Onsanger and Stefan-Maxwell Equations

In summary: Minus sign you are referring to comes from the mistake in equation 1 where you were summing over ##i## when sum is over ##j##: $$c_1 \nabla \bar \mu_1 = K_{11}(v_1 - v_1) + K_{12}(v_2 - v_1) + K_{13}(v_3 - v_1)$$which simplifies to$$c_1 \nabla \bar \mu_1 = K_{12}(v_2 - v_1) + K_{13}(v_3 - v_1)$$which is the same as equation 2.
  • #1
Dario56
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Stefan-Maxwell and Onsanger equations are equations which can be used to calculate mole flux of the component due to different types of gradients. It is assumed that driving forces of mass transfer are in equilibrium with drag forces due to interaction of different types of components.

Stefan-Maxwell equations model diffusion in gases while Onsanger equations model mass transfer in electrolytes where ions can both diffuse and move in the electric field (migration). Both equations can be expressed in the same form, but I will concentrate on the Onsanger type as I'm more interested in it: $$c_i \nabla \bar \mu_i = \sum_j K_{ij} (v_j - v_i) \tag {1}$$

where ##\bar \mu_i## is electrochemical potential of ion ##i##, ##K_{ij}## is interaction coefficient between ions ##i## and ##j##, ##v## is a velocity of the corresponding ion and ##c_i$## is the concentration.

Electrochemical potential gradient takes into account both chemical and electric potential gradient and is therefore sufficient to explain mass transfer in electrolytes.

In the textbook Electrochemical Systems by Newman and Alyea, chapter 12.6: Multicomponent Transport, this equation is written in a bit of a different form on the grounds that in the system of ##n## ions, there are ##n-1## independent velocity differences ##v_j - v_i## or electrochemical potential gradients ## \nabla\bar \mu _i##.

While I understand that there are ##n-1## independent variables previously mentioned, I don't see why does that lead to the equation of this form: $$ c_i \nabla \bar \mu_i = \sum_j M_{ij}(v_j - v_0) \tag {2}$$

where ##v_0## is the velocity of any ion in the solution and ##M_{ij}## is a matrix connected to the interaction coeffiecient matrix ##K_{ij}## and defined as: $$M_{ij} = \begin{cases} K_{ij}, & i \neq j \\ K_{ij} - \sum_k K_{ik}, & i=j \end{cases}$$

Equation (2) is written without a derivation or much explanation and therefore I don't understand how did we get equation (2) from equation (1).
 
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  • #2
Maybe I'm doing something wrong here, but I'm not seeing how this is right. Consider ##n=3## and write out the rhs of equations 1 and 2 explicitly. For equation 1 you get (taking the ##i=1## case WLOG):
$$c_1\nabla\bar{\mu}_1=K_{11}(v_1-v_1)+K_{12}(v_1-v_2)+K_{13}(v_1-v_3)=K_{12}(v_1-v_2)+K_{13}(v_1-v_3)$$
and for equation 2 you get
$$c_1\nabla\bar{\mu}_1=(K_{11}-(K_{11}+K_{12}+K_{13}))(v_1-v_0)+K_{12}(v_1-v_0)+K_{13}(v_1-v_0)=0$$
Did I miss something?
 
  • #3
TeethWhitener said:
Maybe I'm doing something wrong here, but I'm not seeing how this is right. Consider ##n=3## and write out the rhs of equations 1 and 2 explicitly. For equation 1 you get (taking the ##i=1## case WLOG):
$$c_1\nabla\bar{\mu}_1=K_{11}(v_1-v_1)+K_{12}(v_1-v_2)+K_{13}(v_1-v_3)=K_{12}(v_1-v_2)+K_{13}(v_1-v_3)$$
and for equation 2 you get
$$c_1\nabla\bar{\mu}_1=(K_{11}-(K_{11}+K_{12}+K_{13}))(v_1-v_0)+K_{12}(v_1-v_0)+K_{13}(v_1-v_0)=0$$
Did I miss something?
You did. In equation 2 sum goes over ##j## ##(v_j - v_0)##. You read it like is written ##(v_i - v_0)##: $$ c_ 1 \nabla \bar \mu_1 = (K_{11} - (K_{11} + K_{12} + K_{13}))(v_1 - v_0) + K_{12}(v_2 - v_0) + K_{13}(v_3 - v_0)$$
 
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  • #4
Dario56 said:
You did. In equation 2 sum goes over ##j## ##(v_j - v_0)##. You read it like is written ##(v_i - v_0)##: $$ c_ 1 \nabla \bar \mu_1 = (K_{11} - (K_{11} + K_{12} + K_{13}))(v_1 - v_0) + K_{12}(v_2 - v_0) + K_{13}(v_3 - v_0)$$
Ah, yes. Well now there's a sign problem, isn't there?
$$c_ 1 \nabla \bar \mu_1 = -K_{12}(v_1 - v_0) - K_{13}(v_1 - v_0) + K_{12}(v_2 - v_0) + K_{13}(v_3 - v_0)=K_{12}(v_2 - v_1)+ K_{13}(v_3 - v_1)$$
which is the negative of equation 1. Unless of course I missed something else.
But if you correct that with a negative sign (assuming it needs to be corrected), it should be straightforward to see how equation 2 works. The ##i=j## terms will cancel and the ##v_0## terms will also cancel and you'll be left with the same result as equation 1.
 
