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April 3, 2010, 08:53 
increased flowrate decreased in heat rejection?

#1 
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Dear all,
i've a confusion on fluidthermal analysis i'm performing. Using CFX, i find that if i increase the flowrate, the heat taken away by the fluid is decreased, because temperature difference between outlet and inlet have decreased. i think this is not logical. but why does CFX gives me such a result? i used a simple pipe model and the wall of the pipe is heated with heat flux applied. Pipe model is OD=45mm, ID=32mm, length L=100mm. Heated area on pipe is on OD wall, where this area's boundaries are 30mm from both ends of the pipe as shown in attached picture, pipe.jpg. Fluid is WATER. Meshing of fluid is finer than solid. GGI mesh connection is used. i've used both SST and BSL Reynolds Stress turbulence option for the fluid domain, and for both turbulence models with 10L/min and 15L/min flowrates, the 15L/min flowrate resulted in a lower inlet/outlet temperature difference than the 10L/min result. I used the following expression to calculate temperature difference: areaAve(T)@outareaAve(T)@in Then heat taken away by fluid: Qfluid=massFlow()@in*(areaAve(T)@outareaAve(T)@in)*4181.7 where 4181.7 is specific heat capacity of water. Does the results makes sense please? Does higher flowrate rejects less heat? Funny enough, the Wall Heat Transfer Coefficient looks logical where the Wall HTC is higher at higher flowrate. See attached pipe02.jpg (left pic: HTC for 10L/min, right pic: HTC for 15L/min) Thank you for your kind attention. i really really like to read any comments/suggestions/ideas you may have.
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April 3, 2010, 11:27 

#2 
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Dmitry Volkind
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Hello!
If You wish to compare how much heat is taken away at different flow rates, then I think You should better use temperature boundary condition. By specifying heat flux You predetermine the amount of heat that gets into the domain, no matter what the flow is doing. And the heat has nowhere else to go but into the fluid through the wall (I guess solid pipe ends are adiabatic). That means You should get approximately equal results using Your equation. I think, something's really wrong with your solution. Did You use conservation target? Try to get rid of the solid domain  less headache with solid timescales. You can easily calculate wall temperature distribution analytically. 

April 3, 2010, 21:36 

#3 
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Hi dvolkind, thank you for your reply.
i didn't use conservation target. will the default value be enough? how do i set this value please? i don't want to get rid of solid domain, because i would like to use such a CHT model to calculate the heat rejection by water on a more complicated solid model. Using this simple pipe model, i'm trying to get a sense how i should setup my complicated model. may i ask, what do you mean by using temperature boundary condition please? Furthermore, this model was setup following most of the steps in the CFX Tutorial: Conjugate Heat Transfer in a Heating Coil. Where could i have possibly go wrong please? I look forward to hear of comments/suggestions. Thank you all!
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April 4, 2010, 04:21 

#4 
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Usually the default value is enough. You can set this value in
Convergence criteria > Conservation target (if You meant that) By saying temperature boundary condition I meant specifying wall temperature instead of wall heat flux. I'll try to explain myself. Using steadystate model we may consider the heat balance of our entire volume (fluid domain together with solid domain). Steady state presumes that spatial temperature distribution inside the volume does not change in time. It means that the heat inflow is equal to the heat outflow: inflow(inlet) + inflow(cylindrical wall) = outflow(outlet) When You specify the same heat flux every time, you actually fix the value of inflow(cylindrical wall). So, the difference between inlet and outlet heat flow is always equal to the fixed heat flow value at the pipe wall. When You set temperature instead of heat flux at the boundary, then the heat flux value depends of the heat transfer conditions inside the pipe. Thus, You are able to compare the amount of heat taken away at different flow rates. 

April 4, 2010, 07:45 

#5  
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Glenn Horrocks
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Quote:
Dmitry, I think mactech needs the CHT domain for later work so the temperature boundary condition will not help. Quote:


April 4, 2010, 09:35 

#6 
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Quote: "Dmitry, I think mactech needs the CHT domain for later work so the temperature boundary condition will not help."
Why not? I guess he can still apply temperature BC to the outer surface of the solid domain. "i find that if i increase the flowrate, the heat taken away by the fluid is decreased"  wrong and not logical, but decreased temperature difference between outlet and inlet have nothing to do with it (see Mr. Horrocks' last statement). Mr. Machtech001 is interested in heat flow difference [Q(outlet)  Q(inlet)]. But, as I said before, it is fully determined by Q(wall), which is fixed. Thus, if a heat flux BC is used then [Q(outlet)  Q(inlet)]=q(wall)*pi*d*l and does not depend on mass flow rate. That's why Mr. Machtech001 has to use temperature BC to solve for the heat flux. Am I wrong? 

