Modelling of two phase flow in packed bed (continued)

[casualguitar]

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:
$$C_{p, mol} * \frac{1000}{mW} = C_{p, kg}$$

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?

 

[Chestermiller]

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?

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.

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.

 

[Chestermiller]

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)

 

[casualguitar]

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?

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:

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?

 

[Chestermiller]

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.

 

[casualguitar]

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:

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

 

[Chestermiller]

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.

 

[casualguitar]

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

 

[Chestermiller]

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.

 

[casualguitar]

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?

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

 

[Chestermiller]

$\textbf{The overall mass balance for the gas phase:}$
$$\frac{\partial (\epsilon \rho_m)}{\partial t}=-\frac{\partial (\rho_mv_m)}{\partial z}-\sum_{i=1}^{n_c}{\dot{M}_i^"a_s}\tag{1}$$
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.

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.

 

[casualguitar]

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?

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

 

[Chestermiller]

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.

 

[casualguitar]

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

 

[Chestermiller]

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 $(T_z+T_{z-\Delta z})/2$.

 

[casualguitar]

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

 

[Chestermiller]

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.

 

[casualguitar]

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

 

[Chestermiller]

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.

 

[casualguitar]

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?

 

[Chestermiller]

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

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

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

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

 

[casualguitar]

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

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

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

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

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

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

 

[Chestermiller]

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

Why would that be a problem?

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

 

[casualguitar]

Why would that be a problem?

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

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

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

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

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

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

 

[Chestermiller]

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

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

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

 

[casualguitar]

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

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

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

 

[Chestermiller]

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

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

That's already included in their equations.

 

[casualguitar]

That's already included in their equations.

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

 

[Chestermiller]

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

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

 

[casualguitar]

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

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

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

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

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

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

 

[casualguitar]

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

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

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

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

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

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

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

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

 

[Chestermiller]

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

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

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

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

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

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

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

I don't understand.

 

[casualguitar]

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

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

I don't understand.

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

 

[casualguitar]

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

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

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

 

[Chestermiller]

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

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

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

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

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

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

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

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

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

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

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

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

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

 

[casualguitar]

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

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

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

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

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

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

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

 

[Chestermiller]

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

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

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

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

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

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

Sure.

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

 

[casualguitar]

Sure.

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

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

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

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

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

 

[Chestermiller]

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

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

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

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

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

You could define it as a new variable.

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

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

 

[casualguitar]

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

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

You could define it as a new variable.

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

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

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

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

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

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

 

[casualguitar]

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

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

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

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

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

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

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

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

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

 

[Chestermiller]

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

We handle the dispersion differently also.

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

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

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

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

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

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

 

[Chestermiller]

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

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

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

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

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

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

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

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

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

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

 

[casualguitar]

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

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

Also needed are the heat- and mass transfer coefficient correlations

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

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

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

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

No mass flux expressions yet

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

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

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

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

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

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

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

 

[Chestermiller]

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

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

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

No mass flux expressions yet

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

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

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

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

I think we should complete the formulation first.

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

 

[casualguitar]

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

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

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

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

 

[casualguitar]

I think we should complete the formulation first.

Sublimation curve for CO2 (Span and Wagner):

Melting curve for H2O (Span and Wagner):

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

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

 

[Chestermiller]

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

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

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

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

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

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

 

[casualguitar]

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

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

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

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

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

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

 

[Chestermiller]

Mass Transfer Rate for Deposition of CO2 and Water

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

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

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

Thoughts?