Modeling Resistance in Josh’s Separation Data

ByKarl Smith

In Josh’s forthcoming paper about clogging in forward and reverse configurations using 20 nm and 100 nm sepcons, we thought there was reason to discuss how the data agrees with the already existing models of membrane fouling. In this post, Josh discusses some of his efforts to model the data without accounting for the fact that using a centrifuge means the transmembrane pressure changes over time.

What I was curious to see was whether we could watch the transition from one mechanism of fouling to another within the data we already had. Take a look at the graphs below:Screen Shot 2015-04-01 at 8.48.03 PM

What I think we’re seeing is that after 15 seconds, 100 nm particles at 10^9 part/mL reach a high enough concentration that they form a packed bed against the membrane, and change the filtration behavior of the system from one where individual pores being blocked are causing flux decline to one in which the thickness of the packed bed dominates the hydraulic resistance of the system. The same thing seems to happen even more noticeably in the 20 nm particle case at 30 seconds in at 10^12 part/mL. I did some modeling using my sieving model that seems to support this transition based on how the concentration at the membrane surface evolves with time. But using my model is a pretty big hammer for this paper, and we wanted to show the same result with another approach.

I started from a slightly modified version of Darcy’s law that groups the permeability of the membrane, the length, the area, and the viscosity of the fluid all within the single catchall term R_h, which we can think of as the hydraulic resistance.

Q(t) = \frac{\Delta P}{R_h}

\Delta P = \rho g h

Q(t) is flux, with units of m^3/s. rho is the density of water, g is the force due to gravity (690m/s^2 in our small centrifuge when set to 3000 rpm). h is the head height.

\int_{0}^{t}Q(t)dt=\int_{0}^{t}\frac{\rho g h(t)}{R_h}

V_{passed}(t)=\frac{\rho g}{R_h}\int_{0}^{t}h(t)dt

$Latex V_{passed}(t)$ is the volume passed at time t. $Latex V_0 $ is the initial volume in the system.

h(t)=h_{0}(V_{0}-V_{passed})

V_{passed} (t)=\frac{\rho g}{R_h}(h_0 t+c-\int_{0}^{t}V_{passed}(t)dt); c=0

This is identical to the differential equation y'(t) = b t - c y(t) which I solved using an online differential equation software (I could hear my undergraduate physics professors crying). My final result was:

V_{passed}=V_0(1-e^{\frac{-\rho g}{R_h V_0}})

Which at the very least has the correct limiting cases. Next I solve the equation for R_h:

R_h = -\frac{\rho g t}{V_0 ln(-\frac{V_{passed}}{V_0}+1)}

I then plotted Josh’s raw data as R_h \text{vs} t. The results I’ve gotten… well… they aren’t exactly straightforward to interpret. It’s worth thinking about. First, I’ll show the results for the 20 nm nanoparticle cases:

20 10 20 11 20 12 20 13 20 14 20 15

 

Now all together on the same graph:

20 all

Now 100 nm particles:

100 9 100 10 100 11 100 12 100 13

Now all together:

 

 

100 all

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