Aim:-
Give a brief description of why and where a CHT analysis is used.
Maintain the y+ value according to the turbulence model and justify the results.
Calculate the wall/surface heat transfer coefficient on the internal solid surface & show the velocity & temperature contours in appropriate areas.
How would you verify if the HTC predictions from the simulations are right? On what factors does the accuracy of the prediction depend on?
1. Conjugate heat transfer (CHT):
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. Typical applications of this analysis type exist as, but are not limited to, the simulation of heat exchangers, cooling of electronic equipment, and general-purpose cooling and heating systems.
The contemporary conjugate convective heat transfer model was developed after computers came into wide use in order to substitute the empirical relation of proportionality of heat flux to temperature difference with heat transfer coefficient which was the only tool in theoretical heat convection since the times of Newton. 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. The physical processes and solutions of the governing equations are considered separately for each object in two subdomains. Matching conditions for these solutions at the interface provide the distributions of temperature and heat flux along the body–flow interface, eliminating the need for a heat transfer coefficient.
2. Why CHT analysis is used:
Computational Fluid Dynamics (CFD) and more specifically conjugate heat transfer (CHT) analysis can accurately predict heat transfer by simultaneously solving all the relevant solid and flow field heat transfer processes, for example, conduction through solids, free and forced convection in the gases/fluids and thermal radiation.
The CHT approach has an advantage over FE thermal analyses in that wall heat transfer coefficients and their local variations on surfaces are directly calculated within the model rather than based on simplified empirical calculations. The CHT approach, therefore, has benefits for those applications where heat transfer is either non-uniform or difficult to calculate empirically.
3. Where is CHT analysis used:
Starting from simple examples in the 1960s, the conjugate heat transfer methods have become a more powerful tool for modeling and investigating natural phenomena and engineering systems in different areas ranging from aerospace and nuclear reactors to thermal goods treatment and food processing, from complex procedures in medicine to atmosphere/ocean thermal interaction in meteorology, and from relatively simple units to multistage, nonlinear processes. A detailed review of more than 100 examples of conjugate modeling selected from a list of 200 early and modern publications shows that conjugate methods are now used extensively in a wide range of applications. That also is confirmed by numerous results published after this book appearance (2009) that one may see, for example, at the Web of Science. The applications in specific areas of conjugate heat transfer at periodic boundary conditions and in exchanger ducts are considered in two recent books.
Applications:
Effective heat transfer:
Efficiently combining heat transfer in fluids and solids is the key to designing effective coolers, heaters, or heat exchangers.
The fluid usually plays the role of energy carrier on large distances. Forced convection is the most common way to achieve a high heat transfer rate. In some applications, the performances are further improved by combining convection with phase change (for example liquid water to vapor phase change).
Even so, solids are also needed, in particular, to separate fluids in a heat exchanger so that fluids exchange energy without being mixed.
Energy savings:
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… provided they are not in motion. In many situations, gas is preferred to other materials due to its low weight. In any case, it is important to limit the heat transfer by convection, in particular by reducing the natural convection effects. Judicious positioning of walls and the use of small cavities helps to control the natural convection. Applied at the micro-scale, the principle leads to the insulation foam concept where tiny cavities of air (bubbles) are trapped in the foam material (e.g. polyurethane), which combines high insulation performances with lightweight.
4. Heat transfer co-efficient:
The heat transfer coefficient or film coefficient, or film effectiveness, in thermodynamics and in mechanics is the proportionality constant between the heat flux and the thermodynamic driving force for the flow of heat (i.e., the temperature difference, ΔT):
The overall heat transfer rate for combined modes is usually expressed in terms of an overall conductance or heat transfer coefficient, U. In that case, the heat transfer rate is:
{displaystyle {dot {Q}}=hA(T_{2}-T_{1})}
where:
A: the surface area where the heat transfer takes place, m2
T_{2}: temperature of the surrounding fluid, K
T_{1}: temperature of the solid surface, K.
The general definition of the heat transfer coefficient is:
h = frac{q}{Delta T}
where:
q: heat flux, W/m2; i.e., thermal power per unit area, q = dQ'/dA
h: heat transfer coefficient, W/(m2•K)
ΔT: difference in temperature between the solid surface and surrounding fluid area, K
It is used in calculating the heat transfer, typically by convection or phase transition between a fluid and a solid. The heat transfer coefficient has SI units in watts per squared meter kelvin: W/(m2K).
The heat transfer coefficient is the reciprocal of thermal insurance. This is used for building materials (R-value) and for clothing insulation.