April 4, 2010, 10:48 

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Quote:
Quote:
Below are the results of heat taken away by fluid at 10L/min & 15L/min using the expression below: Qfluid=massFlow()@in*(areaAve(T)@outareaAve(T)@in)*4181.7 5528.34W(@10L/min), 5436.39W(@15L/min). Funny enough, it is greater than the 5kW heat flux applied. (area of heat flux applied is 5654.87mm^2, so heat flux applied on this area is 884193.624Wm^2 in CFX.).... why? Flowrate is in kg/s, so Water density in CFX is 997kgm^3, at 15L/min = 0.24925kgs^1; at 10L/min=0.1661667kgs^1 Comments appreciated.
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April 4, 2010, 12:23 

#8 
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Quote: "i can't apply temperature BC, because i want to find out the temperature at the surface with known heat loss (5kW) and fluid flowrate. ... I'm interested to know the heat taken away by the fluid at different flowrate, not heat flow difference between inlet and outlet."
"Heat loss" is the amount of heat taken away by the fluid. If You know it's 5 kW, then what do You solve for? Applying the heat flux BC is the right thing to solve for the surface temperature, but not for heat taken away. Using Your formula (which is actually the heat flow difference between inlet and outlet) is a good way to check the global energy conservation if You don't trust the "conservation target". By the way, it might be significant to consider variable heat capacity. Quote: "5528.34W(@10L/min), 5436.39W(@15L/min). Funny enough, it is greater than the 5kW heat flux applied. (area of heat flux applied is 5654.87mm^2, so heat flux applied on this area is 884193.624Wm^2 in CFX.).... why?" If You are running steady state, then it should be exactly 5 kW (theoretically), for those 5 kW You provide are the only heat source for the fluid. Why it's not exactly 5 kW? I can't say for sure, but I see 4 possible measures to fix it: 1. Make specific heat capacity temperaturedependent. 2. Review convergence criteria. 3. Use double precision. 4. Refine mesh. My point is: solve for heat loss OR for outer wall temperature, not both of them at the same time. 

April 5, 2010, 08:19 

#9 
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Glenn Horrocks
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As Dmitry has said, it looks like the problem is apparent now  mactech's simulation is not fully converged and that is why one run gives 5528W and the other gives 5436W. These are just random numbers within the numerical accuracy of an inadequately converged solution, so comparing them is meaningless. Further convergence should bring that back to 5000W.


April 5, 2010, 21:07 

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Hi Dmitry and Glenn, thank you so much for your replies and comments.
my solution did in fact converged. please see attached graphs. all RMS residuals stopped equal or below 1e5. i've tried to use another model, to try to include air convection BC as well. this is attached as "pipeTH3.jpg". Heat source is the bottom volume of 3kW. Air convection is imposed on the highlighted areas in 'pipeTH3.jpg'. Fluid settings: flowrate=15L/min,inlet temperature=25C. Air convection: HTC=5W/m^2/K, outside temperature=25C. Solution shows convergence to RMS residual=1e5. Heat taken by fluid(water in this case) is 3113.827W. Does it mean that air convection setting is not happening please? Quote:
Look forward to comments soon.
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April 5, 2010, 21:45 

#11  
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And Dimtry's point about what are you trying to solve for is the key point. Either you specify the heat input and get the temperature field as a result, or you specify the temperature inputs and get the heat flow as a result. Reading back through your posts I cannot see what you are actually trying to do. 

April 6, 2010, 03:19 

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Quote:
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Fluid settings: flowrate=15L/min,inlet temperature=25C. Air convection: HTC=5W/m^2/K, outside temperature=25C. Currently, I have the following Domain Imbalance values: HEnergyfluid Domain Imbalance=0.132% TEnergysolid Domain Imbalance=0.1788% TEnergyheatsource Domain Imbalance=0.0639% Checking heat taken away by fluid, Q1= 3113.82W With FINER MESHING on the fluid, I have the following Domain Imbalance values: HEnergyfluid Domain Imbalance=0.1413% TEnergysolid Domain Imbalance=0.2450% TEnergyheatsource Domain Imbalance=0.0547% Checking heat taken away by fluid, Q2= 3303.11W > Q1 how come? What else should I check please? Quote:
Then, I tried imposing temperature BC on the outer wall where it is exposed to air, i.e. temperature=25C, and the heat taken away by fluid, Q3=232.422W. Assuming remaining heat is taken away by the 25C air, do I calculate it through: areaInt(Wall Heat Flux)@outerwall please? Look forward to more discussions/comments please.
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April 6, 2010, 06:29 

