A composite slab consists of one
layer of brick 500 mm thickness and two layers of insulation. The inner layer
of insulation is 100 mm thickness and outer layer is 60 mm thickness. The
thermal conductivity of the brick, inner and outer layer are 15 w/mk, 0.12 w/mk
and 0.082 w/mk respectively. The brick side is exposed to gases at 800 c and
the outer insulation is exposed to ambient air at 30 c, The brick side and air
side heat transfer co-efficient are 300 w/m2k and 150 w/m2k. Find the heat
transfer through this composite slab and inter face temperature.
Preference > thermal > ok.
Pre-processor > Element type > add/edit/delete
> add > select (link > convection 34) > apply > select (link
> 3d conduction 33) > ok > close.
Pre-processor > Real constants >
add/edit/delete > add > select type 2 link 33 > ok > enter cross -
area = 1 sqm > apply > ok.
Pre-processor > Material properties >
material models > thermal > convection or film wet > (enter HF = 300)
> new material > ID = 2 > ok > conductivity > isotropic > KXX
= 15 > ok > new material > ID = 3 > ok > conductivity > isotropic > KXX = 0.12
> ok > new material > ID = 4 > ok > conductivity > isotropic
> KXX = 0.082 > ok > new material > ID = 5 > ok > convection
or film coefficient > HF = 150 > ok > exit.
Pre-processor > modelling > create >
nodes > In active CS > enter the values in the below table and follow the
flow of the problem.
|
X
|
Y
|
Z
|
|
|
0
|
0
|
0
|
APPLY
|
|
0.01
|
0
|
0
|
APPLY
|
|
0.51
|
0
|
0
|
APPLY
|
|
0.61
|
0
|
0
|
APPLY
|
|
0.67
|
0
|
0
|
APPLY
|
|
0.68
|
0
|
0
|
OK
|
Pre-processor > modelling > create >
elements > element attribute > change element type number = 1 link 34
& material number, real constant set number = 1 > ok.
Pre-processor > modelling > create >
elements > auto numbered > thru nodes > select 1 - 2 nodes > ok.
NOTE: Now repeat the above two steps again
as below manner.
·
Now change element type no = 2 link 33 and
material number = 2 > ok > and
join 2 - 3 nodes by using thru nodes.
·
Now
change element type no = 2 link 33 and material number = 3 > ok > and
join 3-4 nodes by using thru nodes.
·
Now
change element type no = 2 link 33 and material number = 4 > ok > and
join 4-5 nodes by using thru nodes.
·
Now
change element type no = 1 link 33 and material number = 5 > ok > and
join 5-6 nodes by using thru nodes.
Solution > define loads > apply >
thermal > temperature > on nodes > select node 1 > ok > temp
> load temp = 800 > apply > select node 6 > ok > temp > load
temp > 30.
Solution > solve > current LS > ok
> close.
General post processor > element type > define table
> add > dof solution > temperature "TEMP" > apply >
add > by sequence > SMISC 1 (HEAT FLOW) > ok > close.
General post processor > list results > element table
data > select both TEMP & SMISC 1 > ok (By this we will get average
temperature of element and heat flow thru element)
General post processor > list results > reaction solution
> heat flow "HEAT" > ok. (We will get heat flow in watts)
General post processor > list results > nodal
solution > select nodal solution > dof solution > nodal temp > ok
(By this we will get the intermediate temp between layers)
NOTE: Plot ctrls > style > size and
shape > display elements (on) > ok. (now again run the nodal solution)
Comments
Post a Comment