Modelling of two phase flow in packed bed (continued)

[Chestermiller]

Hmm I redid the substitution and seem to come up with the same answer. I have multiplied in the negative sign though in this version:
$\epsilon_g\rho_m\frac{\partial y_i}{\partial t} = \phi_{z-\Delta z/2}(\frac{y_{z-\Delta z}- y_z}{\Delta z}) -\dot{M}_{i,z}^"a_s$

$\epsilon_g \rho_mC_{p,g,m}\frac{\partial T_g}{\partial t}= \phi_{z-\Delta z/2}C_{p,z-\Delta z/2}(\frac{T_{z-\Delta z} - T_z}{\Delta z}) -q_{g,I,z}a_s$

These look effectively identical to the equivalent equations we had at this stage in model 1. No good?

This looks better, although you are missing the summation term in the individual species mass balance equation.

 

[casualguitar]

This looks better, although you are missing the summation term in the individual species mass balance equation.

$\epsilon_g\rho_m\frac{\partial y_i}{\partial t} = \phi_{z-\Delta z/2}(\frac{y_{z-\Delta z}- y_z}{\Delta z}) -\dot{M}_{i,z}^"a_s+y_{i,z}\sum_{j=1}^{n_c}\dot{M}_j^"a_s$

Time to get rid of the $\Delta z$ terms and convert density to mass by multiplying the gas/solid mass/heat balances by $A_C\epsilon\Delta z$ and $A_C(1-\epsilon)\Delta z$ respectively? If so I'll do that

I can do this now if this is an appropriate next step?

Edit: Hmm actually this next step seemed to make sense to me when I didn't have the final summation term in the individual mass balance equation. Now however because this is present it doesn't seem like multiplying by $A_C\epsilon\Delta z$ will work for this term?

Edit: If I were to guess I would say that $y_{i,z}*A_C\epsilon\Delta z$ will equal a term that we know or we can calculate quite easily, however I'm not yet sure what this term is

 

[Chestermiller]

$\epsilon_g\rho_m\frac{\partial y_i}{\partial t} = \phi_{z-\Delta z/2}(\frac{y_{z-\Delta z}- y_z}{\Delta z}) -\dot{M}_{i,z}^"a_s+y_{i,z}\sum_{j=1}^{n_c}\dot{M}_j^"a_s$I can do this now if this is an appropriate next step?
Actually, it would just be $A_C\Delta z$ for each equation. This is optional; it makes the model seem much more like actual tanks, but this is really necessary.
Edit: Hmm actually this next step seemed to make sense to me when I didn't have the final summation term in the individual mass balance equation. Now however because this is present it doesn't seem like multiplying by $A_C\Delta z$ will work for this term?

Why would that be a problem?

We need to do more on the mass transfer rate expression that they use. I definitely don't like what they did. I'll get back to that later.

 

[casualguitar]

Why would that be a problem?

We need to do more on the mass transfer rate expression that they use. I definitely don't like what they did. I'll get back to that later.

Well if the guess below is correct then there is no problem.

Edit: If I were to guess I would say that yi,z∗ACϵΔz will equal a term that we know or we can calculate quite easily, however I'm not yet sure what this term is

Actually yes I had forgotten that we make the V/n substitution anyway (as we did in model 1) to get rid of the $\Delta z$ term

Time to get rid of the Δz terms and convert density to mass by multiplying the gas/solid mass/heat balances by ACϵΔz and AC(1−ϵ)Δz respectively? If so I'll do that

An appropriate time to do this? If so I'll do that

 

[Chestermiller]

Well if the guess below is correct then there is no problem.

Actually yes I had forgotten that we make the V/n substitution anyway (as we did in model 1) to get rid of the $\Delta z$ termAn appropriate time to do this? If so I'll do that

I have no problem with this, but like I said above, you should only multiply all the equations by $A_C\Delta z$.

 

[casualguitar]

I have no problem with this, but like I said above, you should only multiply all the equations by $A_C\Delta z$.

Did you have a better alternative in mind? If not I'll do this now

Also why leave out the $\epsilon$ and $1-\epsilon$ terms? Do we not have to account for the space occupied by the gas/solid phases?

 

[Chestermiller]

Did you have a better alternative in mind? If not I'll do this now

Also why leave out the $\epsilon$ and $1-\epsilon$ terms? Do we not have to account for the space occupied by the gas/solid phases?

That's already included in their equations.

 

[casualguitar]

That's already included in their equations.

Whoops. Perfect I'll multiply through by $A_C\Delta z$ then. This looks like it will leave us with a set of equations that are almost ready to be solved. I suppose correlations for the heat transfer coefficients (from Bird et al similar to the last model) and a correlation for the mass deposition rate are needed before we solve. I can use the one they give for now if that's alright, so that we can get some initial results?

 

[Chestermiller]

Whoops. Perfect I'll multiply through by $A_C\Delta z$ then. This looks like it will leave us with a set of equations that are almost ready to be solved. I suppose correlations for the heat transfer coefficients (from Bird et al similar to the last model) and a correlation for the mass deposition rate are needed before we solve. I can use the one they give for now if that's alright, so that we can get some initial results?

As I said, I don't like their mass transfer approach, but that can be replaced later.

