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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…
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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