Modelling of two phase flow in packed bed (continued)

In summary: I don't know actually, but I think you will be right about the CO2 depositing temporarily on the bed. What I thought would happen (assuming a bed colder than the freezing point of CO2) was that the ambient CO2 enriched stream would enter the cold bed and immediately the CO2 at the 'front' of the stream would freeze. The pure air would carry on through the bed. Then the newly entering stream - which is at ambient temperature - would vaporise the frozen CO2, and the vaporised CO2 plus the CO2 'behind' it in the stream would now be frozen/deposited slightly further downstream. This process repeats until you
  • #36
casualguitar said:
This is a plot of the heat capacities of N2,O2,CO2 and H20 versus temperature, for a pressure of 5 bar:
I thought we were going to operate this at 1 bar, like in the paper.
casualguitar said:
View attachment 299082
The N2 and O2 heat capacities are almost identical (I don't think we have discussed the inclusion of O2 in this model however I've included it here). The CO2 heat capacity is also very similar to N2/O2 while in the gas phase. H20 heat capacity is not similar to the others however if N2/O2/CO2 take up a high enough mass fraction of the total mixture then this constant molar heat capacity assumption seems reasonable
Your curve for water vapor is wrong. The molar heat capacity of water vapor is about 36 J/mol-K. We are interested only in the heat capacities in the gas phase for this model. At 1 bar, we are talking about the ideal gas heat capacities which are all functions only of temperature.
casualguitar said:
The CO2 heat capacity will jump up when it transitions to solid phase. This plot however is actually the correlation used for the liquid phase (as thermo does not recognise solid CO2 it uses the liquid correlation instead). We can manually add solid correlations where needed. It doesn't look like we will need one here though given the constant molar heat capacity assumption
We are not assuming constant molar heat capacity. The temperature dependence of the heat capacities should be included in the model. We are only assuming negligible effect of composition (i.e., changes mole fraction).
casualguitar said:
Hmm not an obvious one. Possibly you're looking at the derivative itself? The way they have written them, its really two first derivatives multiplied together, rather than a second derivative which is what it should be
Yes, the first T should not be in there.
casualguitar said:
Will converting to molar basis involve replacing each mass term with molar*mW term? The mW would then cancel out it seems and we would be left with the same equations except assuming a molar basis?
Please tell me you don't think this?
 
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  • #37
Chestermiller said:
I thought we were going to operate this at 1 bar, like in the paper.

Your curve for water vapor is wrong. The molar heat capacity of water vapor is about 36 J/mol-K. We are interested only in the heat capacities in the gas phase for this model. At 1 bar, we are talking about the ideal gas heat capacities which are all functions only of temperature.
Well yes the molar heat capacity of water vapor is about 36 J/mol.K in the vapor phase. Actually I hadn't thought about the phase of water entering the bed. If we are dealing with combustion gases then the temperature will surely be in the vapour range for water. However water won't desublimate like CO2 will at atmospheric pressure so it seems we would have an intermediate liquid phase

Heat capacity including the vapour phase for water:
Screenshot 2022-03-29 at 17.34.40.png

Chestermiller said:
We are not assuming constant molar heat capacity. The temperature dependence of the heat capacities should be included in the model. We are only assuming negligible effect of composition (i.e., changes mole fraction).

Yes, the first T should not be in there.
Hmm ok so we will set up an initial mixture of x,y,z% N2,CO2,H2O (the feed composition) and then calculate all heat capacities (as a temperature dependent property) based on this feed mixture assuming x,y,z stay the same?

Chestermiller said:
Please tell me you don't think this?
I thought it was a possibility. Mass fractions don't convert to mole fractions smoothly like this, however I thought it was reasonable to guess that the conversion would leave us with model equations that look similar to the model equations we have currently on the mass basis i.e we wouldn't be left with terms like the mW at the end
 
  • #38
casualguitar said:
Well yes the molar heat capacity of water vapor is about 36 J/mol.K in the vapor phase. Actually I hadn't thought about the phase of water entering the bed. If we are dealing with combustion gases then the temperature will surely be in the vapour range for water. However water won't desublimate like CO2 will at atmospheric pressure so it seems we would have an intermediate liquid phase
The model assumes that condensed water is all immobilized on the bed. You can see from their model plots that most of the water is released again to the gas phase before the CO2 begins accumulating on the bed (so there is not significant interaction, according to their results).
casualguitar said:
Heat capacity including the vapour phase for water:
View attachment 299093

Hmm ok so we will set up an initial mixture of x,y,z% N2,CO2,H2O (the feed composition) and then calculate all heat capacities (as a temperature dependent property) based on this feed mixture assuming x,y,z stay the same?
In terms of the heat transfer, yes.
casualguitar said:
I thought it was a possibility. Mass fractions don't convert to mole fractions smoothly like this, however I thought it was reasonable to guess that the conversion would leave us with model equations that look similar to the model equations we have currently on the mass basis i.e we wouldn't be left with terms like the mW at the end
I'm not going to convert their equations to moles and mole fractions, which would be too much of a pain in the butt. I'm just going to reformulate the equations in terms of moles and mole fractions from scratch.
 
  • #39
Chestermiller said:
The model assumes that condensed water is all immobilized on the bed. You can see from their model plots that most of the water is released again to the gas phase before the CO2 begins accumulating on the bed (so there is not significant interaction, according to their results).
CO2 will solidify at a temperature below the freezing temperature of the water so we will see solid phase water. So are we saying that liquid and solid phase water are both immobilised on the bed, and that both of these phases would be accounted for in the water mass balance?
Chestermiller said:
I'm not going to convert their equations to moles and mole fractions, which would be too much of a pain in the butt. I'm just going to reformulate the equations in terms of moles and mole fractions from scratch.
Ok sounds good. Genuinely my thoughts were given that the mass fraction terms are kg/kg (units cancel), the equations in their current format would still apply on a molar basis i.e. we can directly apply them as they are, using a mole basis. This is not the case as you have said however I don't yet see why not. Why are the equations not immediately applicable on a mole basis?
 
  • #40
casualguitar said:
CO2 will solidify at a temperature below the freezing temperature of the water so we will see solid phase water. So are we saying that liquid and solid phase water are both immobilised on the bed, and that both of these phases would be accounted for in the water mass balance?
No. See their Fig. 3.
casualguitar said:
Ok sounds good. Genuinely my thoughts were given that the mass fraction terms are kg/kg (units cancel), the equations in their current format would still apply on a molar basis i.e. we can directly apply them as they are, using a mole basis. This is not the case as you have said however I don't yet see why not. Why are the equations not immediately applicable on a mole basis?
Let's see how it plays out.

On a mole basis, the overall mass balance equation for the gas phase will read: $$\frac{\partial (\epsilon \rho_m)}{\partial t}=-\frac{\partial (\rho_mv_m)}{\partial z}-\sum_{i=1}^{n_c}{\dot{M}_i^"a_s}\tag{1}$$where ##\rho_m## is the molar density of the gas $$\rho_m=\frac{P}{RT}\tag{2}$$ ##v_m## is the superficial molar flow rate, and ##\dot{M}_i^"## is the molar deposition rate of species i per unit area of bed. Note from Eqn. 2 that the molar density of the gas (assumed ideal) does not depend on the composition of the gas (i.e., the mole fractions of the various species), but only the absolute temperature.

