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May 12, 2000, 09:49 
Pipe wall temperature

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I have a 2mm thich aluminium pipe that is about 100mm in diameter, about 1m long with numerous bends, curves, variations in diameter etc. Air is blown down at an average of 5m/s and a highly accurate CFD simulation of the fluid conducted (it IS accurate, take that as said). Is there any way of determining the most efficient location of a point source of heat (say, 10W) so as to minimise the temperature that the aluminium pipe at that point reaches from the CFD predicted fluid field alone (e.g. without trial and error thermal simulations done on a nonbuoyant static fluid field). [what I'm after is really: "Is there a relationship between Heat Transfer Coefficient and the Temperature+Fluid field or between Heat Transfer Coefficient and the Fluid field alone, does htc=f(fluid,temperature) or does htc=f(fluid)?"] Fred. 

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May 12, 2000, 10:20 
Re: Pipe wall temperature

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(1). The heat transfer coefficient is derived from the temperature gradient. No temperature gradient, no heat transfer. (2). The temperature gradient comes from the temperature field, which is the solution of the energy equation. (3). The energy equation consists of the diffusion terms (the conduction in fluid), the source terms (the heating element), and the convection terms (coupled through the velocity field). So, in order to solve the energy equation for the temperature field, one needs the velocity field. (4). The velocity field comes from the solution of the continuity equation and the three momentum equations. (5). I am still trying to understand your question. (In some cases, heat transfer coefficient can be related to the skin friction coefficient. But in general, it depends on the energy equation and the momentum equations.)


May 12, 2000, 10:39 
Re: Pipe wall temperature

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Fred,
Chapter 6 of Frank White's "Heat and Mass Transfer" (ISBN 020117099X) has a well laid out definition that will give you a good understanding of convective heat transfer. I'll try to sum it up and hit the high points. The situation you describe is clearly forced convection. In fact, it's probably turbulent since your velocity is 5m/s through a 10cm diameter pipe, although I haven't bothered to calculate a Reynold's number. Typical handbook equations for calculating h require you to first calculate the Nusselt number (which is the ratio of convective heat transfer to conductive heat transfer). Equations for the Nusselt number, for forced convection are functions of the Reynolds number and Prandtl number. The Prandtl number is purely a material property. The Reynolds number is a function of material properties, the tube diameter and the flow velocity. Based on this, the heat transfer coefficient is a function only of the flow. However, (and this is before the pureists jump all over me), there are some caveats. This assumes that the temperature of the fluid is a constant, which it is not, since you have a heat source. Temperature changes will affect the fluid properties, and the Prandtl and Reynolds number. In many cases these will be second order effects, and they can be ignored. In your example, a small 10W heat source will probably not change the temperature of the air by more than a few degrees C. You would therefore be safe assuming constant temperatures, getting air properties for that temperature, and calculating the heat transfer coefficient based only on the flow. If you were dealing with a large temperature change (pick something very large like 1000C), then the changes in fluid properties become very important and can not be ignored. Regards, Alton 

May 12, 2000, 11:01 
Re: Pipe wall temperature

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Alton,
>Based on this, the heat transfer coefficient is a function only of the flow I agree! Though I don't think the changes in fluid properties with T are the sensitive issue. Going back to the original question then ("by looking at just the flow field could you find the best place (lowest resultant wall temperature) for the point source of heat?") you indicate that YES you can. Big important point. There seems to be a contradiation of understanding generally in the literature as to whether HTC is a 'derived' property, something that is calculated 'after' the heat has flowed, or that it is a propert of the fluid, e.g. the fluids' ability to remove heat "should heat ever be released at that point". I'll follow up on John's post where he seemed to capture this contradaition in the last sentence. Cheers Fred. 

