Modelling of two phase flow in packed bed (continued)

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The discussion focuses on modeling the two-phase flow of air through a packed bed, specifically addressing the complexities introduced by CO2 freezing out from the process stream. Participants explore the behavior of CO2 in a colder bed, suggesting that it may deposit temporarily before forming a solid plug, which could complicate the flow dynamics. There is a consensus on the importance of understanding thermodynamics and phase equilibria of the gas mixture before advancing the model, with recommendations to analyze vapor pressures and phase compositions. The impact of varying superficial velocities across different phases is highlighted as a significant factor that could affect the model's accuracy. Overall, the conversation emphasizes a careful and methodical approach to developing the model to account for these complexities.
  • #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.
 
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  • #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.
 
  • #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?
 
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  • #71
casualguitar said:
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.
 
  • #72
Chestermiller said:
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##

casualguitar said:
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
 
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  • #73
casualguitar said:
##\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.
 
  • #74
Chestermiller said:
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.
casualguitar said:
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

casualguitar said:
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
 
  • #75
casualguitar said:
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##.
 
  • #76
Chestermiller said:
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?
 
  • #77
casualguitar said:
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.
 
  • #78
Chestermiller said:
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?
 
  • #79
casualguitar said:
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.
 
  • #80
Chestermiller said:
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
 
  • #81
casualguitar said:
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
 
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  • #82
casualguitar said:
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?
casualguitar said:
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.
 
  • #83
Chestermiller said:
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##

Chestermiller said:
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?
 
  • #84
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
 
  • #85
casualguitar said:
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.
casualguitar said:
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##.

casualguitar said:
##\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##.
 
  • #86
Chestermiller said:
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##?
 
Last edited by a moderator:
  • #87
casualguitar said:
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, $$\dot{m}_j=\dot{m}_{j-1}+\frac{\rho_m}{T}A_C\Delta z\frac{\partial T}{\partial t}-\sum_{i=1}^{n_c}{\dot{M}_{i,j}^"A_S}$$
 
  • #88
Chestermiller said:
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
 
  • #89
casualguitar said:
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}##
casualguitar said:
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##.
casualguitar said:
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.
casualguitar said:
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.
 
  • #90
Chestermiller said:
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:
Screenshot 2022-04-09 at 22.19.21.png

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
 

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