TO BE CONTINUED
 
  • #41
Chestermiller said:
On a mole basis, the overall mass balance equation for the gas phase will read: $$\frac{\partial (\epsilon \rho_m)}{\partial t}=-\frac{\partial (\rho_mv_m)}{\partial z}-\sum_{i=1}^{n_c}{\dot{M}_i^"a_s}\tag{1}$$where ##\rho_m## is the molar density of the gas $$\rho_m=\frac{P}{RT}\tag{2}$$ ##v_m## is the superficial molar flow rate, and ##\dot{M}_i^"## is the molar deposition rate of species i per unit area of bed. Note from Eqn. 2 that the molar density of the gas (assumed ideal) does not depend on the composition of the gas (i.e., the mole fractions of the various species), but only the absolute temperature.

TO BE CONTINUED
Followed fully. No questions on this

If I were to guess I would say that the solid phase mass balance, and the gas and solid energy balance will also keep the same format as the mass basis model equations, so maybe:
$$\frac{\partial \dot{{m}_i}}{\partial t} = {\dot{M}_i^"a_s}\tag{3}$$
$$(\epsilon_g\rho_g C_{p,g} + \rho_s(1-\epsilon_g)C_{p,s})\frac{\partial T}{\partial t} = -\rho_gv_gC_{p,g}\frac{\partial T}{\partial z} + \frac{\partial}{\partial z}(\lambda_{eff}\frac{\partial T}{\partial z}) - \sum_{i=1}^{n_c}{\dot{M}_i^"a_s}\Delta h_i\tag{4} $$

Where ##C_{p,g}## is the molar heat capacity of the gas as a function of temperature only in J/mol.K, and ##\Delta h_i## is the latent heat of vaporisation in J/mol

Edit: ##\Delta h_i## is the latent heat of sublimation or vaporisation
 
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  • #42
casualguitar said:
Followed fully. No questions on this

If I were to guess I would say that the solid phase mass balance, and the gas and solid energy balance will also keep the same format as the mass basis model equations, so maybe:
The species mass balance equation for deposition at the interface should read: $$\frac{\partial M_i}{\partial t} ={\dot{M}_i^"a_s}\tag{3}$$where ##M_i## is the number of moles of the species deposited at the interface between gas and bed.
casualguitar said:
$$(\epsilon_g\rho_m C_{p,g} + \rho_s(1-\epsilon_g)C_{p,s})\frac{\partial T}{\partial t} = -\rho_mv_mC_{p,g}\frac{\partial T}{\partial z} + \frac{\partial}{\partial z}(\lambda_{eff}\frac{\partial T}{\partial z}) - \sum_{i=1}^{n_c}{\dot{M}_i^"a_s}\Delta h_i\tag{4} $$

Where ##C_{p,g}## is the molar heat capacity of the gas as a function of temperature only in J/mol.K, and ##\Delta h_i## is the latent heat of vaporisation in J/mol
In addition, ##\rho_m## is the molar density of the gas (i.e., P/RT) and ##v_m## is the superficial molar velocity.

I will have more to say about the heat balance later, and how it should be expressed in terms of separate heat balances for the gas and bed, with separate temperatures and with heat transfer between the two (like in the tanks model). At least this is my opinion.
 
  • #43
Chestermiller said:
The species mass balance equation for deposition at the interface should read: $$\frac{\partial M_i}{\partial t} ={\dot{M}_i^"a_s}\tag{3}$$where ##M_i## is the number of moles of the species deposited at the interface between gas and bed.

In addition, ##\rho_m## is the molar density of the gas (i.e., P/RT) and ##v_m## is the superficial molar velocity.
Great, fully followed so far then
Chestermiller said:
I will have more to say about the heat balance later, and how it should be expressed in terms of separate heat balances for the gas and bed, with separate temperatures and with heat transfer between the two (like in the tanks model). At least this is my opinion.
Ok so is it reasonable to say that the main two reasons for going with this approach are:
1) It avoids using the simplification that the bed and fluid are at the same temperature
2) It 'aligns' this model with the old one i.e. they will both use a similar 'tanks' approach

Splitting the left side of the heat balance seems straightforward. The right side not so much. The split will possibly look something like this:

$$\epsilon_g\rho_g C_{p,g}\frac{\partial T_f}{\partial t} = -\rho_gv_gC_{p,g}\frac{\partial T_f}{\partial z} - \sum_{i=1}^{n_c}{\dot{M}_i^"a_s}\Delta h_i + Ua_s(T_s-T_f)\tag{5}$$

$$\rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T_s}{\partial t} = \sum_{i=1}^{n_c}{\dot{M}_i^"a_s}\Delta h_i + Ua_s(T_f-T_s)\tag{6}$$
Where U is the gas solid heat transfer coefficient in W/m2.K

My rationale for the above equations being:
1) The fluid will have an energy contribution from the solid now (the last term in the fluid heat balance)
2) The solid will not have a convection term associated with it
3) The solid will have an energy contribution from the fluid (equal and opposite to the fluid-solid energy contribution)
4) The 'phase change' term is an energy transfer between the fluid and solid so it will be present in both equations (and equal and opposite)

I did not know what to do with this term ## \frac{\partial}{\partial z}(\lambda_{eff}\frac{\partial T}{\partial z})##. Previously we lumped this term in with the fluid-solid heat transfer coefficient U, so in the interest of making these equations similar in style to the old model I did that here also
 
  • #44
Individual Species Mass Balance Equation for Gas Phase in Terms of Molar Quantities

$$\epsilon_g\rho_m\frac{\partial y_i}{\partial t}+\rho_m v_m\frac{\partial y_i}{\partial z}=\frac{\partial}{\partial z}\left(\rho_m D_{eff}\frac{\partial y_i}{\partial z}\right) -\dot{M}_i^"a_s$$$$+y_i\sum_{j=1}^{n_c}\dot{M}_j^"a_s\tag{5}$$where ##y_i## is the mole fraction of species i in the gas phase.

I'd now like to talk about how I would change the model if I were in charge (hopefully to improve it). Of course, in this project, you are in charge and the final decisions on all aspects of the work are up to you.

In the heat balance, I would have separate temperatures for the gas and the bed (##T_g## and ##T_b##) and a separate heat balance for each of them. The two heat balances would be linked by the balance for heat flows into and out of the interface: $$q_{g,I}+\sum_{j=1}^{n_c}\dot{M}_j^"\Delta h_j=q_{I,b}\tag{6a}$$where the heat fluxes from the gas to the interface ##q_{g,I}## and from the interface to the solid bed ##q_{I,b}## are given by $$q_{g,I}=U_g(T_g-T_I)\tag{6b}$$ and $$q_{I,b}=U_b(T_I-T_b)\tag{6c}$$with ##T_I## representing the temperature at the interface (deposit), and where ##U_g## and ##U_b## represent the effective heat transfer coefficients between the bulk gas and the interface, and between the interface and the solid bed, respectively. These would be determined by the heat transfer correlations for packed beds from Bird, Stewart, and Lightfoot that we are using in the tanks model.

If we solve Eqns. 6 for the temperature at the interface, we obtain: $$T_I=\frac{\sum_{j=1}^{n_c}\dot{M}_j^"\Delta h_j}{U_g+U_b}+\frac{U_gT_g+U_bT_b}{U_g+U_b}\tag{7}$$From this it follows that the heat flux from the gas to the interface is given by: $$q_{g,I}=-\frac{U_g\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}}{U_g+U_b}+U^*(T_g-T_b)\tag{8}$$ where U* is the overall heat transfer coefficient: $$\frac{1}{U^*}=\frac{1}{U_g}+\frac{1}{U_b}$$
Similarly, the heat flux from the interface to the solid bed is given by: $$q_{I,b}=+\frac{U_b\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}}{U_g+U_b}+U^*(T_g-T_b)\tag{9}$$

Thoughts?
 
