Modelling of two phase flow in packed bed (continued)

[casualguitar]

Updating on an initial attempt to reformulate the sharp front model on a mole basis:

Converting their capture step sharp front equations to a mole basis -

Component mass balance for defrost front:
\begin{equation}
\Phi'_1 \omega'_{i1} = \Phi'_2 \omega'_{i2} + A m'_i v_d
\end{equation}
$\Phi'_1$ and $\Phi'_2$ are the molar flow rates after and before the defrost front, respectively (in moles per second), $\omega'_{i1}$ and $\omega'_{i2}$ are the molar fractions of component 'i' after and before the defrost front, respectively. $A$ is the cross-sectional area of the bed through which the gas flows (in square meters). $m_i$ is the number of moles of component 'i' deposited per unit of bed volume (in moles per cubic meter). $v_d$ is the velocity of the defrost front (in meters per second), calculated as the distance the defrost front travels divided by the time taken.

Component mole balance for frost front:
\begin{equation}
\Phi'_0 \omega'_{i0} = \Phi'_1 \omega'_{i1} + A m'_i v_f
\end{equation}

So we can see here that the molar flow out of the defrost front is equal to the molar flow into the frost front.

The frost and defrost front velocities dont change for our model. I am guessing that we will use the $\Delta z$ value to track the position of the front:
\begin{equation}
v_d = \frac{z_{d,2} - z_{d,1}}{\Delta t}
\end{equation}
\begin{equation}
v_f= \frac{z_{f,2} - z_{f,1}}{\Delta t}
\end{equation}

Energy balances on a mole basis:
\begin{equation}
A_{vd} \left[ \rho_s C_{p,s}' (T_2 - T_1) + m_i' \Delta H' \right] = \Phi_2' (T_2 - T_1) \left( \omega_{i2}' C_{p,i}' + \omega_{j2}' C_{p,j}' \right)
\end{equation}
\begin{equation}
A_{vf} \left[ \rho_s C_{p,s}' (T_1 - T_0) - m_i' \Delta H' \right] = \Phi_0' (T_1 - T_0) \left( \omega_{i0}' C_{p,i}' + \omega_{j0}' C_{p,j}' \right)
\end{equation}

where the heat capacities are in $J/mol.K$

Overall mole balance for each front:
\begin{equation}
\Phi'_0 \omega'_{i0} = \Phi'_1 \omega'_{i1} + A m'_i v_f
\end{equation}

\begin{equation}
\Phi'_0 = \Phi'_1 + A m'_i v_f
\end{equation}

We also have the molar desublimation rate equation which might be useful to calculate $m_i$:
\begin{equation}
M_i'' = k_i\frac{Py_i - p_T}{RT}
\end{equation}

Just one question - have I missed a relation here? It looks like we have 7 equations and 8 unknowns. I'm defining the unknowns as: $T$, the molar fluxes out of the fronts $\Phi_0$ and $\Phi_1$, the mole fractions out of the fronts $\omega_0$ and $\omega_1$, the front velocities $v_d$ and $v_f$, and the amount of solid buildup $m_i$ in mol/m3.

If this is not the intended reformulation just let me know and I can reform these equations

 

[casualguitar]

Initial attempt to generate the sharp front heat balances from our equations:

The heat balance for the gas phase is given by:
\begin{equation}
\epsilon \rho_m C_{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}(k_{eff}\frac{\partial T_g}{\partial z}) - q_{g,I}a_s
\end{equation}
And the heat balance for the packed bed:
\begin{equation}

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

\end{equation}
Assuming an infinite heat transfer coefficient means that Tg = Tb. Also assuming axial dispersion = 0 we can discount the dispersion term.

If we make these assumptions, and add the two heat balances we get:
\begin{equation}

\epsilon \rho_m C_{p,g,m} \frac{\partial T}{\partial t} + \rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T}{\partial t} = - \phi_m C_{p,g,m}\frac{\partial T_g}{\partial z} + q_{I,b}a_s - q_{g,I}a_s

\end{equation}
We know the relationship between $q_{I,b}$ and $q_{g,I}$ so we can sub out $q_{I,b}$ using:
\begin{equation}

q_{g,I} + \sum\limits_{i=1}^{n_c} M_j''\Delta h_j = q_{I,b}

\end{equation}
Subbing in and cancelling:
\begin{equation}

\epsilon \rho_m C_{p,g,m} \frac{\partial T}{\partial t} + \rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T}{\partial t} = - \phi_m C_{p,g,m}\frac{\partial T_g}{\partial z} + \sum\limits_{i=1}^{n_c} M_j''\Delta h_ja_s

