Conjugate heat transfer

ANS 1 :      CHT analysis allows us to simulate the heat tranfer between the solid body and the fluid . These types of simulations are governed by the convection type of heat tranfer . It's done between the interface of the solid body and the fluid domain exchanging thermal energy . 

 

ANS 2 :    

 

# CASE 1 : Baseline Mesh

 

# Geometry 

 

We have extracted the volume of the exhaust pipe and thus , we have

1. Solid volume ( exterior of the pipe )

2. Fluid volume ( extracted volume )

 

 

 

# Mesh :

 

Named selection :

 

1  inlet

2  outlet

3  wall symmetry convection

 

number of elements : 137391

we have used the shared topology , for the conformal mesh .

 

 

 

 

 

 

 

 

# Fluent :

Physics :

Steady state condition

pressure based solver

absolute velocity

k epsilon (RANS ) , turbulent based solver

thermal conditions valid

 

Boundary conditions :

 

inlet velocity : 5 m/s

temperature : 700 k

 

Material :

 

Fluid : air

Solid : aluminium

 

We have changed the thermal conductivity of the wall symmetry convection to ( 20 )

 

 Residuals

 

 

 Contours 

 

The velocity contour is plotted

 

 

 

# Results :

 

Velocity contours with the streamlines is shown in the post results

 

 

 

* Here we can see the mesh being coarse , so the wall heat tranfer coeff. being seen in a very coarse manner.

 

 

 

 

 

 

# CASE 2 :  Refined mesh with inflation layerG

 

# Geometry

 

The geometry that we have used is the same as the above mesh that we have used .

 

# Mesh  

 

Element number : 424262

 

 

 

 

We have provided the inflation layer , of total thickness , total thickness ( 5mm ) 

 

 

 

# Fluent

 

1  Physics

2.  Boudary layer condition

 

We have used the same condition as that of the case 1.

 

# Residuals

 

 

 

# Temperature contour

 

 

 

# Results :

 

1. Velocity contour

 

 

 

2.  Wall Heat transfer  coeff. contour

 

 

 

# CONCLUSION :

As we have seen in the case 1 the mesh was coarse and thus the heat transfer coeff. is coarse and as seen in case 2 we see the smooth line of that represents the heat transfer coiff. . Also , the heat transfer coeff. is maximum where the velocity is  maximum. As , the condition is steady state so as per the mass flow rate is same at the inlet and the outlet thus the velocity is maximum at the outlet.

 

 

ANS 3 : We can verify the that the results are satisfying as ;

 

1.  The residuals are converging 

2.   The simulation is acting as steady state.

3.   As the condition is stedy state thus [ m (in) = m (out) ] , so the velocity at the outlet is maximum.

4.   Thus the heat trnsfer coeff. is maximum ( as the reynolds number is greater).

 

Factors on which the results depend upon are ;

 

1.  solid body which is not the part of the wall symmetry convection are set to have adiabatic conditions ( thus the heat transfer is 0 there )

2.   the velocity of the flow  at the outlet is maximum ( as per steady state there is only 1 outlet ).

3.   thus the reynolds number is greater 

4.   so the heat tranfer coeff is maximum there.

 

 

 


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