  • #5
TeethWhitener said:
Ah, yes. Well now there's a sign problem, isn't there?
$$c_ 1 \nabla \bar \mu_1 = -K_{12}(v_1 - v_0) - K_{13}(v_1 - v_0) + K_{12}(v_2 - v_0) + K_{13}(v_3 - v_0)=K_{12}(v_2 - v_1)+ K_{13}(v_3 - v_1)$$
which is the negative of equation 1. Unless of course I missed something else.
But if you correct that with a negative sign (assuming it needs to be corrected), it should be straightforward to see how equation 2 works. The ##i=j## terms will cancel and the ##v_0## terms will also cancel and you'll be left with the same result as equation 1.
Minus sign you are referring to comes from the mistake in equation 1 where you were summing over ##i## when sum is over ##j##: $$c_1 \nabla \bar \mu_1 = K_{11}(v_1 - v_1) + K_{12}(v_2 - v_1) + K_{13}(v_3 - v_1)$$
 
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  • #6
Ok then, thanks. Do you see how equation 2 comes from equation 1 then?
 
  • #7
TeethWhitener said:
Ok then, thanks. Do you see how equation 2 comes from equation 1 then?
Well, kind of. There is no doubt that equation 2 is correct. I agree with what you wrote about which terms cancel and why. So, I guess I see it, but if you asked me to derive it from equation 1, I'm not sure I would be able to do it.
 
  • #8
TeethWhitener said:
Ok then, thanks. Do you see how equation 2 comes from equation 1 then?
I have an additional question which builds upon what we've already discussed.

Equation 2 can be inverted so as to express velocity differences (flux of the component) as a function of driving forces explicitly. Equation is again only written without going into much detail regarding derivation:

$$ v_j - v_0 = - \sum_{k \neq 0} L_{jk}^0 c_k \nabla \bar \mu_k \text{,}\quad j \neq 0 \tag{3} $$

where $$L^0 = -(M^0)^{-1}$$

##M^0## is a submatrix of ##M## obtained by removing a row and a column corresponding to ion ##0## from ##M##. Also, it is to be noted that ##L^0## is symmetric due to the fact that ##K_{ij}## is symmetric: $$L_{ij}^0 = L_{ji}^0 $$

I can't really figure out how was equation 3 derived from equation 2. It must have something to do with inverting matrices, but I think I miss some piece of math needed to this correctly.
 
  • #9
This is basically just a matrix equation written in component form. In equation 1, you have a matrix ##\mathbf{M}## with components ##M_{ij}## and you’re multiplying it by a column vector ##\mathbf{v}## with components ##v_j-v_0## to get a vector ##\mathbf{c\nabla\bar{\mu}}## with components ##c_i\nabla\bar{\mu}_i##. In matrix notation, equation 1 looks like
$$
\begin{pmatrix}
c_1\nabla\bar{\mu}_1 \\
\vdots \\
c_n\nabla\bar{\mu}_n
\end{pmatrix}
=
\begin{pmatrix}
M_{11} & \cdots & M_{1n} \\
\vdots & \ddots & \vdots \\
M_{n1} & \cdots & M_{nn}
\end{pmatrix}
\begin{pmatrix}
v_1-v_0 \\
\vdots \\
v_n-v_0
\end{pmatrix}
$$
Or in more compact notation:
$$\mathbf{c\nabla\bar{\mu}}=\mathbf{M}\mathbf{v}$$
This equation is easily inverted by left multiplying by the inverse matrix of ##\mathbf{M}## to give
$$\mathbf{M}^{-1}\mathbf{c\nabla\bar{\mu}}=\mathbf{v}$$
Equation 3 is just this equation in component form.
 
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  • #10
TeethWhitener said:
This is basically just a matrix equation written in component form. In equation 1, you have a matrix ##\mathbf{M}## with components ##M_{ij}## and you’re multiplying it by a column vector ##\mathbf{v}## with components ##v_j-v_0## to get a vector ##\mathbf{c\nabla\bar{\mu}}## with components ##c_i\nabla\bar{\mu}_i##. In matrix notation, equation 1 looks like
$$
\begin{pmatrix}
c_1\nabla\bar{\mu}_1 \\
\vdots \\
c_n\nabla\bar{\mu}_n
\end{pmatrix}
=
\begin{pmatrix}
M_{11} & \cdots & M_{1n} \\
\vdots & \ddots & \vdots \\
M_{n1} & \cdots & M_{nn}
\end{pmatrix}
\begin{pmatrix}
v_1-v_0 \\
\vdots \\
v_n-v_0
\end{pmatrix}
$$
Or in more compact notation:
$$\mathbf{c\nabla\bar{\mu}}=\mathbf{M}\mathbf{v}$$
This equation is easily inverted by left multiplying by the inverse matrix of ##\mathbf{M}## to give
$$\mathbf{M}^{-1}\mathbf{c\nabla\bar{\mu}}=\mathbf{v}$$
Equation 3 is just this equation in component form.
Concise and correct. Thank you. I knew it must have something to do with the matrix inversion. I do know this stuff, but for some reason I couldn't figure it out.
 

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