Modelling of two phase flow in packed bed (continued)

[casualguitar]

In addition, one thing I have found is that the $Q_{IB}$ and $Q_{GI}$ values at the left boundary start at a few hundred and trend gradually to zero (as expected), however the internal node values of $Q_{IB}$ and $Q_{GI}$ start at a decimal value and trend to very small numbers (10^-10), suggesting that heat transfer in the inner nodes is much less. Looking into this

 

[Chestermiller]

So using their operating conditions as much as is possible (I can lay out the exact values used if necessary), this is the position vs gas temperature output:

View attachment 304767
And the Tuinier et al equivalent:

View attachment 304768
So they are not that similar in that our one stops at -90C (around the desublimation temperature), and that there is no clear constant temperature section.

One thing I noticed - here's the plot of CO2 solid buildup versus position:View attachment 304769
This is in moles. Notice how the solid builds up at each position and then decreases, suggesting that sublimation is also occurring. But how can this happen if the temperature is below the sublimation temperature? I don't know the exact temperature of sublimation but even if the gas temperature is slightly above that it would still likely be slow desublimation, not like the above

The mechanics for the sublimation/liquefaction pressure are as follows:
View attachment 304770
and for water:
View attachment 304771
The above is just to show what I do outside the temperature bounds. I either set the sublimation/liquefaction pressure equal to 0 or a very large number

It seems odd that the solid buildup would take on a normal trend, while the gas temperature stays below (or on) the sublimation temperature?

EDIT: I mentioned above that there is no clear constant temperature section in the plot, when actually this is probably not true as the constant temperature section would occur at the maximum temperature on this plot. I guess if the temperature went higher (above sublimation temperature) we would see that constant section

You need to compare these on a more common basis. Remember that $\Delta x=L/n$, where n is the total number of tanks. The tank centers are therefore at $\Delta x/2, 3\Delta x/2, 5\Delta x/2,\ etc$. Plot the data at these distances, not tank numbers. There is also a data point at x = 0, of course, corresponding to the inlet conditions.

I'm guessing that you are using 10 tanks for the whole bed. They seem to be using a finer grid, with approximately 3x your resolution. So,, for comparison, please use 30 tanks.

You should also be comparing the mass buildup per unit area of bed, not the molar solid buildup within each tank. Please, for comparison, show the mass buildups per unit bed area within each tank.

I'm definitely not going to go through your coding. If you want to show a logic diagram for how the deposition is calculated, I'll consider it.

What is this all about: "I either set the sublimation/liquefaction pressure equal to 0 or a very large number". I thought we are assuming the total pressure is 1 bar.

 

[casualguitar]

I'm guessing that you are using 10 tanks for the whole bed. They seem to be using a finer grid, with approximately 3x your resolution. So,, for comparison, please use 30 tanks.

I'm currently using 10 tanks yes. Ah ok I did attempt to make the grids comparable, using the below statement from the paper (and knowing that the bed is 300m in length):

I thought that 300mm length divided into 3cm increments would be about 10 'tanks'. Is this incorrect? If so I can switch to 30 but I don't yet see how you got 30

You should also be comparing the mass buildup per unit area of bed, not the molar solid buildup within each tank. Please, for comparison, show the mass buildups per unit bed area within each tank.

I noticed this also yes that they use $kg.m^3$ not just $kg$. I guess multiplying my molar solid buildup value by a tank volume ($dz * A_c$) works here?

I'm definitely not going to go through your coding. If you want to show a logic diagram for how the deposition is calculated, I'll consider it.

Yes that was my intention to show the logic above. I'll make a logic diagram for how this currently works

What is this all about: "I either set the sublimation/liquefaction pressure equal to 0 or a very large number". I thought we are assuming the total pressure is 1 bar.