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  • #45
Lots of questions on this one
Chestermiller said:
Individual Species Mass Balance Equation for Gas Phase in Terms of Molar Quantities

$$\epsilon_g\rho_m\frac{\partial y_i}{\partial t}+\rho_m v_m\frac{\partial y_i}{\partial z}=\frac{\partial}{\partial z}\left(\rho_m D_{eff}\frac{\partial y_i}{\partial z}\right) -\dot{M}_i^"a_s$$$$+y_i\sum_{j=1}^{n_c}\dot{M}_j^"a_s\tag{5}$$where ##y_i## is the mole fraction of species i in the gas phase.
Why is the last term included here? The second last term is the total moles of species i deposited on the bed. The last term seems to be the total amount of all other species deposited on the bed. But if this is the case, why would we have more than one species considered in the same mass balance? Are these not going to be considered in their own mass balances?

I notice here that the convective and diffusive terms have opposing signs. Why is this? Is this a convention?
Chestermiller said:
In the heat balance, I would have separate temperatures for the gas and the bed (##T_g## and ##T_b##) and a separate heat balance for each of them.
Agreed. This would make the formulation more similar to what we had in the previous model, and also avoids the assumption of ##T_g = T_b##
Chestermiller said:
The two heat balances would be linked by the balance for heat flows into and out of the interface: $$q_{g,I}+\sum_{j=1}^{n_c}\dot{M}_j^"\Delta h_j=q_{I,b}\tag{6a}$$where the heat fluxes from the gas to the interface ##q_{g,I}## and from the interface to the solid bed ##q_{I,b}## are given by $$q_{g,I}=U_g(T_g-T_I)\tag{6b}$$ and $$q_{I,b}=U_b(T_I-T_b)\tag{6c}$$with ##T_I## representing the temperature at the interface (deposit), and where ##U_g## and ##U_b## represent the effective heat transfer coefficients between the bulk gas and the interface, and between the interface and the solid bed, respectively.
What is the 'interface' defined as? As far as I understand its the zone around a solid particle in the bed, where the gas changes phase and is deposited as a solid. Is this correct? If so then 6a, 6b and 6c all make sense to me
Chestermiller said:
If we solve Eqns. 6 for the temperature at the interface, we obtain: $$T_I=\frac{\sum_{j=1}^{n_c}\dot{M}_j^"\Delta h_j}{U_g+U_b}+\frac{U_gT_g+U_bT_b}{U_g+U_b}\tag{7}$$From this it follows that the heat flux from the gas to the interface is given by: $$q_{g,I}=-\frac{U_g\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}}{U_g+U_b}+U^*(T_g-T_b)\tag{8}$$ where U* is the overall heat transfer coefficient: $$\frac{1}{U^*}=\frac{1}{U_g}+\frac{1}{U_b}$$
Similarly, the heat flux from the interface to the solid bed is given by: $$q_{I,b}=+\frac{U_b\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}}{U_g+U_b}+U^*(T_g-T_b)\tag{9}$$

Thoughts?
Brilliant, fully followed so far

Thoughts - so there are time and position derivatives clearly present in the species mass balance. Currently we're dealing with heat flux terms ##q_{I,b}## and ##q_{I,g}## on the heat side which are not currently time or space dependent. I'm not sure if this is a natural way to go about developing the model, but it seems to me like we should try to 'put' the time and space derivates in somehow. How would these heat flux terms be tied back into a heat balance that is time and space dependent?

Also - looking at my previous response, I took the approach of manually 'splitting' the solid and gas heat balance. Ignoring the fact that my equations are likely incorrect, is this approach also ok?

Edit: I haven't done the solution out myself for eqs 7,8,9 yet so I'll work through that this evening
 
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  • #46
casualguitar said:
Lots of questions on this one

Why is the last term included here? The second last term is the total moles of species i deposited on the bed. The last term seems to be the total amount of all other species deposited on the bed. But if this is the case, why would we have more than one species considered in the same mass balance?
We discussed this when we talked about the comparison between the divergence form of the species mass balance and this equivalent form of the species mass balance. We noted that the divergence form has only the single species deposition term. This form is obtained by multiplying the overall mass balance by the mole fraction yi, and subtracting the resulting equation from the divergence form of the mass balance. That leads to the term you are referring to. There is nothing wrong with any of these equations.
casualguitar said:
Are these not going to be considered in their own mass balances?
Yes, terms like this are included in all the species mass balances when expressed in this non-divergence form.
casualguitar said:
I notice here that the convective and diffusive terms have opposing signs. Why is this? Is this a convention?
You had no problem with this in the heat balance equation, where the thermal flux is proportional to minus the temperature gradient; why do you have a problem now where the mass flux is proportional to minus the mole fraction (concentration) gradient.
casualguitar said:
Agreed. This would make the formulation more similar to what we had in the previous model, and also avoids the assumption of ##T_g = T_b##

What is the 'interface' defined as? As far as I understand its the zone around a solid particle in the bed, where the gas changes phase and is deposited as a solid. Is this correct? If so then 6a, 6b and 6c all make sense to me
I envision it the same as if the solid and liquid deposition is an extremely thin layer at the solid bed surface, and is the same as if it were just absorbed (vanished) at the bed surface; so its only effect is the release of heat at the interface between the gas and the solid bed.
casualguitar said:
Brilliant, fully followed so far

Thoughts - so there are time and position derivatives clearly present in the species mass balance. Currently we're dealing with heat flux terms ##q_{I,b}## and ##q_{I,g}## on the heat side which are not currently time or space dependent. I'm not sure if this is a natural way to go about developing the model, but it seems to me like we should try to 'put' the time and space derivates in somehow. How would these heat flux terms be tied back into a heat balance that is time and space dependent?
There are no space and time derivatives because the amount of mass involved is negligible. You don't seem to have this same problem in the formulation of the other model when we express the heat flux between the air and the bed as ##U(T_a-T_s)##. The equations I've written here are not the full heat balances. They just give the heat fluxes to the gas and the bed. They must be combined with the accumulation and dispersion terms to give the full heat balances for the gas and the bed. This is what I'll write down next.
casualguitar said:
Also - looking at my previous response, I took the approach of manually 'splitting' the solid and gas heat balance. Ignoring the fact that my equations are likely incorrect, is this approach also ok?
I don't know what you mean.
 
  • #47
Chestermiller said:
We discussed this when we talked about the comparison between the divergence form of the species mass balance and this equivalent form of the species mass balance. We noted that the divergence form has only the single species deposition term. This form is obtained by multiplying the overall mass balance by the mole fraction yi, and subtracting the resulting equation from the divergence form of the mass balance. That leads to the term you are referring to. There is nothing wrong with any of these equations.
My apologies for all the questions. I decided its surely better to start asking those questions with seemingly obvious answers as soon as I see them, to avoid having them compound down the line (I didn't do this for model 1).

I think the current confusion comes down to two things which are the terminology used to reference equations and the rationale for manipulating equations in some cases.

Terminology - For me when we say 'individual component mass balance' (as used in post 27) means the same thing as the 'species mass balance'. Is this correct?

