Calculation of surface product inactivation of particles as a function of particle size

[Ellehcim60]

Product inactivation, Pi for a particle in a hostile short term environment follows the following general formula:

Pi = 100 e-ε X/Tout

Where X is the penetration depth in μm from the particle surface, Tout is the outlet temperature of a dryer in degrees Kelvin and ε is a sensitivity factor (K/μm).

If we keep the dryer outlet temperature, Tout constant when comparing the total inactivation for different particle sizes, the equation becomes simpler:

Pi = 100 e-αX

Where α is again a sensitivity factor (μm-1).

The total product inactivation of a given particle can be approximated by calculating the average inactivation % for a given sphere shell, then step one sphere shell deeper and recalculate the average inactivation % for this shell and continue to do this until the center of the particle have been reached. Then the sum of all the sphere shell inactivation contributions can be calculated and divided with the total sphere/particle volume. This yields an approximation of the total Inactivation % of the entire spherical particle.

The approach is simple like linear integration but applied on spherical shells instead. A large spread sheet can do the job by using say 100 shell approximations for each particle ranging from 5 micron to 100 micron in average particle size in 10 particle size steps. The predicted total inactivation % as a function of the average particle size can then be plotted and a trend function fitted to the 10 data points.

But can the problem be solved mathematically?

 

[Dr.Peterson]

Product inactivation, Pi for a particle in a hostile short term environment follows the following general formula:

Pi = 100 e-ε X/Tout

Where X is the penetration depth in μm from the particle surface, Tout is the outlet temperature of a dryer in degrees Kelvin and ε is a sensitivity factor (K/μm).

If we keep the dryer outlet temperature, Tout constant when comparing the total inactivation for different particle sizes, the equation becomes simpler:

Pi = 100 e-αX

Where α is again a sensitivity factor (μm-1).

The total product inactivation of a given particle can be approximated by calculating the average inactivation % for a given sphere shell, then step one sphere shell deeper and recalculate the average inactivation % for this shell and continue to do this until the center of the particle have been reached. Then the sum of all the sphere shell inactivation contributions can be calculated and divided with the total sphere/particle volume. This yields an approximation of the total Inactivation % of the entire spherical particle.

The approach is simple like linear integration but applied on spherical shells instead. A large spread sheet can do the job by using say 100 shell approximations for each particle ranging from 5 micron to 100 micron in average particle size in 10 particle size steps. The predicted total inactivation % as a function of the average particle size can then be plotted and a trend function fitted to the 10 data points.

But can the problem be solved mathematically?

I'm not sure how to interpret the equations. Is the first intended to mean this? [math]P_i=100e^\frac{-\epsilon X}{T_{out}}[/math]
If so, you should have typed "P_i = 100 e^{-ε X/T_out}", or "P_i = 100 e^{-ε X/T_out}", or something like that. If this is not quite what you mean, please correct it.

 

[Ellehcim60]

Dear Dr.Peterson,

Thank you for finding my copy/paste error! - Sorry for the confusion!

Yes, you are absolutely right - The equation should be:

Pi = 100 e^(-ε X/Tout}

and the next equation:

Pi = 100 e^(- αX}

 

[Dr.Peterson]

The approach is simple like linear integration but applied on spherical shells instead. A large spread sheet can do the job by using say 100 shell approximations for each particle ranging from 5 micron to 100 micron in average particle size in 10 particle size steps. The predicted total inactivation % as a function of the average particle size can then be plotted and a trend function fitted to the 10 data points.

But can the problem be solved mathematically?

I think all you are asking for is to integrate in spherical coordinates. Is that something you are familiar with, at least a little?

I'm not sure I follow the details of what you are doing enough to actually try to carry this out, as I don't quite know how particle size, depth, etc. fit together. Is there a reference you could provide to give us more background?

 

[Ellehcim60]

Yes that is also my own conclusion. The formula should most likely be:

Pi = e^(-a(R-X)), where R is the radius of the particle/sphere

To solve this the equation will be a triple integration of the above formula (dx, dy, dz), with limits from -R to +R I guess

 

[Dr.Peterson]

Yes that is also my own conclusion. The formula should most likely be:

Pi = e^(-a(R-X)), where R is the radius of the particle/sphere

To solve this the equation will be a triple integration of the above formula (dx, dy, dz), with limits from -R to +R I guess

I take it you don't know what spherical coordinates are? And you seem to have changed the definition of X without saying what it is. In fact, you've managed to avoid answering any of my questions.

What I'm lacking, before I can try an integration, is the details about what everything means. That's why I asked for a reference; searching didn't turn up anything like this. Without knowing exactly what "product inactivation" means, and so on, I really can't commit myself to any answer.

But if Pi is a quantity that applies to each point within a sphere of radius R, and you want to find the average of this per unit volume, then you would integrate something like this: [math]\int e^{-\alpha(R-r)} dV = \int_0^{2\pi}\int_0^\pi\int_0^R e^{-\alpha(R-r)} r^2\sin(\theta)dr d\theta d\phi[/math]
The average would be that integral divided by the volume of the sphere.

 

[Ellehcim60]

I take it you don't know what spherical coordinates are? And you seem to have changed the definition of X without saying what it is. In fact, you've managed to avoid answering any of my questions.

What I'm lacking, before I can try an integration, is the details about what everything means. That's why I asked for a reference; searching didn't turn up anything like this. Without knowing exactly what "product inactivation" means, and so on, I really can't commit myself to any answer.

