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  3. Week 4 - CHT Analysis on Exhaust port

Week 4 - CHT Analysis on Exhaust port

Conjugate Heat Transfer The Conjugate Heat Transfer (CHT) analysis type allows for the simulation of heat transfer between solid and fluid domains by exchanging thermal energy at the interfaces between them Conjugate refers to the fact that both conduction and convection contribute to the…

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  • Yogessvaran T

    updated on 27 Sep 2022

Conjugate Heat Transfer

The Conjugate Heat Transfer (CHT) analysis type allows for the simulation of heat transfer between solid and fluid domains

by exchanging thermal energy at the interfaces between them

Conjugate refers to the fact that both conduction and convection contribute to the transfer of heat.

This model, based on a strictly mathematically stated problem, describes the heat transfer between a body and a fluid

flowing over or inside it as a result of the interaction of two objects

Heat Transfer in Solids

In most cases, heat transfer in solids, if only due to conduction, is described by Fourier’s law defining the conductive heat

flux, q, proportional to the temperature gradient: .

For a transient calculation,

Heat Transfer in Fluids

Due to the fluid motion, three contributions to the heat equation are included:

The transport of fluid implies energy transport too, which appears in the heat equation as the convective contribution.

Depending on the thermal properties on the fluid and on the flow regime, either the convective or the conductive heat

transfer can dominate.

The viscous effects of the fluid flow produce fluid heating. This term is often neglected, nevertheless, its contribution is

noticeable for fast flow in viscous fluids.

As soon as a fluid density is temperature-dependent, a pressure work term contributes to the heat equation. This accounts

for the well-known effect that, for example, compressing air produces heat.

Accounting for these contributions, in addition to conduction, results in the following transient heat equation for the

temperature field in a fluid:

 

Applications of CHT analysis

Efficiently combining heat transfer in fluids and solids is the key to designing effective coolers, heaters, or heat exchangers.

Heat transfer in fluids and solids can also be combined to minimize heat losses in various devices. Because most gases

(especially at low pressure) have small thermal conductivities, they can be used as thermal insulators only if   they are not in

motion

In heat engines parts such as in exhaust port, Cylinder or wherever the heat transfer occur. This calculation helps in design

and selection of material properties based on the output from the simulation

CHT analysis is done on Computer mother boards to analyse the amount of heat transfer occur and to counter act accordingly

Boundary layer theory

Suppose fluid flows over a thin plate. Under no-slip condition the velocity of fluid is zero at boundary and keeps increasing as

we move away from the boundary layer. In a laminar flow the variation is linear but when it comes to the turbulent flow case

it can be divided into three layers as the and they are as follows

Inner layer or Viscous sublayer which is very close to the boundary layer,Here the viscous forces dominates

A buffer layer or defect layer where large scale turbulent eddy shear dominates

Overlap layer or log layer where the velocity profile shows logarithametic variation

 

Y+ Value

 Here y plus values are selected in accordance with the turbulence model we choose to solve the problem with. So particular

turbulance model have its own range of y plus values to get more accurate results

Yplus> 500 for the outer layer

30<Yplus<500 for log low region(High reynolds number formulation hence K-epsilon is used)

5<Yplus<30 for the buffer layer

Yplus<5 for viscous sublayer(low reynolds number formulation hence K-w or SST is used)

Yplus calcutaion using Matlab

 

clear all
close all
clc

Re = 67915.39
rho = 1.225
U = 6
Yplus = 30
Mu = 1.7965e-5;

Cf = 0.079*Re^(-0.25)
Tw = 0.5*(Cf*rho*U^2);
Ut = (Tw/rho)^0.5
dy = (Yplus*Mu)/(rho*Ut)

The obtained result was 0.001485 m 

Calculation of Heat transfer coeffecient

The properties of air in current problem is as follows

Density , ρ">ρ

 

= 1.225 kg/m^3

Viscosity = 1.7894e-5

Dynamic viscosity,μ">μ

 

= 1.7965e-5 kg/ms

Specific heat at constant pressure,Cp = 1.006e3 j/kgk

Conductivity , k = 0.0242 

Prandtl number for air at sea level = 0.746

Pipe diametre = 0.166 m

Velocity of stream = 6 m/s

1)Reynolds number 

Re=ρvdμ">Re=ρvdμ

 

=1.225 x 6 x 0.166/1.796e-5

= 67915.39

Re > 10,000

so it is a turbulant flow

2) Nusselt number

Nu = 0.023Re0.8Pr0.4">0.023Re0.8Pr0.4

 

0.023 x 67915.39^0.8 x 0.746^0.4

= 150.105

3) Calculation of heat transfer 

Nu = h/(k/l)

h = Nu x (k/l)

= 150.105 x 0.0242/0.166

=21.882 w/m^3k

Geometry

Duplicate  edges were removed by selecting repair extra edges tool from Repair

Then volume extract option from prepare is selected to extract fluid volume

Share topology between solid and volume volume is enabled(pink circles ensures the share topology is enabled)

Then created a new component with both volumes(solid and fluid)

Mesh

Here we will be setting up with different cases

Boundaries were named as inlet,outlet,Outer-wall convection(solid volume), and the plates were named as adiabatic wall

which refers it does not allow any heat transfer

The edge sizing for the input and output area is set to 36 divisions

Case 1

 

 

Case 2

 

10 inflation layers were given

Y+ value is set to 30 and hence the inflation layer thickness is set to 1.485 mm as first layer thickness

The mesh size is given as 50mm

 

Boundary conditions

Inlet velocity is set to 6 m/s

Inlet Temperature is set to 700k

On the outer wall the thermal condition is set to Convection, Heat transfer coeffecient to 20 w/m^2k, and free stream

temperature as 300k

Outlet temperature intial condition, Temperature = 300k

Setup

For both the cases same setup is followed

Equation for energy is selected for involving the temperature into the calculation

The viscous model is set to K-epsilon invloving 2 equations as our reynolds number exeeds 10000 we can say that it is a

turbulent flow.

Materials is set to air and aluminium for fluid and solid respectively

I've defined to plot area weighted y-plus values , Nusselt number and heat transfer coeffecient

In the reference values I,ve set the values as follows

Coupled solution method is used

And then for solving I,ve given 300 iteration with hybrid initialisation

Results

Case 1(element size - 150 mm)

Velocity contour

Temperature Contour

 

Case 2 (refined - Element size is set to 50mm)

Velocity Contour

Temperature contour

 

Observations

Solution convergancy can easily be detected by looking at outlet temperature plot

The obtained y=plus values is in range of 30-500, hence this approach is valid

It was observed that the velocity is higher at a point in bend(red point in velocity contour) of outlet pipe and since the

velocity is higher there the heat transfer rate will also be higher at that point

The wall heat transfer coefficient is specificallly made for walls only

At the temperature contour the temperature at the walls is less due to the fact that convective heat transfer has occured

There was no much of difference spotted comparing both cases, there was only a little change in the values obtaine

 

 

  HTC Nusselt number Outlet temparature Error in HTC
Case 1 (k- epsilon(standard)) 23.73867 162.8355 682.91 8.48
Case 2 (k-Epsilon (standard)) 23.419348 160.64511 683.12 7..025
Analatical 21.882 150.105   0

 

Using the above data the calculation done can be validated as the values are closer to the analatical values

Conclusion

HTC analysis was conducted

Plotted area weighted average on HTC, Y-plus ,nusselt number and outlet temperature

Velocity and Temperature contours were obtained

HTC was calculated analaticalliy and was used to validate the obtained results

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