Rationale - so the overall mass balance tracks the total amount of gas, and the total amount of solid. We took this overall mass balance and multiplied it by ##y_i##. Then we took the divergence form of the mass balance, which only deals with a single component, and subtracted the overall mass balance from this, which leads to equation 5. Two questions I think will fully clear this up:
1) What is the advantage of subtracting one from the other? (Or why do this?)
2) Does that last term in equation 5 have any physical meaning, or does it just arise as a result of doing the above?

Chestermiller said:
You had no problem with this in the heat balance equation, where the thermal flux is proportional to minus the temperature gradient; why do you have a problem now where the mass flux is proportional to minus the mole fraction (concentration) gradient.
Ah ok. I was trying to visualise an element where you have an accumulation of mass over time (term 1 in eq.5), mass entering due to convection (term 2), mass entering due to diffusion (term 3), and mass leaving the gas phase due to deposition (term 4). I can absolutely see that mass flux would be proportional to minus the concentration gradient. Does this also mean that the convective term is proportional to minus the diffusion term?

Chestermiller said:
I envision it the same as if the solid and liquid deposition is an extremely thin layer at the solid bed surface, and is the same as if it were just absorbed (vanished) at the bed surface; so its only effect is the release of heat at the interface between the gas and the solid bed.
Got it

Chestermiller said:
There are no space and time derivatives because the amount of mass involved is negligible. You don't seem to have this same problem in the formulation of the other model when we express the heat flux between the air and the bed as ##U(T_a-T_s)##. The equations I've written here are not the full heat balances. They just give the heat fluxes to the gas and the bed. They must be combined with the accumulation and dispersion terms to give the full heat balances for the gas and the bed. This is what I'll write down next.
Yes I didn't have this same problem because we had a time derivative on the LHS and also had used the method of lines for the spatial domain so in effect we did have position and time built in.
Ah understood, yes when they heat fluxes are combined with the accumulation/dispersion term I think this will sit with me a lot better.

Actually also I don't have your understanding of the road ahead i.e. where the model is going to next (in this case finishing off the heat balance seems to be the next step but I would not have known this myself). I think having this vision or 'map' of the road ahead in a sense would make each step feel a lot more natural as I would know why its happening. So because I don't have this (I think this will come with time!) I am much better able to connect the dots of the model in reverse i.e. I can see why you make a certain step after the next step is made in the model.

So yes I think finishing out the heat balance out will possibly clear up the rationale behind the steps we have taken so far leading up to it, if this makes sense
 
  • #48
casualguitar said:
My apologies for all the questions. I decided its surely better to start asking those questions with seemingly obvious answers as soon as I see them, to avoid having them compound down the line (I didn't do this for model 1).

I think the current confusion comes down to two things which are the terminology used to reference equations and the rationale for manipulating equations in some cases.

Terminology - For me when we say 'individual component mass balance' (as used in post 27) means the same thing as the 'species mass balance'. Is this correct?
yes
casualguitar said:
Rationale - so the overall mass balance tracks the total amount of gas, and the total amount of solid. We took this overall mass balance and multiplied it by ##y_i##. Then we took the divergence form of the mass balance, which only deals with a single component, and subtracted the overall mass balance from this, which leads to equation 5. Two questions I think will fully clear this up:
1) What is the advantage of subtracting one from the other? (Or why do this?)
Because the original derivation of the species mass balance resulted in the divergence form of the equation which did not give the partial derivative of the mole fraction with respect to time directly. By manipulating to the non-divergence form, we get the partial derivative of the mole fraction with respect to time directy.
casualguitar said:
2) Does that last term in equation 5 have any physical meaning, or does it just arise as a result of doing the above?
In my judgment, the latter.
casualguitar said:
Ah ok. I was trying to visualise an element where you have an accumulation of mass over time (term 1 in eq.5), mass entering due to convection (term 2), mass entering due to diffusion (term 3), and mass leaving the gas phase due to deposition (term 4). I can absolutely see that mass flux would be proportional to minus the concentration gradient. Does this also mean that the convective term is proportional to minus the diffusion term?
No. You can see that there are other terms in the equation as well.
casualguitar said:
Got itYes I didn't have this same problem because we had a time derivative on the LHS and also had used the method of lines for the spatial domain so in effect we did have position and time built in.
Ah understood, yes when they heat fluxes are combined with the accumulation/dispersion term I think this will sit with me a lot better.

Actually also I don't have your understanding of the road ahead i.e. where the model is going to next (in this case finishing off the heat balance seems to be the next step but I would not have known this myself). I think having this vision or 'map' of the road ahead in a sense would make each step feel a lot more natural as I would know why its happening. So because I don't have this (I think this will come with time!) I am much better able to connect the dots of the model in reverse i.e. I can see why you make a certain step after the next step is made in the model.

So yes I think finishing out the heat balance out will possibly clear up the rationale behind the steps we have taken so far leading up to it, if this makes sense
Yes. I will try to do that tomorrow. There are several additional improvements to what they did that I have in mind.
 
  • #49
Chestermiller said:
yes

Because the original derivation of the species mass balance resulted in the divergence form of the equation which did not give the partial derivative of the mole fraction with respect to time directly. By manipulating to the non-divergence form, we get the partial derivative of the mole fraction with respect to time directy.
Great. Completely understood. No further questions

Chestermiller said:
No. You can see that there are other terms in the equation as well.
Whoops yes 'proportional' wasn't the right word. What I meant was, does the convective term increasing strictly mean that the diffusive term decreases?
Chestermiller said:
Yes. I will try to do that tomorrow. There are several additional improvements to what they did that I have in mind.
Yes you mentioned there is an overwhelming advantage to working on a mole basis? Why are we converting to a mole basis?
 
  • #50
casualguitar said:
Great. Completely understood. No further questionsWhoops yes 'proportional' wasn't the right word. What I meant was, does the convective term increasing strictly mean that the diffusive term decreases?
No. Why would you think this?
casualguitar said:
Yes you mentioned there is an overwhelming advantage to working on a mole basis? Why are we converting to a mole basis?
How do the differences species molar heat capacities compare with the differences in species mass heat capacities? How does the variation of overall molar heat capacity with species concentrations compare the the overall mass heat capacity with species concentrations? How does the variation of molar density with species concentrations compare with the variation of mass density with species concentrations?
 
  • #51
Chestermiller said:
How do the differences species molar heat capacities compare with the differences in species mass heat capacities?
Going from molar (mol) to mass (kg) the relationship would be:
$$C_{p, mol} * \frac{1000}{mW} = C_{p, kg}$$
Chestermiller said:
How does the variation of overall molar heat capacity with species concentrations compare the the overall mass heat capacity with species concentrations?
This is the same except instead we would have the mole fraction weighted average molecular weight. So they are related by ##\frac{1000}{MW_{avg}}##

So the conversion between kg and moles seems to be straightforward. I don't know for sure but I guess the advantage is to do with the calculation of mixture or average properties being more convenient on a mole basis, which is the case i.e. mixture heat capacity, density, etc?
 
  • #52
casualguitar said:
Going from molar (mol) to mass (kg) the relationship would be:
$$C_{p, mol} * \frac{1000}{mW} = C_{p, kg}$$
So, if the molar heat capacities are nearly equal, and the molecular weights are very different, what does this tell you about whether you need to more seriously consider the differences in the mass heat capacities?
casualguitar said:
This is the same except instead we would have the mole fraction weighted average molecular weight. So they are related by ##\frac{1000}{MW_{avg}}##
The consequence is that the mole fraction weighted overall heat capacity varies with species concentration much less than the mass fraction weighted overall heat capacity.
casualguitar said:
So the conversion between kg and moles seems to be straightforward. I don't know for sure but I guess the advantage is to do with the calculation of mixture or average properties being more convenient on a mole basis, which is the case i.e. mixture heat capacity, density, etc?
What is the effect of species concentrations on the molar density vs the mass density? And consider the mass transfer rates between the gas and the solid which are driven by the total pressure times mole fraction rather than times mass fraction.
 
