Particle Capture on Pores

[latexpage]

Last summer, Tejas and I spent a few days beating our heads against the wall because I was trying to develop a mathematical model of particle capture. Specifically, I was trying to understand what amount of transmembrane pressure we would need to hold a particle on (or in, as we are a politically correct, non-discrimination laboratory) a pore. After much frustration, we were finally able to develop a model for a rigid, spherical particle that relates the transmembrane pressure to the velocity of the top fluid. This post is a summer later than it should have been, but now it is starting to become more relevant, so here goes a lot of math.

Figure 1: Force balance diagram of a particle captured on a pore, showing how the downstream drag force [math] F_d [/math], transmembrane force [math] F_D [/math], and the surface normal forces [math] F_{N,1}[/math] and [math] F_{N,2}[/math] balance each other to cause the particle to remain on the pore even in tangential flow.

This model is built around balancing two major forces that are acting on the particle as it is going through the capture process. The first force is the downstream drag force that is acting on the particle. We know from fluid mechanics that for a spherical particle in free stream flow, the drag force acting the particle is given by Stokes’ Law:

\begin{equation}

F_d = 6 \pi \mu R V

\end{equation}

where [math] \mu [/math] is the fluid viscosity, [math] R [/math] is the particle radius, and [math] V [/math] is the fluid velocity. This drag force is meant to be applied in the case of free stream flow, meaning that the whole sphere is exposed to the flow. However, in a capture event part of the sphere will be below the flow. Therefore, Stokes’ drag represents the maximum drag that will act on the particle. To balance this downstream drag, we now consider the transmembrane force that is acting on the particle, or the force that is holding it on the pore. If we account for the all the forces, then we must consider both gravity and the pressure that will act on the particle over the area of the pore. This downward force is given as:

\begin{equation}

F_D = F_g + F_{TMP}

\end{equation}

or

\begin{equation}

F_D = m g + \Delta P * A_{pore}

\end{equation}

where [math] m [/math] is the mass of the particle, [math] g [/math] is the gravitational acceleration, and [math] \Delta P [/math] is the transmembrane pressure. The importance of the gravitational force is determined by the size of the particle and for a smaller particle, the force will be negligible compared to the force generated by the transmembrane pressure. However, we will retain the gravitational term in our analysis for the sake of completeness.

As the particle comes into contact with the pore, there are other components that come into play. Namely, these components are the backwards normal forces between the surface of the particle and the corners of the pore. As the particle rests on the pore, it applies a normal force to the corners at the point of contact. This can be considered as a reaction force that pushes the particle away ( given by [math] F_{N,1} [/math]) and a reaction force that provides resistance to the particle moving (given by [math] F_{N,2} [/math]). These forces contribute in the x-direction and the y-direction as:

\begin{equation}

F_{N1} = F_{N1} \sin{\theta} + F_{N1} \cos{\theta}

\end{equation}

and

\begin{equation}

F_{N2} = F_{N2} \sin{\theta} + F_{N2} \cos{\theta}

\end{equation}

where theta is the contact angle between the particle and the surface of the membrane. At first glance, this contact angle appears to be difficult to determine, but it is actually very straightforward and is defined as:

\begin{equation}

\theta = \arctan{\frac{\sigma}{\delta}}

\end{equation}

where [math] \sigma [/math] is the radius of the pore and [math] \delta [/math] is the height of the centerline of the particle above the surface, or

\begin{equation}

\delta = R – \sigma

\end{equation}

which will give us a value of 0 for a particle that is the same diameter as the pore. Using (4) and (5), we can then balance the drag force and the transmembrane force with the normal force, arriving at:

\begin{equation}

F_d = F_{N1} \sin{\theta} – F_{N2} \sin{\theta}

\end{equation}

and

\begin{equation}

F_D = F_{N1} \cos{\theta} + F_{N2} \cos{\theta}

\end{equation}

which will be helpful later. The final components that we need to define are the torques. The downstream torque is given as:

\begin{equation}

T_d = F_d \delta + F_{N1} R

\end{equation}

which accounts for the downstream drag provided by Stokes’ Law and the reaction force provided by the corner of the pore. The transmembrane torque, which is the torque acting to keep the particle in place, is given as:

\begin{equation}

T_D = F_D \sigma

\end{equation}

where the only force to consider is the transmembrane force keeping the particle in place. In order for the particle to remain on the pore, the torques acting on it must be balanced:

\begin{equation}

T_D = T_d

\end{equation}

or

\begin{equation}

F_D \sigma = F_d \delta + F_{N1} R

\end{equation}

which is the condition where the pressure is exactly equal to what is necessary to prevent the particle from being pulled downstream. In order to solve this equation, we know expressions for two of the forces, but the third force – [math] F_{N1} [/math] – is unknown. However, we can solve for this normal force using (7) and (8). From these, we can find that the normal forces are given as:

\begin{equation}

F_{N1} = \frac{F_d \csc{\theta} + F_D \sec{\theta}}{2}

\end{equation}

and

\begin{equation}

F_{N2} = \frac{F_D \sec{\theta} – F_d \csc{\theta}}{2}

\end{equation}

which can then be used in the torque balance to solve for the pressure. We can solve the balance for when the transmembrane torque is greater than the downstream torque, which gives:

\begin{equation}

T_D > T_d

\end{equation}

or

\begin{equation}

F_D \sigma > F_d \delta + F_{N1} R

\end{equation}

which, with the expressions for the forces plugged in, gives:

\begin{equation}

(m g + \Delta P A_p) \sigma > 6 \pi \mu R V \delta + \frac{(m g + \Delta P A_p) \sec{\theta} + (6 \pi \mu R V) \csc{\theta}}{2} R

\end{equation}

where [math] A_p [/math] is the area of the pore. Plugging in for [math] \sigma [/math], which we will define for convenience as:

\begin{equation}

\sigma = \frac{d_{pore}}{2}

\end{equation}

and rearranging the equation so that we are left with only the transmembrane pressure on the left hand side of the equation, we arrive at:

\begin{equation}

\Delta P > \frac{6 \pi \mu R V (2 \delta – R \csc{\theta}) – m g (d_p – R \sec{\theta})}{(A_p d_p – A_p R \sec{\theta})}

\end{equation}

where [math] d_p [/math] is the pore diameter. This equation represents the required minimum transmembrane pressure to hold a spherical particle particle of radius [math] R [/math] on a pore of diameter [math] d_p [/math] for a given downstream velocity [math] V [/math]. For a rigid particle, the pressure can be this value or higher, but for a deformable particle, considerations must be made as to the pressure at which the particle would deform and the system would have to be maintained very carefully in that range or particles would pass through the pore.

Using Equation (20), we can generate a Matlab script to plot the required pressure particles of various diameter for a 50 nm diameter pore at a flow rate of 10 μL/min. This results in Figure 2.

Figure 2: Required transmembrane pressure for a single pump system to hold particles of varying diameter on a 50 nm pore in a flow rate of 10 μL/min.

While these are relatively small transmembrane pressures, we can determine if they are actually being generated in our system. By using the Hagen-Poiseuille equation, we can plot the transmembrane pressure as a function of downstream flow rate, which is given in Figure 3.

Figure 3: Transmembrane pressure generated as a function of the flow rate. Note that even for a flow rate of 50 µL/min we are not even reaching the pressure necessary to hold a 60 nm particle on the pore.

This analysis is very important as it demonstrates that if we are using a membrane with 50 nm pores, we can essentially only capture particles that are the same size as the pores with a single pump system for a flow rate of 10 µL/min. For particles larger than 50 nm, we would need a flow rate higher than 50 µL/min, which then breaks our condition of a low Peclet number. This conclusion is fundamental to demonstrating the need for two pumps to generate sufficient transmembrane pressure to capture particles larger than the pores while still maintaining a low Peclet number system.

 

I have attached Tejas’s and my original work, updated to account for an error, so that if you wish you can see each step in the process we went through to generate this equation.

 

Particle Capture Full Derivation