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Nusselt number 4.36 validation in a tube in constant heat flux situation |
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May 2, 2012, 09:00 |
Nusselt number 4.36 validation in a tube in constant heat flux situation
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New Member
Felix
Join Date: Feb 2012
Location: Qingdao,China
Posts: 5
Rep Power: 14 |
Hi,guys.
According to classic heat transfer knowledge ,the Nusselt number in a fully developed tube with constant heat flux should be 4.36 . CFD code like fluent is so powerful ,so I want to validate the problem in fluent .And if I model right ,I should get the right Nusselt Number. But maybe there is something wrong in my model process , I just can not get the Nusselt number as described above . These is the process below. according to experience :the length of entry region l=0.05Re×Pr。 Material is water : and I set Re=100. Re=Rho×velocity×D/viscousity. then I get inlet veloctiy=0.1m/s. Pr(30Celsius-60Celsius)is about 3 .then we get the entry length l=0.05×100×3=0.15m. I build a cylinder in Gambit with the dimension below: D=0.1m,L=0.3m. Boundary condition: inlet velocity :v=0.1m/s .T=300K. outflow in outlet。 heat flux of wall =q=400W/m^2. the conductivity of water :lamda=0.6. I use the 3D single precise solver to solve the problem. Set the residual to 10-4. At the same time ,I set a temperature monitor point in the middle point at the tube . The calculation of Nusselt number: Nu=h*D/Lamda. h=q/DletaT=q/(Twall-Tref). Twall is the temperature of wall of the tube ,Tref is the average temperature of the section I choose in the fully developed region. I create a section in fully developed region . plot temperature coutour of this section ,and I get max temperature in the wall is 300.948K . The difficulty comes as I want to calculate the average temperature of the section I created. According to heat transfer textbook ,Tref=Tm=(Intergral of(u×T)in seciton )/(pi×(D/2)^2×Veloctiy)。 The result of the section is 0.002341306。So Tref=Tm=0.002341341/(3.1415926×0.05^2×0.1)=298.2555K. Then h=q/(Twall-Tref)=400/(300.948-298.2555 )k=148.560817084494 w/(m^2.k). At last ,I get Nu=h×D/Lamda=148.560817084494*0.01/0.6=2.4760136180749 != 4.36. I don't know what was wrong with the above process .Can somebody so kind to point the error in my process ? |
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