simulation lab - m.e thermal (converted1)

Simulation Laboratory

M.E.Thermal Engineering] 5 | GCE, Salem - 11

Ex.No : 01STEADY STATE CONDUCTION IN SOLIDS

Date:

Aim: To find the Temperature Distribution of the given solid by using ANSYS 10.1

Problem Description:

1) Length of the solid is 1cm.2) Height of the solid is 1cm.

Boundary conditions:

Top is maintained Problem Description:


1) Top is maintained at 500C.2) Other sides are maintained at 100C.3) Thermal Conductivity (K) of the material is 10 W/mc

Figure:



PROCEDURE:STARTING ANSYS: Click on ANSYS 10.1 in the programs menu

MODELLING THE STRUCTURE:1. Go to the ANSYS main menu2. Create geometry.

Pre-processor >Modeling >create>Areas>Rectangle>By 2 corners>X=0,Y=0.Width =1,Height =1

ELEMENT PROPERTIESSELECTING ELEMENT TYPE:

Define the type of elementPre-processor >element Type >Add //edit/Delete...>click `Add` >select Thermal mass

solid, Quad 4 Node 55

Element material properties Pre processor > Material props> Material Modes > Thermal > conductivity>Isotropic > KXX =10 (Thermal conductivity)



MESHINGDIVIDING THE CHANNEL INTO ELEMENTS:

Mesh Size Preprocessor> Meshing > size cntrls > Manual size > Areas > All Areas > 0.05 MeshMesh

Pre processor > Meshing > Mesh > Areas > Free > Pick AllBOUNDARY CONDITIONS AND CONSTRAINTS

Define Analysis Type

Solution > Analysis Type > New Analysis > steady state

Apply ConstraintsSolution > Define Loads > Apply> Thermal > Temperature > On Lines

Apply temperature as per the figure

SOLUTION

Solve the system

Solution > solve > current LS>Ok

POST PROCESSING

Plot the Temperature distribution

General post proc > Plot results > Contour lot > Nodal Solu… > DOF solution,Temperature TEMP.



TEMPERATURE DISTRIBUTION

RESULTS:

Thus the temperature distribution of the given solid is determined.

Ex.No : 02STEADY STATE CONDUCTION IN SOLIDS

Date:

Aim:

To find the temperature distribution of the given solid by using ANSYS 10.1


1) Length of the solid cylinder is 5cm.2) Diameter of the solid cylinder is 2cm.

Boundary condition:

1) Wall temperature is 20 C.2) Base of the rod is 320 C.3) Convection Co-efficient of the wall is 100 W/m2 C4) Rod tip is insulated.

Figure:



PROCEDURE:

STARTING ANSYS:

Click on ANSYS 10.1 in the programs menu

MODELLING THE STRUCTURE:

1) Go to the ANSYS Main Menu2) Create geometry

Pre-processor > Modeling > create > Volume > Cylinder > Solid Cylinder > Radius=1 , Depth = 5.

ELEMENT PROPEERTIES

SELECTING ELEMENT TYPE:

Define the Type of element

Preprocessor > Element type > Add/Edit/Delete...> click `Add` > select ThermalMass Solid, 20 nodes 90.

Material properties

Preprocessor > Material Props > Material models > Thermal Conductivity >Isotropic > kxx =50 (Thermal conductivity)

MESHING

DIVIDING THE SHANNEL INTO ELEMENTS:

Mesh size



Pre-processor > Meshing > Size cntrls > Manual size > Areas > All Areas > 0.05

Mesh

Pre processor > Meshing > Mesh > Volume > Free > Pick All

BOUNDARY CONDITIONS AND CONSTRAINTS

Analysis Type


Constraints

Solution > Define Loads > Apply

Thermal > Temperature > On lines

Fill the window in as shown to constraint the side to constant temperature of 320


Thermal > Convection > On areas

Fill the window in as shown to constrain the side to film coefficient of 100 and bulk Temperature of 20.(As shown in Figure)



Fill the window in as shown to constrain the side to film coefficient of 0 and bulkTemperature of 0. .(As shown in Figure)

SOLUTION

Solve the system

Solution > solve > Current LS

SOLVE

POST PROCESSING



Plot the temperature distribution

General Postproc > Plot Results > Contour plot > Nodal Solu… > DOF solution,Temperature TEMP




VECTOR PLOT:

Now go to General Postproc > Plot Results > Vector Plot > Predefined and selectVelocity.



RESULTS:

Thus the temperature distribution of the given solid cylinder is determined.

