(ANSYS APDL THERMAL) Composite Slab

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)