Ambiguity in the droplet temperature and diameter.

[H. M. Iqbal Mahmud]

 

 

Hi Sikanen / All,

 

I was trying to see the change of temperature and diameter of a single droplet due to evaporation. Using read_prt5.m file, i plotted the time vs temperatue and time vs droplet diameter. However, those graphs showed very peculiar result. As the ambient temperature is 60 and droplet temperature is 20, water particle should be evaporated form the droplet; as a result, the droplet diameter should be reduced, and also the temperature of droplet should rise as it is absorbing heat from ambient.

 

Does anyone know the reason of this ambiguity?

 

Here, I have uploaded the output graphs.

 

The input file is as below.

 

 

&HEAD CHID='single_droplet', TITLE='Water Evaporation Test, SVN $Revision: 7031 $' /

 

&MESH IJK=20,20,20, XB=0.0,2.0,0.0,2.0,0.0,3.0 /

 

&TIME T_END=1.5/

 

&MISC SURF_DEFAULT='WALL', TMPA=60.0, GVEC=0,0,-9.8, RADIATION=.FALSE./

 

&SURF ID='WALL', ADIABATIC=.TRUE., COLOR='SILVER' /

 

&PART ID='water drops', WATER=.TRUE., QUANTITIES(1:3)='DROPLET_DIAMETER','DROPLET_TEMPERATURE','DROPLET_VELOCITY', SAMPLING_FACTOR=1 /

 

&INIT PART_ID='water drops',XYZ=1.0,1.0,2.9, NUMBER_INITIAL_DROPLETS=1,DIAMETER=200, MASS_PER_TIME=0.00000523,TEMPERATURE=20.0/

 

&VENT MB='ZMIN', SURF_ID='OPEN' /

&VENT MB='ZMAX', SURF_ID='OPEN' /

 

&TAIL /

 

 

Regards

[Topi Sikanen]

Your simulation time is 1.5 seconds. The droplet temperature at the
beginning of simulation is 20 degrees celsius. How much water do you
expect to evaporate in 1.5 seconds ?

The droplet evaporation model is described on page 70 of the FDS
Technical reference guide. You can calculate the evaporation rate from
a droplet by hand. My guess is that the droplet is simply not hot
enough to evaporate any mass. Also note that the humidity is probably
not zero.

[Randy McDermott]

The FDS-SMV/Verification/Sprinklers_and_Sprays/water_evaporation_2.fds test case is the example to follow.  Elizabeth Blanchard set this up and the results are compared with a single droplet experiment in the Verification Guide.

[H. M. Iqbal Mahmud]

Hi Sikanen,
Thanks for your reply. Your logic is authentic. However, i will increase the simulation time and also the height of the room and then will see the effect.

Hi Randy,
Thanks for the link. It seems to be very useful. However, I am not clear about a command in this file.
That is-

&INIT PART_ID='water drops', MASS_PER_VOLUME=0.01, NUMBER_INITIAL_DROPLETS=10000 /

As, the droplet size is 200 micron and number of droplet is 10000, the total mass of water droplet is 4.189E-5 kg. The volume of the room is 1cubic meter.
Therefore, MASS_PER_VOLUME should be 4.189E-5 kg/m3.

Would you please let me know that what its actual meaning?

Kind regards

Iqbal

[dr_jfloyd]

It is not feasible to model every droplet that a sprinkler injects.  A typical sprinkler is injecting hundreds of thousands to millions of drops per second.  Instead each droplet in FDS represents multiple droplets. So in your case above each of those 10000 initial droplets actually represents about 240 droplets.  FDS does all the heat transfer and movement based on the single droplet's diameter and temperature and then multiplies the effect by a weighting factor.

[H. M. Iqbal Mahmud]

Hi Floyd / Sikanen,

Thanks for the useful information. Yes, you are right that it is not feasible to model every droplet from a sprinkler injects.  However, still i am not clear about the use of the input parameter of MASS_PER_VOLUME or MASS_PER_TIME.