  • #53
Separate Heat Balances for Gas and Bed:

$$\epsilon_g \rho_mC_{p,g,m}\frac{\partial T_g}{\partial t}=-\phi_m C_{p,g,m}\frac{\partial T_g}{\partial z}+\frac{\partial}{\partial z}\left(k_{eff}\frac{\partial T_g}{\partial z}\right)-q_{g,I}a_s $$
and
$$\rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T_b}{dt}=q_{I,b}a_s$$where ##\rho_m## is the molar density of the gas (##=P/RT_g##), ##C_{p,g,m}## is the. molar heat capacity of the gas, ##\phi_m## is the superficial molar flux of the gas (##=\rho_mv_m##), ##k_{eff}## is the "effective" axial conductivity }(including dispersion) in the gas, and where ##q_{g,I}## and ##q_{I,b}## are the heat fluxes between the gas and the deposit interface and between the deposit interface and the solid bed, respectively as derived in post #44:

$$q_{g,I}=-\frac{U_gq^*}{U_g+U_b}+U^*(T_g-T_b)\tag{8}$$ and $$q_{I,b}=+\frac{U_bq^*}{U_g+U_b}+U^*(T_g-T_b)\tag{9}$$with $$q^*=\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}$$ and where U* is the overall heat transfer coefficient: $$\frac{1}{U^*}=\frac{1}{U_g}+\frac{1}{U_b}$$
If we add the two separate heat balances together and set ##T_g=T_b=T## (assuming extremely high heat transfer coefficients), we end up with the lumped heat balance in the Tuinier paper.

OK so far? (There are additional steps coming)
 
  • #54
Chestermiller said:
The consequence is that the mole fraction weighted overall heat capacity varies with species concentration much less than the mass fraction weighted overall heat capacity.
So the assumption of properties (heat capacity and density) being independent of composition is more accurate when using a molar basis (when compared to mass basis), due to the lower variation w.r.t composition?
Chestermiller said:
Separate Heat Balances for Gas and Bed:

$$\epsilon_g \rho_mC_{p,g,m}\frac{\partial T_g}{\partial t}=-\phi_m C_{p,g,m}\frac{\partial T_g}{\partial z}+\frac{\partial}{\partial z}\left(k_{eff}\frac{\partial T_g}{\partial z}\right)-q_{g,I}a_s $$
and
$$\rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T_b}{dt}=q_{I,b}a_s$$where ##\rho_m## is the molar density of the gas (##=P/RT_g##), ##C_{p,g,m}## is the. molar heat capacity of the gas, ##\phi_m## is the superficial molar flux of the gas (##=\rho_mv_m##), ##k_{eff}## is the "effective" axial conductivity }(including dispersion) in the gas, and where ##q_{g,I}## and ##q_{I,b}## are the heat fluxes between the gas and the deposit interface and between the deposit interface and the solid bed, respectively as derived in post #44:

$$q_{g,I}=-\frac{U_gq^*}{U_g+U_b}+U^*(T_g-T_b)\tag{8}$$ and $$q_{I,b}=+\frac{U_bq^*}{U_g+U_b}+U^*(T_g-T_b)\tag{9}$$with $$q^*=\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}$$ and where U* is the overall heat transfer coefficient: $$\frac{1}{U^*}=\frac{1}{U_g}+\frac{1}{U_b}$$
If we add the two separate heat balances together and set ##T_g=T_b=T## (assuming extremely high heat transfer coefficients), we end up with the lumped heat balance in the Tuinier paper.

OK so far? (There are additional steps coming)
Yep makes sense. No questions on this bit. The heat balances are starting to look very similar to the heat balances from model 1.

Just one possible sign error:
IMG_1821.JPG

After checking that this model is the same as the Tuinier model once ##T_g=T_b=T##, I found that the solid deposition term was positive (rather than negative as is it is in the Tuinier model). Is there an error in what I have done, or is it possible that the definition of q* should be a negative term i.e. $$q^*=-\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}$$

Besides this I am very much ok so far on this

Edit: Also, the next step I would guess is to discretise the spatial domain using the method of lines?
 
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  • #55
casualguitar said:
So the assumption of properties (heat capacity and density) being independent of composition is more accurate when using a molar basis (when compared to mass basis), due to the lower variation w.r.t composition?

Yep makes sense. No questions on this bit. The heat balances are starting to look very similar to the heat balances from model 1.

Just one possible sign error:
View attachment 299374
After checking that this model is the same as the Tuinier model once ##T_g=T_b=T##, I found that the solid deposition term was positive (rather than negative as is it is in the Tuinier model). Is there an error in what I have done, or is it possible that the definition of q* should be a negative term i.e. $$q^*=-\sum_{j=1}^{n_c}{\dot{M}_j^"\Delta h_j}$$
It's got to be a typo in the Tuinier paper. If mass is building up at the interface, then heat is being released.
 
  • #56
Chestermiller said:
It's got to be a typo in the Tuinier paper. If mass is building up at the interface, then heat is being released.
After looking at model 1 (post #104), it looks like this model is at about the right stage to implement a finite difference scheme. I don't see a problem with setting up the grid similar to last time. This is your comment about that from the previous post:
Chestermiller said:
The finite difference scheme I am recommending employs a finite difference grid with spacing Δx, grid cell boundaries at (x=0, Δx, 2Δx...), and grid cell centers at (x=Δx/2, 3Δx/2, 5Δx/2, ...). The mass flux into the bed ϕ0 is known at the cell boundary x = 0, and we calculate the mass flux at all the other cell boundaries. On the other hand, we calculate the fluid temperatures, enthalpies, and densities at the cell centers.

Based on this description, and employing the method of lines, the 2nd order accurate finite difference representations of the mass balance and heat balance equations for the fluid are expressed as:
However one difference is that we now have the ##k_{eff}## term rather than the dispersion coefficient ##D## as we had in model 1, which was approximated to ##ul##, and ##l## was then further approximated to ##l=\frac{\Delta x}{2}##. You did mention that the ##k_{eff}## term will include dispersion. Was this a suggestion that we will go the same route as model 1 here and approximate using the dispersion parameter? If so I can attempt the FD scheme
 
  • #57
casualguitar said:
After looking at model 1 (post #104), it looks like this model is at about the right stage to implement a finite difference scheme. I don't see a problem with setting up the grid similar to last time. This is your comment about that from the previous post:

However one difference is that we now have the ##k_{eff}## term rather than the dispersion coefficient ##D## as we had in model 1, which was approximated to ##ul##, and ##l## was then further approximated to ##l=\frac{\Delta x}{2}##. You did mention that the ##k_{eff}## term will include dispersion. Was this a suggestion that we will go the same route as model 1 here and approximate using the dispersion parameter? If so I can attempt the FD scheme
Yes, this is exactly what I had in mind. Let's see what you come up with. I have worked this out already.
 
