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Old   November 24, 1998, 15:27
Default a problem with Boundary condition
  #1
M Rad
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I have a very simple CFD problem but dont know how to find a boundary condition.

It is a very simple diffusion and convection problem, however, the boundary conditions seems to be a problem for me.

Consider the flow of air inside a rectangular duct. Only one of the four faces of the duct is heated. To find the temperature profile inside the duct, I was going to use the Finite difference method to solve the governing differential equation (steady state). However I dont know, how to define the boundary condition where air exits the duct. We have an equation that total heat added inside the duct is equal to increase in outlet enthalpy, however this only gives a Summation equation.

Thanks, M_Rad
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Old   November 24, 1998, 15:48
Default my correct email
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M Rad
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Sorry my email address is wrong in the original message.

the correct one: MRAD@hotpop.com
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Old   November 24, 1998, 15:51
Default Re: a problem with Boundary condition
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Abdul Aziz Jaafar
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Dear Mehdi

If the problem is incompressible steady flow in certain length of pipe, you could set boundary condition at the exit fixed pressure boundary, fixed mass flux or derivative boundary conditions for static/total enthalphy/temperature.

As you are solving this problem numerically, it certainly have very small error in overall energy balance as the solution converge.

Hope this help A. Aziz Jaafar
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Old   November 24, 1998, 18:51
Default Re: a problem with Boundary condition
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M Rad
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It is actually a simpler version of looking for both Temp. and velocity. just looking for Temp. profile, assuming uniform velocity profile, and I dont have the Temp. or its derivative at the exit.

regards
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Old   November 25, 1998, 03:38
Default Re: a problem with Boundary condition
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Abdul Aziz Jaafar
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Dear Mehdi

What about zero derivative for temperature

Hope this help

A. Aziz Jaafar
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Old   November 25, 1998, 14:25
Default Re: a problem with Boundary condition
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Adrin Gharakhani
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>just looking for Temp. profile,assuming uniform velocity profile.

Do I understand you correctly that you have uniform velocity in the entire duct? If so, there are two separate issues.

1- uniform velocity implies slip velocity BC on the walls - and I assume you have walls since you said flow in a duct. This is physically impossible, and unless you are trying to verify the code for something, you are going to get meaningless results for your temperature no matter what BC's you use.

2- if you are not concerned with physics, then a uniform velocity profile simplifies the problem significantly to the point that you won't even need CFD for this. You should be able to get an (semi?) analytic solution for this.

Adrin Gharakhani
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Old   November 25, 1998, 16:19
Default Re: a problem with Boundary condition
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M Rad
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I dont see any problem with taking a zero derivative for T except, It is the same condition as for the three walls with no heat.

Is'nt it a problem? (for three non heated walls the BC is a Zero drivative, and there is no air flow: While for the exit there is air flow)

many tanks again
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Old   November 25, 1998, 16:29
Default Re: a problem with Boundary condition
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M Rad
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Thanks, But for a moment, forget about what is happening inside the duct. Even if we try to solve both governing equations for the momentum and heat, we still need a BC for the exit air temperature.

What BC for Temp. we can consider for exit air? (no matter how to look at the problem)

best regards M.Rad
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Old   November 25, 1998, 17:43
Default Re: a problem with Boundary condition
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Adrin Gharakhani
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>What BC for Temp. we can consider for exit air? (no matter how to look at the problem)

Well, if you can assume that the duct is long enough so that it reaches fully developed thermal profile, you can use the zero temperature gradient in the axial direction at the exit.

Remember that your problem is not unique... The correct choice of the exit boundary condition is a problem, be it for the momentum or the energy equations. You must assign a BC to solve the problem and you have a handful of options; the gradient BC being one. You could also assign a temperature profile at the exit (based on some sort of physical analysis of the exit condition, or based on available experimental data for the particular problem)

Adrin Gharakhani
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Old   November 25, 1998, 17:54
Default Re: a problem with Boundary condition
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Adrin Gharakhani
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>I dont see any problem with taking a zero derivative for T except, It is the same condition as for the three walls with no heat.

If you have a duct flow, then there are _only_ 2 walls, one inlet and one exit (which are not walls). If you claim there are 3 walls, then please be specific about the inflow condition and the type of problem you are trying to solve (unless you have cavity-type problem).

In case of a 2-wall problem, you can assign a temperature profile for the inlet and a zero-gradient BC for the exit (and the 2 walls) and the problem will be well-posed.

But I have to take issue with your overall approach to solving this problem. It seems that you are looking for "correct" BC's, but care little about the physics!! You have to remember that the BC's are part of the solution package, different BC's with the same PDE will predict totally different flow phenomena. You can't talk about finding a "valid" BC in abstract terms.

Adrin Gharakhani
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Old   November 27, 1998, 07:31
Default Re: a problem with Boundary condition
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Abdul Aziz Jaafar
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Dear Mehdi

I think you are right that the zero derivative at the exit is same way that you define zero derivative at the wall.

I hope if I am wrong someone could correct it for me. As far as I understood, when zero derivative for temperature is set at the wall, this mean that this is an adiabatic wall, or no heat through it(it does not mean unheated wall).

I think the exit temperature then will be the same as the inlet temperature. Provided that the flow is stationary, the flow remain stationary if no body forces is presence.

Hope this help

A. Aziz Jaafar

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Old   November 27, 1998, 12:07
Default Re: a problem with Boundary condition
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M Rad
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You are totally right that the ["when zero derivative for temperature is set at the wall, this mean that this is an adiabatic wall, or no heat through it(it does not mean unheated wall)]. OK

However, I did not understand the following!~!

[I think the exit temperature then will be the same as the inlet temperature. Provided that the flow is stationary, the flow remain stationary if no body forces is presence.]

If it is stationary, yes the temp. profile is symetric, however there is flow in and out from the duct (someone forces it and we assume a constant uniform flow in the duct).

Now the only thing we know is that total heat added by coduction to the fluid is equal to increase in the DelT out from the system. But it does not give a simple BC. the equation simplifies to a SUMATION form, and also I am not sure if it is an independent equation.

Thanks again
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Old   November 27, 1998, 12:49
Default Re: a problem with Boundary condition
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Abdul Aziz Jaafar
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Dear Mehdi

I am glad that we agree in the matter.

Again if I am wrong please do correct my understanding.

If the problem is in adiabatic system, you expect changes in temperature, provided that no frictional heating occur in the system. If there is frictional effect then you can find changes in your temperature at the exit. Normally this could happen when the Eckert number is higher.

Assume that the frictional effect can be neglected, there is no effect of the flow on the temperature (isothermal flow)

The motion in the duct is created by the pressure gardient ( ex. pump), friction ( as in Coutte flow ) or body forces such as bouyancy for example.

Yes you are right that the total heat will equal to the net increase in the enthalpy. Do not worry about the overall energy balance.

If you still not performing any computation yet. Just set the constant amount of heat flux through the walls of the duct and set zero derivative at the exit and let the computer calculate for you for a simple start. Then we can see any changes in the exit temperature profile compare to the inlet.

Hope this help
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