My understanding is, whatever FDS do, input parameter should match each other.
As i said before - with input command of 200-micron drop dia and 10000 numbers of droplets, the MASS_PER_VOLUME should be equal to 4.189E-5. But, in the input command it is written as MASS_PER_VOLUME=0.01? Is it not contradictory?

I think, i have misunderstood something. Would you please make me clear about this command?

I have the same question about MASS_PER_TIME input command.  Would you please have a look at my calculation which i asked earlier in my another post with title - "About the terminal_velocity_dt=1_0.fds and read_prt5.m script"? Randy advised me to look at your post, but i am still in dark with the use of those input commands. The link of the post is -

http://groups.google.com/group/fds-smv/browse_thread/thread/41c6cf90749d20df/c83d853d3d706293?hl=en&lnk=gst&q=About+the+terminal_velocity_dt%3D1_0.fds+and+read_prt5.m+script#c83d853d3d706293

You will find my question in my last post in this link.

I would be very glad to get a clear idea from anyone of you.

Kind regards

Iqbal

[dr_jfloyd]

Tracking every drop produced by a sprinkler is costly in terms of time and memory.  Rather than do this, we limit the total number of droplets.  Instead each drop in FDS is weighted to behave like more than one drop.  That weighting factor accounts for the difference in mass.  As a completely made up example:

You set MASS_PER_VOLUME to be 100 kg and set INIT to put in 1 droplet that weights 1 kg.  To obtain the total MASS_PER_VOLUME of 100 kg, the single 1 kg droplet is given a weighting factor of 100.  100 x 1 kg = 100 kg.

In this example evaporation and movement would be computed once for the single droplet and the effects would be multiplied by 100 (the weighting factor). 

The same thing is happening in your inputs only the weighting factor is whatever it needs to be to get the mass to balance.

[H. M. Iqbal Mahmud]

Hi Floyd,

A lot of thanks to you for your kind patience to make me understand those commands. I think, i have understood the meaning of those commands.

Just for a cross check, I am explaining it in my own way. Please, have a look to it that i am right or not.

In my case -

&INIT PART_ID='water drops', MASS_PER_VOLUME=0.01, DIAMETER=200., NUMBER_INITIAL_DROPLETS=10000 /

That means, FDS will simulate 10000 drops with diameter 200 micron. The mass of 10000 drops of 200 micron diameter is 4.189E-5 kg. The total mass of water in the volume is MASS_PER_VOLUME=0.01. Therefore, to get the total effect of the total mass of water, FDS use a weighting factor, 0.01/4.189E-5 = 238.72. Am I right?


____________________________________________________________________


On the same way, for MASS_PER_TIME:

&PART ID='embers', SAMPLING_FACTOR=1, EVAPORATE=.FALSE., DENSITY=1000, DIAMETER=3000, QUANTITIES(1:2)='DROPLET DIAMETER','DROPLET TEMPERATURE' /
&INIT XB=9.8,10.0,2.4,2.7,1.3,1.7, PART_ID='embers' NUMBER_INITIAL_DROPLETS=20, MASS_PER_TIME=3.64E-4, DT_INSERT=0.005/

Droplet diameter is 3000 micron. So, mass of one droplet is = (1/6) x pi x D3 x rho = 1.41372E-5 kg.

 

NUMBER_INITIAL_DROPLETS=20, and DT_INSERT=0.005. So, number of drops per second = 20 / 0.005 = 4000. Therefore, mass of water droplet per time = 1.41372E-5 x 4000 = 5.65E-2 Kg/s.

 

That means, FDS will simulate 5.65E-2 Kg/s of water with 3000 micron diameter and 20 number of droplets inserted in every 0.005 seconds.

Total mass of water per time is MASS_PER_TIME=3.64E-4 kg/s. Therefore, to get the total effect, FDS will use a weighting factor, 3.64E-4/5.65E-2 =0.0065. Am I right?

