+1} \left( x \right) + J_{n-1} \left( x \right) = \frac{2n}{x} J_{n} \left( x \right)$$

So those last undetermined constants are: $$ C_i = { {\frac{J_{1} \left( \lambda_{i} \right)}{\lambda_{i}^{3}} } \over {\frac{1}{2} \left( J_{1}(\lambda_{i}) \right)^2 }} = \frac{2}{\lambda_{i}^{3} J_{1}(\lambda_{i})} $$

Putting it all back together: $$ \phi (\xi, \tau) = \frac{1}{4} \left( 1 - \xi^2 \right) - 2 \sum \limits_{i=1}^{\infty} { {J_{0}(\lambda_{i} \xi)} \over { \lambda_{i}^{3} J_{1}(\lambda_{i}) }} \exp \left( - \lambda_{i} \tau \right) $$

Then back into original variables $$ v_{z}(r,t) = - \frac{\Delta P \cdot R^2}{4 \mu L} \left[ 1 - \left(\frac{r}{R}\right)^2 - 8 \sum \limits_{i=1}^{\infty} { {J_{0}(\frac{\lambda_{i} r}{R})} \over { \lambda_{i}^{3} J_{1}(\lambda_{i}) }} \exp \left( - \frac{\lambda_{i} \mu t}{\rho R^2} \right) \right] $$ Where the $ \lambda_{i} $ are the roots of the 0th Bessel function.