Yes the total pressure is 1 bar however the sublimation rate is dependent on the sublimation pressure at a given temperature. Above the critical temperature I set the sublimation/liquefaction pressure equal to a very large number to stop any desublimation happening

 

[Chestermiller]

I'm currently using 10 tanks yes. Ah ok I did attempt to make the grids comparable, using the below statement from the paper (and knowing that the bed is 300m in length):View attachment 304777
I thought that 300mm length divided into 3cm increments would be about 10 'tanks'. Is this incorrect? If so I can switch to 30 but I don't yet see how you got 30

OK. I was confused. I missed the point that their total length of bed was 30 cm. However, the locations of their thermocouples do not necessarily correspond to the grid spacing in their model. I would suggest using more tanks in your calculations. Their graphs seem to suggest that the used a finer resolution than $\Delta x = 3\ cm.$. I suggest 30 tanks, so that $\Delta x = 1 \ cm$ and so that the center of the first tank is at x = 0.5 cm and the center of the last tank is at 29.5 cm.

The locations where the temperatures are changing substantially in their model do not seem to correspond to where they are changing substantially in your model. All the variation seems to be happening closer to the inlet in your model. Is there a scaling problem on time?

I noticed this also yes that they use $kg.m^3$ not just $kg$. I guess multiplying my molar solid buildup value by a tank volume ($dz * A_c$) works here?

I'll let you work this geometric conversion out. But please provide the rationale and equations for the conversion that you develop. It is not simply Adz.

 

[Chestermiller]

On second thought, surface area to volume ratio of a particle is $$\frac{4\pi r_p^2}{\frac{4}{3}\pi r_p^3}=\frac{3}{r_p}=\frac{6}{d_p}$$The particle volume per bed column volume is $1-\epsilon$. The column volume per tank is $A_c\Delta z$. So, the available deposition area per tank is $$\frac{6}{d_p}(1-\epsilon)A_c\Delta z$$

 

[casualguitar]

I suggest 30 tanks, so that Δx=1 cm and so that the center of the first tank is at x = 0.5 cm and the center of the last tank is at 29.5 cm

Change made

The locations where the temperatures are changing substantially in their model do not seem to correspond to where they are changing substantially in your model. All the variation seems to be happening closer to the inlet in your model. Is there a scaling problem on time?

The variation happening close to the inlet is something I noticed also and I did spot something yesterday on that (my post #246 from yesterday above, I'll quote it here):

In addition, one thing I have found is that the $Q_{IB}$ and $Q_{GI}$ values at the left boundary start at a few hundred and trend gradually to zero (as expected), however the internal node values of $Q_{IB}$ and $Q_{GI}$ start at a much smaller decimal value and trend to very small numbers (10^-10), suggesting that heat transfer in the inner nodes is much less.

i.e. the variation is happening closer to the inlet. Possibly suggesting I've set up the boundary conditions incorrectly

Also I don't think there's a time scaling problem. I checked the simulation length time and it equals the length of the time array for the solution, meaning that there is a 1:1 matching

I'll let you work this geometric conversion out. But please provide the rationale and equations for the conversion that you develop. It is not simply Adz.

Will do

 

[casualguitar]

On second thought, surface area to volume ratio of a particle is $$\frac{4\pi r_p^2}{\frac{4}{3}\pi r_p^3}=\frac{3}{r_p}=\frac{6}{d_p}$$The particle volume per bed column volume is $1-\epsilon$. The column volume per tank is $A_c\Delta z$. So, the available deposition area per tank is $$\frac{6}{d_p}(1-\epsilon)A_c\Delta z$$

The units here seem to be $mol.m^2$ whereas the units in Tuinier et al are $kg.m^3$. Is $m^2$ right here?

Also, if the above is correct then I've potentially made a mistake elsewhere -

The solid phase mass balance is:

And I have defined $a_s$ (the specific surface area), as $\frac{6}{d_p}(1-\epsilon)A_c\Delta z$. Is this incorrect? Everywhere we have $a_s$ in the model equations I have used the above

Looking into the 'all the variation happening at the inlet' issue now

 

[Chestermiller]

The units here seem to be $mol.m^2$ whereas the units in Tuinier et al are $kg.m^3$. Is $m^2$ right here?

Remember, we decided to work in terms of moles rather than mass. This is signified by using M rather than m. Also, the subscript "I" signifies that it is averaged over tank i.

Also, if the above is correct then I've potentially made a mistake elsewhere -

The solid phase mass balance is:
View attachment 304803

In our development, $a_s$ is the surface area available for deposition in tank i.