Ex.No : 03STEADY STATE RADIATION IN SOLIDS

Date:

Aim :



1. Length of the strip is 2cm.2. Height of the strip is 1cm.

Boundary condition:

1. Left and Right side temperature is 900 C and Emissive is 0.7.2. Convection Co-efficient of the top is 100W/m2C and Bulk temperature is 50C.3. Stephen Boltze` man constant is 5.67e-8W/m2K4.

Material properties:

1. Thermal conductivity is 3W/m C.2. Density is 1600 kg/m3.3. Specific heat is 800 J/kg C.



Figure:

PROCEDURE:

STARTING ANSYS:



1. Go to the ANSYS Main Menu2. Create geometry

Pre-processor > Modeling > create > Areas > Rectangle > By 2 Corners > X =0, Y=0,Width=2, Height = 1.

ELEMENT PROPEERTIES



Preprocessor > Element type > Add/Edit/Delete...> click `Add` > select ThermalMaterial properties

Preprocessor > Material Props > Material models > Thermal >Conductivity>Isotropic > kxx =3 (Thermal conductivity)

Specific heat =800Density =1600.

MESHING




Mesh size


Mesh

Pre processor > Meshing > Mesh > Areas > Free > Pick All


Analysis Type


Constraints


Thermal > Temperature > On lines

Fill the window in as shown to constraint the side to constant temperature of 900


Thermal > Convection > Online

Fill the window in as shown to constrain the side to film coefficient of 100 and bulk Temperature of 50.


Thermal > Radiation > Online

Fill the window in as shown to constrain the side to a emissivity of 0.7 and enclosure no is2.

Preprocessor > Radiation opts > solution opt

Fill the window in as shown to constrain the side to a Stephen Boltze man constant of 5.67e-8.7 and enclosure no is 2.

Main menu > Radiation Opt > Matrix method > Other settings

Fill the window in as shown to constrain the side to a Stephen Boltze man constant of 5.67e-8.7 And type of geometry is 2D.



SOLUTION

Solve the system

Solution > solve > Current LS

SOLVE

POST PROCESSING


General postproc > Plot Results > contour plot > Nodal Solu… > DOF solution,Temperature TEMP




VECTOR PLOT:

Now go to General Postproc > Plot Results > Vector Plot > Predefined and select

Thermal flux



RESULTS:

Thus the temperature distribution and thermal flux of the given strip is determined.

Ex.No : 04COMBINED CONDUCTION AND CONVECTION IN SOLIDS

Date:

Aim:

To find the Temperature distribution of the given solid by using ANSYS 10.1


Dimensions:

1. Length of the fin is 8 cm.2. Width of the fin is 2cm.

Boundary Conditions:

1. Wall temperature is 20 C2. Base of the rod is 300 C3. Convection Co-efficient of the wall is 40 W/m2C.

Figure



PROCEDURE;

STARTING ANSYS:

Click on ANSYS 10.1 in the program menu.


1) Go to the ANSYS main menu.2) Create geometry.

Preprocessor > Modeling > create > Areas > Rectangle > By 2 corners > X=0, Y=0,Width =8, Height=2.

ELEMENT PROPERTIES

SELECTING ELEMENT TYPE

Define the type of element

Preprocessor > Element Type > Add/Edit/Delete >click `Add` > select

Thermal Mass Solid, Quad 4Node 55

Element material properties

Preprocessor> Material props > Material Models > Thermal conductivity>Isotropic > KXX = 1 (Thermal conductivity)

MESHING



DIVIDING THE CHANNEL INTO ELEMENTS:

Mesh Size

Preprocessor > Meshing > size cntrls > Manual size > Areas > All areas > 0.05 Mesh

Preprocessor > Meshing > Mesh > Areas > Free > Pick all



Solution > Analysis Type > New Analysis > steay state Apply Constraints


Thermal > convection> Online lines

Fill the window in as shown to constrain the strip to a film coefficient of 40 and bulktemperature of 20.

SOLUTION

Solve the system

Solution > solve > current LS

SOLVE



POST PROCESSING


General postproc > Plot results > Contour Plot > Nodal Solu…> DOF solution,Temperature TEMP




RESULT:

Thus the temperature distribution of the given Strip is determined.

Ex.No : 05CONDUCTION HEAT TRANSFER IN A PLATE

Date:

Aim :

To solve the 2-D heat conduction problem below, using ANSYS. And find the



Temperature distribution.