\end{equation}

Assuming an infinite heat transfer coefficient means that all co2 in a spatial element is either sublimated or desublimated depending on the temperature (this happens instantly). So $\sum M''_i a_s$ is actually replaced with $m_i Av$, where A is the CSA and v is the velocity (of the front):
\begin{equation}

\epsilon \rho_m C_{p,g,m} \frac{\partial T}{\partial t} + \rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T}{\partial t} = - \phi_m C_{p,g,m}\frac{\partial T_g}{\partial z} + m_i Av\Delta h_j

\end{equation}
At this point I'm not sure how to finish this out. If we let subscript 0 denote before the defrost front, subscript 1 denote between the defrost and frost fronts, and subscript 2 denote after the frost front, we can say that $\frac{\partial T_g}{\partial z}$ = $T_1 - T_0$ for defrost and $\frac{\partial T_g}{\partial z}$ = $T_2 - T_1$ for frost.

I can recreate the RHS of their heat balance for the defrost front for example knowing that:
\begin{equation}

- \phi_1 C_{p,g,m} = \phi_m (C_{p,i}y_i + C_{p,j}y_j)(T_1 - T_0)

\end{equation}
That leaves the LHS as below, which I am not sure how to simplify to what Tuinier has:
\begin{equation}

\epsilon \rho_m C_{p,g,m} \frac{\partial T}{\partial t} + \rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T}{\partial t} + m_i Av_d\Delta h_i = \phi_m (C_{p,i}y_i + C_{p,j}y_j)(T_1 - T_0)

\end{equation}
They seem to have factored out $Av_d$ but we only have one velocity term on the LHS. Are we going to integrate the time partial derivatives here or do we know the value of this term?

For reference, their LHS is = $Av_d(\rho_s C_{p,s}(T_1-T_0) - m_i\Delta h_i)$, and their RHS is the same as ours

 

[Chestermiller]

$$\epsilon \rho_m C_{p,g,m} \frac{\partial T}{\partial t} + \rho_s(1-\epsilon_g)C_{p,s}\frac{\partial T}{\partial t} = - \phi_m C_{p,g,m}\frac{\partial T_g}{\partial z} + m_i Av\Delta h_j$$

I need to think about this some more. If the deposition term were not there, we Ould have $$\frac{\partial T}{\partial t}+\frac{\phi_mC_{p,g,m}}{\epsilon \rho C_{pig,m}+\rho_s(1-\epsilon)C_{p,s}}\frac{\partial T}{\partial x}=0$$

 

[casualguitar]

I need to think about this some more. If the deposition term were not there, we Ould have $$\frac{\partial T}{\partial t}+\frac{\phi_mC_{p,g,m}}{\epsilon \rho C_{pig,m}+\rho_s(1-\epsilon)C_{p,s}}\frac{\partial T}{\partial x}=0$$

Interesting. I assume this form is useful because it is a form of the advection equation, where everything before the spatial partial derivative is the velocity of the front. This equation seems to apply when we're outside the defrost/frost zone (as $m_i$ is zero)

If we did have the deposition term (which is the case in between the frost and defrost fronts):
\begin{equation}
\frac{\partial T}{\partial t}+\frac{\phi_mC_{p,g,m}}{\epsilon \rho C_{pig,m}+\rho_s(1-\epsilon)C_{p,s}}\frac{\partial T}{\partial x}= m_i Av\Delta h_j
\end{equation}
We get a first-order linear non-homogeneous PDE, which is the same as you had with an additional source term:
\begin{equation}
\frac{\partial T}{\partial t}+v\frac{\partial T}{\partial x}= m_i Av\Delta h_j
\end{equation}
If we're going the road of using a step function then I suppose the general solution is below but it doesn't seem useful:
\begin{equation}
T(x,t) = f(x-vt) + \frac{m_i Av\Delta h_j}{v}
\end{equation}
Can we equate the velocity in the deposition term and the velocity term in the advection equation defined by you above?