And I have defined $a_s$ (the specific surface area), as $\frac{6}{d_p}(1-\epsilon)A_c\Delta z$. Is this incorrect? Everywhere we have $a_s$ in the model equations I have used the above

In our development, this is correct to use. Of course, you have to be able to convert from our notation to theirs to compare the results. Are you not able to do this?

 

[casualguitar]

Remember, we decided to work in terms of moles rather than mass. This is signified by using M rather than m. Also, the subscript "I" signifies that it is averaged over tank i.

Yes apologies for the confusion I'm ok with the kg and mol difference however their plot here seems to have m3 in the units whereas ours has $m^2$ instead

In our development, this is correct to use. Of course, you have to be able to convert from our notation to theirs to compare the results. Are you not able to do this?

So I'm not fully clear on how to convert between yet. $m^2$ makes sense to me as we're looking for the amount of solid available for deposition. Why would they use $m^3$ here?

 

[casualguitar]

In addition, the difference in $Q_{GI}$ values stems from the $\Delta T$ between gas and bed at the inlet being higher than the delta T for the internal nodes:

I'm not sure why this happens yet. I would have expected the blue curve to be similar to the orange and green curves. Looking into it

 

[casualguitar]

Final update (apologies). So the reason that there is more variation at the inlet is this -

The gas phase heat balance is this:

When setting up this equation, I have one equation for the boundary tank (which I think is tank zero), and one equation for the other tanks.

$\dot{m}_{j-1}$ is the inlet molar flow rate for tank zero (boundary tank) and $T_{j-1}$ is the inlet flow temperature.

For the rest of the system I just take the outlet flow and temperature from the previous tank and use this for the $j-1$ values.

So plotting $T_{j-1} - T_j$ for all tanks gives:

which shows that there is a much larger temperature change occurring at the boundary always. Is this expected?

 

[Chestermiller]

Yes apologies for the confusion I'm ok with the kg and mol difference however their plot here seems to have m3 in the units whereas ours has $m^2$ instead

View attachment 304805

So I'm not fully clear on how to convert between yet. $m^2$ makes sense to me as we're looking for the amount of solid available for deposition. Why would they use $m^3$ here?

Yes. My bad. Their $m_i$ is supposed to be mass per unit of bed volume (actually column volume). So their $a_s$ is equal to $\frac{6}{d_p}(1-\epsilon)$, and is the deposition surface area per unit volume of column. Our $M_i$ is supposed to be total moles of species deposited on solid surface in tank and our $M_i^"$ is the moles of species deposited per unit time per unit of deposition surface in tank. So, $$M_i=\frac{1000m_iA_c\Delta z}{W_i}$$and$$M_i^"=\frac{1000m_i^"A_c\Delta z}{W_i}$$where $W_i$ is the molecular weight of the species.

 

[Chestermiller]

Final update (apologies). So the reason that there is more variation at the inlet is this -

The gas phase heat balance is this:
View attachment 304808
When setting up this equation, I have one equation for the boundary tank (which I think is tank zero), and one equation for the other tanks.

$\dot{m}_{j-1}$ is the inlet molar flow rate for tank zero (boundary tank) and $T_{j-1}$ is the inlet flow temperature.

For the rest of the system I just take the outlet flow and temperature from the previous tank and use this for the $j-1$ values.

So plotting $T_{j-1} - T_j$ for all tanks gives:
View attachment 304809
which shows that there is a much larger temperature change occurring at the boundary always. Is this expected?

What is your exact equation for tank 1?

 

[casualguitar]

What is your exact equation for tank 1?

