Answer to Question #108965 in Calculus for azike

Question

integrate between limits 0 and d50〖1/(√2π *dp* ln GSD) exp[-〖{ln(dp )-ln(MMD)}〗^2/(2〖{ln(GSD)}〗^2 )]ddp 〗.

Note: the above model is a particle collection efficiency model that is used for calculating particle collection efficiency of mechanical cyclones. The upper limit is zero (0) while the lower limit is d50. The square root covers only 2pi.

After solving please test result with known variables given as MMD=19; GSD=1.4; d50=8.25; Expected result=99.3%. Also test MMD=13; GSD=1.7; d50=4.85; Expected result =96.8%.

Thank you very much as l am very confident that you will be able to help me solve this problem that have been giving me headache over the years. Thanks as l expect your reply.

Answer 1

It is necessary to calculate the integral

$I=\int\limits_{d50}^0\left(\frac{\exp\left({-\displaystyle\frac{\left(\ln dp-\ln MMD\right)^2}{2\cdot\ln^2GSD}}\right)}{\sqrt{2\pi}\cdot dp\cdot\ln GSD}\right)d(dp)$

Note : Your constants are too inconvenient, therefore, to calculate this integral, we introduce simple variables, and at the end we will write the answer through your variables.

$d50=x,\quad dp=y,\quad MMD=a,\quad GSD=b$

Yet we need such a thing as the error function:

$erf(x)=\frac{2}{\sqrt{\pi}}\cdot\int\limits_0^xe^{-t^2}dt\\[0.3cm] 1-erf(x)=\frac{2}{\sqrt{\pi}}\cdot\int\limits_x^{\infty}e^{-t^2}dt\\[0.3cm] erf(-x)=-erf(x)$

This is not an elementary function that can only be calculated using a computer.

( More information : https://en.wikipedia.org/wiki/Error_function )

In our case,

$I=\int\limits_{x}^0\left(\frac{\exp\left({-\displaystyle\frac{\left(\ln y-\ln a\right)^2}{2\cdot\ln^2b}}\right)}{\sqrt{2\pi}\cdot y\cdot\ln b}\right)d(y)=\\[0.3cm] \int\limits_{x}^0\left(\frac{\exp\left({-\left(\displaystyle\frac{\ln y-\ln a}{\sqrt{2}\cdot\ln b}\right)^2}\right)}{\sqrt{2\pi}\cdot y\cdot\ln b}\right)d(y)$

We introduce the change of variable

$t=\frac{\ln y-\ln a}{\sqrt{2}\cdot\ln b}\longrightarrow\sqrt{2}\cdot t\cdot\ln b=\ln\left(\frac{y}{a}\right)\\[0.3cm] e^{\sqrt{2}\cdot t\cdot\ln b}=\frac{y}{a}\longrightarrow\boxed{y=a\cdot e^{\sqrt{2}\cdot t\cdot\ln b}}\\[0.3cm] \boxed{dy=a\cdot e^{\sqrt{2}\cdot t\cdot\ln b}\cdot\sqrt{2}\cdot\ln b\cdot dt}\\[0.3cm] y=0=a\cdot e^{\sqrt{2}\cdot t\cdot\ln b}\to\boxed{t=-\infty}\\[0.3cm] y=x=a\cdot e^{\sqrt{2}\cdot t\cdot\ln b}\to\boxed{t=B\equiv\frac{\ln x-\ln a}{\sqrt{2}\cdot\ln b}}$

We substitute all this into the original integral

$I=\int\limits_{B}^{-\infty}\left(\frac{e^{-t^2}}{\sqrt{2\pi}\cdot a\cdot e^{\sqrt{2}\cdot t\cdot\ln b}\cdot\ln b}\right)\cdot\left(a\cdot e^{\sqrt{2}\cdot t\cdot\ln b}\cdot\sqrt{2}\cdot\ln b\right)dt=\\[0.3cm] =\frac{1}{2}\cdot\frac{2}{\sqrt{\pi}}\cdot\int\limits_B^{-\infty}e^{-t^2}dt=[t=-k]=\\[0.3cm]=\frac{1}{2}\cdot\frac{2}{\sqrt{\pi}}\cdot\int\limits_{-B}^{\infty}e^{-m^2}(-dm)=-\frac{1}{2}\cdot\underbrace{\frac{2}{\sqrt{\pi}}\cdot\int\limits_{-B}^{\infty}e^{-m^2}dm}_{1-erf(-B)}\\[0.3cm] \boxed{I=-\frac{erf(B)+1}{2}}$

In the given variables, the answer looks like this

$B=\frac{\ln x-\ln a}{\sqrt{2}\cdot\ln b}=\frac{\ln(d50)-\ln MMD}{\sqrt{2}\cdot\ln GSD}\\[0.3cm] \boxed{I=-\frac{1}{2}\left(1+erf\left(\frac{\ln(d50)-\ln MMD}{\sqrt{2}\cdot\ln GSD}\right)\right)}$

Check this result using the suggested values.

1) $MMD=19; GSD=1.4; d50=8.25;\text{Expected result}=0.993$

$I=-\frac{1}{2}\left(1+erf\left(\frac{\ln(8.25)-\ln 19}{\sqrt{2}\cdot\ln 1.4}\right)\right)\approx-0.00658\\[0.3cm] 1+I=1-0.00658=0.99342$

2) $MMD=13; GSD=1.7; d50=4.85; \text{Expected result} =0.968$

$I=-\frac{1}{2}\left(1+erf\left(\frac{\ln(4.85)-\ln 13}{\sqrt{2}\cdot\ln 1.7}\right)\right)\approx-0.031576\\[0.3cm] 1+I=1-0.031576=0.968424$

Hint : I don’t know what the problem is, but the indicated integral itself does not give a result that is declared as a test, but the value $(1-I)$ is very well consistent with the results. Maybe you inaccurately formulated the problem?