 

[casualguitar]

As I said, I don't like their mass transfer approach, but that can be replaced later.

Ideal. These final model equations seem to be extremely close to the ones you developed in model 1. I'll post them this evening (currently away from my pc). Is there anything else (besides the mass transfer approach and the the heat transfer correlations) that should to be done before I can solve these equations in code?

One other question to do with the $\frac{dm_j}{dt}=\frac{V}{n}\frac{d\rho_j}{dt}=\frac{V}{n}\left(\frac{d\rho}{dh}\right)_j\frac{dh_j}{dt}\tag{4b}$ term as we had it in the previous post.

Since we are using temperature rather than enthalpy I guess we can just replace the enthalpy derivatives with equivalent temperature derivatives here. This $\frac{d\rho}{dh}$ term, or $\frac{d\rho}{dT}$ in this model is computationally expensive to calculate using the library I previously used. I was hoping we could swap this out in part i.e. assume it is zero for the solid regions or something similar to make it computationally less expensive. I could also use some other literature to calculate this term for CO2.

As I say I'm just pointing this out now. I think because the thermo library doesn't do solid phase CO2, we could run into a bit of a mess if we attempt the same thing we did previously.

Anyway I'll copy the updated model equations here this evening

 

[casualguitar]

Ideal. These final model equations seem to be extremely close to the ones you developed in model 1. I'll post them this evening (currently away from my pc):

Multiplying the spatially discretised model equations by $A_C\Delta z$:

Individual species mass balance:
$m_{m,j}\frac{\partial y_i}{\partial t} = \dot{m}_{j-1}(y_{j-1} - y_j) - \dot{M}_{i,j}''a_sA_C\Delta z + y_{i,j}*A_C\Delta z\sum_{j=1}^{n_c}\dot{M}_j^"a_s$

Individual species heat balance:
$m_{m,j}c_{p,j}\frac{\partial T_g}{\partial t} = \dot{m}_{j-1}c_{p,j-1}(T_{j-1} - T_j) - q_{g,I,j}a_sA_C\Delta z$

Overall mass balance to the gas phase:
$\frac{\partial m}{\partial t} = m_{j-1} - m_j -A_C\Delta z\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"a_s}$

And then the two equations that do not have spatial derivatives, the species mass balance for deposition at the interface, and the heat balance to the bed:
$\frac{\partial M_i}{\partial t} ={\dot{M}_i^"a_s}$
$\rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T_b}{dt}=q_{I,b}a_s$

I have left $A_C\Delta z$ in three of the terms that I did not know how to simplify. I could sub in V/n here but I would guess there is a better way to get rid of the $A_C\Delta z$ term?

Edit: I also have started re-reading the model from the beginning (cleaning up our comments into a word document), and reading back over the earlier comments with the new knowledge of how these early equations were developed into the equations we now have. A lot of the smaller bits that didn't really make sense (like how they got their individual species mass balance) now do make sense
Edit 2: Also quite funny (in my view) that I seem to respond with 'I fully understand your post' quite often and then continue to misunderstand what you've said for a number of further posts. Apologies about that, will aim to do better

 

[Chestermiller]

Multiplying the spatially discretised model equations by $A_C\Delta z$:

Individual species mass balance:
$m_{m,j}\frac{\partial y_i}{\partial t} = \dot{m}_{j-1}(y_{j-1} - y_j) - \dot{M}_{i,j}''a_sA_C\Delta z + y_{i,j}*A_C\Delta z\sum_{j=1}^{n_c}\dot{M}_j^"a_s$

Individual species heat balance:
$m_{m,j}c_{p,j}\frac{\partial T_g}{\partial t} = \dot{m}_{j-1}c_{p,j-1}(T_{j-1} - T_j) - q_{g,I,j}a_sA_C\Delta z$

Overall mass balance to the gas phase:
$\frac{\partial m}{\partial t} = m_{j-1} - m_j -A_C\Delta z\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"a_s}$

And then the two equations that do not have spatial derivatives, the species mass balance for deposition at the interface, and the heat balance to the bed:
$\frac{\partial M_i}{\partial t} ={\dot{M}_i^"a_s}$
$\rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T_b}{dt}=q_{I,b}a_s$

Why didn't you multiply these equations by $A_C\Delta z$ also?

I have left $A_C\Delta z$ in three of the terms that I did not know how to simplify. I could sub in V/n here but I would guess there is a better way to get rid of the $A_C\Delta z$ term?

I don't understand.

 

[casualguitar]

Why didn't you multiply these equations by $A_C\Delta z$ also?

Yes I should have however I didn't really know what to do with the term once it was multiplied in:
$\frac{\partial M_i}{\partial t} ={\dot{M}_i^"a_s}A_C\Delta z$
$\rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T_b}{dt}=q_{I,b}a_sA_C\Delta z$

I don't understand.

In the previous model, the final model equations did not have any $\Delta z$ or $A_c$ term in them. We were able to sub them out for a term with physical meaning i.e. these substitutions $m_j=\rho_jA_C\epsilon \Delta x=\rho_j(V/n)$, or $\dot{m_j}=\phi_{x+\Delta x/2}A_C\epsilon$, whereas here we seem to be left with $\Delta z$ and $A_c$ terms in the final equations? Do we have equivalent substations in these cases, or will we have $\Delta zA_c$ terms in the final model equations?