Problem description:Rectangular plate based on the specified temperatures on the plate edges, and the plate

Dimensions. Note: Thermal conductivity of the plate, Kxx=401 W/(m-K)

Procedure:

Define element type:

Preprocessor -> Element Type -> Add/Edit/Delete> Thermal Solid> Quad 4node 55

Define Material Properties:

Preprocessor -> Material Properties -> Material Models> Thermal



ModelingPreprocessor -> Modeling-> Create -> Areas -> Rectangle -> By DimensionsFill in the fields, (X1, X2) as 0, 10 & (Y1, Y2) as 0, 20 and then click “OK”.

Specify mesh density controls.

We will specify numbers of element divisions along lines. Choose: Preprocessor -> -Meshing- Size Controls -> Manual Size -> Lines-> Picked Lines

The picking menu (below left) appears. On the graphics window, click on the bottomHorizontal line (this is one of the 10 meter lines), to highlight it. Then, click “OK” in thePicking menu. Then, the “Element Size” menu (below right) appears. Enter “10” for“NDIV”, as shown, then click “OK”. Now, repeat the above process to specify 20 divisionsalong either of the vertical lines.

Mesh the rectangle to create nodes and elements.Preprocessor -> Meshing -> Mesh -> Areas -> Mapped -> 3 or 4 Sided

A picking menu appears. Select “Pick All”. The rectangle will be meshed.

Solution:

Apply temperatures around the edges:Solution -> Define Loads-> Apply -> Thermal-> Temperature -> On Lines

A picking menu appears. Highlight the two vertical lines (the 20 meter lines), which have atemperature of 100 C, then click on “OK” in the picking menu. The box on the next page Appears. Highlight “TEMP” for “DOFs to be constrained”, and enter “100” for “VALUE”.Repeat the above process to apply the 100 C temperature to the bottom horizontal line, but inthis case, choose “Yes” for “Apply TEMP to endpoints?” Repeat the process once more, toapply the 200 C temperature to the top horizontal line, but in this case, choose “No” for“Apply TEMP to endpoints?”

Now, to address the fact that two corners do not have a specified temperature, as anApproximation, we will set the temperature at these to corners to 150 C.

Solution -> Define Loads-> Apply -> Thermal-> Temperature -> On Key points

A picking menu appears. Note that there are four “key points” in the model, one at each



corner of the large rectangular area. Click on the upper two corners, at the intersections of the100 C and 200 C lines. When these corner “key points” are highlighted, choose “OK” in thepicking menu, and the following box appears: Click on “TEMP” for “DOFs to beconstrained” and enter “150” for VALUE, then click “OK”.

Solve the problem:

Solution ->-Solve -> Current LS

Post processing:Plot the temperature distribution:

General Post proc -> Plot Results -> Contour Plot-> Nodal SolutionThe box below appears. Click on “DOF solution”, then “Temperature”, and then click “OK”.

List the temperatures at the middle of the plate:Utility Menu > Select > EntitiesSelect Nodes and By Location from the scroll down menus. Click X coordinates and type 5into the input box , then click OK. Then go to

General Post proc -> List Results -> Nodal SolutionIn the box that appears, click on “DOF Solution” and “Temperature”, as shown, then click“OK”.

Write down the results:

SI no Node number Temperature SI no Node number Temperature

1 11

2 12

3 134 145 15

6 16



7 17

8 18

9 19

10 20



Plot Results




Result:

Thus the given 2-d heat conduction problem is solved using ansys. and the temperature

distribution within the rectangular plate, based on the specified temperatures on the plate edges, and the

plate dimensions was found.

Ex.No : 06 THERMAL ANALYSIS USING MIXED BOUNDARY CONDITIONS



Date:

Aim: To solve the 2-D heat conduction and convection problem below using ANSYS, and find the temperature

distribution.


The Mixed Convection/Conduction/Insulated Boundary Conditions Example is constrained as shown in thefollowing figure (Note that the section is assumed to be infinitely long):

Procedure:Create geometry

Preprocessor > Modeling > Create > Areas > Rectangle > By 2 Corners > X=0, Y=0,Width=1, Height=1Define the Type of Element

Preprocessor > Element Type > Add/Edit/Delete... > click 'Add' > Select Thermal Mass Solid, Quad 4Node55ET, 1, PLANE55

Material Properties



Preprocessor > Material Props > Material Models > Thermal > Conductivity > Isotropic >KXX = 10This will specify a thermal conductivity of 10 W/m*C.Element

Mesh SizePreprocessor > Meshing > Size Cntrls > Manual Size > Areas > All Areas > 0.05

MeshPreprocessor > Meshing > Mesh > Areas > Free > Pick All

Solution Phase: Assigning Loads and SolvingDefine Analysis Type

Solution > Analysis Type > New Analysis > Steady-State

Apply Conduction ConstraintsIn this Problem, all 2 sides of the block have fixed temperatures, while convection occurs on the other 2 sides.

