I have laminar flow in a tube. Consider the tube to be 0.2 m long and with an average velocity of 0.05 m/s. The analytical expression for my transfer function is: $E(t)= \frac{\tau^2}{2*t^3}$ for $t$>=$\frac{\tau }{2}$ and $E(t)=0$ for $t < \frac {\tau }{2}$. $\tau$ is the mean residence time. In this case: $\tau$=0.2m/0.05m/s=4 s. I want to convolute this with an exponential equation: $E_2=(1-exp(\frac{-t}{2.55}))$. This equation descripes the magnetization of a particle in a static magnetic field.

I want to get the average magnetization of the particles at the outlet of the tube. So I thought I would do the following: $E_{out}(t)=E(t)*E2(t)$. I would then take the value at t=4s. I have the results from CFD which gives the following: CFD result, the red line is the magnetic field and the green line is $E_{out}(t)$. I'm interested in the value at x=0,2m which is 0.7942.

My results of the convolution are totally different:Convolution result

What am I doing wrong? I'm a little bit confused. Has anyone an idea of how to approach this problem.

Best regards,


  • $\begingroup$ The RTD of a laminar flow reactor is E(t) = tau^2/(2*t^3) for t larger or equal than tau/2, isn't it? $\endgroup$
    – Toulousain
    Feb 19 '18 at 14:41
  • $\begingroup$ You are right. In my program I used the right equation it was just the typo here. $\endgroup$
    – Gesetzt
    Feb 20 '18 at 15:27

The average magnetization exiting the tube in steady state should just be the product not the convolution. $$\int_0^\infty E(t)\, E2(t) \, dt$$ $$\int_{\frac{\tau}2}^\infty \left(1-e^{-t\frac{t}{k}}\right)\, \frac{\tau^2}{2\,t^3} \, dt$$ $$-\frac{\tau^2 Ei\left(-\frac{\tau}{2 k}\right)}{4 k^2} - \frac{\tau \, e^{-\frac{\tau}{2 k}}}{2 k} + e^{-\frac{\tau}{2 k}}-1$$ Where $Ei$ is the Exponential Integral

This looks like it comes out to a bit less than your CFD results: Plot of fraction of flow magnetized

The x axis in this case is the average residence time which should be proportional to the length along the tube, making our plots comparable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.