 

[casualguitar]

Just reading back over progress to date I have one other question (which I think you have already answered but I want to ask it in another way so maybe the reasoning will become clear to me).

We initially had the overall mass balance and the divergence form of the mass balance. For me it looks like the divergence form is immediately in a useful format. However we did not use it in that base format. Instead we multiplied the overall mass balance by $y_i$ and subtracted this from the divergence form of the mass balance. So the question would be - why can we not use the divergence form in its base form i.e. why do we need to subtract the overall mass balance*y_i from it?

You mentioned that the divergence form only has the single species deposition term. Is that the reason for this manipulation? To get the multi species deposition term $y_i\sum_{j=1}^{n_c}\dot{M}_j^"a_s$ in there? If so, why do we need this term? I suppose I am confused as to why the individual species mass balance deals with more than one species in the one equation

 

[Chestermiller]

Multiplying the spatially discretised model equations by $A_C\Delta z$:

Individual species mass balance:
$m_{m,j}\frac{\partial y_i}{\partial t} = \dot{m}_{j-1}(y_{j-1} - y_j) - \dot{M}_{i,j}''a_sA_C\Delta z + y_{i,j}*A_C\Delta z\sum_{j=1}^{n_c}\dot{M}_j^"a_s$

Individual species heat balance:
$m_{m,j}c_{p,j}\frac{\partial T_g}{\partial t} = \dot{m}_{j-1}c_{p,j-1}(T_{j-1} - T_j) - q_{g,I,j}a_sA_C\Delta z$

Overall mass balance to the gas phase:
$\frac{\partial m}{\partial t} = m_{j-1} - m_j -A_C\Delta z\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"a_s}$

The above three equations involve moles, right, not mass.

And then the two equations that do not have spatial derivatives, the species mass balance for deposition at the interface, and the heat balance to the bed:
$\frac{\partial M_i}{\partial t} ={\dot{M}_i^"a_s}$

Multiplying this equation by $A_C\Delta z$ gives:

$A_C\Delta z\frac{\partial M_i}{\partial t} ={\dot{M}_i^"A_C\Delta za_s}$

where $A_C\Delta zM_i$ are the number of moles of deposit between axial locations $z-\Delta z/2$ and $z+\Delta z/2$, and $A_C\Delta za_s$ is the surface area of packing between axial locations $z-\Delta z/2$ and $z+\Delta z/2$.

$\rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T_b}{dt}=q_{I,b}a_s$

Multiplying this equation by $A\Delta z$ gives:

$\rho_s(1-\epsilon_g)A\Delta z C_{p,s}\frac{\partial T_b}{dt}=q_{I,b}A_C\Delta za_s$

where $\rho_s(1-\epsilon_g)A\Delta z$ represents the mass of packing between axial locations $z-\Delta z/2$ and $z+\Delta z/2$.

 

[casualguitar]

The above three equations involve moles, right, not mass.

Yes exactly, typo by me there. Also typo in forgetting to multiply the LHS of those equations by $A\Delta z$.

If we let:
$$A_s = A_C\Delta za_s$$
$$M_s = \rho_s(1-\epsilon_g)A\Delta z$$
$$N_i = A_C\Delta zM_i$$

Then we get:
$$A_C\Delta z\frac{\partial M_i}{\partial t} ={\dot{M}_i^"A_S}$$
and
$$M_sC_{p,s}\frac{\partial T_b}{dt}=q_{I,b}A_S$$

I suppose we could sub in $N_i$ on the LHS in the first equation but I don't think this is useful.

For the remaining equations, using the above substitutions we can get to:
$$m_{m,j}\frac{\partial y_i}{\partial t} = \dot{m}_{j-1}(y_{j-1} - y_j) - \dot{M}_{i,j}''A_S + y_{i,j}\sum_{j=1}^{n_c}\dot{M}_j^"A_S$$
$$m_{m,j}c_{p,j}\frac{\partial T_g}{\partial t} = \dot{m}_{j-1}c_{p,j-1}(T_{j-1} - T_j) - q_{g,I,j}A_S$$
and
$$\frac{\partial m}{\partial t} = \dot{m}_{j-1} - \dot{m}_j -\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"A_S}$$

If the above substitutions are ok, then the species mass balance for deposition at the interface is the only remaining equation with a $A_C\Delta z$ term. We could sub in $N_i$ and this $\frac{\partial N_i}{\partial t}$ term would be the total moles of species i deposited between $z-\Delta z/2$ and $z+\Delta z/2$?

 

[Chestermiller]

Yes exactly, typo by me there. Also typo in forgetting to multiply the LHS of those equations by $A\Delta z$.

If we let:
$$A_s = A_C\Delta za_s$$
$$M_s = \rho_s(1-\epsilon_g)A\Delta z$$
$$N_i = A_C\Delta zM_i$$

Then we get:
$$A_C\Delta z\frac{\partial M_i}{\partial t} ={\dot{M}_i^"A_S}$$
and
$$M_sC_{p,s}\frac{\partial T_b}{dt}=q_{I,b}A_S$$

I suppose we could sub in $N_i$ on the LHS in the first equation but I don't think this is useful.

