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Steady State Heat Transfer through a Composite Slab in ANSYS APDL Software

Steady State Heat Transfer through a Composite Slab in ANSYS APDL Software:The main aim of  experiment:Steady State Heat Transfer through a Composite Slab is to determine the amount of heat flow through the nodes and the interface temperatures or Nodal Temperatures under the application of Temperature.Here,in this experiment,the composite slab consists of 3 materials having their thermal conductivities K1,K2 & K3.One end of the composite slab is exposed to External temperature and the other end is exposed to ambient air.The complete Numerical is placed below.

In order to analyse the Steady State Heat Transfer through a Composite Slab in ANSYS Software,we require 4 steps.

I.Preferences
II.Preprocessor
III.Solution
IV.General Post Processor.
The Numerical on Steady State Heat Transfer through a Composite Slab was shown below…
Numerical:

Steady State Heat Transfer through a Composite Slab

A  Composite slab consists of one layer of brick 500mm thick ad two layers of insulation. The inner layer of insulation is 100mm thick and  outer layer is 60mm thick.The thermal conductivitie’s of the brick, inner and outer layers are 15W/mK,0.12W/mK and 0.082W/mK resp.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 the air side heat transfer coefficient are 300W/m2k and 150W/m2k resp.Find the heat transfer through this composite slab and the interface temperature?

Procedure:

I.PREFERENCES:

Thermal—> h-method—> OK.

II.PREPROCESSOR

1.Element Type:

  • Add—> link—> convection34—> apply—> ok
  • Add—> link—> 3d conduction—> apply—> ok

As only conduction and convection taking place and that’s the reason,the two links are generated.

2.Real Constants:

Add/edit—>ok—>area=1m2

3.Material Properties:

Material 1:

Material Models—>Thermal—>convection(or film coefficient)—>HF=300W/m2k

Material 2:

Material—>no.2—> Conductivity—>Isotropic—>Kxx=15w/mk-OK.

Material 3:

Material—>no.3—> Conductivity—>Isotropic—>Kxx=0.12W/mk-OK.

Material 4:

Material—>no.4—> Conductivity—>Isotropic—>Kxx=0.082W/mk-OK.

Material 5:

Material no.5—>Convection or film coeff. =150w/m2k-OK.

4.MODELING:

Create –nodes-in active CS—>Now,create the length of the slab by providing length in the form of nodes..

Consider: 0.01m a small thickness taken for the convection element.

Vertices:

  • 1(0,0,0,),
  • 2(0.01,0,0)  ; here a small thickness of 0.01 m has taken for the convection element.
  • 3(0.01+0.5,0,0)=3(0.51,0,0)
  • 4(0.51+0.1,0,0)=4(0.61,0,0)
  • 5(0.61+0.06,0,0)=5(0.67,0,0)
  • 6(0.67+0.01,0,0)=6(0.68,0,0)

5.Elements:

Here we need to concentrate on two features.1.Element attributes and 2.auto numbered

Element 1:

  • Element attributes—>link 34(for convection)—>( Material no.1)—> Real const.no.1-OK.
  • Auto numbered—>Through Nodes—>pick the nodes 1&2 –OK.

Element 2:

  • Element attributes—>link 33(for conduction)—>( Material no.2)—> Real const.no.1-OK.
  • Auto numbered—>Through Nodes—>pick the nodes 2&3 –OK.

Element 3:

  • Element attributes—>link 33(for conduction)—>( Material no.3)—> Real const.no.1-OK.
  • Auto numbered—>Through Nodes—>pick the nodes 3&4 –OK.

Element 4:

  • Element attributes—>link 33(for conduction)—>( Material no.4)—> Real const.no.1-OK.
  • Auto numbered—>Through Nodes—>pick the nodes 4&5 –OK.

Element 5:

  • Element attributes—>link 34(for convection)—>( Material no.5)—> Real const.no.1-OK.
  • Auto numbered—>Through Nodes—>pick the nodes 5&6 –OK.

6.Define loads:

  • Apply—>Thermal—>Temperature—>on nodes—>temp=800°C—>OK.
  • Now pick the last node-6th node—>30°C—>OK.

III.SOLUTION:

  • Analysis Type—> New Analysis—> Steady-State—> OK.
  • Solve—> Current LS—> OK.

IV.GENERAL POST PROCESSOR

  • Element table—>Define Table—>DOF solution—>Temp—>OK.
  • Element table—>Define Table—>By sequence no.-SMISC,1.—>OK.

List results:

  • Element Table Data—>(Temp&SMISC)—>ok (we will get a box)—>Pick the Values from that box.

Reaction solutions:

  • Heat flow—>heat—>OK.

Heat flow @node 1 is 478.74 and @node6 is 478.74

Thertefore,

heat flow=478.74W

Figure:

  • Plot controls—>style—>size and shape—>display of element(ON)-OK.
  • Plot results-Contour plot-Nodal solution-DOF solution-Nodal temp-ok.

Nodal solutions:To get Interface Temperatures from T1 to T6,go to

-DOF solution-nodal temp-ok.

NodesTemperature
1 T1 
2 T2
 3 T3
4 T4
5 T5
6T6

This is the nodal temp. or the interface temp’s b/w layers of the composite slab.

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