[Antanas]

Originally Posted by techtuner

Look, in Celsius difference between 2.6 C and 0.7 C in % with average liquid temperature about 28C is very high. In other word it isn't correct to present temperature in %.

I'd used double presision solver in computations and the same result for temperature was obtained.

In my case Qwall isn't high enough. By this reason dT is low. That's why I shouldn't consider natural convection and inconsistency of properties in a liquid that may affect on the heat transfer coefficient.
But low dT isn't the reason of the fault in heat transfer coefficient that may observed due to numerical errors. That's because of accuracy of the numerical simulation (double) is much higher than observed error.

If you use 'Wall Heat Transfer Coefficient' variable to calculate Nu number you'll get value that is more close to what you want (I got 785 w/o 'tbulk for htc' and 440 with 'tbulk for htc' = 300 K).

[techtuner]

Originally Posted by Antanas

If you use 'Wall Heat Transfer Coefficient' variable to calculate Nu number you'll get value that is more close to what you want (I got 785 w/o 'tbulk for htc' and 440 with 'tbulk for htc' = 300 K).

Wall Heat Transfer Coefficient in CFD Post is different parameter from the traditional Wall Heat Transfer Coefficient in a pipe. In my case it is useless.

In CFX alpha computed as alpha=Qwall/(Twall-T1stLayer) - without 'tbulk for htc' and alpha=Qwall/(Twall-TbulkForHTC) with determined T bulk for HTC.
Here the T1stLayer is temperature of the first node near the wall, TbulkForHTC is 'tbulk for htc'=300 K.

For industrial facilities much more interested parameter is Wall Heat Transfer Coefficient that determined as alpha=Qwall/(TwallAve-TLiquid).
Sometimes difference in Wall Heat Transfer Coefficient in 20-30% is critical. In case of 300% error CFX looks like completely useless CFD code for convective heat transfer simulation opposite to Fluent or Star-CCM+.

[Antanas]

Originally Posted by techtuner

Wall Heat Transfer Coefficient in CFD Post is different parameter from the traditional Wall Heat Transfer Coefficient in a pipe. In my case it is useless.

In CFX alpha computed as alpha=Qwall/(Twall-T1stLayer) - without 'tbulk for htc' and alpha=Qwall/(Twall-TbulkForHTC) with determined T bulk for HTC.
Here the T1stLayer is temperature of the first node near the wall, TbulkForHTC is 'tbulk for htc'=300 K.

For industrial facilities much more interested parameter is Wall Heat Transfer Coefficient that determined as alpha=Qwall/(TwallAve-TLiquid).
Sometimes difference in Wall Heat Transfer Coefficient in 20-30% is critical. In case of 300% error CFX looks like completely useless CFD code for convective heat transfer simulation opposite to Fluent or Star-CCM+.

Ok. Then make additional variable for Tw and check 'boundary only field'. Set it to be equal Temperature in domain options. Then use it in CFD post as Twall in your expression for Nu. I got 596.756 for Nu using this approach. Also it may be useful to read sections 2.7.5.6.-2.7.5.9 in CFX Modelling Guide.

LIBRARY: 
  ADDITIONAL VARIABLE: Tw
    Boundary Only Field = On
    Option = Definition
    Tensor Type = SCALAR
    Units = [K]
    Variable Type = Unspecified
  END
END

FLOW: Flow Analysis 1
  DOMAIN: HeXe
    FLUID MODELS: 
      ADDITIONAL VARIABLE: Tw
        Additional Variable Value = T
        Option = Algebraic Equation
      END
    END
  END
END

[techtuner]

Originally Posted by Antanas

Ok. Then make additional variable for Tw and check 'boundary only field'. Set it to be equal Temperature in domain options. Then use it in CFD post as Twall in your expression for Nu. I got 596.756 for Nu using this approach. Also it may be useful to read sections 2.7.5.6.-2.7.5.9 in CFX Modelling Guide.

LIBRARY: 
  ADDITIONAL VARIABLE: Tw
    Boundary Only Field = On
    Option = Definition
    Tensor Type = SCALAR
    Units = [K]
    Variable Type = Unspecified
  END
END

FLOW: Flow Analysis 1
  DOMAIN: HeXe
    FLUID MODELS: 
      ADDITIONAL VARIABLE: Tw
        Additional Variable Value = T
        Option = Algebraic Equation
      END
    END
  END
END

Thank you for your reply.

I have got Nu~572 (sidewall temperature averaging procedure was Twall=areaAve(Tw)@Sidewall) in case of coarse mesh (y+~45). Theoretical one value is Nu=575.