By tank 1 (if we're talking about the boundary tank), then my exact equation gas phase heat balance is:
$$\frac{\partial T_g}{\partial t} = \frac{\dot{m}_{in}C_{p}(T_{in} - T_1) - q_{g,I}A_s}{m_1C_{p}}$$
where $\dot{m}_{in}$ is the inlet molar flow and $T_{in}$ is the temperature of the inlet flow, and $m_1$ is the molar holdup in tank 1

Is this reasonable?

and here is the general equation:
$$\frac{\partial T_g}{\partial t} = \frac{\dot{m}_{j-1}C_{p}(T_{j-1} - T_j) - q_{g,I}A_s}{m_1C_{p}}$$

 

[Chestermiller]

By tank 1 (if we're talking about the boundary tank), then my exact equation gas phase heat balance is:
$$\frac{\partial T_g}{\partial t} = \frac{\dot{m}_{in}C_{p}(T_{in} - T_1) - q_{g,I}A_s}{m_1C_{p}}$$
where $\dot{m}_{in}$ is the inlet molar flow and $T_{in}$ is the temperature of the inlet flow, and $m_1$ is the molar holdup in tank 1

Is this reasonable?

and here is the general equation:
$$\frac{\partial T_g}{\partial t} = \frac{\dot{m}_{j-1}C_{p}(T_{j-1} - T_j) - q_{g,I}A_s}{m_1C_{p}}$$

Looks OK

 

[casualguitar]

Looks OK

So the boundary condition equation is set up ok. This graph doesn't seem intuitive though ($T_{j-1}$ - $T_j$ for all tanks). This delta T discrepancy is driving the higher rate of variation at the inlet:

Maybe one of the heat transfer terms driving heat transfer axially through the bed is low? The $Q_GI$ term at the inlet is much higher than $Q_GI$ for the other tanks also. I don't think this is to be expected.

I've checked the molar desublimation rates across the bed and they seem ok everywhere.

I messed around with the heat of vaporisations/desublimations just on the off chance they were incorrect. Currently the heat of desublimation and vaporisation are 26000 and 40650 J/mol respectively. If I just divide them both by 1000 (no particular reason other than maybe the current units are wrong), then the gas temperature levels off at a higher temperature. Still too low, but higher. Suggesting maybe that this is a value error rather than a desublimation mechanics error?

Plot of gas temperature with the heat of vaporisation/desublimation values divided by 1000:

Original values:

 

[casualguitar]

Slightly more interestingly, if I divide those heat of vaporisation/desublimation values by 1000 again then we get a relatively normal looking profile (note these temperatures are degrees C):

This might suggest that I haven't divided by 1000 somewhere where I should have. 1000 would come up when converting from kg to mol so maybe I haven't done that somewhere? Complete guess but will check

 

[Chestermiller]

Have you checked what the literature gives for these heats?

i’n having trouble understanding what seems wrong.

 

[Chestermiller]

If it wouldn't be too much trouble, could you please plot up the 30 tank results of temperature vs distance and CO2 deposited vs distance (with time as a secondary parameter) using the units in the Turnier paper so that we can compare directly with their paper. Thanks.

 

[Chestermiller]

By tank 1 (if we're talking about the boundary tank), then my exact equation gas phase heat balance is:
$$\frac{\partial T_g}{\partial t} = \frac{\dot{m}_{in}C_{p}(T_{in} - T_1) - q_{g,I}A_s}{m_1C_{p}}$$
where $\dot{m}_{in}$ is the inlet molar flow and $T_{in}$ is the temperature of the inlet flow, and $m_1$ is the molar holdup in tank 1

Is this reasonable?

and here is the general equation:
$$\frac{\partial T_g}{\partial t} = \frac{\dot{m}_{j-1}C_{p}(T_{j-1} - T_j) - q_{g,I}A_s}{m_1C_{p}}$$

In these equations, we use $m_j$ to represent the total number of moles of all species in tank j, and $\dot{m}_{j}$ to represent the total number of moles per unit time exiting tank j and entering tank j-1.

 

[Chestermiller]

Yes. My bad. Their $m_i$ is supposed to be mass per unit of bed volume (actually column volume). So their $a_s$ is equal to $\frac{6}{d_p}(1-\epsilon)$, and is the deposition surface area per unit volume of column. Our $M_i$ is supposed to be total moles of species deposited on solid surface in tank and our $M_i^"$ is the moles of species deposited per unit time per unit of deposition surface in tank. So, $$M_i=\frac{1000m_iA_c\Delta z}{W_i}$$and$$M_i^"=\frac{1000m_i^"A_c\Delta z}{W_i}$$where $W_i$ is the molecular weight of the species.