May 12, 2000, 11:07 
Re: Pipe wall temperature

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John,
Your point (1) indicates that the efficiency by which heat is removed from a surface can ONLY be determined 'AFTER' the heat has flowed. Intuitively I would think that if the fluid if flowing at 100m/s the HTC would be higher than if the fluid flowed at 5m/s (I'm ignoring the coupling between HTC and T due to buoyancy), nothing to do with temperature. The question is essentially: "Is HTC is a 'derived' property, something that is calculated 'after' the heat has flowed, or is it a property of the fluid, e.g. the fluids' ability to remove heat "should heat ever be released at that point". Your last sentence seems to indicate both. Does the equation Q = HTC A (Twall Tambient) capture the 2 mechanisms that govern efficient heat removal, a low ambient temperature and a high HTC, or is HTC a function of the fluid temperature??? Fred. 

May 12, 2000, 11:51 
Re: Pipe wall temperature

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If you look at handbook or text book type equations dealing with convection, you can clearly see that it depends on what type of convection you're dealing with. For problems that are clearly forced convection, the Nusselt number is a function of only the Prandtl number (material property) and Reynolds number (a non dimentional representation of the flow velocity). The heat transfer coefficient can be calculated from a flow only calculation, assuming temperature dependent material properties are not important, and applied to a seperate heat transfer problem. I've actually done this, and it yields reasonably accurate solutions. The key here is that the heat flux depends on the temperature difference, but the heat transfer coefficient is not temperature dependent.
For problems that are clearly natural convection, there are a different set of equations that can be used to calculate the Nusselt number. These still involve the Prandtl number, but they use the Grashof number instead of the Reynolds number. The Grashof number is a nondimensional number that related to the change in a fluid's density with changes in temperature. It really only has meaning in the context of a heat transfer calculation. The situation that the hand book equations will not help you solve is one of mixed convection where both forced and natural convection act. It's hard to come up with an example, but picture some heat producing machinery in an industrial plant. There is an HVAC system that is circulating air through the room, however, the bulk velocities are not large (or the workers would complain about the wind). The machines make enough heat to generate bouyant flows around them. To calculate the temperature of the machines, you could use the free convection equations, but they would under predict the amount of heat transfered. You could use the forced convection equations, but what's the right Reynolds number to use? If you use one based on the bulk flow velocity, you'll probably also under predict the heat transfer. The only option is to consider natural and forced convection at the same time, and the best way to do that is with a coupled simulation. The situation you describe is clearly one of forced convection. The HTC can be obtained from the flow field and later applied to a heat transfer analysis. Regards, Alton 

May 12, 2000, 12:29 
Re: Pipe wall temperature

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Alton,
>The situation you describe is clearly one of forced convection. The HTC can be obtained from the flow field and later applied to a heat transfer analysis. It was purposly posed as forced convection. So what 'feature' of the flow filed could be used to determine the location of maximum potential HTC. skin friction? Fred. 

May 12, 2000, 15:03 
Re: Pipe wall temperature

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(1). It is very hard to understand your question and problem. (2). I am going to rephrase it and then try to answer it in my way first. (3). A heat source is placed inside a thin aluminum pipe 10cm in diameter. Since the pipe is not a straight one ( complex shape ), the heat source effect is not going to stay along the centerline, when the flow in the pipe is moving at 10m/sec. (4). The heat source will affect the wall temperature in certain ways in general, and it will produce hot spot somewhere on the wall surface. (5). If I only have the flow field information (the temperature field is not available), is it possible to identify the wall surface hot spot as a function of the location of the heat source? (6). The answer is : Yes, from the location of the heat source, you can identify the hot spot location on the wall surface, without solving the energy equation and the temperature field. From the velocity field, do the fluid particle tracing in the postprocessor from the heat source location.(diffusion and body force not included) If the trajectory stays in the center of the pipe (away from the wall), you are not going to get hot spot on the wall. Otherwise, if it moves closer to the wall at certain location, you will likely to get hot spot there on the wall. (7). The heat transfer coefficient is a global parameter useful for certain similar types of problems. The heat transfer coefficient is usually expressed in terms of other global parameters, and the local temperature or difference. It is useful only if the types of problem are closely related, and the primary variables can be easily identified. Once the empirical formula is established, it can be used in engineering design and analysis. The primary requirement is that the flow field and the temperature field must remain similar across the range of the parameters. In a complex flow field associated with complex geometry, a slight change in a parameter can easily change the downstream flow separation pattern. Then it would be very difficult to obtain useful heat transfer coefficient formula. It is similar to creating a formula to estimate the weight of a person based on his height. It is possible only if they are similar in shape, bone structure, etc.. For a single case, the heat transfer coefficient for that case is unique. This is also true for a person in terms of his weight and height. So, you will get the same Q , either from the heat transfer coefficient formula or from the wall interface temperature gradient calculation (solve energy equation). Beyond the single case , it is an engineering formula only.( a formula to cover a range of similar problems) (8). So, the heat transfer coefficient is practical for simple problems only, where the flow field and the temperature field remain similar over certain range of parameters.