Solution > Define Loads > Apply > Thermal > Temperature > On LinesSelect the top line of the block and constrain it to a constant value of 500 C Using the same method, constrain the left side of the block to a constant value of 100 C

Apply Convection Boundary ConditionsSolution > Define Loads > Apply > Thermal > Convection > On Lines

Select the right side of the block.Fill in the window. This will specify a convection of 10 W/m2*C and an ambient temperature of 100 degrees Celsius. Note that VALJ and VAL2J have been left blank. This is because we have uniform convection across the line.Apply Insulated Boundary Conditions

Solution > Define Loads > Apply > Thermal > Convection > On Lines

Select the bottom of the block.Enter a constant Film coefficient (VALI) of 0. This will eliminate Convection through the side, thereby modeling an insulated wall. Note: you do not need to enter a Bulk (Or

ambient) temperature.

Solve the SystemSolution > Solve > Current LSSOLVE



Post processing:Viewing the ResultsPlot Temperature

General Post proc > Plot Results > Contour Plot > Nodal Solu ... > DOF solution, Temperature



TEMP

List the temperatures at the middle of the plate:

Utility Menu > Select > Entities

Select Nodes and By Location from the scroll down menus. Click X coordinates and type0.5 into the input box , then click OK. Then go to

General Postproc -> List Results -> Nodal Solution

In the box that appears, click on “DOF Solution” and “Temperature”, as shown, then click“OK”.

Write down the results

SI no Node number Temperature SI no Node number Temperature

1 11

2 12

3 13

4 14

5 15

6 16

7 17

8 18

9 19

10 20



POST PROCESSING


General postproc > Plot results > Contour Plot > Nodal Solu…> DOF solution,Temperature TEMP



Result:

Thus the given 2-D heat conduction and convection problem is solved using ANSYS.

And the temperature distribution within the component based on the specified temperatures

on the edges, and the dimensions, was found.



Ex.No : 07LAMINAR FLOW OVER FLAT PLATE

Date:

Aim:

To find the Temperature Distribution of the Laminar flow over flat plate.


Dimensions:

1) Length of the plate is 1 m long.2) The flow area is 1 m * .25 m. .


1) The velocity of the air at infinite distance from the plate is 0.5 m/s.2) Atmospheric pressure is assumed on all faces except the face where velocity is

input into the system.

Figure:



PROCEDURESTARTING ANSYS

1) Click on ANSYS 10.1 in the programs menuMODELING THE STRUCTURE: Click

Preprocessor>Modeling>Create>Areas>Rectangle->By2 Corners.

The width is 1 and the height is .25. Starting position is at (0,0).

The modeling of the problem is done

ELEMENT PROPERTIES

SELECTING ELEMENT TYPE:ClickPreprocessor>Element Type>Add/Edit/Delete...> Flotran CFD > Flotran 141

MESHING


Go to Preprocessor>Meshing>SizeCntrls>ManualSize>Lines>Picked Lines.

Select thetop and bottom lines (horizontal lines of the rectangle).In the window that comes up type

50 in the field for 'No. of element divisions'. Now Click OK.

Go to



Preprocessor>Meshing>SizeCntrls>ManualSize>Lines>Picked Lines.

Select the two vertical lines of the rectangle. In the window that comes up type 100 in the field for 'No. of

element divisions' and type 10 in the field for 'Spacing ratio'Now click OK

Go toPreprocessor>Meshing>Size Cntrls>ManualSize>Lines>Flip Bias. Select the left

line of the rectangle and click OK.

Now go to

Preprocessor>Meshing>Mesh>Areas>Free

Fluid Properties (air):

Go to Preprocessor>Flotran Set Up>Fluid Properties.

On the box, shown below, set the first 2 input fields to Air-Si. Then click on OK.


The boundary conditions in this problem are an imposed velocity over the plate, and

the no-slip condition on the plate itself.

To apply the imposed velocity ,go to

Preprocessor>Loads>DefineLoads>Apply>Fluid CFD>Velocity>On lines.

Pick the left edge and the top edge of the rectangle and Click OK. The following window comes up.

Enter0.5 in the VX field and0 in the VY and VZ fields. Make sure'Apply toendpoints' is set to Yes.

Then click OK. This number corresponds to the velocity of 0.5 meter per second of air flowing from

The left side and the assumption that thevelocity is equal to 0.5 far from the plate..

