Week 4 - CHT Analysis on Exhaust port
AIM:
- The aim of this assignment is to perform a Conjugate Heat Transfer analysis and simulate flow inside an Exhaust port.
OBJECTIVE:
The objectives of this assignment are:
- To give a brief description about Conjugate Heat Transfer (CHT) analysis and where and why it is used.
- To perform CHT analysis on Exhaust Port.
- To select a proper turbulence model and maintain the correct Y plus value.
- To compute the Velocity, Temperature, Heat transfer coefficient contours and Y plus plots and other required contours and plots to show the results of the CHT analysis on the Exhaust Port.
- To validate the result and verify if the HTC predictions from the simulations are right or not.
- To state on what factors does the accuracy of the prediction depend on.
INTRODUCTION:
CONJUGATE HEAT TRANSFER ANALYSIS:
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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. Efficiently combining heat transfer in fluids and solids is the key to designing effective coolers, heaters, or heat exchangers.
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The fluid usually plays the role of energy carrier on large distances. Forced convection is the most common way to achieve high heat transfer rate. In some applications, the performances are further improved by combining convection with phase change (for example liquid water to vapour 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.
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The temperature field and the heat flux are continuous at the fluid/solid interface. However, the temperature field can rapidly vary in a fluid in motion: close to the solid, the fluid temperature is close to the solid temperature, and far from the interface, the fluid temperature is close to the inlet or ambient fluid temperature. The distance where the fluid temperature varies from the solid temperature to the fluid bulk temperature is called the thermal boundary layer. The thermal boundary layer size and the momentum boundary layer relative size is reflected by the Prandtl number. For the Prandtl number to equal 1, thermal and momentum boundary layer thicknesses need to be the same. A thicker momentum layer would result in a Prandtl number larger than 1. Conversely, a Prandtl number smaller than 1 would indicate that the momentum boundary layer is thinner than the thermal boundary layer. The Prandtl number for air at atmospheric pressure and at 20°C is 0.7. That is because for air, the momentum and thermal boundary layer have similar size, while the momentum boundary layer is slightly thinner than the thermal boundary layer. For water at 20°C, the Prandtl number is about 7. So, in water, the temperature changes close to a wall are sharper than the velocity change.
Heat transfer in solids and heat transfer in fluids are combined in the majority of applications. This is because fluids flow around solids or between solid walls, and because solids are usually immersed in a fluid. An accurate description of heat transfer modes, material properties, flow regimes, and geometrical configurations enables the analysis of temperature fields and heat transfer. Such a description is also the starting point for a numerical simulation that can be used to predict conjugate heat transfer effects or to test different configurations in order, for example, to improve thermal performances of a given application.
APPLICATIONS:
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Computational Fluid Dynamics (CFD) and more specifically conjugate heat transfer (CHT) analysis can accurately predict heat transfer by simultaneous 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.
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The CHT approach has an advantage over FE thermal analysis 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.
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The heat transfer projects have covered a wide range of aerodynamic and hydrodynamic industrial applications but projects typically had objectives such as:
- Temperature distribution analysis.
- Heat flux rate analysis – calculation of wall heat flux rates or heat transfer coefficients.
- Visualization of heat transfer features for improved understanding of system flow phenomena.
- Identification of localized ‘hot spots or heat sinks in checking against specified requirements.
- Evaluating variation in temperature peaks for different operating conditions/ failure modes.
- Assessing insulation or heat shield effectiveness and optimizing of material selections.
- Coupling CHT with FE analysis to evaluate thermally induced stresses, expansion or fatigue.
- Temperature field analysis
- CFD thermal analysis
- Heat transfer coefficient analysis
- Heat transfer rate analysis
METHODOLOGY:
CASE-1:
- Y+ value = 1
- First layer thickness = 0.00005 m
- Body element size = 0.01 m
- Total number of elements = 339674
GEOMETRY:
MESHING:
- For this problem we will be using K-Omega SST model.
- The y plus range for it is 0-5 in order to calculate the First layer thickness.
- By using Y plus of 1 we get the first layer thickness as 0.00005 m.
CALCULATING Y PLUS:
Inlet diameter of the pipe = 0.16m.
Density of air = 1.225 Kg/m^3.
Velocity of the fluid = 5 m/s.