May 12, 2000, 16:49 
Re: Pipe wall temperature

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i've had this discussion about the heat transfer coefficient before. h is not a fluid property it depends on the flow, the wall temperature and the materials (solid and fluid) in contact with one another. in forced convection which this seems to be case here the heat transfer coefficient depends only on the fluid flow so if you have an accurate CFD analysis then you can calculate h at all points on the relevant surface with no need to know anything about the temperature of the surface. there are a few caveats though. tpically if the fluid is laminar the correlations for the nusselt number depend on if the wall is at constant temp or specified heat flux and often you need to know in what range the temperature or heat flux are which implies an iterative calculation. for turbulent flows if you assume forced convection then usually the wall condition is unimportant. but since you aren't using correlations here this is probably irrelevant to you. remember though that the assumption of natural convection is only an approximation. it gets better with higher heat transfer coefficients (which usually translates to higher reynold's numbers). the question often arises as to when you should assume forced convection there are several thing sto take into account. (1) is the ratio of fluid buoyancy to fluid inertia high (Rayleigh number)? it is not appropriate to just look at the reynolds number since the fluid might have a low prandtl number (like a liquid metal) in which case buoyancy can be significant even in a high Re flow.(2) the thermal capacity of the body. if the body has a low thermal capacity (or if there is low or no internal heat generation) then the Rayleigh and Grashoff numbers can vary significantly during heat tansfer


May 12, 2000, 22:42 
Re: Pipe wall temperature

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(1). The heat transfer coefficient is defined for a group of similar problems, at the interface wall having the wall temperature Twall and a reference temperature Tambinet. (2). For the equation to be useful with the given h (can be a function of other parameters), the problem must be similar to the original group of problems. (3). So, the concept of the heat transfer coefficient is for the problem as a whole. And the h must be derived from Q/(A*(TwallTambient)) for that group of problems first, through experimental data, or cfd calculations. (4). Once h is obtained this way, it can then be used to evaluate the heat flux Q at the wall for a similar problem (only),in a similar way.(5). So, if you can find a problem with similar nature and conditions, then it is all right to apply the formula to compute the heat flux Q. Otherwise, you will have to run the test or perform the cfd calculations to determine the Q.


May 15, 2000, 03:45 
Re: Pipe wall temperature

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Clifford,
Thanks, good answer. I too believe that a knowledge of the Temperature field is not required prior to deciding the optimimun location (lowest resultant Twall) for the point source of heat on the outside of the pipe for this particular stylised case. I still don't know exactly what parameter of the flow field could be used to infer the variation of the HTC field...?? I still think there are fundementally 2 mechanisms responsible for the efficient removal of heat by air from a solid surface. There is Ta, the lower the ambient air temperature the lower the wall temperature. Then there is the HTC itself. I think of this more as the air's ability to 'scour' heat from the surface, it is this that should be a function of the flow field. All empricial correlations deduce HTC 'after the heat has already flowed' thus Nu contains both mechanisms (this is why Nu is highest near the entrance to a fixed Twall duct, because Ta is lowest, it has not been preheated). So if [Q = HTC A (TwallTambient)] how could HTC (or at least its variation within a CFD solved flow field) be derived from the flow field? What about the Stanton Number? Fred. 