To set the no-slip condition on the plate, repeat the above procedure but this time set

the Velocity to ZERO for the bottom line of the rectangle. (VX=VY=0). This time

make sure 'Apply to endpoints' is set to Yes.

Now Atmospheric Pressure must be set for the right side of the rectangle.



To do this, select

Preprocessor>Loads>Define Loads>Apply>FluidCFD>Pressure DOF>On lines.

Click the right line and then OK. The following window will now appear.

Enter 0 for the constant pressure value for these faces and click OK. This sets the pressure to Atmospheric.

SOLUTION

Go to ANSYS Main Menu>Solution>Flotran Set Up>Execution Ctrl.

Go to Solution>Run FLOTRAN.

POST-PROCESSING


General Postproc>Plot Results>Contour Plot>Nodal Solution.



VECTOR PLOT

Now go to General Postproc>Plot Results>Vector Plot>Predefined andselect Velocity. Enter a scale factor (VRATIO) of 0.4. The vector plot looks like this:



VECTOR PLOT

RESULT :

Thus the temperature distribution of the laminar flow over the flat plate is determined.



Ex.No : 08COMPOSITE WALL

Date:

Aim:

To find the Temperature distribution of the given composite wall by using ANSYS 10.1


Dimensions:

1) Length = 3 m

2) Width = 3 m

3) Thickness of each Layer = 1 m

Boundary Conditions:

1) The left side of the block has a constant temperature of 400 K.

2) The right side of the block has convection (h=20 W/m-K ; T= 300 K)

3) The Al section generates heat at a rate of 200 W/m3

4) The He section absorbs heat at a rate of 175 W/m3

Figure



PROCEDURE:

STARTING ANSYS:MODELING THE STRUCTURE:

1) Go to the ANSYS main menu2) Create geometry

Preprocessor > Modeling > Create > Areas > Arbitrary > Through KPs . Length = 3m,Width =3m.

ELEMENT PROPERTIES

Aluminum(1stlayer): KAl= 235 W/m*K

Helium(2ndlayer): KHe= 0.1513 W/m*K

Copper(3rdlayer): KCu= 400 W/m*



1) Now that we’ve defined what material ANSYS will be analyzing, we have to definehow ANSYS should analyze our block.

2) Click Preprocessor>Element Type>Add/Edit/Delete... In the 'Element Types'window that opens click on Add... The following window opens:

3) Type 1 in the Element Type reference number.

4) Click on Thermal Mass>Solid and select Quad 8node 77. Click OK. Close the'Element Types' window.

5) Now we have selected Element Type 1 to be a Thermal Solid 8node Element.

6) This finishes the section defining how the part is to be analyzed.

7) Now we have selected Element Type 1 to be a Thermal Solid 8node Element.

8) This finishes the section defining how the part is to be analyzed.


Define the Type of Element

Preprocessor > Element Type > Add/Edit/Delete…>click Add > Thermal Mass >Solid > select Quad8node 77.




Preprocessor > Material props> Material models > Thermal > conductivity >Isotropic.

1) Fill in 235 for Thermal conductivity. Click OK. This is the Thermal Conductivity of Al.

2) Now repeat the steps of clicking Thermal>Conductivity>Isotropic and then Defining the Thermal Conductivityas 0.1513 for the Model 2.

3) You have now defined the k value of Helium.

4) Define the last section and this time use K = 400. This is the Thermal Conductivity of Copper.

MESHING


1) Mesh size

Preprocessor > Meshing > Size controls > Manual size > Lines > All Lines. >0.05

Now go to Preprocessor>Meshing>Mesh Attributes>Default Attributes

2) Mesh

Preprocessor > Meshing > Mesh > Areas >Free

3) A popup window will appear on the left hand side of the screen. This window allows you to select the area to

be meshed.

4) Choose the 1starea and then click OK in the pop-up window. This both meshes the area and defines it as

Material 1. Material 1 (as you recall from before) was set to Aluminum originally by defining the k value of the

material as 235 W/m*K.

5) Now return to Preprocessor>Meshing>Mesh Attributes>Default Attributes. This time, select Material Number2 from the dropdown menu and click OK.

6) Once the pop-up window appears, select the middle layer and click OK.

7) Repeat this process of defining each layer as a different material for Material 3 and mesh it so that all threelayers have been meshed.

8) The block should now look like this when you are done meshing: (if you choose fit in the pan zoom rotatedialog)



BOUNDARY CONDITIONS AND CONSTRAINTS:

Constraints

Apply constant Temperature

Preprocessor > Loads > Define Loads > Apply > Thermal >Temperature > Online.