There are numerous methods for calculating the heat transfer coefficient in different heat transfer modes, different fluids, flow regimes, and under different thermohydraulic conditions. Often it can be estimated by dividing the thermal conductivity of the convection fluid by a length scale. The heat transfer coefficient is often calculated from the Nusselt number (a dimensionless number). There are also online calculators available specifically for Heat-transfer fluid applications. Experimental assessment of the heat transfer coefficient poses some challenges especially when small fluxes are to be measured.
Geometry setup:
Load the model into the SpaceClaim and change the units to meters
Remove the duplicate edges in the model
Then extract the volume in the model and name the inner volume as fluid and outer volume as solid volume and remove the empty components
Enable Share topology of the main component so that the mesh and physics information can be shared and conformal mesh can be achieved.
Lastly, to achieve conjugate heat transfer between the fluid and wall we have to choose share prep under the workbench tool so that the heat transfer is captured throughout the wall from inlet to outlet.
Mesh setup:
After the geometry setup is done in the SpaceClaim open the mesh
select the faces and name them accordingly
Create a name selection for fluid volume so that the inflation layer can be achieved on that body since the conjugate heat transfer is between the fluid volume to the solid wall.
Create a body sizing on the outer wall convection region to achieve mesh refinement and accurate results can be captured at such regions.
Fluent setup:
After updating the mesh open fluent setup, select a pressure-based solver using the steady-state time simulation and absolute velocity formulation under the general settings.
Under physics choose the energy option since we are dealing with temperature properties.
Input the Boundary values at the inlet, 5 m/s as inlet velocity at 700k temperature. Set the outer wall convection boundary to convection type and input a constant value for heat transfer coefficient as 20
Under materials, properties choose air as the fluid material and aluminum as the solid material
Enter the reference values as per the geometry and inlet boundary conditions accordingly.
Initialize by using the hybrid initialization, and the simulations for 400 iterations as it is optimum to achieve a converged solution.
CFD post:
We can calculate the wall heat transfer coefficient by inserting a plane along xy plane
Similarly, Vector and streamline contours can be created by using the insert vector or streamline option.
Case 1: Baseline simulation
element size =150mm
sizing element size=50mm
no. of elements=247571
viscous model=k-epsilon
inlet velocity =5m/s
inlet temperature=700k
HT co-efficient=20W/m^2-k
Mesh details:
Temparature distribution:
Residual plot
surface heat transfer co-efficient
wall heat transfer coefficient
Temperature and velocity contours
velocity streamlined plot
Case 2 Refined mesh
element size =10mm
sizing element size=10mm
no. of elements=336493
viscous model=k-epsilon
inlet velocity =5m/s
inlet temperature=700k
HT co-efficient=20W/m^2-k
Mesh details:
Temparature distribution:
Residual plot
surface heat transfer co-efficient
wall heat transfer coefficient
Temperature and velocity contours
velocity streamlined plot
Case 3 Refined mesh
element size =100mm
sizing element size=20mm
no. of elements=362257
viscous model=k-omega
inlet velocity =5m/s
inlet temperature=700k
HT co-efficient=20W/m^2-k
Temparature distribution:
Residual plot
surface heat transfer co-efficient
wall heat transfer coefficient
Temperature and velocity contours
velocity streamlined plot
conclusion:
cases viscous model no. of elements
surface HT coefficient
in(W/m^2-K)
wall HT coefficient in(W/m^2-K)
1 k-epsilon 247571 31.674 211
2 k-epsilon(refined) 336493 28.58 191.9
By comparing 3 cases using the k omega model we can get more accurate because we are dealing with temperature
Wall heat co-efficient is more in the k-omega model as compared to k epsilon
There is no change in the temperature for all cases.
Y plus should be calculated depending upon the solvers.
How would you verify if the HTC predictions from the simulations are right? On what factors does the accuracy of the prediction depend on?
According to Bernoulli's mass conservation principle mass is conserved in our model which is clearly explained below.
Let us assume the flow rate from each inlet as Q1, Q2, Q3, Q4 respectively, and Q5 be the flow rate(outflow) from the exhaust manifold.
As mass is neither created nor destroyed, mass flow rates are also balanced as below
Q1+Q2+Q3+Q4=Q5Q1+Q2+Q3+Q4=Q5 -------(1)
we have Q=ρ.A.VQ=ρ.A.V -------(2)
now the equation 1 becomes
A1.V1+A2.V2+A3.V3+A4.V4==A5.V5
as it is an incompressible flow to compensate for the mass flow, velocity is increased, and the overall mass flow rate is conserved.
As velocities are high, the Reynolds number at the outlet manifold will be high.
Nu=f(Re, pr) for forced convection applications,
Nu=hD/k
from the above equations
Nu∝Re,
Nu∝h,
hence, Re∝h
From our simulations, we have observed velocities are high at bend section of outlet manifold hence higher Reynolds number which has high Heat transfer coefficients.
Therefore predictions from our simulations can be considered correct.