For the remaining equations, using the above substitutions we can get to:
$$m_{m,j}\frac{\partial y_i}{\partial t} = \dot{m}_{j-1}(y_{j-1} - y_j) - \dot{M}_{i,j}''A_S + y_{i,j}\sum_{j=1}^{n_c}\dot{M}_j^"A_S$$
$$m_{m,j}c_{p,j}\frac{\partial T_g}{\partial t} = \dot{m}_{j-1}c_{p,j-1}(T_{j-1} - T_j) - q_{g,I,j}A_S$$
and
$$\frac{\partial m}{\partial t} = m_{j-1} - m_j -\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"A_S}$$

If the above substitutions are ok, then the species mass balance for deposition at the interface is the only remaining equation with a $A_C\Delta z$ term. We could sub in $N_i$ and this $\frac{\partial N_i}{\partial t}$ term would be the total moles of species i deposited between $z-\Delta z/2$ and $z+\Delta z/2$?

Sure.

Also, $m=\frac{P}{RT}A_C\Delta z$, so $$\frac{\partial m}{\partial t}=-\frac{\rho_m}{T}A_C\Delta z\frac{\partial T}{\partial t}$$So, $$\dot{m}_j=\dot{m}_{j-1}+\frac{\rho_m}{T}A_C\Delta z\frac{\partial T}{\partial t}-\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"A_S}$$

 

[casualguitar]

Sure.

Also, $m=\frac{P}{RT}A_C\Delta z$, so $$\frac{\partial m}{\partial t}=-\frac{\rho_m}{T}A_C\Delta z\frac{\partial T}{\partial t}$$So, $$\dot{m}_j=\dot{m}_{j-1}+\frac{\rho_m}{T}A_C\Delta z\frac{\partial T}{\partial t}-\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"A_S}$$

Ok that's interesting, so we can avoid using thermo to calculate derivatives completely here since no such derivatives show up in this model (like the $\frac{\partial \rho}{\partial H}$ derivative in the last model, or equivalent here)

Could we also sub in $m_m$ for $\rho_mA_C\Delta z$ in both equations i.e. the holdup mass?

The last remaining equation with an $A_C\Delta z$ term is $A_C\Delta z\frac{\partial M_i}{\partial t} ={\dot{M}_i^"A_S}$. Is there a substitution that could be made here to sub out the $A_C\Delta z$ term?

If you don't mind before we move any further I'd like to read back over everything we have done so far on this model to be sure I follow what we've done so far. I think it mostly makes sense however I'd like to clear up the last few questions before we progress just so I'm not trying to build on anything I don't understand

 

[Chestermiller]

Ok that's interesting, so we can avoid using thermo to calculate derivatives completely here since no such derivatives show up in this model (like the $\frac{\partial \rho}{\partial H}$ derivative in the last model, or equivalent here)

For an ideal gas, this is just $\frac{1}{C_p}\frac{\partial \rho}{\partial T}$

Could we also sub in $m_m$ for $\rho_mA_C\Delta z$ in both equations i.e. the holdup mass?

The holdup moles of gas is $\rho_m\epsilon A_C\Delta z$.

The last remaining equation with an $A_C\Delta z$ term is $A_C\Delta z\frac{\partial M_i}{\partial t} ={\dot{M}_i^"A_S}$. Is there a substitution that could be made here to sub out the $A_C\Delta z$ term?

You could define it as a new variable.

If you don't mind before we move any further I'd like to read back over everything we have done so far on this model to be sure I follow what we've done so far. I think it mostly makes sense however I'd like to clear up the last few questions before we progress just so I'm not trying to build on anything I don't understand

I think this is a good idea. The most important part of model development is the formulation of the equations, which represents the translation of the physical and chemical mechanisms involved into the language of mathematics. In my judgment, it is very important to spend lots of time "playing" with the model equations to help decide what is the most favorable forms to work with.

 

[casualguitar]

For an ideal gas, this is just $\frac{1}{C_p}\frac{\partial \rho}{\partial T}$

The holdup moles of gas is $\rho_m\epsilon A_C\Delta z$.

You could define it as a new variable.

I think this is a good idea. The most important part of model development is the formulation of the equations, which represents the translation of the physical and chemical mechanisms involved into the language of mathematics. In my judgment, it is very important to spend lots of time "playing" with the model equations to help decide what is the most favorable forms to work with.

After reading through the model, I think these questions sum up the bits I don't understand (not really all that much):
-The model is similar to the Tuinier model, however there are some differences. Is it fair to say these are the main differences, or have I missed any?
1) no assumption of $T_g = T_b$
2) molar rather than mass balances used
3) The formulation for the rate of mass deposition (will be different)

- I've asked this next one already but I want to ask it again in a different way to hopefully make it clear what exactly is happening here:
I see that the overall gas phase mass balance wasn't useful as it did not have the variation of mole fraction w.r.t time in it. So we multiplied in the mole fraction term. Then we added this equation to species mass balance to obtain the divergence form of the mass balance:

What is the advantage of solving this form of the mass balance over the component mass balance that Tuinier provided (our species mass balance)? As both equations seem to have the mole fraction time derivative. Do you see my confusion there?