So. The error of wall heat transfer coefficient computation was connected with the sidewall temperature computation.

It don't clear for me why averaged sidewall temperature for additional variable Tw different from averaged sidewall temperature for Temperature field. I have read Modelling guide 2.7.5.6-2.7.5.9 but there is no answer.

[Antanas]

Originally Posted by techtuner

Thank you for your reply.

I have got Nu~572 (sidewall temperature averaging procedure was Twall=areaAve(Tw)@Sidewall) in case of coarse mesh (y+~45). Theoretical one value is Nu=575.

So. The error of wall heat transfer coefficient computation was connected with the sidewall temperature computation.

It don't clear for me why averaged sidewall temperature for additional variable Tw different from averaged sidewall temperature for Temperature field. I have read Modelling guide 2.7.5.6-2.7.5.9 but there is no answer.

Because with 'Boundary Only Field' Tw contains hybrid (boundary) values of Temperature. When you calculate something using CEL it uses conservative values. In this case areaAve(T)@Sidewall is the same as areaAve(Wall Adjacent Temperature)@Sidewall. In section 2.7.5.6 of Modeling guide it is said that CFX uses hybrid value of T at wall to calculate Wall Heat Flux, WHTC an so on. And it calculates WHTC correctly.

[ghorrocks]

Nice work Antanas. Kudos for that one.

[techtuner]

Originally Posted by Antanas

Because with 'Boundary Only Field' Tw contains hybrid (boundary) values of Temperature. When you calculate something using CEL it uses conservative values. In this case areaAve(T)@Sidewall is the same as areaAve(Wall Adjacent Temperature)@Sidewall. In section 2.7.5.6 of Modeling guide it is said that CFX uses hybrid value of T at wall to calculate Wall Heat Flux, WHTC an so on. And it calculates WHTC correctly.

Now it's clear. Thank you!

[leasken]

Originally Posted by Antanas

Ok. Then make additional variable for Tw and check 'boundary only field'. Set it to be equal Temperature in domain options. Then use it in CFD post as Twall in your expression for Nu. I got 596.756 for Nu using this approach. Also it may be useful to read sections 2.7.5.6.-2.7.5.9 in CFX Modelling Guide.

LIBRARY: 
  ADDITIONAL VARIABLE: Tw
    Boundary Only Field = On
    Option = Definition
    Tensor Type = SCALAR
    Units = [K]
    Variable Type = Unspecified
  END
END

FLOW: Flow Analysis 1
  DOMAIN: HeXe
    FLUID MODELS: 
      ADDITIONAL VARIABLE: Tw
        Additional Variable Value = T
        Option = Algebraic Equation
      END
    END
  END
END

Would someone mind explaining this more carefully? I have tried to add an Additional Variable, but cannot select the Boundary Only Field.
Trying to edit the variable in the command editor results in

"The parameter "Boundary Only Field" is present in the object "/LIBRARY/ADDITIONAL VARIABLE:Tw" but it is not physically valid."

What am I doing wrong?

Edit: Found the problem. I had not enabled Beta features. Now it works fine. Thanks for the posts!

[leasken]

Hm. This super variable did not help me..

So I hope that some of you guys will have some ideas of where I am thinking wrong. I have a square duct with a specified heat flux from two opposite walls.

The fluid is air as an Ideal gas. The duct is approximately 20 hydraulic diameters long with a Reynolds number of 80 000.

The mesh has a y+ of ~0.5 and turbulence model is k-omega sst with re-attachement modification (not specifically needed for this smooth channel case, but comes in handy when adding for instance ribs to the walls).

The inlet has a boundary condition of fully developed flow profile with an average velocity of 16 m/s and a constant temperature of 293.15 K. The outlet has relative pressure 0 Pa and the heated walls a heat flux of 1000 W/m^2. The unheated walls are considered to be adiabatic. The reference pressure is set to 1 atm.
Nusselt number integrated from area averaged values on spanwise sections.
Nu = Qwall*Dh/(k*(Twall-Tfluid))

Tfluid is the average temperature of the temperatures at the inlet and outlet, Dh hydraulic diameter, k average conductivity from inlet-outlet and Twall the temperature at the wall.

Results:
Dittus-Boelter: Nu ~175
CFX: Nu ~120

When I use the above mentioned additional variable I get almost the same temperatures on the wall. areaAve(Temperature)@wall1 = 342.351 K and areaAve(Tw)@wall1 = 342.539 K.