  • #58
Chestermiller said:
Yes, this is exactly what I had in mind. Let's see what you come up with. I have worked this out already.
Just to confirm before I give this a go, we're dealing with four core model equations

##\textbf{Individual species mass balance for gas phase in terms of molar quantities}:##
$$\epsilon_g\rho_m\frac{\partial y_i}{\partial t}+\rho_m v_m\frac{\partial y_i}{\partial z}=\frac{\partial}{\partial z}\left(\rho_m D_{eff}\frac{\partial y_i}{\partial z}\right) -\dot{M}_i^"a_s\tag{5}$$

##\textbf{The species mass balance for deposition at the interface:}##
$$\frac{\partial M_i}{\partial t} ={\dot{M}_i^"a_s}\tag{3}$$

##\textbf{Individual species heat balance for gas phase in terms of molar quantities:}##
$$\epsilon_g \rho_mC_{p,g,m}\frac{\partial T_g}{\partial t}=-\phi_m C_{p,g,m}\frac{\partial T_g}{\partial z}+\frac{\partial}{\partial z}\left(k_{eff}\frac{\partial T_g}{\partial z}\right)-q_{g,I}a_s\tag{10}$$

##\textbf{Individual species heat balance for bed:}##
$$\rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T_b}{dt}=q_{I,b}a_s\tag{11}$$

The heat flux term will also be involved in the differencing scheme but besides that, these are the 4 core model equations? That gives us a mass and heat balance for both the gas and the bed
 
  • #59
casualguitar said:
Just to confirm before I give this a go, we're dealing with four core model equations

##\textbf{Individual species mass balance for gas phase in terms of molar quantities}:##
$$\epsilon_g\rho_m\frac{\partial y_i}{\partial t}+\rho_m v_m\frac{\partial y_i}{\partial z}=\frac{\partial}{\partial z}\left(\rho_m D_{eff}\frac{\partial y_i}{\partial z}\right) -\dot{M}_i^"a_s\tag{5}$$

##\textbf{The species mass balance for deposition at the interface:}##
$$\frac{\partial M_i}{\partial t} ={\dot{M}_i^"a_s}\tag{3}$$

##\textbf{Individual species heat balance for gas phase in terms of molar quantities:}##
$$\epsilon_g \rho_mC_{p,g,m}\frac{\partial T_g}{\partial t}=-\phi_m C_{p,g,m}\frac{\partial T_g}{\partial z}+\frac{\partial}{\partial z}\left(k_{eff}\frac{\partial T_g}{\partial z}\right)-q_{g,I}a_s\tag{10}$$

##\textbf{Individual species heat balance for bed:}##
$$\rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T_b}{dt}=q_{I,b}a_s\tag{11}$$

The heat flux term will also be involved in the differencing scheme but besides that, these are the 4 core model equations? That gives us a mass and heat balance for both the gas and the bed
In Eqn. 10, I would rewrite the dispersion terms as
$$\frac{\partial}{\partial z}\left(k_{eff}\frac{\partial T_g}{\partial z}\right)=\frac{\partial}{\partial z}\left(\rho_mC_{p,g,m}D_{eff,T}\frac{\partial T_g}{\partial z}\right)$$$$=\frac{\partial}{\partial z}\left(\rho_mC_{p,g,m}(v_ml_T)\frac{\partial T_g}{\partial z}\right)=\frac{\partial}{\partial z}\left(l_T\ \phi_m \ C_{p,g,m}\frac{\partial T_g}{\partial z}\right)$$
where ##l_T## is the thermal dispersion length. I would do something similar for the dispersion term in the species mass balance equation.

An equation that is missing from this list is the overall mass (molar) balance equation on the gas.
 
  • #60
casualguitar said:
Just to confirm before I give this a go, we're dealing with four core model equations

##\textbf{Individual species mass balance for gas phase in terms of molar quantities}:##
$$\epsilon_g\rho_m\frac{\partial y_i}{\partial t}+\rho_m v_m\frac{\partial y_i}{\partial z}=\frac{\partial}{\partial z}\left(\rho_m D_{eff}\frac{\partial y_i}{\partial z}\right) -\dot{M}_i^"a_s\tag{5}$$

##\textbf{The species mass balance for deposition at the interface:}##
$$\frac{\partial M_i}{\partial t} ={\dot{M}_i^"a_s}\tag{3}$$

##\textbf{Individual species heat balance for gas phase in terms of molar quantities:}##
$$\epsilon_g \rho_mC_{p,g,m}\frac{\partial T_g}{\partial t}=-\phi_m C_{p,g,m}\frac{\partial T_g}{\partial z}+\frac{\partial}{\partial z}\left(k_{eff}\frac{\partial T_g}{\partial z}\right)-q_{g,I}a_s\tag{10}$$

##\textbf{Individual species heat balance for bed:}##
$$\rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T_b}{dt}=q_{I,b}a_s\tag{11}$$

The heat flux term will also be involved in the differencing scheme but besides that, these are the 4 core model equations? That gives us a mass and heat balance for both the gas and the bed
##\textbf{The overall mass balance for the gas phase:}##
$$\frac{\partial (\epsilon \rho_m)}{\partial t}=-\frac{\partial (\rho_mv_m)}{\partial z}-\sum_{i=1}^{n_c}{\dot{M}_i^"a_s}\tag{1}$$
The rewriting of equation 10 makes sense to me. Just one question on the equations we have. If we have the solid and gas species mass balances, what additional information does having the overall mass balance for the gas phase give us i.e. why would we need this also?

Chestermiller said:
I would do something similar for the dispersion term in the species mass balance equation.
Can we reasonably let ##D_{eff} = v_mL_T## in the mass balance equation also? Effectively saying that ##D_{eff}## has the same value in the mass and heat balances
 
  • #61
casualguitar said:
##\textbf{The overall mass balance for the gas phase:}##
$$\frac{\partial (\epsilon \rho_m)}{\partial t}=-\frac{\partial (\rho_mv_m)}{\partial z}-\sum_{i=1}^{n_c}{\dot{M}_i^"a_s}\tag{1}$$
The rewriting of equation 10 makes sense to me. Just one question on the equations we have. If we have the solid and gas species mass balances, what additional information does having the overall mass balance for the gas phase give us i.e. why would we need this also?
We need to use the overall mass balance to get the values of ##\phi_m=\rho_mv_m## at the cell boundaries (for use in the other balance equations). This is what we do in model 1.
casualguitar said:
Can we reasonably let ##D_{eff} = v_mL_T## in the mass balance equation also? Effectively saying that ##D_{eff}## has the same value in the mass and heat balances
I would be inclined to say that the two l's are the same because the dispersion is essentially mechanical (dominated by axial mixing). You could use different values if you desire, but, if they are the same and equal to ##\Delta x/2##, the resulting simplification is very attractive.

I'm interested in seeing how you discretize the equations with respect to z.
 
  • #62
The finite difference scheme employs a finite difference grid with spacing ##\Delta x##, grid cell boundaries at ##(z=0,\ \Delta z,\ 2\Delta z...)##, and grid cell centers at ##(z=\Delta z/2,\ 3\Delta z/2,\ 5\Delta z/2,\ ...)##. The mass flux into the bed ##\phi_0## is known at the cell boundary z = 0, and we calculate the mass flux at all the other cell boundaries. On the other hand, we calculate the fluid temperatures, enthalpies, and densities at the cell centers.