Lastly what is the physical significance of the last term if any?
$m_{m,j}\frac{\partial y_i}{\partial t} = \dot{m}_{j-1}(y_{j-1} - y_j) - \dot{M}_{i,j}''A_S + y_{i,j}\sum_{j=1}^{n_c}\dot{M}_j^"A_S$

Besides this no further questions I am good to proceed. I'll write out the up to date model equations with relevant substitutions now

 

[casualguitar]

Besides this no further questions I am good to proceed. I'll write out the up to date model equations with the relevant substitutions now

So the equations we're solving:
The gas phase mole balance:
$m_{m,j}\frac{\partial y_{i,j}}{\partial t} = \dot{m}_{j-1}(y_{j-1} - y_j) - \dot{M}_{i,j}''A_{S,j} + y_{i,j}\sum_{j=1}^{n_c}\dot{M}_j^"A_S$

The gas phase heat balance:
$m_{m,j}c_{p,j}\frac{\partial T_{g,j}}{\partial t} = \dot{m}_{j-1}c_{p,j-1}(T_{j-1} - T_j) - q_{g,I,j}A_{S,j}$

Bed heat balance:
$M_{s,j}C_{p,s,j}\frac{\partial T_{b,j}}{dt}=q_{I,b,j}A_{S,j}$

Solid phase mass balance:
$A_C\Delta z\frac{\partial M_{i,j}}{\partial t} ={\dot{M}_{i,j}^"A_{S,j}}$

Variation of gas phase mass holdup w.r.t. time:
$\frac{\partial m_j}{\partial t} = \dot{m}_{j-1} - \dot{m}_j -\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"A_{S,j}}$
$\frac{\partial m_j}{\partial t}=-\frac{\rho_{m,j}}{T_j}A_C\Delta z\frac{\partial T_{g,j}}{\partial t}$

Mass flow out of a tank:
$\dot{m}_j=\dot{m}_{j-1}+\frac{\rho_{m,j}}{T}A_C\Delta z\frac{\partial T}{\partial t}-\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"A_{S,j}}$

Where:
$A_s = A_C\Delta za_s$
$M_s = \rho_s(1-\epsilon_g)A\Delta z$
$m_j=\rho_jA_C\epsilon \Delta x=\rho_j(V/n)$
$\dot{m_j}=\phi_{x+\Delta x/2}A_C\epsilon$
$A_{S,j}=A/n$ where n is the number of tanks
$M_{S,j}=M/n$

Do these equations look ok? In your view are they in a final 'solvable' form?

 

[Chestermiller]

After reading through the model, I think these questions sum up the bits I don't understand (not really all that much):
-The model is similar to the Tuinier model, however there are some differences. Is it fair to say these are the main differences, or have I missed any?
1) no assumption of $T_g = T_b$
2) molar rather than mass balances used
3) The formulation for the rate of mass deposition (will be different)

We handle the dispersion differently also.

- I've asked this next one already but I want to ask it again in a different way to hopefully make it clear what exactly is happening here:
I see that the overall gas phase mass balance wasn't useful as it did not have the variation of mole fraction w.r.t time in it. So we multiplied in the mole fraction term. Then we added this equation to species mass balance to obtain the divergence form of the mass balance:
View attachment 299658

No. The divergence form was the original form of the individual species mass balance equation that they derived (not shown in their formulation). So to get the time derivatives of the mass fractions, they multiplied the overall mass balance equation by species mass fraction and subtracted.

What is the advantage of solving this form of the mass balance over the component mass balance that Tuinier provided (our species mass balance)? As both equations seem to have the mole fraction time derivative. Do you see my confusion there?

I am not recommending solving using the divergence form. In the newly derived finite difference equations, we have used the non-divergence form (material derivative form).

Lastly what is the physical significance of the last term if any?
$m_{m,j}\frac{\partial y_i}{\partial t} = \dot{m}_{j-1}(y_{j-1} - y_j) - \dot{M}_{i,j}''A_S + y_{i,j}\sum_{j=1}^{n_c}\dot{M}_j^"A_S$

I suppose we could figure out a physical significance to it (based on the material derivative frame of reference of an observer moving with the mass- or molar average velocity), but I don't think it is really worth it.

 

[Chestermiller]

So the equations we're solving:
The gas phase mole balance:
$m_{m,j}\frac{\partial y_{i,j}}{\partial t} = \dot{m}_{j-1}(y_{j-1} - y_j) - \dot{M}_{i,j}''A_{S,j} + y_{i,j}\sum_{j=1}^{n_c}\dot{M}_j^"A_S$

The gas phase heat balance:
$m_{m,j}c_{p,j}\frac{\partial T_{g,j}}{\partial t} = \dot{m}_{j-1}c_{p,j-1}(T_{j-1} - T_j) - q_{g,I,j}A_{S,j}$

Bed heat balance:
$M_{s,j}C_{p,s,j}\frac{\partial T_{b,j}}{dt}=q_{I,b,j}A_{S,j}$

Solid phase mass balance:
$A_C\Delta z\frac{\partial M_{i,j}}{\partial t} ={\dot{M}_{i,j}^"A_{S,j}}$

Variation of gas phase mass holdup w.r.t. time:
$\frac{\partial m_j}{\partial t} = \dot{m}_{j-1} - \dot{m}_j -\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"A_{S,j}}$
$\frac{\partial m_j}{\partial t}=-\frac{\rho_{m,j}}{T_j}A_C\Delta z\frac{\partial T_{g,j}}{\partial t}$

These previous two equations are not used in the computer model. They are combined in the equation below to give the flow out of each tank.