Any ideas of what to test next? Should I expect the Nusselt number to be around the Nu of Dittus-Boelter even though I only have two heated walls out of four?

[techtuner]

Originally Posted by leasken

Hm. This super variable did not help me..

So I hope that some of you guys will have some ideas of where I am thinking wrong. I have a square duct with a specified heat flux from two opposite walls.

The fluid is air as an Ideal gas. The duct is approximately 20 hydraulic diameters long with a Reynolds number of 80 000.

The mesh has a y+ of ~0.5 and turbulence model is k-omega sst with re-attachement modification (not specifically needed for this smooth channel case, but comes in handy when adding for instance ribs to the walls).

The inlet has a boundary condition of fully developed flow profile with an average velocity of 16 m/s and a constant temperature of 293.15 K. The outlet has relative pressure 0 Pa and the heated walls a heat flux of 1000 W/m^2. The unheated walls are considered to be adiabatic. The reference pressure is set to 1 atm.
Nusselt number integrated from area averaged values on spanwise sections.
Nu = Qwall*Dh/(k*(Twall-Tfluid))

Tfluid is the average temperature of the temperatures at the inlet and outlet, Dh hydraulic diameter, k average conductivity from inlet-outlet and Twall the temperature at the wall.

Results:
Dittus-Boelter: Nu ~175
CFX: Nu ~120

When I use the above mentioned additional variable I get almost the same temperatures on the wall. areaAve(Temperature)@wall1 = 342.351 K and areaAve(Tw)@wall1 = 342.539 K.


Any ideas of what to test next? Should I expect the Nusselt number to be around the Nu of Dittus-Boelter even though I only have two heated walls out of four?

Nusselt number that you were obtained by CFX is pretty accurate from my point of view. There it is two main reasons.
1. You have only 2 out of 4 heated walls. That means you should to obtain twice lower Nu number;
2. You didn't have stabilization of thermal field at the inlet of the pipe. This condition lead to incresing of heat transfer coefficient in comparison to stabilized flow. And don't forget that Dittus-Boelter equation is applicatiable only for the flow with stabilized heat transfer.

So, try to compute heat transfer in long pipe (approximately 100 gauges) with uniform wall heating. Then try to compute averaged by perimeter of the pipe wall Nu number in cross section with longitudinal coordinate equal to 80 gauges. I'm shure, you will obtain pretty good coincidence of Nu number with value that you were computed by Dittus-Boelter equation.

You have been obtained low difference between temperatures, computed by common Temperature variable and temperature, that was recorded in Additional Variable, due to low Pr number of air and low value of y+. Try to compute with y+=50, difference will be much higher. And don't worry about quality of computation at high y+. Just beleive in near wall functions )))

[leasken]

Originally Posted by techtuner

Nusselt number that you were obtained by CFX is pretty accurate from my point of view. There it is two main reasons.
1. You have only 2 out of 4 heated walls. That means you should to obtain twice lower Nu number;
2. You didn't have stabilization of thermal field at the inlet of the pipe. This condition lead to incresing of heat transfer coefficient in comparison to stabilized flow. And don't forget that Dittus-Boelter equation is applicatiable only for the flow with stabilized heat transfer.

So, try to compute heat transfer in long pipe (approximately 100 gauges) with uniform wall heating. Then try to compute averaged by perimeter of the pipe wall Nu number in cross section with longitudinal coordinate equal to 80 gauges. I'm shure, you will obtain pretty good coincidence of Nu number with value that you were computed by Dittus-Boelter equation.

You have been obtained low difference between temperatures, computed by common Temperature variable and temperature, that was recorded in Additional Variable, due to low Pr number of air and low value of y+. Try to compute with y+=50, difference will be much higher. And don't worry about quality of computation at high y+. Just beleive in near wall functions )))

Thanks for the response!

I made a major error when constructing the mesh and with a new mesh methodology I obtain Nusselt values that are +- 10% of the DB-correlation. The mesh was poorly constructed from my side, eventhough the y+ was below 1, the transition from inflation layers to the bulk mesh was not sufficiently balanced.

I deem these results as accurate. However, I agree with you regarding the thermal field and that is of course contributing to the results being a bit inaccurate.

And regarding the Nusselt number. I do not agree with you that I should expect it to be halved compared to a uniformly heated channel. I have read in literature and made simulations that suggests that the Nusselt number is not highly dependent on the wall temperature. And the DB-correlation requires no temperature input, which is also pointing towards a Nusselt number being more or less the same, for a fixed geometry, whatever the temperature boundary conditions are.