Dynamic Viscosity = 1.7894*10^(-5) Kg/(m-s)
Reynold's Number = `(rho*V*D)/mu`
= 1.225*5*0.16/1.7894*10^(-5)
= 54766.96 = 5.48 * 10^(4)
Skin Friction Coefficient `C_f` = 0.079 * Re^(-0.25) (For internal flows)
= 0.005164
Wall Shear Stress `tau_w` = `(1/2)*C_f*rho*V^2`
= 0.079075
Shear Velocity `U_T` = `sqrt(tau/rho)`
= 0.254
Y+ = `(rho*U_T*triangley_1)/mu`
where,
`triangley_1`= First Layer Thickness
For Y+ value of 1 by substituting the above values we get:
First layer thickness = 0.00005
Also we can calculate the same by another method:
- Now we can do the meshing by adding the Inflation layer between the solid and fluid interface and give the first layer thickness as calculated:
SETUP:
- Turbulence Model: K-Omega SST model
- Fluid: Air
- Solid: Aluminum
- Inlet Velocity: 5m/s
- Inlet Temperature: 700K
- Outer layer heat transfer coefficient (Convection): 20 W/m^2-K
RESULTS:
RESIDUALS:
SURFACE HEAT TRANSFER COEFFICIENT:
Y PLUS PLOT:
TEMPERATURE CONTOUR:
TEMPERATURE AT OUTLET:
VELOCITY STREAMLINE:
VELOCITY AT OUTLET:
SURFACE HEAT TRANSFER COEFFICIENT:
WALL HEAT TRANSFER COEFFICIENT:
CASE-2:
- Y+ value = 0.8
- First layer thickness = 0.00004 m
- Body element size = 0.01 m
- Total number of elements = 339885
GEOMETRY:
MESHING:
- For this problem we will be using K-Omega SST model.
- The y plus range for it is 0-5 in order to calculate the First layer thickness.
- By using Y plus of 0.8 we get the first layer thickness as 0.00004 m.
CALCULATING Y PLUS:
Inlet diameter of the pipe = 0.16m.
Density of air = 1.225 Kg/m^3.
Velocity of the fluid = 5 m/s.
Dynamic Viscosity = 1.7894*10^(-5) Kg/(m-s)
Reynold's Number = `(rho*V*D)/mu`
= 1.225*5*0.16/1.7894*10^(-5)
= 54766.96 = 5.48 * 10^(4)
Skin Friction Coefficient `C_f` = 0.079 * Re^(-0.25) (For internal flows)
= 0.005164
Wall Shear Stress `tau_w` = `(1/2)*C_f*rho*V^2`
= 0.079075
Shear Velocity `U_T` = `sqrt(tau/rho)`
= 0.254
Y+ = `(rho*U_T*triangley_1)/mu`
where,
`triangley_1`= First Layer Thickness
For Y+ value of 0.8 by substituting the above values we get:
First layer thickness = 0.00004
Also we can calculate the same by another method:
- Now we can do the meshing by adding the Inflation layer between the solid and fluid interface and give the first layer thickness as calculated:
SETUP:
- Turbulence Model: K-Omega SST model
- Fluid: Air
- Solid: Aluminum
- Inlet Velocity: 5m/s
- Inlet Temperature: 700K
- Outer layer heat transfer coefficient (Convection): 20 W/m^2-K
RESULTS:
RESIDUALS:
SURFACE HEAT TRANSFER COEFFICIENT:
Y PLUS PLOT:
TEMPERATURE CONTOUR:
TEMPERATURE AT OUTLET:
VELOCITY STREAMLINE:
VELOCITY AT OUTLET:
SURFACE HEAT TRANSFER COEFFICIENT:
WALL HEAT TRANSFER COEFFICIENT:
CASE-3:
- Y+ value = 0.4
- First layer thickness = 0.00002 m
- Body element size = 0.007 m
- Total number of elements = 479767
GEOMETRY:
MESHING:
- For this problem we will be using K-Omega SST model.
- The y plus range for it is 0-5 in order to calculate the First layer thickness.
- By using Y plus of 0.4 we get the first layer thickness as 0.00002 m.
CALCULATING Y PLUS:
Inlet diameter of the pipe = 0.16m.
Density of air = 1.225 Kg/m^3.