#13 
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Dmitry Volkind
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Quote: "Is it enough to monitor HEnergy of fluid and TEnergy of solid domain imbalances please?"
Yes, should be enough. Quote: "Heat source is the bottom volume of 3kW" Why do You model the bottom volume? Can't it be raplaced with heat flux? Try to make your model as small as possible. You can't model the whole universe anyway) Quote: "Q2= 3303.11W > Q1 how come?" No idea at all. Quote: "I would like to do the former, i.e. I have heat input and get temperature field, but imposing air convection since outer wall of solid is exposed to air." Are You sure You want to do that? You definitely don't need it with circular pipe. But If You wish to do the same thing on a complex geometry, then don't forget to use bouyancy model. By the way, using walls as outer boundaries for air is incorrect if natural convection is modelled. areaInt(Wall Heat Flux)@outerwall  OK, still I wouldn't make it a wall. 

April 6, 2010, 07:11 

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Thanks for your reply Dmitry.
Quote:
how should i model natural air convection please?
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April 6, 2010, 07:38 

#15 
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Quote: "Why use buoyancy model?"
Because it's a little easier than setting momentum source manually. "how should i model natural air convection please?" 1. Enable bouyancy, set bouancy ref. temperature and gravity. 2. Move air domain boundaries further away from the pipe. 3. Change outer nonsleep walls to freeslip walls/inlets/outlets/openings. My advice is to model heat transfer for water and air separately and connect both models with evaluated HTC's (but do not use standard HTC fron the .outfile). Solve water first, specifying a reasonable value for air HTC. Then solve air, specifying the HTC You got from water solution. Then solve water again with corrected air HTC and so on, until [HTC(solved)HTC(specified)]/min[HTC(solved),HTC(specified)]=desired tolerance, %. 

April 6, 2010, 18:50 

#16 
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Glenn Horrocks
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I think at this time it is a good idea to ask mactech what he/she is trying to achieve with this model. Dmitry's comments about modelling the convective boundary may or may not be a good idea, it depends on what you are trying to achieve.
So what are you looking for with this simulation? 

April 6, 2010, 20:51 

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To answer Glenn's question above, i've attached a picture, cfxOutput.jpg, of a simple model that represents the eventual complex geometry. It is a cylindrical model that can accommodate an electrical machine in the inner space. In this cylindrical housing, there is an internal cooling channel with ribs to increase the effective surface area for heat convection. The electrical machine in the housing will have losses. Losses are applied to the model as heat flux. I am trying to obtain a design that is to use water that flows in the channel to take away all the losses. Outer surfaces of the cylindrical housing is air. Does buoyancy modelling matter in such a case please? I started off by following most of the setup in the CHT Tutorial of a Heating coil. no mention of buoyancy modeling. How then can i improve the air convection modeling please? i look forward to hear any comments/suggestions you may have.
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April 7, 2010, 04:09 

#18 
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This time it's correct to apply heat flux to determine water flow rate required. Heat flow has a choice now: it might be carried away with both water and air.
Quote: "Does buoyancy modelling matter in such a case please?" Bouyancy does matter on the outer surface, if natural air convection is being modelled. As for the water domain, bouyancy may or may not matter. Basically, the contribution of natural convection to heat transfer is judged by Grashof and Reynolds numbers. Their correlation yields viscous, gravitational or viscousgravitational flow regime. Depending on the regime, You may or may not have to consider bouancy. Quote: "I started off by following most of the setup in the CHT Tutorial of a Heating coil. no mention of buoyancy modeling." Due to high flow rate and low thermal expansivity of water natural convection doesn't matter in this case. Or might be modelling error, but it's OK for tutorial purpose. If interested, calculate Re and Gr for that tutorial device and check which regime is that. Quote: "How then can i improve the air convection modeling please?" If that thing in the picture is Your actual geometry, then use empirical formula for HTC (Nu = F (Re, Gr, Pr, etc.)) for a cylinder, it should be accurate enough. The HTC value calculated for pure natural air convection should lie somewhere between 1 and 20 kW/(m^2*K). And keep in mind that HTC strongly depends on temperature difference for natural convection. 

April 7, 2010, 07:35 

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April 7, 2010, 09:26 

#20 
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Bouyancy is not a boundary condition, it's a fluid model option. By saying "Bouyancy does matter on the outer surface" I presumed that air domain surrounding the device is modelled. I also said that You don't really need to model natural air convection for that particular device in the picture You provided, for its shape is primitive and You may use empiric formula for HTC.


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