Mass flow out of a tank:
$\dot{m}_j=\dot{m}_{j-1}+\frac{\rho_{m,j}}{T}A_C\Delta z\frac{\partial T}{\partial t}-\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"A_{S,j}}$

Where:
$A_s = A_C\Delta za_s$
$M_s = \rho_s(1-\epsilon_g)A\Delta z$
$m_j=\rho_jA_C\epsilon \Delta x=\rho_j(V/n)$
$\dot{m_j}=\phi_{x+\Delta x/2}A_C\epsilon$
$A_{S,j}=A/n$ where n is the number of tanks
$M_{S,j}=M/n$

Do these equations look ok? In your view are they in a final 'solvable' form?

As best I can tell, these equations look OK. However they are not the complete formulation. Also needed are the heat- and mass transfer coefficient correlations, the expressions for the heat- and mass fluxes to the gas and bed, and the temperature at the solid deposit (interface between the gas and bed) which is used to calculate the partial pressure of the depositing species.

 

[casualguitar]

These previous two equations are not used in the computer model. They are combined in the equation below to give the flow out of each tank.

As best I can tell, these equations look OK. However they are not the complete formulation

Also needed are the heat- and mass transfer coefficient correlations

$\frac{1}{U^*}=\frac{1}{U_g}+\frac{1}{U_b}$
So we need correlations for $U_b$ and $U_g$. We know how to get these so that should be ok

No mass transfer coefficient correlations yet. We could go with their fudge factor approach for now until a basic model in code has been set up

the expressions for the heat- and mass fluxes to the gas and bed

Heat fluxes:
$q_{g,I}=-\frac{U_g\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}}{U_g+U_b}+U^*(T_g-T_b)$
$q_{I,b}=+\frac{U_b\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}}{U_g+U_b}+U^*(T_g-T_b)$

No mass flux expressions yet

the temperature at the solid deposit (interface between the gas and bed)

$T_I=\frac{\sum_{j=1}^{n_c}\dot{M}_j^"\Delta h_j}{U_g+U_b}+\frac{U_gT_g+U_bT_b}{U_g+U_b}$

which is used to calculate the partial pressure of the depositing species.

So we also need a way to calculate the partial pressure of a species. Thermo should be able to do this I'll check. Tuinier provides a saturation pressure correlation for CO2 but not water so I suppose we will need to get an equivalent solidification curve for water here also

So to sum up what other things are needed:
- $U_b$ and $U_g$ correlations from Bird et al
- Mass transfer coefficient correlations
- Expressions for mass flux to the gas and bed
- Partial pressure correlations

I guess Tuinier have provided the desublimation curve for #CO_2##. We might also need an equivalent solidification (liquid to solid) curve for water?

Do you think it is better to discuss the computational flow i.e. how to solve this system next, or should we get correlations for the above first?

 

[Chestermiller]

$\frac{1}{U^*}=\frac{1}{U_g}+\frac{1}{U_b}$
So we need correlations for $U_b$ and $U_g$. We know how to get these so that should be ok

No mass transfer coefficient correlations yet. We could go with their fudge factor approach for now until a basic model in code has been set up

Heat fluxes:
$q_{g,I}=-\frac{U_g\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}}{U_g+U_b}+U^*(T_g-T_b)$
$q_{I,b}=+\frac{U_b\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}}{U_g+U_b}+U^*(T_g-T_b)$

No mass flux expressions yet

$T_I=\frac{\sum_{j=1}^{n_c}\dot{M}_j^"\Delta h_j}{U_g+U_b}+\frac{U_gT_g+U_bT_b}{U_g+U_b}$So we also need a way to calculate the partial pressure of a species. Thermo should be able to do this I'll check. Tuinier provides a saturation pressure correlation for CO2 but not water so I suppose we will need to get an equivalent solidification curve for water here also

So to sum up what other things are needed:
- $U_b$ and $U_g$ correlations from Bird et al
- Mass transfer coefficient correlations
- Expressions for mass flux to the gas and bed
- Partial pressure correlations

I guess Tuinier have provided the desublimation curve for #CO_2##. We might also need an equivalent solidification (liquid to solid) curve for water?

Do you think it is better to discuss the computational flow i.e. how to solve this system next, or should we get correlations for the above first?

I think we should complete the formulation first.

BTW, I meant to say "equilibrium vapor pressures," not "partial pressures"

 

[casualguitar]

BTW, I meant to say "equilibrium vapor pressures," not "partial pressures"

Yes I thought so, however I have a history of being wrong so far

The paper provides this curve for $CO_2$. None for water though so I suppose I can just get an equivalent liquid-solid curve to add to the $CO_2$ vapour-solid curve they give

I think I can make a good attempt at $U_g$ and $U_b$, and I can get the water equilibrium vapour pressure curve. How does starting with the mass transfer correlations/expressions sound? Seems like the most difficult of the bunch

 

[casualguitar]

I think we should complete the formulation first.