Enter 400 in the popup window as the set temperature for the left edge of the first section:

Apply Convection

Preprocessor > Loads > Define Loads > Apply > Thermal >Convection > Online

Apply Heat Generation and Heat Absorption

Preprocessor > Loads > Define Loads > Apply > Thermal >Heat General > Areas

Enter 200 W/m3for the generation and the hit ok.

Repeat this step for the second area but input -175 W/m3to imply absorption.

SOLUTION

Solve the system

Solution > Solve > Current LS

SOLVE



POST PROCESSING

Now go to General Post processing > Plot Results >Contour Plot > Nodal

Solution.> DOF solution > Temperature.




VECTOR PLOT:

Now go to General Postproc > Plot Results > Vector Plot > Predefined and select

Thermal flux



RESULT:

Thus the temperature distribution of the composite wall is determined.



Ex.No : 09STEADY STATE CONVECTION IN FLUIDS

Date:

Aim:

To plot the velocity distribution of the flow of air over an infinite cylinder.


Dimensions:

1) The cylinder is .25m in diameter.

2) Because of symmetry, the problem is simplifies so that half of the cylinder is

computed. The results are assumed to be the same on the other side of the x axis

(axis of symmetry).

3) The flow area is 2m by 0.5m. This arbitrary size serves to set up the boundary

conditions of flow over a cylinder.

4) The velocity of the air at infinite distance from the plate is 5 mm/s (.005m/s)



Figure:

PROCEDURE:

STARTING ANSYS:





preprocessor > Modeling >Create > Areas > Rectangle > By 2 corners.

The width is 2 and the height is .5. Starting position is at (0,0).

To create the circle

preprocessor > Modeling > Create > Areas >Circle > Solid Circle. The radius is .125 and the center is

at (0.5,0)

To subtract the circular area from the rectangle click

Preprocessor > Modeling >Operate > Booleans > Subtract > Areas.

Select the rectangle and click OK.

Next select the circle and circle OK. You will be left with this area.

ELEMENT PROPERTIES


Preprocessor > Element Type > Add /Edit/Delete… >Flotran CFD>2D Flotran 141

MESHING


Preprocessor > Meshing >Size cntrls > Manual size > Lines > Picked Lines.

In the window that comes up type .02 in the `Element edge length` box. Now go to

Pre processor > Meshing > Mesh >Areas > Free.

Click the area and OK.



FLUID PROPERTIES (AIR):

Preprocessor > Flotran set up > Fluid properties.

On the box, shown below, set the first two input fields to Air –SI. Then click on OK


To make loading easier, go to Plot> Lines on the main file menu.

Go to

Pre processor >Loads > Define Loads > Apply >Fluid CFD > Velocity >On lines.

pick the left and top edge of the rectangle and click OK. Enter 0.005 in the VX field and 0

in the VY. Leave VZ blank. then click OK. This number corresponds to the velocity of 0.005 meter per econd of air

flowing from the leftside:

To account for symmetry, the normal component to the axis of symmetry is set

zero. To set this go to

Preprocessor>Loads>Define Loads > Apply > FluidCFD > Velocity > on lines.

And select the bottom line of the rectangle. Enter 0 inthe VY field. Leave VX and VZ blank.

Repeat the above and set the velocity to ZERO for the 2 lines that make up thehalf – circle. (VX =

VY=0,VZ can be left blank). This is due to the no –slip condition along the cylinder.

The last step is to apply atmospheric pressure to the outlet of the fluid region. Go to

Preprocessor > Loads > Define Loads > Apply >Fluid CFD > PressureDOF > On Lines and

select the right most vertical line and click OK. In the Window that appears, enter 0 for the constant pressure value

and click OK.

SOLUTION

Go to ANSYS main menu > Solution > Flotran set up > Execution Cntrl. The following window appears. Change the first input value to 1000, as shown. No other changes are

needed. Click OK.

Go to Solution > Run FLOTRAN.



POST PROCESSING

1) Plotting the velocity distribution.

2) Go to General post proc > Read Results > Last set.

3) Then go to General Postproc > Plot Results > Contour plot > Nodal solution

select VSUM and click OK



VECTOR PLOT:

NOW GO TO General Post proc > Plot Results > vector Plot > Predefined and

select velocity. The vector plot looks like this.



VECTOR PLOT

RESULTS:

Thus the velocity distribution of a flow of air over an infinite cylinder is computed andPlotted.



