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So after posting this question and reading all your comments I would like to make this new question (update).
If you consider the three equations presented here: $$\frac{\partial \rho}{\partial t} +\frac{\partial (\rho u)}{\partial z}=0$$ $$\frac{\partial C_i}{\partial t} +\frac{\partial (u C_i)}{\partial z}=Reactionrate$$ $$\rho=MW_{mean}\frac{P}{RgT}, MW_{mean}=\sum^{N_{comp}}_{i=1}(Y_i*mw_i), Y_i=\frac{C_i}{\sum^{N_{comp}}_{i=1}{C_i}}$$ continuity equation, mass balance for each species & an equation to calculate density, what would the discretized form (in $z$ direction, with backwards finite differences) be? I mean, if we consider (for solution purposes) that the variables are $C_i$, $u$ & $\rho$ (Case_1 a, b & c) then it is totally different than considering that the variables are $C_i$, $(uC_i)$, $\rho$ & $(\rho u)$ (Case_2)(as mentioned by Wolfgang Bangerth at the previous question, the term $\frac{\partial (\rho u)}{\partial z}$ implies conservation of quantity $(\rho u)$ in direction $z$, so should we treat this quantity like one?). Next I present the discretized form of both this cases (please correct me if I am wrong, or more precisely, tell me which is the correct one):
Case_1a
$$\frac{\partial \rho_j}{\partial t} +\frac{(\rho_j u_j - \rho_{j-1} u_{j-1})}{\Delta z}=0$$ $$\frac{\partial C_{i,j}}{\partial t} +\frac{(u_j C_{i,j} -u_{j-1} C_{i,j-1})}{\Delta z}=Reactionrate$$ $$\rho_j=MW_{mean,j}\frac{P}{RgT}, MW_{mean,j}=\sum^{N_{comp}}_{i=1}(Y_{i,j}*mw_i), Y_{i,j}=\frac{C_{i,j}}{\sum^{N_{comp}}_{i=1}{C_{i,j}}}$$

Case_1b
$$\frac{\partial \rho_j}{\partial t} +\rho_j\frac{(u_j - u_{j-1})}{\Delta z} +u_j\frac{(\rho_j - \rho_{j-1})}{\Delta z}=0$$ $$\frac{\partial C_{i,j}}{\partial t} +u_j\frac{(C_{i,j} - C_{i,j-1})}{\Delta z} +C_{i,j}\frac{(u_j - u_{j-1})}{\Delta z}=Reactionrate$$ $$\rho_j=MW_{mean,j}\frac{P}{RgT}, MW_{mean,j}=\sum^{N_{comp}}_{i=1}(Y_{i,j}*mw_i), Y_{i,j}=\frac{C_{i,j}}{\sum^{N_{comp}}_{i=1}{C_{i,j}}}$$

Case_2
$$\frac{\partial \rho_j}{\partial t} +\frac{(\rho u)_j - (\rho u)_{j-1}}{\Delta z}=0$$

$$\frac{\partial C_{i,j}}{\partial t} +\frac{(u C_i)_j - (u C_i)_{j-1}}{\Delta z}=Reactionrate$$

$$\rho_j=MW_{mean,j}\frac{P}{RgT}, MW_{mean,j}=\sum^{N_{comp}}_{i=1}(Y_{i,j}*mw_i), Y_{i,j}=\frac{C_{i,j}}{\sum^{N_{comp}}_{i=1}{C_{i,j}}}$$

First of all, should there be any difference detween the results in Case_1a & Case_1b given the same initial $(t=0)$ and boundary $(z=0)$ conditions? What is the correct formulation (mathematically speaking)? Do you think that this set of equations has one unique solution? Because I am almost sure that it does not. What assumption could justify a change like this:
Case_1c $$\frac{\partial \rho_j}{\partial t} +\rho_j\frac{(u_j - u_{j-1})}{\Delta z}=0$$ $$\frac{\partial C_{i,j}}{\partial t} +u_j\frac{(C_{i,j} - C_{i,j-1})}{\Delta z}=Reactionrate$$ $$\rho_j=MW_{mean,j}\frac{P}{RgT}, MW_{mean,j}=\sum^{N_{comp}}_{i=1}(Y_{i,j}*mw_i), Y_{i,j}=\frac{C_{i,j}}{\sum^{N_{comp}}_{i=1}{C_{i,j}}}$$ which I believe that does have one uniquie solution?
And referring to the Case_2, this way we have four(*) unknown variables ($C_i$, $(uC_i)$, $\rho$ & $(\rho u)$) which means that we need one more equation? Would an equation like: $$\frac{(uC_i)_j}{C_{i,j}}=\frac{(\rho u)_j}{\rho_j}$$ work? and with given initial and boundary conditions provide one and only one solution to this problem?

*actually the variables are more than four. If $N_c=Number of Components$ then the variables are $N_c*C_i+N_c*(uC_i)+\rho+(\rho u)$ times the number of discrete points in z direction but the use of the last equation should give the same number of equations and variables along with the boundary conditions.

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  • $\begingroup$ Can anybody answer? Or give me a hint? $\endgroup$ – ASK22 Feb 11 '16 at 9:52

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