Sublimation curve for CO2 (Span and Wagner):

Melting curve for H2O (Span and Wagner):

If we need the vapour-liquid curve for CO2 we could take this from Span and Wagner also just so we would have alignment between the VL and VS curves

I will make an attempt at doing the $U_g$ and $U_b$ correlations this evening. I guess this will be the the same process as the previous model? Just adjusting for gas/solid parameters as needed

 

[Chestermiller]

Sublimation curve for CO2 (Span and Wagner):
View attachment 299816

What's wrong with using the CO2 vapor pressure from Tuinier et al?

Melting curve for H2O (Span and Wagner):
View attachment 299817

If we need the vapour-liquid curve for CO2 we could take this from Span and Wagner also just so we would have alignment between the VL and VS curves

I will make an attempt at doing the $U_g$ and $U_b$ correlations this evening. I guess this will be the the same process as the previous model? Just adjusting for gas/solid parameters as needed

Yes, the BSL correlation for the gas, and the asymptotic (long time) equation for the solid particles.

 

[casualguitar]

What's wrong with using the CO2 vapor pressure from Tuinier et al?

Yes, the BSL correlation for the gas, and the asymptotic (long time) equation for the solid particles.

Nothing I just thought it would be nice to have both fits come from the same source Span and Wagner. Either one is fine with me

Ok great. I'll make an attempt at these this evening then

By any chance could we talk about the computational flow in short before doing the mass transfer correlations? I think given the multiple equations involved it will take some time for me to understand this

For reference this is the computational flow for the previous model:

 

[Chestermiller]

Mass Transfer Rate for Deposition of CO2 and Water

The following is still somewhat approximate, but is a significant improvement to the mass transfer rate approach used by Tuinier et al:
$$\dot{M}_i^"=k_i\left(\frac{Py_i-p_i(T_I)}{RT_I}\right)\tag{1}$$where $\dot{M}_i^"$ is the molar flux of species per unit area of surface, P is the total pressure, $y_i$ is the mole fraction of species in the bulk gas, $p_i(T_I)$ is the equilibrium vapor pressure of species at the interface temperature $T_I$, and $k_i$ is the mass transfer coefficient (units of m/s) of species for deposition from the gas.

The mass transfer coefficient $k_i$ is related to the local Sherwood number for mass transfer by $$Sh_{loc,i}=\frac{k_iD_p}{(1-\epsilon_g)D_i\psi}$$where $D_i$ is the diffusion coefficient of species in N2 and $\psi$ is the packing particle shape factor (equal to 1.0 for spherical particles and 0.92 for cylindrical particles). By Reynolds analogy, the Sherwood number $Sh_{loc,i}$ is the same function of Reynold number and Schmidt number as the Nussult number is as a function of Reynolds number and Prantdl number, respectively.

An additional constraint on Eqn. 1 is that $\dot{M}_i^"$ must be zero if the number of moles of deposit per unit area of surface is zero and the right hand side of Eqn. 1 is negative. This means that the number of moles per unit area of deposit on the surface of the particles can never go negative.

Thoughts?

 

[casualguitar]

Mass Transfer Rate for Deposition of CO2 and Water

The following is still somewhat approximate, but is a significant improvement to the mass transfer rate approach used by Tuinier et al:
$$\dot{M}_i^"=k_i\left(\frac{Py_i-p_i(T_I)}{RT_I}\right)\tag{1}$$where $\dot{M}_i^"$ is the molar flux of species per unit area of surface, P is the total pressure, $y_i$ is the mole fraction of species in the bulk gas, $p_i(T_I)$ is the equilibrium vapor pressure of species at the interface temperature $T_I$, and $k_i$ is the mass transfer coefficient (units of m/s) of species for deposition from the gas.

Ok then similar to the Tuinier et al model you're saying that mass transfer rate will be proportional to the difference between the partial pressure of the species and the equilibrium vapour pressure of the species

Where does equation 1 come from? Is this an equation you derived or is it an established equation?

The mass transfer coefficient ki is related to the local Sherwood number for mass transfer by Shloc,i=kiDp(1−ϵg)Diψwhere Di is the diffusion coefficient of species in N2 and ψ is the packing particle shape factor (equal to 1.0 for spherical particles and 0.92 for cylindrical particles).

Got it, so I guess we will be able to find Sherwood number correlations for packed beds in Bird et al that we can use here to solve for the mass transfer coefficient

By Reynolds analogy, the Sherwood number Shloc,i is the same function of Reynold number and Schmidt number as the Nussult number is as a function of Reynolds number and Prantdl number, respectively.

So if the Nusselts number tells you how convection or conduction dominant the heat transfer is, the Sherwood number tells you how convection or diffusion dominant the mass transfer is?

An additional constraint on Eqn. 1 is that M˙i" must be zero if the number of moles of deposit per unit area of surface is zero and the right hand side of Eqn. 1 is negative. This means that the number of moles per unit area of deposit on the surface of the particles can never go negative.

Got it

This just leaves us with the $U_g$ and $U_b$ correlations then? On this now

 

[Chestermiller]

Ok then similar to the Tuinier et al model you're saying that mass transfer rate will be proportional to the difference between the partial pressure of the species and the equilibrium vapour pressure of the species

Where does equation 1 come from? Is this an equation you derived or is it an established equation?

