Week 4 - CHT Analysis on Exhaust port

1. Give a brief description of why and where a CHT analysis is used.
2. Maintain the y+ value according to the turbulence model and justify the results.
3. Calculate the wall/surface heat transfer coefficient on the internal solid surface & show the velocity & temperature contours in appropriate areas.
4. How would you verify if the HTC predictions from the simulations are right? On what factors does the accuracy of the prediction depend on?

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:

1. 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.
2. 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.
3. 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 , `rho` = 1.225 kg/m^3
• Viscosity = 1.7894e-5
• Dynamic viscosity,`mu` = 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 = (rhovd)/mu`

=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 = `0.023Re^0.8Pr^0.4`

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 obtained

• 	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

References

Conjugate Heat Transfer Analysis | Analysis Types | SimScale

Conjugate Heat Transfer | COMSOL Blog