User:Brian P. Josey/Notebook/2010/05/27

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Theory Behind Model

I am working through some numbers today to help show that my "salad dressing" idea is feasible. But before I calculate some results, I want to justify my reasoning, and leave it open to comment and criticism. Essentially, my plan is to look at the surface tension at the interface of water and an oil and show that the pressure from the surface tension exerts enough force on the ferritin to keep it from moving into the oil. After I show that, I want to then calculate the velocity that a water droplet, with a given radius and number of ferritin would move under the force of a magnetic field. From there, I will have a set of values that I can use to experimentally check my calculations.

Interfacial Tension and Pressure

Interfacial tension is the term for the surface tension that one fluid has when it is in a second fluid. So while surface tension in water sitting in the air is called "surface tension" the surface tension for water sitting in oil is called "interfacial tension." Because of this, I'll use the terms interchangeably.

In a drop of liquid, there exists a difference in pressure between the surrounding medium and inside the drop. This difference is given by the formula:

\Delta P= P_i-P_o= \frac {2 \gamma}{R}

  • ΔP is the difference in pressure between the inside, Pi, and the outside, Po
  • γ is the surface tension, and
  • R is the radius of the droplet

This implies that the pressure inside the droplet is always higher than the pressure outside of the drop, and that it only depends on the surface tension, or interfacial tension, and the size of the droplet. A small drop has a greater change in pressure than a larger one. For a bubble, this factor becomes 4 times the surface tension over the radius, as opposed to two times.

From this pressure, I can calculate the minimum force necessary to pull an object through the interface. To do this, I determine the force as a function of the pressure and surface area of the object I am considering. In this case, I choose to use the area of a circle to measure the surface area of ferritin. The reason for this is that the force from the surface tension acts normal to the surface of the ferritin. At the maximum force, this covers half of the surface of ferritin, and there is some cancellation of the force that will reduce down to a term that is equal to the force on a circular surface with the same radius as ferritin.

So, the area A of a circle with a radius r is:

 A = \pi r^{2} \,

From the definition of pressure,

 P = \frac {F}{A}

I get force as a function of area:

 F = P A \,

Plugging the area of a circle and the pressure from surface tension, I ultimately get:

 F = \frac {2 \gamma \pi r^{2}} {R}

  • F is the maximum force I can exert on ferritin before it leaves the droplet
  • r is the radius of ferritin, and
  • R is the radius of the droplet.

Unfortunately, as of writing this, I have been unsuccessful in finding numerical values for γ for the interfaces of water with different oils. However, because force is indirectly proportional to the radius of the droplet, and the droplets in a flow cell will be very small, it is safe too assume that the surface tension force is significantly larger than the magnetic force. However, when I do find some numbers for interfacial tension, I want to double check this, for the sake of consistency.

Velocity of Water Droplet

As I've already discussed a couple of weeks ago, the magnitude of the terminal terminal velocity of a spherical object moving through a medium can be calculated by

 V = \frac {F} {6 \pi \mu R}

  • V is the speed
  • F is the force acting on the drop
  • μ is the viscosity of the medium
  • R is the radius of the drop

However, the force in this situation is a discrete multiple of the force each individual ferritin feels when subjected to an external magnetic field. So by replacing the total force, F, with the number of ferritin in the droplet, n, by the individual force acting on each ferritin, Ff I get the final equation for the velocity of the water droplet:

 V = \frac {n F_f} {6 \pi \mu R}

I have to also consider the possibility that the water droplet will change it's shape, and in turn it's maximum velocity through the oil, but I will write this off as a small concern. The reason is that the speeds that I am thinking of, on the order of μm/s, will not put too great of a strain on the shape of the droplet. Also, water will not readily mix with the oil, and I doubt the effect would be noticeable. When an air bubble in soda rises, which is a similar event, it maintains its ruffly circular shape throughout the liquid. Another thing that I need to remember, is that viscosity and surface tension are functions of temperature, and as the temperature increases, they both decrease as their fluidity increases. So if the liquid is significantly heated while in the microscope, it could potentially change the result. But, looking ahead I doubt this will be a significant factor, as the changes in water at least are slight between 0 and 100°C.

Some Basic Calculations

I calculated the force required to remove a single ferritin protein from a water droplet. Because I am still searching for values of interfacial tension, I had to settle on a water droplet resting in room temperature air. At this point, water has a surface tension of 72.8 mN/m, or 7280 fN/nm. Using this I generated the following graph that represents the data for droplets of water ranging in 100 nm to 5000 nm, or 0.1 μm to 5μm, in radius.

As expected, for small radii, the force was very large, and gradually lightened up as the radius increased. From the graph, a force on the order of magnitude of 100 to 10,000 fN is necessary to pull a single ferritin out of a water drop and into the air. This is much larger than the forces that I calculated, which were on the order of 0.1 to 10 fN. It is then safe to conclude that it would be impossible to pull ferritin out of water with the magnets I have.

A more important result, however, is that I created a new formula for the terminal velocity that is based on the number of ferritin in a sample. The sample that I have been using has a density of 56 mg/mL of ferritin in water. With a mass of 440 kDa (u), this gives a total number of 7.661 * 1022 ferritin proteins per cubic meter, or the more useful 7.661 * 10-5 ferritin proteins per cubic nanometer. Then given the equation above, I substituted in for the number of ferritin this number multiplied by the volume of a circle with radius R. This gives me a final equation of:

 V = \frac {2 \rho R^{2} F_f} {9 \mu}


  • ρ is the number of ferritin per unit volume
  • R is the radius of the drop
  • Ff is the individual force on a ferritin
  • μ is the viscosity

The main advantage of this new function is that it assumes that the number of ferritin is consistent across all given values of R, and calls for a constant density of ferritin. It is much simpler than the first approach I was attempting, where I was trying to find the number of ferritin in the sample by using sphere stacking and the force the exert on each other. From this I calculated the terminal velocity for several different radii again ranging from 100 nm to 5 μm. Here is a graph of the values I determined:

These velocities were calculated for a water droplet moving through mineral oil, which has a viscosity of 50 cP. The velocity grows quickly, reaching 100 μm/s for the 5 μm radius droplet, which makes physical sense. The number of ferritin, and in turn the force, goes as a product of the volume, while Stoke's drag goes as a product of the radius. I then calculated the terminal velocity for several different water droplets suspended in different liquids.

View/Edit Spreadsheet

Here the velocity is measure in μm/s, and all droplets are 1μm in radius. The viscosity is given in two sets of units, the standard cP, centiPoise, and the convenient, fN*s/um*nm. Of course, the velocities vary widely as do the viscosities.

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