If this explanation is ok, then i have understood the meaning of those input commands.

Thanks again.

Kind regards

[Elizabeth Blanchard]

Dear Iqbal,

For answering to your email you sent me on my pro box, I confirm what
Jason said (that's why I report few of his comments):

FDS uses an Eulerian-Lagrangian approach to simulate evaporating
droplets. This means that trajectories and evaporations of water
droplets are individually tracked. However, FDS (like some other
codes) cannot model every droplet, only a limited number of droplets
is tracked. Thus, each droplet in the calculation is assumed to
represent many others of similar size and trajectory. The droplets
tracked have to constitute a representative sample of the entire
spray. The weighting factor of each droplet accounts for the
difference in mass and the size distribution.

I you e-mail, you are speaking about the verification case
water_evaporation_2.fds but you describe the case
water_evaporation_1.fds. These cases are quite different since one
involves a sample of droplets and the other only one droplet.

In the case water_evaporation_1.fds, MASS_PER_VOLUME is set to be 0.01
kg/m3 in the cubic domain 0.0 1.0 0.0 1.0 0.0 1.0. The corresponding
water mass in that domain is equal to 0.01 kg. In this case, we
planned to simulate a sample of droplets. Since we cannot simulate all
the droplets, we decided to simulate only 10000 droplets. If you
assume they have the same size (200 µm), they weight 4.18879E-05 kg.
To obtain the total MASS_PER_VOLUME, the ratio between the real mass
and the simulated mass is around 239.

In the case water_evaporation_2.fds, MASS_PER_VOLUME is set to be
523.6 kg/m3 in the cubic domain 0.2 0.3 0.2 0.3 0.2 0.3. The
corresponding water mass in that domain is equal to 0.5236 kg. In this
case, we planned to only simulate one single droplet. For reaching
this objective, we set its diameter to be 100000µm. If you do the
ratio between its mass and the simulated mass, you get that a ratio
equal to 1.

Elizabeth

[H. M. Iqbal Mahmud]

Hi Elizabeth,

I am happy to hear from you. Yes, you are right. Now, I am clear about the function of MASS_PER_VOLUME command.

However, I have a question about the test case water_evaporation_2.fds. Here, you have simulated 1 droplet with 100000µm of 0.5236 kg. But, there is a effect of droplet size.
Suppose, if i simulate number of droplets (say, 100) in the same domain with droplet size 200µm whose total weight is 0.5236 kg, then certainly the evaporation rate, change of temperature and droplet diameter etc will be different, is not it?

Other thing is that, if i set the MASS_PER_VOLUME and droplet diameter to a fixed value and change the initial droplet number, then the evaporation rate, droplet temperature and diameter also change. That means, though the total weight of water and droplet diameter is fixed in a domain, it has an effect on how many droplets are simulating. But, I think, it should not be. Because, the total weight of water and droplet diameter is fixed.

Thanks again for your kind information.

Regards

[dr_jfloyd]

The goal of that case is to get the steady state temperature and show that we conserve energy and mass and that we get the correct end state. Ideally diameter should not impact the end state only the time it takes, but on the coarse grids used to make it run fast small drops with fast evaporation rates result in larger errors from the underlying numerics. To minimize this error the large diameter is used.

[dr_jfloyd]

[dr_jfloyd]

In terms of droplets numbers, we evaluate each drop one at a time updating the mass fraction and temperature if the gas after each drop. So if we have one super drop per cell, all of its subdrops will evaporate. If we have many super drops per cell we may reach vapor saturation before all drops are evaluated.

[Victor Kim]

Why, that was very useful and interesting.
Thanks

On Nov 2, 1:31 am, "H. M. Iqbal Mahmud" <hmiqbalmah...@gmail.com>
wrote:

>  Droplet Diameter.jpg
> 19KViewDownload
>
>  Droplet Temperature.jpg
> 24KViewDownload