Ex.No : 10STEADY STATE RADIATION IN SOLID CYLINDER

Date:

Aim:

To find the Temperature Distribution of the given solid cylinder by using ANSYS 10.1


1. Radius of the solid is 1m.2. Height of the solid is 1m.


1. Top side is maintained at 700°C & emissivity 0.82. Bottom side is maintained at 500°C & emissivity 0.43. Circumference is maintained at 400°C & emissivity 1(Black body)

Figure:

T=700°C & ε=0.8

T=500°C T=500°C

T=500°C & ε=0.4

PROCEDURE:



STARTING ANSYS:



1. Go to the ANSYS main menu2. Create geometry.

Pre-processor >Modelling >create>Volume>Cylinder>Solid cylinder>X=0,Y=0.Radius =1, Height =1

ELEMENT PROPERTIES



Pre-processor >element Type >Add //edit/Delete...>click `Add` >select Thermal masssolid, 20 node 90.


Pre processor > Material props> Material Modes > Thermal > conductivity>Isotropic > KXX =10 ( Thermal conductivity)

MESHING


Mesh Size

Preprocessor> Meshing > size cntrls > Manual size > Areas > All Areas > 0.05 Mesh

Pre processor > Meshing > Mesh > Volume > Free > Pick All






Apply Constraints

Solution > Define Loads > Apply>Thermal > Temperature > On Area

Apply Temperature as per the figure

Solution > Define Loads > Apply>Thermal > Radiation > On Area

Apply Emissivity as per the figure and enclosure no is 1 for all area.

Preprocessor > Radiation opts > solution opt

Fill the window in as shown to constrain the side to a Stephen Boltze man constant of 5.67e-8.7 And enclosure no is 1.

Main menu > Radiation Opt > Matrix method > other settings

Fill the window in as shown to constrain the side to a Stephen Boltze man constant of 5.67e-8.7 And type of geometry is 2D.

SOLUTION

Solve the system

Solution > solve > current LS > OK



POST PROCESSING

Plot the Temperature distribution

General post proc > Plot results > Contour lot > Nodal Solu… > DOF solution,Temperature TEMP.




RESULTS:

Thus the temperature distribution of the given solid is determined.

Ex.No : 11COMBINED STEADY STATE CONDUCTION AND CONVECTION IN FIN

Date:

Aim:





1) Length of the fin is 1 m.2) Height of the fin is 0.5m3) Width of the fin is 5 m

Boundary condition:

1) Wall temperature is 200 C.2) Ambient air is 40 C.3) Convection Co-efficient of the wall is 50 W/m2 C4) Thermal conductivity 200 W/m C

Figure:



PROCEDURE:

STARTING ANSYS:



1) Go to the ANSYS Main Menu2) Create geometry

Pre-processor > Modelling > create > Volume > Block > By 2 corner & Z > Length=1, Height=0.5 , Depth = 5.

ELEMENT PROPEERTIES



Preprocessor > Element type > Add/Edit/Delete...> click `Add` > select ThermalMass Solid, 20 node 90.

Material properties

Preprocessor > Material Props > Material models > Thermal Conductivity >Isotropic > kxx =200 (Thermal conductivity)

MESHING


Mesh size


Mesh

Pre processor > Meshing > Mesh > volume > Free > Pick All




Analysis Type


Constrints


Thermal > Temperature > On Area

Apply temperature as per the figure



Apply film coefficient of 50 and bulk temperature of 40 in the remaining sides’

SOLUTION

Solve the system

Solution > solve > Current LS > ok

SOLVE



POST PROCESSING


General postproc > Plot Results > Contour plot > Nodal Solu… > DOF solution,Temperature TEMP




VECTOR PLOT:

Now go to General PostProc > Plot Resuots > Vector Plot > Predefined and selectvelocity.



RESULTS:

Thus the temperature distribution and thermal flux of the given fin is determined.

Ex.No : 12STEADY STATE CONVECTION IN COMPOSITE HOLLOW CYLINDER

Date:

Aim:

To find the temperature distribution of the given composite hollow cylinder by using ANSYS 10.1




1. Inner diameter of the cylinder 0.025m2. Outer diameter of the cylinder 0.038m3. Inner diameter of the cylinder (insulator) 0.038m4. Outer diameter of the cylinder (insulator) 0.058m5. Depth 2m

Boundary condition:

6. Hot gas flow inside the cylinder 330 C & h=400 W/m2 C7. Outer surface of the insulator is at 30C & h=60 W/m2 C

Figure:

PROCEDURE:

STARTING ANSYS:





Go to the ANSYS Main Menu>Create geometry

Pre-processor >Modelling >create>Volume>Cylinder>Hollow cylinder> Rad1 =0.025, Rad2 =0.038,Depth=2