On second thought, I think we should let $A_s$ represent the deposition area available within each tank: $$A_s=\frac{6}{d_p}(1-\epsilon)A\Delta z$$and we should let $M_i^"$ represent the molar deposition rate per unit deposition area available in tank "I":$$M_i^"=\frac{1000m_i^"}{W_i}$$So $$\frac{dM_i}{dt}=M_i^"A_s$$

 

[casualguitar]

Have you checked what the literature gives for these heats?

Yes and the values I had originally are the literature values

i’n having trouble understanding what seems wrong.

So the last two plots in post #261 show the temperature levelling off at a temperature that is below the inlet temperature (100C). I was pointing out in post #262 that if the heat of vaporisation/sublimation is multiplied by a factor of about 1000 then we get the correct 'temperature level off'. Obviously we can't just multiply by 1000 though so I thought maybe a factor of 1000 was missing somewhere else (possibly I hadn't converted between kg and mol properly somewhere). Just a thought though. But more importantly it shows that the desublimation/vaporisation mechanics aren't fully broken in that the gas temperature can go beyond the desublimation/vaporisation temperature

If it wouldn't be too much trouble, could you please plot up the 30 tank results of temperature vs distance and CO2 deposited vs distance (with time as a secondary parameter) using the units in the Turnier paper so that we can compare directly with their paper. Thanks.

Can do

In these equations, we use mj to represent the total number of moles of all species in tank j, and m˙j to represent the total number of moles per unit time exiting tank j and entering tank j-1.

Should this read 'entering tank j+1'?

On second thought, I think we should let $A_s$ represent the deposition area available within each tank: $$A_s=\frac{6}{d_p}(1-\epsilon)A\Delta z$$and we should let $M_i^"$ represent the molar deposition rate per unit deposition area available in tank "I":$$M_i^"=\frac{1000m_i^"}{W_i}$$So $$\frac{dM_i}{dt}=M_i^"A_s$$

Understood. Actually this was the representation I was using. So to convert between their CO2 solid buildup plots all that is needed is to multiply the mole values by $\frac{W_i}{1000*A_C*dz}$?

Doing those plots now

 

[casualguitar]

Here are the plots of position vs Mass of solid CO2 build up and position versus gas temperature (for a selection of times):

Some notes:
- The CO2 mass buildup is in the general ballpark of the Tuinier et al paper. The max buildup I see (about 120 kg/m3) is about double what the Tuinier et al paper shows

- Secondly, as you said earlier the activity in the above plots seems to be concentrated at the inlet much more so than the Tuinier et al plots (unless I'm reading it incorrectly)

- Lastly the gas temperature seems to max out at about -90C which is not the case in the Tuinier paper. We initially thought this was to do with sublimation mechanics but now that we can fix this by 'dividing the heat of vaporisation/sublimation by 1000', maybe this isn't the case

I'll take a look for forgotten 'divisions by 1000'. If there's any other useful plots I could do just let me know

 

[Chestermiller]

Yes and the values I had originally are the literature values

So the last two plots in post #261 show the temperature levelling off at a temperature that is below the inlet temperature (100C). I was pointing out in post #262 that if the heat of vaporisation/sublimation is multiplied by a factor of about 1000 then we get the correct 'temperature level off'. Obviously we can't just multiply by 1000 though so I thought maybe a factor of 1000 was missing somewhere else (possibly I hadn't converted between kg and mol properly somewhere). Just a thought though. But more importantly it shows that the desublimation/vaporisation mechanics aren't fully broken in that the gas temperature can go beyond the desublimation/vaporisation temperatureCan do

Should this read 'entering tank j+1'?

Yes. My mistake.

Understood. Actually this was the representation I was using. So to convert between their CO2 solid buildup plots all that is needed is to multiply the mole values by $\frac{W_i}{1000*A_C*dz}$?

Doing those plots now

Yes. Another mistake of mine. I misread the units as kg/m^2

 

[casualguitar]

Yes. My mistake.

