【讨论】对流换热系数跟理论解不匹配？

ainite

为了验证热边界层理论的Pohlhausen解，作了一个小例子，但结果中对流换热系数跟边界层解差了几倍，不知原因何在？

算例如下图所示，并附上求解命令流：

/PREP7
ET,1,FLUID141
RECTNG,,6,,1
LESIZE,2, , ,20,10
LESIZE,4, , ,20,.1
LESIZE,1, , ,80
LESIZE,3, , ,80
AMESH,1

/SOL
DL,4,,TEMP,20,1
DL,1,,TEMP,100,1
DL,4,,VX,1,1
DL,4,,VY,0,1
DL,3,,VY,0,1
DL,1,,VX,0,1
DL,1,,VY,0,1
DL,2,,PRES,0,1
FLDATA1,SOLU,TEMP,1
FLDATA1,SOLU,TURB,1
FLDATA2,ITER,EXEC,100
FLDATA5,OUTP,HFLM,1
FLDATA8,NOMI,DENS,1000
FLDATA8,NOMI,VISC,.01
FLDATA8,NOMI,COND,1
FLDATA8,NOMI,SPHT,100
SOLVE
FINISH

/POST1
SET,LAST
PATH,pathA,2,30,20
PPATH,1,1
PPATH,2,2
PDEF,,HFLM,,AVG
/PBC,PATH,,0
PLPATH,HFLM
FINISH

ainite

速度对强制对流换热系数影响很大，当流速从1m/s变为2m/s时，换热系数结果对比如图：

nwpuyl

欢迎ainite大师来捧场
希望能带动本版的CFD讨论

iomega

based on your data, I assume:

a=1W/m-K/1000kg/m^3*100J/kg/K= 1e-5m^2/s
mu=0.01/1000kg/m^3=1e-5 m^2/s
Pr=mu/a = 1
Re_x=U*x/mu=x*1e5

For laminar external flow through the constant T plate   (Pohlhausen solution)
Nu_x=0.332*Pr^1/3*Re_x^1/2=h_x*x/k
h_x=- 0.332*Re_x^2/x*k= 0.332*sqrt(1e5*x)/x
The curve is even a little lower than your ANSYS simulation. (from -190 to -50  W/m^2-K)
I just curious about the format of your 理论解...

ainite

原来是输入错误，flotran里面输入的是动力黏度，不是运动黏度，唉！

谢谢楼上这位iomega

shaoyun1987

05年就已经是这个版面的版主了，版主，你现在依旧很犀利，赞美