 

[casualguitar]

Hi Chet, I just wanted to clear up the wording on the constant dispersion length assumption - are we saying the first or second statement below?
1) the dispersion length $l$ is assumed to be equal to $\frac{\Delta x}{2}$
2) the spatial increment $\Delta x$ is assumed to be equal to $2l$, where $l$ is the dispersion length

I realise that these are very similar. In our discussion it seems that you're saying its "let $\Delta x$ = $2l$", but when we get to this point in our discretisation scheme:
\begin{flalign*}
\phi\frac{\partial y}{\partial z} &= \frac{1}{2}[\phi_{z+\Delta z/2}(\frac{y_{z+\Delta z - y_z}}{\Delta z}) &\\
&\phantom{=}+ \phi_{z-\Delta z/2}(\frac{y_z-y_{z-\Delta z}}{\Delta z})] &\\
\phi C_p\frac{\partial T}{\partial z} &= \frac{1}{2}[\phi_{z+\Delta z/2}C_{p,z+\Delta z/2}(\frac{T_{z+\Delta z} - T_z}{\Delta z}) &\\
&\phantom{=}+ \phi_{z-\Delta z/2}C_{p,z-\Delta z/2}(\frac{T_z-T_{z-\Delta z}}{\Delta z})] &\\
l\frac{\partial}{\partial z}(\phi_m\frac{\partial y}{\partial z}) &= \frac{l}{\Delta z}[\phi_{z+\Delta z/2}(\frac{y_{z+\Delta z} - y_z}{\Delta z}) &\\
&\phantom{=}- \phi_{z-\Delta z/2}(\frac{y_z-y_{z-\Delta z}}{\Delta z})] &\\
l\frac{\partial}{\partial z}(\phi_m C_{p,g,m}\frac{\partial T}{\partial z}) &= \frac{l}{\Delta z}[\phi_{z+\Delta z/2}C_{p,z+\Delta z/2}(\frac{T_{z+\Delta z}-T_z}{\Delta z}) &\\
&\phantom{=}- \phi_{z-\Delta z/2}C_{p,z-\Delta z/2}(\frac{T_z-T_{z-\Delta z}}{\Delta z})] &
\end{flalign*}

We sub in $l$ = $\frac{\Delta x}{2}$ and we dont sub in $2l$ for $\Delta x$:
\begin{equation*}
\epsilon \rho_m C_{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
\end{equation*}
\begin{equation*}
\epsilon\frac{\partial \rho_m}{\partial t} = \frac{\phi_{z-\Delta z/2} - \phi_{z+\Delta z/2}}{\Delta z} - \sum\limits_{i=1}^{n_c} M_{j,z}'a_s
\end{equation*}
\begin{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}) - M_i''a_s + y_{i,z}\sum\limits_{i=1}^{n_c} M_{j,z}''a_s
\end{equation*}

Is it true then to say that we are actually saying "assume $l$ = $\Delta x$/2", and not "assume $\Delta x$ = 2*l?

Apologies for the verbose question

 

[Chestermiller]

Interesting. I assume this form is useful because it is a form of the advection equation, where everything before the spatial partial derivative is the velocity of the front. This equation seems to apply when we're outside the defrost/frost zone (as $m_i$ is zero)

If we did have the deposition term (which is the case in between the frost and defrost fronts):
\begin{equation}
\frac{\partial T}{\partial t}+\frac{\phi_mC_{p,g,m}}{\epsilon \rho C_{pig,m}+\rho_s(1-\epsilon)C_{p,s}}\frac{\partial T}{\partial x}= m_i Av\Delta h_j
\end{equation}
We get a first-order linear non-homogeneous PDE, which is the same as you had with an additional source term:
\begin{equation}
\frac{\partial T}{\partial t}+v\frac{\partial T}{\partial x}= m_i Av\Delta h_j
\end{equation}
If we're going the road of using a step function then I suppose the general solution is below but it doesn't seem useful:
\begin{equation}
T(x,t) = f(x-vt) + \frac{m_i Av\Delta h_j}{v}
\end{equation}
Can we equate the velocity in the deposition term and the velocity term in the advection equation defined by you above?

I'm not so sure this is exactly right. I'm struggling with this, and am having trouble getting the mental concentration to do this right. I'm usually much better at formulation than this.

 

[casualguitar]

I'm not so sure this is exactly right. I'm struggling with this, and am having trouble getting the mental concentration to do this right. I'm usually much better at formulation than this.

As a side point, I'm not exactly sure if the time required here is worth what it would give us? It strengthens the case for saying the model is 'validated' yes, however we have already validated this model at least somewhat with the Tuinier experimental data

 

[Chestermiller]

As a side point, I'm not exactly sure if the time required here is worth what it would give us? It strengthens the case for saying the model is 'validated' yes, however we have already validated this model at least somewhat with the Tuinier experimental data

I have a different perspective. It looks like the simple sharp front model tells 90 % of the story, except for a small amount of dispersion at the fronts. It certainly would be much simpler to do the real modeling and design with the sharp front model, and the results would be easier to interpret physically.