May 15, 2000, 14:42 
Re: Pipe wall temperature

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in forced convection problem you can calculate the heat transfer coefficient at the surface of the body. h = k*dT/dn where k is the conductivity of the fluid and dT/dn is the derivative of the temperature in the direction normal to the wall. as for implementing this you can check out a CFD book like Hirsch, Hoffman or Tannehill Anderson and Pletcher. of course in your solution you can't use an adiabatic wall so usually you impose a specified wall temperature. many CFD analyses don't pay much attention to thermal bc usually the adiabatic one is used. if possible a conjugate heattransfer solution would be best if you could do it.


May 15, 2000, 14:48 
i messed up sorry

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what i meant to say was that since Q=k*dT/dn=h(TwallTref) then his found by changing the usbject of the formula to h (i hope didn't mess up this time.) also you have to figure out your choice for Tref. (Q is heat transferred per unit area)


May 16, 2000, 01:34 
Re: i messed up sorry

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Hold on Clifford, now we're back to using the thermal field to calculate (derive) h!! I thought we agreed that you could just use the flow field in this case.
Actually I was very recently told the solution to my problem . It does involve the Stanton number and a generalised form of the 'TaylorPrandtl analogy'. H can be calculated without a T in sight (just a speed and a tauw). Fred. 

May 16, 2000, 09:41 
Re: i messed up sorry

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(1). If the temperature field solution is similar to the velocity field , one might be able to derive the temperature related global parameters from the velocity related parameters. This is possible for very simple cases, such as flow over a flat plate (check out some heat transfer books). (2). In most cases, the local heat transfer coefficient is not related to the local skin friction coefficient.


May 16, 2000, 10:02 
Re: i messed up sorry

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(2). In most cases, the local heat transfer coefficient is not related to the local skin friction coefficient.
Well that's as about contentious as you can get! I think that most CFD codes that do conjugate heat transfer use the Stanton number approach to calculate a local HTC. How else do you think they do it? The HTC has to be calculated as part of a CFD solution, not derived. So how is it done? Fred. 

May 16, 2000, 13:46 
Re: i messed up sorry

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i wrote before: Q=k*dT/dn=h(TwallTref) then his found by changing the usbject of the formula to h (i hope didn't mess up this time.) also you have to figure out your choice for Tref. (Q is heat transferred per unit area)
you don't need anything but the flowfield solution because k is the thermal conductivity of the fluid dT/dn is the normal temperature gradient of the fluid at the surface, Twall is also given by the solution and Tref is specified by the analyst. for an incompressible flow the solution is straighforward because the energy equation can be solved a posteriori ie the continuity and momentum equations are solved by whatever technique you like (eg pressure based) then the temperature field in the fluid alone is calculated via the energy equation. typically this is done iteratively because the thermal wall bc is not often easy to specify (i'm sure John can give more details i can only speak from a theoretical point of view). 

May 16, 2000, 13:52 
Re: Does it matter how it was derived, calculated, or used?

