have a particle of barium ... (Part II)
"Iuuuu?!? Anyone ?!?!" Wednesday, September 10, 2008
Most Common Pick 3 Numbers
[Particle Barium in Brownian motion in a chemical trail.]
Una critica che spesso viene mossa ai modellini dagli sciachimisti (in quei pochi casi in cui riescano a muovere delle critiche che non siano "Il tuo modello matematico non mi piace!") consiste nell'asserire che siano sostanzialmente deterministici e che quindi non vadano bene per modellare fenomeni "caotici" (quando loro stessi non hanno la più pallida idea di come e perchè si definisca "caotico" un sistema). Ecco dunque che spinti da questa critica e dall'esigenza di calcolare il mean free path per una particella di aerosol/particolato dobbiamo iniziare a tenere conto anche di fenomeni aleatori come il moto Browniano.
Quello che finora ho sentito circa le dimensioni del particolato è più o meno classificabile come:
1) Piccolissimo
2) Al di sotto del millimetro (Dr.s Staninger)
etc.
Nessuna indicazione precisa circa le dimensioni. Insomma una nanoparticella per essere definita tale ha un diametro di 1-100 nm (nanometri). A queste dimensioni occorre iniziare a tener conto dello Slip Correction Factor, perchè ci si avvicina al regime di transizione (Numero Kn di Knudsen pari a circa 1). Considerando che l'aria è composta prevalentemente di azoto e che questo ha un raggio di Van der Waals pari a circa 155 pm (picometri, 10^(-12) m), e che le particelle in esame hanno un raggio equivalente di circa 1 nm, c'è un rapporto di quasi uno a mille ed continuare ad usare l'approssimazione continua senza le dovute corrections can be risky.
The goal that we aim to conduct this analysis is the "complication" of the formula was derived in the first part , and the derivation of a report for the calculation of the mean free path and diffusion coefficient to be used in formulas more complicated, that the diffusion model with the Laplacian instead of the Brownian motion and detailed descriptions of each individual particle interactions with nearby particles.
To achieve our evil purposes (of course, misinform, confuse and obscure, as well as conquer the world) we use a very simple equation:
m_ {p} \\ frac {d \\ bold {v}} {dt} =- \\ frac {6 \\ pi \\ mu R_ {p}} {C_ {c}} \\ bold {v} + m_ {p} \\ cdot \\ bold {a} with R p
the equivalent radius of the particle, m p
mass of the particle , to Brownian acceleration, C c
slip correction factor, μ the viscosity of the medium in which the particle is immersed and of course v the particle's velocity in space. The equation in question is also well known in statistical physics, both known to have a proper name of the equation:
Langevin equation.
At first glance it seems very quiet equation, only thing is that the variables in bold (v Langevin equation.
ed
a ), sono in realtà delle variabili aleatorie, e l'equazione in questione in realtà un'equazione differenziale stocastica. Nulla di complicato (almeno concettualmente parlando) dal momento che l'accelerazione accelerazione Browniana è una variabile "casuale" a media nulla, che appunto contribuisce ad un moto discontinuo e casuale della particella in esame. Ora dal momento che essere spinti in una direzione è equiprobabile all'essere spinti nella direzione opposta, si suppone che vi sia la proprietà di isotropia del movimento, ossia che l'accelerazione non abbia una direzione preferita. Forti di queste conoscenze, introduciamo una variabile ausiliaria