But if Pi is a quantity that applies to each point within a sphere of radius R, and you want to find the average of this per unit volume, then you would integrate something like this: [math]\int e^{-\alpha(R-r)} dV = \int_0^{2\pi}\int_0^\pi\int_0^R e^{-\alpha(R-r)} r^2\sin(\theta)dr d\theta d\phi[/math]
The average would be that integral divided by the volume of the sphere.

I am so sorry "Dr.Peterson"!

I have not deliberately tried to be arrogant or in any way tried to offend you :-(

This is my first post and I have a full time busy job and I am most likely 6-9 time zones east of you - assuming you live in the US.

If the spherical particle contains a protein and the particles are emerging from a drying droplet inside a drying chamber, the proteins closest to the particle surface will be damaged/inactivated/denaturated the most, but the degree of damage quickly drops when you move away from the surface towards the core.

The sensitivity factor applies to different proteins. Low values means high sensitivity typical for linear proteins without cysteine bridges. High values means low sensitivity typical for proteins with many cysteine bridges and a developed 3-D structure which are much less prone for inactivation.

The variable X is thus the distance from the center as you have already realized and thus X will go from 0 to R, the radius of the spherical particle. This is the reason I modified the formula from "X" to "(R-X)". The 100 is just a matter of measuring the inactivation in %, so it doesn't have to be part of the equation to be solved. And therefore moving to spherical coordinates it makes perfect sense to substitute the variable X with r.

And you are right: I am not familiar with triple integration using spherical coordinates, but eager to learn - what can I say? - I am an engineer by trade and we like to solve problems using numerical integration, so the huge spread sheet solution works fine. But I am also curious as I off course know a mathematical solution to the problem is possible - Just not something I have done much of the last 35 years.

If you search the internet you won't find anything about "surface product inactivation of particles as a function of particle size" since this is something my own research have documented for the products I am working with.

The relevance of solving this problem both by numerical integration and by math, is to use this for a "Scale-down justification Analysis" of a drying process.

In pilot scale this drying process can only yield small particles (10 micron) which will have a high degree of product inactivation, compared to production scale which will be able to yield much larger particles (80 micron) which will have much less product inactivation according to the function: Pi (r).

The formula and theory will thus yield an estimation of how much product inactivation we should expect in pilot scale. And then furthermore yield an estimate of how much less product inactivation we then will expect in production scale.

By documenting this in both reality and in theory will prove that product inactivation is not a scalable product quality attribute, which means it cannot be fully tested/confirmed in laboratory or pilot scale, but will have to be proven in production scale.

It also gives a good reason NOT to abandon the drying technology just because experiments yields disappointing levels of product inactivation in pilot scale as it is merely a consequence of the pilot scale drying equipment's inability to make the large particles expected in production scale (due to smaller drying chambers and thus shorter particle residence time in the drying chamber).

I hope the above explanation offers some clarity to what the concept/equation entails.

Have a great evening and thank you for you response!

Kind regards

Ellehchim60

NB: Below is the extract of the numerical integration carried out using an Excel sheet:

 

[Dr.Peterson]

I have not deliberately tried to be arrogant or in any way tried to offend you :-(

Arrogant is not the word (perhaps that's what I've been) -- just not seeing our perspective as outsiders lacking information. It's easy for any of us to assume everyone knows the things we know; that's a common failing of teachers, but can work both ways.

You've now given us more than enough background information, which is good.

The next step is to carry out the integration; I did the hard part, in a sense, in providing the integral you need. I'm not sure whether you're saying you also haven't done integration in a long time, or if you could do the next step. (If necessary, you could just use WolframAlpha or some other program to do it for you.)

The link I included has an answer, but I haven't checked it yet.

 

[Ellehcim60]

Arrogant is not the word (perhaps that's what I've been) -- just not seeing our perspective as outsiders lacking information. It's easy for any of us to assume everyone knows the things we know; that's a common failing of teachers, but can work both ways.

You've now given us more than enough background information, which is good.

The next step is to carry out the integration; I did the hard part, in a sense, in providing the integral you need. I'm not sure whether you're saying you also haven't done integration in a long time, or if you could do the next step. (If necessary, you could just use WolframAlpha or some other program to do it for you.)

The link I included has an answer, but I haven't checked it yet.

Thank you so much!

The link you provided yielded the follwing solution:


Which I will now compare with the numerical solution from my (engineering) spread sheet

I will revert with a comparison asap.

Thank you again! and have a great weekend!

Kind regards

Ellehcim60

 

[Ellehcim60]

Arrogant is not the word (perhaps that's what I've been) -- just not seeing our perspective as outsiders lacking information. It's easy for any of us to assume everyone knows the things we know; that's a common failing of teachers, but can work both ways.

You've now given us more than enough background information, which is good.

The next step is to carry out the integration; I did the hard part, in a sense, in providing the integral you need. I'm not sure whether you're saying you also haven't done integration in a long time, or if you could do the next step. (If necessary, you could just use WolframAlpha or some other program to do it for you.)

The link I included has an answer, but I haven't checked it yet.

Dear "Dr.Peterson",
I have now used the same sensitivity factors and ranges of particle sizes when applying the mathematical solution obtained from WolframAlpha and when compared with my numerical solution to the problem, the two sets of inactivation estimates are the same for the naked eye

Again thank you for your support finding the solution I was looking for!

Have a great weekend!