Backing conditions

[DRAKE JONES]

Hi,

       BACKING=’INSULATED’ prevents any heat loss from the backside of the obstruction and the backside is considered to be perfectly insulated. However I found that the spread of fire at the back of my obstruction where the boundary condition is insulated. As per my boundary conditions there should not be any fire at the back right? Or am I missing something here?. TIA

[dr_jfloyd]

When you apply a SURF_ID to an OBST, that SURF is applied to every wall cell exactly as defined on the SURF. In your case by using an insulated SURF_IF you are telling FDS that every exposed wall cell is your combustible material backed by insulation. See 11.3.1 Structure of Solid Boundaries in the User's Guide.

[DRAKE JONES]

Yeah I completely agree.  In your case by using an insulated SURF_IF you are telling FDS that every exposed wall cell is your combustible material backed by insulation. But my question is why there is a fire spread at the insulated area?

[dr_jfloyd]

Did you read 11.3.1?

If you have an OBST that you have given a SURF_ID with an IGNITION_TEMPERATURE and an HRRPUA then EVERY wall cell on that OBST will burn.  SURF_ID means apply the boundary condition to EVERY wall cell of the OBST. On EVERY cell the SURF_ID is applied with the first layer as the exposed layer for that wall cell. 

[Hannah 11]

Only the wall cells burn, what about the interior cells? 

Please have a look at the following case. 

I have an HRRPUA curve obtained for a material of  thickness 8 cm. I have an obstruction of thickness 8 cm. I use a grid cell of size, say 2 cm, which is very much less than 8 cm obstruction (obviously the obstruction is divided into more number of cells). I specify HRRPUA model and an ignition temperature. This means that only the cells on the surface of the obstruction will begin to burn on reaching the wall temperature of ignition temperature. What happens to the cells at the bottom? which face of the cell has to reach the ignition temperature?.

[Hannah 11]

What difference it is going to create when the BACKING=INSULATED is used on the SURF_ID? 
 BACKING=INSULATED- Doesn't it mean that the whole obstruction is backed by an insulation as per user guide-The expression BACKING=’INSULATED’ on the SURF line prevents any heat loss from the back side of the material. Use of this condition means that you do not have to specify properties of the inner insulating material because it is assumed to be perfectly insulated. 

I want the backside of my obstruction to be insulated but as per your statement, every wall cell is backed by insulation which is applicable to the wall cell on the top of my obstruction. If I am going wrong here, please correct me. Thanks in advance.

[dr_jfloyd]

SURF defines how FDS treats a wall cell from the exposed thickness to the other side. Say I have a 10 cm grid can the following OBST and SURF

&SURF ID='My SURF', THICKNESS=0.01, MATL_ID='My MATL', BACKING='Insulated'/

&OBST XB=0.4,0.5,0.4,0.5,0.4,0.5,SURF_ID='My SURF'/

Since the OBST is 10 cm x 10 cm x 10 cm and the grid is 10 cm, this OBST will occupy one grid cell and there will be six wall cells. EVERY wall cell will be treated as layer of My MATL that is 1 cm thick and backed by insulation. FDS will treat the top wall cell as 1 cm of My MATL backed by insulation, and FDS will treat the bottom wall cell as 1 cm of My MATL backed by insulation. If this is the behavior you want, then there isn't necessarily anything wrong with that. However, if you are trying to model a layer of foam ontop of a layer of insulation, then this is not correct. You would need to assign a different SURF for the bottom of the OBST so FDS does not treat the exposed wall as if it can burn.

[Hannah 11]

Thank you so much. Is this the case with the interior cells also (considering the case wherein there are multiple cells in an obstruction)?

Supposing, I assign a HRRPUA to that Obstruction with an ignition temperature. This means that all the six surfaces begin to burn as the HRRPUA specified after reaching the ign. temperature. Am I right?

 

And also, I tried two different ignition scenarios of my material using HRRPUA model

1. with the backing condition as insulated

2. with no backing condition

However I could see from the results that the heat release rates in both the cases are same. So I concluded that since I have already prescribed the rate at which my material should burn and also the heat feedback is not adjusted in HRRPUA model, the HRR remains the same in both the cases. Is the conclusion right? 