Pre-processor >Modelling >create>Volume>Cylinder>Hollow cylinder> Rad1 =0.038, Rad2 =0.058, Depth=2

Pre-processor >Modelling >operate>booleans >glue>areas

ELEMENT PROPEERTIES



Preprocessor > Element type > Add/Edit/Delete...> click `Add` > select ThermalMaterial properties

Preprocessor>Material Props>Materialmodels1>Thermal>Conductivity>Isotropic >kxx =15

Preprocessor>Material Props>Materialmodels2>Thermal>Conductivity>Isotropic >kxx =0.2(Thermal conductivity)

MESHING


Mesh size


Mesh

Pre processor > Meshing > Mesh > Volume> Free > Pick All


Analysis Type




Constraints

Solution > Define Loads > Apply>Thermal > Convection > on areas

Apply as per the diagram (Bulk Temperature = 330, film coefficient =400)

For Insulator

Solution > Define Loads > Apply>Thermal > Convection > On areas Apply as per the diagram (Bulk Temperature = 30, film coefficient =60)

SOLUTION

Solve the system

Solution > solve > Current LS> Ok

SOLVE

POST PROCESSING






POST PROCESSING






VECTOR PLOT:


Now go to General PostProc > Plot Results > Vector Plot > Predefined and Thermal Flux



RESULTS:

Thus the temperature distribution of the given composite hollow cylinder is determined.



Ex.No : 13COMBINED CONDUCTION AND CONVECTION IN TRIANGULAR FIN

Date:

Aim:

To find the Temperature distribution of the given triangular fin by using ANSYS 10.1


1. Length of the fin is 0.04 m.2. Side of the fin is 0.005m.3. Wall temperature is 400 C4. Convection Co-efficient of the wall is 90 W/m2C5. Bulk Temperature 50 C

Figure

PROCEDURE;



STARTING ANSYS:




Preprocessor > Modeling > create > Volume> Prism > By Side Lengths> Z=0.04, Side=0.005

ELEMENT PROPERTIES




Thermal Mass Solid, 20 node90



MESHING


Mesh Size

Preprocessor > Meshing > size cntrls > Manual size > Areas > All areas > 0.05

Mesh



Preprocessor > Meshing > Mesh > Volume > Free > Pick all



Solution > Analysis Type > New Analysis > steay state

Apply Constraints

Solution > Define Loads > Apply Thermal > Temperature > Online Areas

Apply as per the drawing

Solution > Define Loads > Apply Thermal > Convection> Online Areas

Apply as per the figure (film coefficient of 90 and bulk temperature of 50.)

SOLUTION

Solve the system


SOLVE

POST PROCESSING




General postproc > Plot results > Contour Plot > Nodal Solu…> DOF solution ,Temperature TEMP



VECTOR PLOT:


General PostProc > Plot Results > Vector Plot > Predefined and Thermal Flux



RESULT:




Ex.No : 14RADIATION HEAT TRANSFER IN TRIANGULAR FIN

Date:

Aim:

To find the Temperature distribution of the given triangular strip by using ANSYS 10.1


1. Radius of the strip is 1 m.2. Angle of the strip is 60.3. Wall temperature at the sides are 900 C and 400 C respectively4. Emissivities of the sides are 0.8 for both sides.5. Heat transfer at the bottom side is zero6. K=30 W/m C

Figure



PROCEDURE;STARTING ANSYS:




Preprocessor > Modeling > create > Area> Polygon >Triangle> X=0,Y=0, Radius=1,Theta=60Preprocessor > Modeling > create > Area> Circle >Solid circle> X=0,Y=0, Radius=0.5

ELEMENT PROPERTIES




Thermal Mass Solid, Quad node55



MESHING


Mesh Size

Preprocessor > Meshing > size cntrls > Manual size > Areas > All areas > 0.05

Mesh



Preprocessor > Meshing > Mesh > Areas > Free > Pick all





Solution > Analysis Type > New Analysis > steay state

Apply Constraints

Solution > Define Loads > Apply >Thermal > Temperature > Online

Apply as per the drawing

Solution > Define Loads > Apply >Thermal > radiation> Online

Apply as per the figure (Emissivity)

SOLUTION

Solve the system


SOLVE



POST PROCESSING


General postproc > Plot results > Contour Plot > Nodal Solu…> DOF solution ,Temperature TEMP




VECTOR PLOT:


General PostProc > Plot Results > Vector Plot > Predefined and Thermal Flux



RESULT:


simulation lab - m.e thermal (converted1)

Documents