r
che ci dia la posizione della particella. Since the motion of the particle in question is isotropic
= 0 (r will also mean nothing, always from the speech that the Brownian motion has no preferred direction). che ci dia la posizione della particella. Since the motion of the particle in question is isotropic
remember that the average of a function f
of a random variable x continuous (or, equivalently, its expected value) is defined as:
Now, dividing the original equation for the mass and calculating the expected value of the variables, you get \\ frac {d} {dt} \\ langle \\ bold { r} \\ cdot \\ bold {v} \\ rangle =- \\ frac {1} {\\ tau} \\ langle \\ bold {r} \\ cdot \\ bold {v} \\ rangle + \\ frac {3KT} {m_ {p}}
Where τ is defined as:
Where τ is defined as:
\\ tau = \\ frac {C_ {c} m_ {p}} {6 \\ pi \\ mu R_ {p}}
integrating the equation normally (in the variable ), we get:
\\ langle \\ bold {r} \\ cdot \\ bold {v} \\ rangle = \\ frac {3KT \\ tau} {m_ {p}} + \\ langle \\ bold {r} _ {0} \\ cdot \\ bold {v} _ {0} \\ rangle e ^ {- \\ frac {t} {\\ tau}}
hours noting that:
\\ langle \\ bold {r} \\ cdot \\ bold {v } \\ rangle = \\ langle \\ bold {r} \\ cdot \\ frac {\\ bold {r}} {dt} \\ rangle = \\ frac {1} {2} \\ frac {d} {dt} \\ langle r ^ {2 } \\ rangle
the final equation becomes: \\ frac {1} {2} \\ frac {d} {dt} \\ langle r ^ {2} \\ rangle = \\ frac {3KT \\ tau} {m_ {p}} + \\ langle \\ bold {r} _ {0} \\ cdot \\ bold {v} _ {0} \\ rangle e ^ {- \\ frac {t} {\\ tau}}
asymptotically (but we saw that the actual time settling of small particles to the longitudinal axis speed is very quick), we can happily omit the exponential term in dt and integrate, obtaining:
\\ langle r ^ {2} \\ rangle = \\ frac {3KT \\ tau} {m_ { p}} t = \\ frac {kTC_ {c}} {\\ pi \\ mu R_ {p}} t
Also we have that for isotropy: \\ frac {1} {3} \\ langle r ^ {2} \\ rangle = \\ langle x ^ {2} \\ rangle = \\ langle y ^ {2} \\ rangle = \\ langle z ^ {2} \\ rangle
and therefore: \\ langle x ^ {2} \\ rangle = \\ langle y ^ {2} \\ rangle = \\ langle z ^ {2} \\ rangle = \\ frac {kTC_ {c}} {3 \\ pi \\ mu R_ {p}} t
This result, which Einstein has come in another way, was also confirmed experimentally. It follows that the distance traversed mean square of the particle is directly proportional to the time that has suffered the Brownian motion.
But now it hits anything with the coefficient of diffusion?
Let us write the equation di diffusione semplice in 3 dimensioni:
This result, which Einstein has come in another way, was also confirmed experimentally. It follows that the distance traversed mean square of the particle is directly proportional to the time that has suffered the Brownian motion.
But now it hits anything with the coefficient of diffusion?
Let us write the equation di diffusione semplice in 3 dimensioni:
\frac{\partial N(x,y,z,t)}{\partial t}=D\cdot\nabla^{2}N(x,y,z,t) 
dove l'operatore laplaciano di una funzione è definito come: \nabla^{2}f(x,y,z,t) = \frac{\partial^{2} f(x,y,z,t)}{\partial x^{2}}+\frac{\partial^{2} f(x,y,z,t)}{\partial y^{2}}+\frac{\partial^{2} f(x,y,z,t)}{\partial z^{2}}
mentre la funzione N fornisce il numero di particelle che si sta muovendo di moto Browniano. Possiamo adesso considerare la diffusività dovuta al moto Browniano as a macroscopic phenomenon. Now suppose we find N 0
particles in the plane (y, z) and suppose that N does not depend on y or z.