Yes. Another mistake of mine. I misread the units as kg/m^2

Those changes had been incorporated anyway. The plots above (post #268) are the plots you mentioned earlier and use as much of the Tuinier data as possible. In places I had to make assumptions (like picking very high $U_g$ and $U_b$ values to approximate their infinite gas-solid heat transfer coefficient). Unless I've read the plots incorrectly they're not very similar yet. I'm looking into the possibility that I've forgotten to convert between kg and mol somewhere along the way

Maybe I've misunderstood again but this plot (Fig 7 in Tuinier) doesn't seem to make sense):

The x-axis is time in seconds and the temperature is the outlet temperature of the packed bed. Does this not say that their temperature reaches a maximum at roughly -90C also? But surely this is at odds with Fig 6 and 5 which show the temperature going above this. Or have I misunderstood? Because fig 7 looks a lot like our time vs Tg plot (besides the time scale)

 

[Chestermiller]

Those changes had been incorporated anyway. The plots above (post #268) are the plots you mentioned earlier and use as much of the Tuinier data as possible. In places I had to make assumptions (like picking very high $U_g$ and $U_b$ values to approximate their infinite gas-solid heat transfer coefficient). Unless I've read the plots incorrectly they're not very similar yet. I'm looking into the possibility that I've forgotten to convert between kg and mol somewhere along the way

Those new graphs in #268 are encouraging to me. Some questions:

1. Why didn't you show temperatures from -90 to 100 C?

2. Although they use infinite U's and we use finite U's, our U's are still pretty high, as evidenced by the very small differences between the gas temperature and the bed temperature. What do our results look like using our correlations.

3. There seems to be a time scaling issue here. Are you sure you are showing the results at the correct times? Are you using fixed time interval, or having the integrator spit out results at specified times? The time scaling factor seems to be something like 10x.

Maybe I've misunderstood again but this plot (Fig 7 in Tuinier) doesn't seem to make sense):
View attachment 304867
The x-axis is time in seconds and the temperature is the outlet temperature of the packed bed. Does this not say that their temperature reaches a maximum at roughly -90C also? But surely this is at odds with Fig 6 and 5 which show the temperature going above this. Or have I misunderstood? Because fig 7 looks a lot like our time vs Tg plot (besides the time scale)

It looks like the operating conditions for Fig.7 were a little different than for figs. 5 & 6. I wouldn't worry too much about this.

 

[casualguitar]

1. Why didn't you show temperatures from -90 to 100 C?

Because the temperature only rises to about -90C for some reason. I can run the simulation for much longer times but the temperature will max out at -90C

Although they use infinite U's and we use finite U's, our U's are still pretty high, as evidenced by the very small differences between the gas temperature and the bed temperature. What do our results look like using our correlations.

I haven't added in these correlations yet given the above issue. I think we agreed earlier its better to fix that first rather than add the U correlations in and then fix. But if necessary I can add these

3. There seems to be a time scaling issue here. Are you sure you are showing the results at the correct times? Are you using fixed time interval, or having the integrator spit out results at specified times? The time scaling factor seems to be something like 10x.

I think so anyway. So I pass t_eval to the integrator with is an array of times that I would like the solution stored at. I just store the time at each second interval. I also checked the length of the time array output by the integrator and it is the same length as the simulation time, surely indicating that there is no time scaling issue?

 

[Chestermiller]

Because the temperature only rises to about -90C for some reason. I can run the simulation for much longer times but the temperature will max out at -90C

How can they not get higher than -90 C at short distances from the inlet? The stream coming in is at 100 C.

I haven't added in these correlations yet given the above issue. I think we agreed earlier its better to fix that first rather than add the U correlations in and then fix. But if necessary I can add these

Yes, please turn them on and see what we get.

I think so anyway. So I pass t_eval to the integrator with is an array of times that I would like the solution stored at. I just store the time at each second interval. I also checked the length of the time array output by the integrator and it is the same length as the simulation time, surely indicating that there is no time scaling issue?
View attachment 304868

We seem to be losing a factor of 10 elsewhere then. It's hard to imagine where, unless it is the individual tank holdup time (molar holdup divided by molar flow rate).

 

[casualguitar]

How can they not get higher than -90 C at short distances from the inlet? The stream coming in is at 100 C.