These are the individual species mass and heat balances, and the overall mass balance to the gas phase, the equations with spatial derivatives. Its just the first step though (haven't subbed out ##l## yet). All ok so far?
IMG_1833.JPG

IMG_1834.JPG


If these three equations look ok I'll let ##l = \Delta z/2## in the morning. Quite a long train journey ahead tomorrow, so plenty time for some calculations

Note: I have written x in the FD scheme. I should have used z. I will rewrite this in the next iteration
 
  • #63
casualguitar said:
The finite difference scheme employs a finite difference grid with spacing ##\Delta x##, grid cell boundaries at ##(z=0,\ \Delta z,\ 2\Delta z...)##, and grid cell centers at ##(z=\Delta z/2,\ 3\Delta z/2,\ 5\Delta z/2,\ ...)##. The mass flux into the bed ##\phi_0## is known at the cell boundary z = 0, and we calculate the mass flux at all the other cell boundaries. On the other hand, we calculate the fluid temperatures, enthalpies, and densities at the cell centers.

These are the individual species mass and heat balances, and the overall mass balance to the gas phase, the equations with spatial derivatives. Its just the first step though (haven't subbed out ##l## yet). All ok so far?
View attachment 299408
View attachment 299409

If these three equations look ok I'll let ##l = \Delta z/2## in the morning. Quite a long train journey ahead tomorrow, so plenty time for some calculations

Note: I have written x in the FD scheme. I should have used z. I will rewrite this in the next iteration
I can't read your handwritten version. Please just present the results using LaTex. Thanx.
 
  • #64
Chestermiller said:
I can't read your handwritten version. Please just present the results using LaTex. Thanx.
Here are the three equations with spatial derivatives (I have not subbed ##l## out yet or simplified, this is just the first run through).

Note: Given that these all deal with species i only I have left the i subscript out of this version for brevity

Individual species mass balance:
##\epsilon_g\rho_m\frac{\partial y_i}{\partial t} + \phi_m\frac{y_z - y_{z-\Delta z}}{2\Delta z} = +l\frac{\phi_{z+\Delta z/2}(y_{z+\Delta z}-y_z)-\phi_{z-\Delta z/2}(y_z-y_{z-\Delta z})}{(\Delta z)^2} - \dot{M}_z^"a_s\tag{1}##

Individual species heat balance:
##\epsilon_g\rho_mc_p\frac{\partial T_g}{\partial t} = - \phi_mc_p\frac{T_{g,z} - T_{g,z-\Delta z}}{2\Delta z} +lc_p\frac{\phi_{z+\Delta z/2}(T_{g,z+\Delta z}-y_z)-\phi_{z-\Delta z/2}(T_{g,z}-T_{g,z-\Delta z})}{(\Delta z)^2} - \dot{M}_z^"a_s - q_{g,I,z}a_s\tag{2}##

Individual species heat balance:
##\epsilon_g\frac{\partial \rho_m}{\partial t} = \frac{\phi_{z-\Delta z/2} - \phi_{z+\Delta z/2}}{\Delta z} -\sum_{i=1}^{n_c}{\dot{M}_z^"a_s}\tag{1}##

If these look alright to you I'll sub in for ##l## and simplify
 
  • #65
casualguitar said:
Here are the three equations with spatial derivatives (I have not subbed ##l## out yet or simplified, this is just the first run through).

Note: Given that these all deal with species i only I have left the i subscript out of this version for brevity

Individual species mass balance:
##\epsilon_g\rho_m\frac{\partial y_i}{\partial t} + \phi_m\frac{y_z - y_{z-\Delta z}}{2\Delta z} = +l\frac{\phi_{z+\Delta z/2}(y_{z+\Delta z}-y_z)-\phi_{z-\Delta z/2}(y_z-y_{z-\Delta z})}{(\Delta z)^2} - \dot{M}_z^"a_s\tag{1}##

Individual species heat balance:
##\epsilon_g\rho_mc_p\frac{\partial T_g}{\partial t} = - \phi_mc_p\frac{T_{g,z} - T_{g,z-\Delta z}}{2\Delta z} +lc_p\frac{\phi_{z+\Delta z/2}(T_{g,z+\Delta z}-y_z)-\phi_{z-\Delta z/2}(T_{g,z}-T_{g,z-\Delta z})}{(\Delta z)^2} - \dot{M}_z^"a_s - q_{g,I,z}a_s\tag{2}##

Individual species heat balance:
##\epsilon_g\frac{\partial \rho_m}{\partial t} = \frac{\phi_{z-\Delta z/2} - \phi_{z+\Delta z/2}}{\Delta z} -\sum_{i=1}^{n_c}{\dot{M}_z^"a_s}\tag{1}##

If these look alright to you I'll sub in for ##l## and simplify
For these terms, I think you meant to write the following 2nd order approximations:
$$\phi \frac{\partial y}{\partial z}=\phi_z\left[\frac{y_{z+\Delta z}-y_{z-\Delta z}}{2\Delta z}\right]$$
$$\phi C_p \frac{\partial T}{\partial z}=\phi_zC_{p,z}\left[\frac{T_{z+\Delta z}-T_{z-\Delta z}}{2\Delta z}\right]$$
The 2nd order finite difference approximations I am recommending lead to much more attractive and compelling results in the final finite difference equations:
$$\phi \frac{\partial y}{\partial z}=\frac{1}{2}\left[\phi_{z+\Delta z/2}\left(\frac{y_{z+\Delta z}-y_z}{\Delta z}\right)+\phi_{z-\Delta z/2}\left(\frac{y_z-y_{z-\Delta z}}{\Delta z}\right)\right]$$

$$\phi C_p \frac{\partial T}{\partial z}=\frac{1}{2}\left[\phi_{z+\Delta z/2}C_{p,z+\Delta z/2}\left(\frac{T_{z+\Delta z}-T_z}{\Delta z}\right)+\phi_{z-\Delta z/2}C_{p,z-\Delta z/2}\left(\frac{T_z-T_{z-\Delta z}}{\Delta z}\right)\right]$$

$$l\frac{\partial}{\partial z}\left(\phi\frac{\partial y}{\partial z}\right)=\frac{l}{\Delta z}\left[\phi_{z+\Delta z/2}\left(\frac{y_{z+\Delta z}-y_z}{\Delta z}\right)-\phi_{z-\Delta z/2}\left(\frac{y_z-y_{z-\Delta z}}{\Delta z}\right)\right]$$

$$l\frac{\partial}{\partial z}\left(\phi C_p\frac{\partial T}{\partial z}\right)=\frac{l}{\Delta z}\left[\phi_{z+\Delta z/2}C_{p,z+\Delta z/2}\left(\frac{T_{z+\Delta z}-T_z}{\Delta z}\right)-\phi_{z-\Delta z/2}C_{p,z-\Delta z/2}\left(\frac{T_z-T_{z-\Delta z}}{\Delta z}\right)\right]$$
If we are considering the temperature dependence of heat capacity in the analysis,, then ##C_{p,z+\Delta z/2}## refers to the heat capacity at ##(T_z+T_{z+\Delta z})/2## and ##C_{p,z-\Delta z/2}## refers to the heat capacity at ##(T_z+T_{z-\Delta z})/2##.
 