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(1). If there is a heat source outside a pipe, then one needs to solve three regions, namely, (a). the flow field outside the pipe, including the heat source, (b). the heat conduction inside the pipe wall, and (c). the convection inside the pipe. (2). The unknowns are the Twall,inside and the Twall,outside. (3). With the wall temperatures known, the heat conduction inside the pipe wall can be computed, and this include the heat flux,Q. (4). The Twall,outside can be used as the boundary condition to solve the external flow problem. And it will provide the temperature field along with the wall temperature gradient at the outer pipe wall. From there, one can compute the heat flux,Q also. (5). Then, inside the pipe, the Twall,inside can be used as the wall boundary condition to obtain the temperature field distribution. In the same way, the temperature gradient at the inner wall can be computed from the temperature distribution and provide the heat flux,Q also. (6). At steady state, there must be only one value for the heat flux,Q. And this constraint can be used to adjust the wall temperatures, until the steady state is reached. (7). With the final wall temperatures distributions known ( as a function of the surface position ), there are three distinct and well defined problems, namely, the condution inside the pipe wall, the external heat transfer, and the internal heat transfer in the pipe flow. And each will have its own unique temperature distributions. And it is obtained from the NavierStokes equation and energy equation. (inside the pipe wall, the convection is zero) There is no Heat transfer coefficients involved in the calculations. (8). But, if you don't want to solve the flow field and the temperature field inside and outside the pipe , you can formulate it in a different way. That is, Q=h,inside_wall*A*(Tref,insideTwall,inside), and Q=h,outside_wall*A*(Tref,outsideTwall,outside). The unknowns are still the Twall,inside and the Twall,outside. (9). The problem can also be solved if h,inside and h,outside functions are known. In general, the heat transfer coefficient is not known and must be obtained in a separate exercise through testing or cfd calculations. (10). In case, you don't have the exact formula for h for your specific problem, you can still borrow an existing formula and assume that it is applicable to your problem. As long as you are happy with the results, it is really doesn't matter how you solve the problem, or how the heat transfer coefficients are computed (or derived). If you borrow the wrong formula from a wrong problem, then the results will likely be wrong.


May 16, 2000, 15:04 
Re: Does it matter how it was derived, calculated, or used?

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John, i don't think i'm understanding his physical layout but i'm not sure it's important. if the situation is one of forced convection then all he needs is the flowfield and since the problem is incompressible we can ignore the energy equation isn't that so? as a result the heat transfer coefficient can be calculated from the solution. the only difficulty i can forsee if he's using a time marching method which includes the energy equation is specifying the thermal bc. as you correctly pointed out this is to be done iteratively. as a general comment i think most CFD up til now has given heat transfer a short thrift this is due (i think) to the emphasis on aerodynamic applications where people just assume an adiabatic wall as it makes no difference because they usually don't care about heat transfer. heat transfer calculations bring up the thorny issue of bc. of course as you said if you use a true conjugate solver there is no thermal bc at the wall since the thermal field is continuous. however most people like the two code approach where a CFD code is used for the fluid and an FEM code for the solid. my conceptual approach which i've thought of on my own is to start the calculation with CFD solution with an adiabatic wall (or some other initial guess on the wall heat flux) this will give wall temperatures which can be used by the FEM calculation as a bc. this will result in new wall heat fluxes which are different from the previous ones. then the CFD calc is repeated to give new wall temps etc until the heat flux converges. of course it's not easy to figure out how much to converge each CFD solution (and FEM solution if an iterative technique is used there eg conjugate gradient) each time. it may well be that a good route would be to converge the CFD solution to about 2 to 4 orders of magnitude and then perform the FEM calc every 1 to 5 timesteps subsequently (or do it every 1 to 5 timesteps from the very start) since a fully/highly converged FEM calc is about as quick as a single iteration of the CFD code (i'm assuming a lot here) then the total calculation would probably be only 2 to 3 times as long as a similar CFD only job (2D, in 3D it'd probably be more). i realise this approach is probably not new, i suppose it's the way most people do it. surprisingly, while conjugate heat transfer is something almost every CFD code vendor claims their code can do with ease i haven't seen many papers on the subject. what is your feedback?


May 16, 2000, 18:10 
Re: Does it matter how it was derived, calculated, or used?

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(1). I will have to read your message several times before I can make comments on it. (2). The use of the heat transfer coefficient derived directly from the velocity field or the skin friction coefficient, I think, is possible when you have similar solutions between the thermal field and the velocity field. (3). So, pick some problems, and plot the surface skin friction and the heat transfer on the surface. In the real world problems, I don't think there is a direct relationship between the two, that is my impression. (4). I know that in some simple cases, it is possible to obtain the heat transfer coefficient from the velocity field alone. What I am saying is that, you are free to use that assumption as long as the results to your problem is acceptable and accurate. My feeling is sometimes the heat transfer on the surface is different from the skin friction on the surface. So, it depends on the specific problem you are trying to solve.


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