Convective Heat Transfer Validation
 

Hello,
Just trying a validation of forced convective heat transfer of a flat plate. And comparing to a hand calculation.

The problem is a flat plate at a temperature of 60C with air at 25C and flowing at 35ms-1 Then trying to work out the convective heat transfer.

The Nusselt Correlation I am comparing it to is the Laminar-Turbulent Average formula
Nu=(Pr^1/3(0.037(Re^0.8)-871)

With a length of 0.75m I make the reynold number to be 1.53x10^6

Using a film temp of 42.5C I work out that the nusselt number works out to be 2152.37
The heat transfer coefficient to be 79.03 W/m^2K
And assuiming an area of 0.00075 due to a 1mm wide cfd I calculate 2.07W

I set up a box as an air domain with length 0.75m and width 1mm and plenty of height. I set the sides as free slip walls. I used the SST solver with no transitional turbulence. When I used the function calculator I use area int with wall heat flux on the hot surface and it gives me a figure of 2.50w

I have noticed odd behaviour with the temperature plot on the hot surface right near the inlet (see pictures)

Any light on what this may be, I believe that the CFD is predicting Turbulent flow way before the reynolds number reaches 5x10^5 but not sure why it would do this.

Mesh is one element thick with tiny thickness towards hot surface.

Thanks Callum

You are not using a transitional turbulence model, so the turbulence model is assuming the whole thing is turbulent. If you want to capture the effect of the turbulence transition you need to add a turbulence transition model. But be aware that the gamma-theta model in CFX is tuned for aerofoil results and I think it unlikely it will work well on this type of flow. You probably will have to define the turbulence transition point yourself.

What is odd about your temperature plot? You appear to be flowing air through a narrow duct, so the air rapidly assumes the hot plate temperature and no heat transfer occurs after that as there is no temperature difference. Your text appears to be talking about heat transfer into an infinite fluid, so you are going to have to model a much bigger depth of fluid to approximate an infinite fluid.

Thanks for the reply,

What I essentially am trying to acheive here is to replicate the average laminar turbulent heat transfer equation on a flat plate. Is it possible to set the model to be laminar until the reynlods number is high enough for the flow to turn to turbulent? My current understanding is I can set the solver to assume only Laminar flow, or only turbulent flow. When I tried to use the gamma theta model wich in my understanding will account for laminar flow, the heat transfer increased. I assume you are saying also that the depth of the domain must be increased (i.e. in the direction of the boundary layer and not in the width) to acheive accurate results.

A slight side note I am wondering how CFX solves conduction. As I understand the equation to be Q=k/s*A*(t2-t1), how is it that I can input an initial temperature on a solid, with a thickness and a thermal conductivity but without a final temperature and yet the solver calculates conduction ?

Thank you

What is "laminar turbulent heat transfer equation"?

Yes, using the turbulence transition model. As I previously mentioned I suspect the built in gamma-theta model will not work in this case so I doubt it will help much. In your case you will probably need the specified intermittency model where you can define the laminar and turbulent regions manually.

If you are modelling a flat plate in an infinite fluid then yes, you need enough fluid such that it acts as an infinite fluid.

I don't understand your final sentence. CFX solves the momentum, pressure and heat equations (and other equations if relevant) on a 3D mesh, and this can capture effects like heat flux through conducting solids. I don't understand what you are actually asking.

Sorry for the late response,

What Im essentially trying to do is to assess the accuracy of CFX solving Conductive heat transfer, by using a solid domain with a fluid domain on top, can I use the area integral of the wall heat flux of the fluid side of the domain to give a power of convective heat transfer. Then use that number for Q in Q=k/s*A*(T2-T1) assuming that in a steady state the conservation energy should mean the convective and conducive heat transfre is equal.

This gives me a very slightly different temperature than the CFX does when I use the average domain side 2 temperature on the function calculator. Can I use this difference to calculate the accuracy to which CFX calculates conduction.

I dont understand how if CFX uses the conduction equation Q=k/s*A*(T2-T1) I can set up a 3d solid in cfx and input a temperature on the bottom surface (T2). I understand that CFX has values for A, K, S and T2 but if it has no values for T1 and Q how can it solve this equation.

Thanks

I don't understand your question.

The approach CFX uses to solve the equations is in the theory documentation. In short, for a laminar flow (I assume this is laminar); the heat flow in each element is determined by the heat equation across that element with the addition of convective heat in the fluid region, and at the solid/fluid boundary the element face on the boundary for the fluid element and the solid element are set to be the same.