It is an established equation using the mass transfer coefficient analogous to the heat transfer coefficient. See chapters 17 and forward in BSL, Treybel (Mass Transfer Operations), and Chemical Engineers' Handbook

Got it, so I guess we will be able to find Sherwood number correlations for packed beds in Bird et al that we can use here to solve for the mass transfer coefficient

Like I said, you use the same BSL correlation for Sh in terms of Re and Sc that you use for heat transfer Nu in terms of Re and Pr.

So if the Nusselts number tells you how convection or conduction dominant the heat transfer is, the Sherwood number tells you how convection or diffusion dominant the mass transfer is?

Nu is for convective heat transfer and Sh is for convective mass transfer.

Got it

This just leaves us with the $U_g$ and $U_b$ correlations then? On this now

I recommend using the same BSL correlations that we used in the other model.

 

[casualguitar]

Like I said, you use the same BSL correlation for Sh in terms of Re and Sc that you use for heat transfer Nu in terms of Re and Pr.

Got it, I found where it does that in BSL also

I recommend using the same BSL correlations that we used in the other model.

Ok interesting so this flow as done in the previous model:

applies for both $U_g$ and $U_b$ (using the relevant solid and fluid parameters for $U_b$ and $U_g$ respectively). We can also swap out the Prandtl number for the Schmidt number and then this flow applies for the mass transfer coefficient $k_i$ also? If so this makes the coding efficient

Where $Sc = \frac{\mu}{\rho D}$. The equivalent $k_g$ term from model 1 was found via the thermo library assuming $k_g$ was temperature and pressure dependent. I don't think thermo will give the mass diffusivity so we will need a correlation for $D$ here also. I guess BSL will have this?

Note:
Just looking at the correlation used for $U$ in the previous model, its not clear to me how to apply this separately to $U_g$ and $U_b$ in this model. It seems that we will only need to use the fact that $T_g$ does not equal $T_b$. Is it correct to say that the only difference between the $U_g$ and $U_b$ models is that the $U_g$ parameters are evaluated at $T_g$ and the $U_b$ parameters are evaluated at $T_b$?

Or if the above is not correct then maybe we should go with the approach below (outlined in BSL ch.14) where you take the average temperature so $U_b$ parameters would be evaluated at $\frac{(TI+TB)}{2}$ and $U_g$ parameters at $\frac{(TI+TG)}{2}$?

 

[casualguitar]

It is an established equation using the mass transfer coefficient analogous to the heat transfer coefficient. See chapters 17 and forward in BSL, Treybel (Mass Transfer Operations), and Chemical Engineers' Handbook

Like I said, you use the same BSL correlation for Sh in terms of Re and Sc that you use for heat transfer Nu in terms of Re and Pr.

Nu is for convective heat transfer and Sh is for convective mass transfer.

I recommend using the same BSL correlations that we used in the other model.

Just one other question - I'll be using the previous HTC functions I made for this (plus a new one for the Sc number). I think there is a bug in one of my functions though. Do you know what the typical ranges for the Pr, Re, Nu, Sc numbers and U in the previous model (model 1) would be? Just to help with the debug!

Also I believe you previously commented on my sleeping patterns (early on in model 1). I see you're also awake at a quite early time of 4am! An early bird or a late night owl?

Edit:
Lastly, we had the lumping of the convective and conductive terms in the last model for U:

Does the same lumping apply for the convective and diffusive mass transfer? Meaning that we would get something like:
$$k_i = \frac{1}{\frac{1}{k_{fs}} + \frac{dp/\beta}{D}}$$

 

[Chestermiller]

Got it, I found where it does that in BSL also

Ok interesting so this flow as done in the previous model:
View attachment 299928
applies for both $U_g$ and $U_b$ (using the relevant solid and fluid parameters for $U_b$ and $U_g$ respectively). We can also swap out the Prandtl number for the Schmidt number and then this flow applies for the mass transfer coefficient $k_i$ also? If so this makes the coding efficient

I can't read your handwriting again. Please use Latex or provide a post # from the previous thread.

Where $Sc = \frac{\mu}{\rho D}$. The equivalent $k_g$ term from model 1 was found via the thermo library assuming $k_g$ was temperature and pressure dependent. I don't think thermo will give the mass diffusivity so we will need a correlation for $D$ here also. I guess BSL will have this?

Just google "diffusivity of CO2 (and H2O)" in Nitrogen. That should be good enough. BSL will have H2O probably.

Note:
Just looking at the correlation used for $U$ in the previous model, its not clear to me how to apply this separately to $U_g$ and $U_b$ in this model. It seems that we will only need to use the fact that $T_g$ does not equal $T_b$. Is it correct to say that the only difference between the $U_g$ and $U_b$ models is that the $U_g$ parameters are evaluated at $T_g$ and the $U_b$ parameters are evaluated at $T_b$?

This is not going to be exact. I would use T at the interface.

Or if the above is not correct then maybe we should go with the approach below (outlined in BSL ch.14) where you take the average temperature so $U_b$ parameters would be evaluated at $\frac{(TI+TB)}{2}$ and $U_g$ parameters at $\frac{(TI+TG)}{2}$?

View attachment 299929

This would be adequate too. Like I said, it won't be exact. This is a judgment call.