[dr_jfloyd]

If you have an OBST that burns away, every exposed wall cell is treated the same as the initial exposed wall cells. 

If you have an OBST with a SURF with an ignition temperature, each wall cell will ignite and burn at the HRRPUA when its temperature reaches the ignition temperature.

If a surface is sufficiently thick, the time to reach the ignition temperature will be independent of the backside boundary condition. It takes time for a thermal wave to move through a solid object. You may wish to review your heat transfer and the concept of thermal penetration time.

A good way to learn how FDS works is to make simple test cases. Also study the various verification files. There is a whole series of verification cases that deal with the ignition and burning of solid objects. 

[Alberto Silingardi]

Good evening Dr. Floyd,

I have carefully read chapter 11.3 I think I understand most of the concepts except one so I am posting here. 

In short, I do not understand the following wordings from 11.3.3 back side boundary conditions where it says:

1) By default, if the obstruction is less than or equal to one cell thick, then the innermost layer will be exposed to the air temperature on the back side.

2) If the obstruction is on the boundary of the domain or is more than one cell thick, then it is assumed to back up to an air gap at ambient temperature.

At first, I couldn't understand the difference between "air temperature" and "air gap at ambient temperature". 

Secondly, are we considering the OBST thickness or the MATL-related thickness? 

Must the OBST thickness be the sum of the MATL thicknesses? In my understanding is not mandatory because the conduction 1D solver just uses the MATL thickness to compute heat transfer.

I've checked out the code of the insulated_steel_plate.fds, also I looked at other examples in the Verification Guide but I am afraid I have only complicated my understanding. Does the Verification contain a particular scenario linked to 11.3.3 User Guide?

I would like to grasp the fundamentals to be able to understand how I should model MATL_ID and SURF for larger-scale models. (e.g. wooden pallets taking fire). 

Might you reference any example or text that could help? Should I deep dive into the mathematics of Conduction to get to the point? (I've just opened Drysdale 2 edition as suggested). 

kindest regards

[Kevin McGrattan]

What version of FDS are you using? Some of the guidance that you cite has changed. 

If your obstruction is thin, like a plate, the back side is assumed to be EXPOSED to the gas behind it. If the obstruction abuts an external boundary, the back side is assumed to be ambient air. 

[Alberto Silingardi]

Thanks for the reply.

According to cmdFDS: FDS6.7.7-0-gfe0d4ef38-release
I'm reading and I've quoted the User Guide "NIST Special Publication 1019 Sixth Edition" and Verification Guide "Special Publication 1018-2"

I have a hard time understanding whether you and the manual are talking about the size of the geometric OBST or the size of the MATL thickness.

is it correct to say that the geometric dimension of the OBST is independent of the 1d conduction calculations?

On Wed, Jan 3, 2024 at 5:14 PM Kevin McGrattan <mcgr...@gmail.com> wrote:

What version of FDS are you using? Some of the guidance that you cite has changed. 

If your obstruction is thin, like a plate, the back side is assumed to be EXPOSED to the gas behind it. If the obstruction abuts an external boundary, the back side is assumed to be ambient air. 

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[Kevin McGrattan]

You should be using the latest version of FDS, 6.8.0, because we are often asked these kinds of questions and we try to clarify things in the manual. 

In short, the THICKNESS on the SURF line is most important. The 1-D heat conduction calculation is done based on the THICKNESS, not the dimensions of the OBST.

[Alberto Silingardi]

Thanks. I will understand if it is better to continue studying on 6.8.0, I downloaded this one to be used with an Italian commercial software that also implements EVAC.

However the guidance I cited should be the latest according to my understanding.

On Wed, Jan 3, 2024 at 7:13 PM Kevin McGrattan <mcgr...@gmail.com> wrote:

You should be using the latest version of FDS, 6.8.0, because we are often asked these kinds of questions and we try to clarify things in the manual. 

In short, the THICKNESS on the SURF line is most important. The 1-D heat conduction calculation is done based on the THICKNESS, not the dimensions of the OBST.

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