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  • #66
Chestermiller said:
For these terms, I think you meant to write the following 2nd order approximations:
$$\phi \frac{\partial y}{\partial z}=\phi_z\left[\frac{y_{z+\Delta z}-y_{z-\Delta z}}{2\Delta z}\right]$$
$$\phi C_p \frac{\partial T}{\partial z}=\phi_zC_{p,z}\left[\frac{T_{z+\Delta z}-T_{z-\Delta z}}{2\Delta z}\right]$$
The 2nd order finite difference approximations I am recommending lead to much more attractive and compelling results in the final finite difference equations:
$$\phi \frac{\partial y}{\partial z}=\frac{1}{2}\left[\phi_{z+\Delta z/2}\left(\frac{y_{z+\Delta z}-y_z}{\Delta z}\right)+\phi_{z-\Delta z/2}\left(\frac{y_z-y_{z-\Delta z}}{\Delta z}\right)\right]$$

$$\phi C_p \frac{\partial T}{\partial z}=\frac{1}{2}\left[\phi_{z+\Delta z/2}C_{p,z+\Delta z/2}\left(\frac{T_{z+\Delta z}-T_z}{\Delta z}\right)+\phi_{z-\Delta z/2}C_{p,z-\Delta z/2}\left(\frac{T_z-T_{z-\Delta z}}{\Delta z}\right)\right]$$

$$l\frac{\partial}{\partial z}\left(\phi\frac{\partial y}{\partial z}\right)=\frac{l}{\Delta z}\left[\phi_{z+\Delta z/2}\left(\frac{y_{z+\Delta z}-y_z}{\Delta z}\right)-\phi_{z-\Delta z/2}\left(\frac{y_z-y_{z-\Delta z}}{\Delta z}\right)\right]$$

$$l\frac{\partial}{\partial z}\left(\phi C_p\frac{\partial T}{\partial z}\right)=\frac{l}{\Delta z}\left[\phi_{z+\Delta z/2}C_{p,z+\Delta z/2}\left(\frac{T_{z+\Delta z}-T_z}{\Delta z}\right)-\phi_{z-\Delta z/2}C_{p,z-\Delta z/2}\left(\frac{T_z-T_{z-\Delta z}}{\Delta z}\right)\right]$$
If we are considering the temperature dependence of heat capacity in the analysis,, then ##C_{p,z+\Delta z/2}## refers to the heat capacity at ##(T_z+T_{z+\Delta z})/2## and ##C_{p,z-\Delta z/2}## refers to the heat capacity at ##(T_z+T_{z-\Delta z})/2##.
Whoops yes that is what I meant

For the 2nd order finite difference approximations you are recommending, why choose these over the first ones? Is this an experience related thing i.e. you know in advance that these will simplify down to favourable results when compared to the first equations?

I'll do the ##l = \frac{\Delta z}{2}## algebra this evening (up since 4.30 am though so there is a chance I'll do this first thing tomorrow instead). Actually today I was building some of the physical system that will be used to run air liquefaction/CO2 solidification experiments
 
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  • #67
casualguitar said:
Whoops yes that is what I meant

For the 2nd order finite difference approximations you are recommending, why choose these over the first ones? Is this an experience related thing i.e. you know in advance that these will simplify down to favourable results when compared to the first equations?
No. I agonized over this for several days. I tried several alternatives, and this one simplified down to a form almost identical to model 1 tanks formulation.
 
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  • #68
In this case (and for model 1) it seems that leaving ##l## equal to ##\Delta z/a## where a is the denominator of the factored out fraction will lead to a lot of cancelling out ,and in this case only leaving us with upwind parameters of the one we're solving for

Letting ##l = \frac{\Delta z}{2}## and using some placeholder variables to simplify the algebra
$$a = \phi_{z+\Delta z/2}(\frac{y_{z+\Delta z}-y_z}{\Delta z})$$
$$b = \phi_{z-\Delta z/2}(\frac{y_z-y_{z-\Delta z}}{\Delta z})$$
$$c = \phi_{z+\Delta z/2}C_{p,z+\Delta z/2}(\frac{T_{z+\Delta z}-T_z}{\Delta z})$$
$$d = \phi_{z-\Delta z/2}C_{p,z-\Delta z/2}(\frac{T_z-T_{z-\Delta z}}{\Delta z})$$

Looking at those terms, we would want ##a## and ##c## to 'cancel out' because they are the terms with downwind parameters

Individual species gas phase mass balance:
##\epsilon_g\rho_m\frac{\partial y_i}{\partial t} = - b -\dot{M}_i^"a_s##
##\epsilon_g\rho_m\frac{\partial y_i}{\partial t} = - \phi_{z-\Delta z/2}(\frac{y_z-y_{z-\Delta z}}{\Delta z}) -\dot{M}_{i,z}^"a_s##

Individual species gas phase heat balance:
##\epsilon_g \rho_mC_{p,g,m}\frac{\partial T_g}{\partial t}= -d -q_{g,I}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-T_{z-\Delta z}}{\Delta z}) -q_{g,I,z}a_s##

The last term to discretise spatially would be the overall mass balance for the gas phase. This is the current discretised equation:
##\epsilon\frac{\partial \rho_m}{\partial t}=\frac{\phi_{z-\Delta z/2} - \phi_{z+\Delta z/2}}{\Delta z}-\sum_{i=1}^{n_c}{\dot{M}_{i,z}^"a_s}##

This is the only equation with a 'downwind' parameter ##\phi_{z+\Delta z/2}##. I think we can make use of this parameter to calculate the flow out of a bed (##m_j## in the previous model)? After checking the previous model yes this is what you did there

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
 
  • #69
casualguitar said:
In this case (and for model 1) it seems that leaving ##l## equal to ##\Delta z/a## where a is the denominator of the factored out fraction will lead to a lot of cancelling out ,and in this case only leaving us with upwind parameters of the one we're solving for

Letting ##l = \frac{\Delta z}{2}## and using some placeholder variables to simplify the algebra
$$a = \phi_{z+\Delta z/2}(\frac{y_{z+\Delta z}-y_z}{\Delta z})$$
$$b = \phi_{z-\Delta z/2}(\frac{y_z-y_{z-\Delta z}}{\Delta z})$$
$$c = \phi_{z+\Delta z/2}C_{p,z+\Delta z/2}(\frac{T_{z+\Delta z}-T_z}{\Delta z})$$
$$d = \phi_{z-\Delta z/2}C_{p,z-\Delta z/2}(\frac{T_z-T_{z-\Delta z}}{\Delta z})$$

Looking at those terms, we would want ##a## and ##c## to 'cancel out' because they are the terms with downwind parameters

Individual species gas phase mass balance:
##\epsilon_g\rho_m\frac{\partial y_i}{\partial t} = - b -\dot{M}_i^"a_s##
##\epsilon_g\rho_m\frac{\partial y_i}{\partial t} = - \phi_{z-\Delta z/2}(\frac{y_z-y_{z-\Delta z}}{\Delta z}) -\dot{M}_{i,z}^"a_s##

Individual species gas phase heat balance:
##\epsilon_g \rho_mC_{p,g,m}\frac{\partial T_g}{\partial t}= -d -q_{g,I}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-T_{z-\Delta z}}{\Delta z}) -q_{g,I,z}a_s##

The last term to discretise spatially would be the overall mass balance for the gas phase. This is the current discretised equation:
##\epsilon\frac{\partial \rho_m}{\partial t}=\frac{\phi_{z-\Delta z/2} - \phi_{z+\Delta z/2}}{\Delta z}-\sum_{i=1}^{n_c}{\dot{M}_{i,z}^"a_s}##

This is the only equation with a 'downwind' parameter ##\phi_{z+\Delta z/2}##. I think we can make use of this parameter to calculate the flow out of a bed (##m_j## in the previous model)? After checking the previous model yes this is what you did there

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
Please check the signs.
 
  • #70
Chestermiller said:
Please check the signs.
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?
 
Last edited:

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