Exactly. However I mentioned yesterday if we scale down the heats of vaporisation/sublimation the temperature does go above -90C and reaches 100C. Obviously we can't just change those values though. But in effect by scaling these down we're actually scaling down $Q_{GI}$ and $Q_{IB}$ because these are a function of the heats of vaporisation/sublimation. I'm not sure why scaling down these values let's the temperature increase up to the inlet temperature but it does for some reason. As you say we're missing a factor of 10 somewhere (or possibly a factor greater than 10)

Yes, please turn them on and see what we get.

Will do. This will take a bit of time

We seem to be losing a factor of 10 elsewhere then. It's hard to imagine where, unless it is the individual tank holdup time (molar holdup divided by molar flow rate).

I agree there is a factor of 10 (or greater) lost somewhere, I'll add those $U_g$ values in anyway and see what happens

 

[casualguitar]

I found one potential factor of 10 loss relating to the inlet molar flow

The Tuinier et al paper uses 0.27 kg/m2.s. I converted this to mol/m2.s (x1000/mW)

Now to convert to molar flow I was multiplying by $A_C$, but is this correct? Or should I be multiplying by the actual flow area which I think would be A_c * epsilon? Including a voidage term min the inlet flow rate term makes a difference in how 'spread out' the variation is.

It doesn't help the temperature reach the inlet temperature unfortunately though but its a start

 

[Chestermiller]

No. This is always given based on the column area Ac.

 

[Chestermiller]

What do you get for the molar flow rate? Assuming a uniform temperature of 300K and a pressure of 1 atm, what do you get for the gas holdup in moles?

 

[casualguitar]

What do you get for the molar flow rate? Assuming a uniform temperature of 300K and a pressure of 1 atm, what do you get for the gas holdup in moles?

I was about to comment in relation to this. Printing the reynolds numbers at each position results in values around Re = 100, which are possibly quite low

For a molar flux of 0.27kg/m2.s as used in Tuinier et al, I get a molar flow of 0.0085mol/s, which gives a gas holdup of 0.0012 mol at a temperature of 300K and 1 atm. Does this seem reasonable?

Also I've implemented the variable $U_g$ values. They do change the plots slightly (not too much). Also printing out these $U_g$ values gives values in the range of about 20 to 60 W/m2.K

Here are the plots for non-constant $U_g$:

Edit: I missed the molar flow rate question.

So the inlet flow is 0.0085mol/s. In the first tank this actually increases up to 0.01 mol/s and stays at about this value throughout the simulation. In the other tanks it also does an initial jump up to 0.01mol/s but then gradually tails off down to about 0.005 mol/s which is more what I would have expected, indicating I've possibly set this up incorrectly at the boundary

 

[Chestermiller]

I was about to comment in relation to this. Printing the reynolds numbers at each position results in values around Re = 100, which are possibly quite low

For a molar flux of 0.27kg/m2.s as used in Tuinier et al, I get a molar flow of 0.0085mol/s, which gives a gas holdup of 0.0012 mol at a temperature of 300K and 1 atm. Does this seem reasonable?

The molar flow rate seems reasonable, but not the holdup. I get a superficial column volume of 289 cc, and, assuming a void fraction of 0.32, I get a column void volume of 92.4 cc = 0.0924 liters. From the ideal gas law at 1 atm and 300 K, I get a molar density of 0.041 moles/liter. So, for a molar holdup, I get 0.0038 moles. So for a mean residence time of the gas, I get 0.0038/.0085 = 0.45 seconds. This is way less than the times we are seeing in our calculations for a temperature wave to travel through the bed, so the solid bed must be having a substantial effect of the speed of temperature waves through the column. Apparently, in our model, the waves are slowed much more than in the Turnier model.

 

[casualguitar]

This is way less than the times we are seeing in our calculations for a temperature wave to travel through the bed, so the solid bed must be having a substantial effect of the speed of temperature waves through the column. Apparently, in our model, the waves are slowed much more than in the Turnier model.

I've calculated the molar holdup by saying that:

$$m_j = \frac{P}{RT_g}A_C*\delta z*\epsilon$$

This is way less than the times we are seeing in our calculations for a temperature wave to travel through the bed,

If these times were increased, is it correct to say that we would see the temperature variation 'spread out' across the bed rather than be concentrated at the inlet?