Plane 75 axisymmetric-harmonic thermal solid

jinqinghai

!!! Plane 75 axisymmetric-harmonic thermal solid
fini
/clear
/filname,fin
/units,si

/prep7
antype,trans
et,1,75
mp,kxx,1,237
mp,dens,1,2702
mp,c,1,903
mp,hf,1,150
mp,kyy,1,237

rectng,0,0.002,0,0.04
lesize,1,,,2
lesize,2,,,8
eshape,2
amesh,1
eplot
/solu
outres,,all  !!! No!! /outres,,all
!!outres,,last
outpr,,all
sfa,1,2,conv,150,25
sfa,1,3,conv,0,25
tunif,25
d,1,temp,85
d,2,temp,85
d,3,temp,85

autots,on
time,0.01
time,8  !8s
nsubst,80,80,40! process 80 steps
NEQIT,20

!trnopt,full
solve
FILE,'file','rst','.'  !!!!
finish

/post26
STORE,MERGE   !!!!

nsol,2,1,temp,,t1
nsol,3,17,temp,,t2
nsol,4,12,temp,,t3

/axlab,y,temp
plvar,2,3,4
fini

jinqinghai

anyone provides explanations to the APDL in the column of ANSYS--材料与接触分析. he/she will get 1-2 credit points surely.

Explainning it with mathmatical  method will be rewarded 2-3 credit points respectively.

iomega

hehe, this problem is kind of little interesting:
I am not familiar with plane 55 element, but after reading the code, my understanding is that you have an aluminum rod with diameter of 4mm and length of 40mm. One end applies 85C and another end is adiabatic. the surface is convective to air with h of 150W/m^2-K. It's a typical fin problem.

Some conclusions from physics that can explain your plot:

(1) temperature in cross-section is uniform due to smaller thermal resistance compared to that of the convection.

(2) thermal diffusion time = L^2/alf = 0.04^2/ (237/(2702*903))= 16.5s
where L is the length =0.04m and alf is the thermal diffusivity of the Al.
This means it takes around 16s for heat to diffuse to the end of the rod.
As can be judged by the attached figure (1): use 80s time and the temperature is close to 80C around 17s.

(3) thermal healing length  L_h=  sqrt ( k*A / (h*P)) = 0.04m
where A is the cross-section area = Pi*r^2 and P is the perimeter = 2*Pi*r. Because L_h is the same as rod length L and the end is adiabatic, when reach steady-state, everywhere in rod should be 80C for two reasons:
a. heat can't conduct along the rod becasue of the adiabatic condition at the end.
b. thermal resistance of the rod in radius direction is much smaller than that of the air convection.
This again can be seen in Fig (1) when the time is long enough.

(3) if the rod lengh L >> L_h, when reachs steady-state, the end with adiabatic conditon is always 25C, becasue the heat already dissipates to the air as it conducts along the rod. In this case, the steady-state temperature distribution will looks like  T( x ) = 85*exp(-x/L_h).

Such kind of "fin" problems is very helpful for understanding the physics of the heat conduction. That's why all the heat transfer book start with the "fin" problem.

One interesting thing is that fourier conduction is diffusion in nature, assuming the speed of thermal wave is infinite, that's why you will see temperature at the end of the rod will rise immediately after t>0 ( which is not correct).

The analytical soultion for this problem does exist - I have solved such problem before with heat flux input at one end. The analytical solution probably gives more physical insight.

jinqinghai

perfect explanation! but I can't award you with my deep pity. well, hereinafter I attached mathmatical solution as a response on your answer.

jinqinghai

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