multiplying both sides of the equation x ² and above for making the integral in dx from - ∞ to + ∞, we get:
\\ int \\ limits_ {- \\ infty} ^ {+ \\ infty x ^ {2} } \\ frac {\\ partial N} {\\ partial t} dx = \\ int \\ limits_ {- \\ infty} ^ {+ \\ infty} x ^ {2} D \\ frac {\\ partial ^ 2} {N} {\\ partial x ^ {2}} dx
now, the first member to full equality in practice is nothing more than the expected value of x ² (once brought out the differential at), since we started with N particles
0, N 0
particles remain, therefore: multiplying both sides of the equation x ² and above for making the integral in dx from - ∞ to + ∞, we get:
\\ int \\ limits_ {- \\ infty} ^ {+ \\ infty x ^ {2} } \\ frac {\\ partial N} {\\ partial t} dx = \\ int \\ limits_ {- \\ infty} ^ {+ \\ infty} x ^ {2} D \\ frac {\\ partial ^ 2} {N} {\\ partial x ^ {2}} dx
now, the first member to full equality in practice is nothing more than the expected value of x ² (once brought out the differential at), since we started with N particles
0, N 0
\\ int \\ limits_ {- \\ infty} ^ {+ \\ infty} x ^ {2} \\ frac {\\ partial N} {\\ partial t} dx = \\ frac {\\ partial} {\\ partial t} \\ int \\ limits_ {- \\ infty} ^ {+ \\ infty} x ^ {2} = N_ {0} Ndx \\ frac {\\ partial \\ langle x ^ {2} \\ rangle} {\\ partial t} Now the second integral is a bit 'hanged more to resolve, but it is Vuorio with relative ease using the integration parties, and recalling that: \\ int g (x) \\ frac {d ^ {2} f (x)} {dx ^ {2}} dx = G (x) \\ frac {df (x)} {dx} - \\ left (\\ frac {dg (x)} {dx} f (x) - \\ int f (x) \\ frac {d ^ {2} g (x)} {dx ^ 2} {dx} \\ right) Applying this simple rule to our full comes out that:
\\ int \\ limits_ {- \\ infty} ^ {+ \\ infty} x ^ {2} D \\ frac {\\ partial ^ 2} {N} {\\ partial x ^ {2}} dx = D \\ left (\\ left [x ^ {2} \\ frac {\\ partial} {N} {\\ partial x} \\ right] _ {- \\ infty} ^ {+ \\ infty} - \\ left [2xN \\ right] _ {- \\ infty} ^ {+ \\ infty} + \\ int \\ limits_ {- \\ infty} ^ {+ \\ 2Ndx infty} \\ right)
the first two terms are zero, while the last term with the integral is precisely equal to 2N 0 .
So now having developed the integrals can match the two results, obtaining: < r >
\\ frac {\\ partial {\\ langle x ^ {2} \\ rangle}} {\\ partial t} = 2D integrating dt, we get:
\\ langle x ^ {2} \\ rangle = 2DT \\ int \\ limits_ {- \\ infty} ^ {+ \\ infty} x ^ {2} D \\ frac {\\ partial ^ 2} {N} {\\ partial x ^ {2}} dx = D \\ left (\\ left [x ^ {2} \\ frac {\\ partial} {N} {\\ partial x} \\ right] _ {- \\ infty} ^ {+ \\ infty} - \\ left [2xN \\ right] _ {- \\ infty} ^ {+ \\ infty} + \\ int \\ limits_ {- \\ infty} ^ {+ \\ 2Ndx infty} \\ right)
the first two terms are zero, while the last term with the integral is precisely equal to 2N 0 .
So now having developed the integrals can match the two results, obtaining: < r >
\\ frac {\\ partial {\\ langle x ^ {2} \\ rangle}} {\\ partial t} = 2D integrating dt, we get:
Equating this report with the report obtained by the initial model with the stochastic differential equation, we get what is called in the literature report Stokes-Einstein
-Sutherland (unless the factor C
c)
D = \\ frac {kT C_ {c}} {6 \\ pi \\ mu R_ {p}}
where k is the Boltzmann constant, R p
is the equivalent radius of the particle, C c < r·v > slip correction factor, T temperature and μ
viscosity of the medium that surrounds the particle.
Now at last we got a realistic diffusion coefficient can be used in future models to partial derivatives.
Now just to get an idea of \u200b\u200bthe forces acting on particles very small, just replace the numeretti in previous reports and calculate the distance traveled by diffusion or gravitational settling, well for particles with equivalent diameter of 0.01
μ
m lo spostamento dovuto alla diffusione è circa 1000 volte più grande dello spostamento dovuto alla forza di gravità. Tuttavia facendosi due conti, ci si rende conto che per particelle che non siano di dimensioni comparabili con le molecole del mezzo in cui sono immerse, la diffusione non dà certo un gran contributo al trasporto.
Ora però due parole sul "mean free path". Il movimento di una particella è caratterizzato dalla velocità termica media:
La teoria cinetica dei gas ci dice che questa può essere legata alla diffusività come segue: D = \\ frac {1} {2} \\ bar {c} _ {p} \\ lambda_ {p}
report using the Stokes-Einstein-Sutherland gives the following expression for the free path average of a particle particle in a gas:
\\ lambda_ {p} = \\ frac {C_ {6} {c} \\ mu} \\ sqrt {\\ frac {2 \\ rho k T R_ {p}} {3}}
With the obvious meaning of the parameters.
We then derived a way to determine the mean free path and diffusion coefficient.
Nella terza parte , tratteremo più in dettaglio la viscosità, i diametri equivalenti e la relazione dinamica generale per una particella singola.
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