Velocity of the fluid = 5 m/s.
Dynamic Viscosity = 1.7894*10^(-5) Kg/(m-s)
Reynold's Number = `(rho*V*D)/mu`
= 1.225*5*0.16/1.7894*10^(-5)
= 54766.96 = 5.48 * 10^(4)
Skin Friction Coefficient `C_f` = 0.079 * Re^(-0.25) (For internal flows)
= 0.005164
Wall Shear Stress `tau_w` = `(1/2)*C_f*rho*V^2`
= 0.079075
Shear Velocity `U_T` = `sqrt(tau/rho)`
= 0.254
Y+ = `(rho*U_T*triangley_1)/mu`
where,
`triangley_1`= First Layer Thickness
For Y+ value of 0.4 by substituting the above values we get:
First layer thickness = 0.00002
Also we can calculate the same by another method:
- Now we can do the meshing by adding the Inflation layer between the solid and fluid interface and give the first layer thickness as calculated:
SETUP:
- Turbulence Model: K-Omega SST model
- Fluid: Air
- Solid: Aluminum
- Inlet Velocity: 5m/s
- Inlet Temperature: 700K
- Outer layer heat transfer coefficient (Convection): 20 W/m^2-K
RESULTS:
RESIDUALS:
SURFACE HEAT TRANSFER COEFFICIENT:
Y PLUS PLOT:
TEMPERATURE CONTOUR:
TEMPERATURE AT OUTLET:
VELOCITY STREAMLINE:
VELOCITY AT OUTLET:
SURFACE HEAT TRANSFER COEFFICIENT:
WALL HEAT TRANSFER COEFFICIENT:
TABULATION:
| SL.NO | Y+ VALUE | FIRST LAYER THICKNESS (mm) | NO. OF MESH ELEMENTS | SURFACE HEAT TRANSFER COEFFICIENT (W/m^2-K) |
| 01. | 1.0 | 0.00005 | 339647 | 20.821365 |
| 02. | 0.8 | 0.00004 | 339885 | 20.60224 |
| 03. | 0.4 | 0.00002 | 479767 | 19.809142 |
VALIDATION OF THE HEAT TRANSFER COEFFICIENT PREDICTIONS FROM THE SIMULATION:
- The heat transfer coefficient obtained from the simulation can be verified manually by calculating the heat transfer coefficient and then validating with the numerical data obtained.
- To calculate the Heat transfer coefficient analytically first we need to calculate the Reynold's number at that particular position by knowing the diameter and velocity of the place we want to calculate.
- For the temperature conditions and the fluid we use we have a certain Prandtl number, and also the conductive heat transfer coefficient of the solid material.
- Then we can use the emperical formula for Nusselt Number, Reynold's number and Prandtl number to calculate the value of the Nusselt number.
- The emperical formula between them is `Nu = 0.664*Re^0.5*Pr^0.33`
- Then we can use the Nusselt number formula to calculate the Heat transfer coefficient i.e.
- `Nu = (h*l)/k`
- `h = (Nu*k)/l`
- From the above calculations we can evaluate the value of the heat transfer coefficient and validate with our numerical data.
- The accuracy of the heat transfer coefficient value depends upon the correct Y+ plus value that we use according to the turbulence models.
CONCLUSION:
- From the above experiment we can conclude that the correct of Y+ value will give the accurate heat transfer coefficient.
- From the temperature contour we can notice that the temperature at the outlet side is more than the inlet side.
- This happens because of conservation of mass i.e. the mass flow rate of fluid exiting out of the port should be equal to the sum of all mass flow rate of the fluid coming in from the inlet side.
- Conservation of mass = `m_o = m_(i1) + m_(i2) + m_(i3) + m_(i4)`
- Due to the conservation of mass the velocity near the outlet side will be more than the inlet side and this can be clearly seen from the velocity plots.
- Due to high velocity at that position the Reynold's number will also be high which in turn increases the Nusselt number and thus the value of heat transfer coefficient near the outlet region increases.
- This increase in the value of the surface heat transfer coefficient at the outlet region can be observed from the contour provided above.
- To validate the results of the heat transfer coefficient obtained we have performed the simulation by refining the mess and can be observed that the value of the heat transfer coefficient remains almost same.
- From the above Wall Y plus plots it can be clearly observed that the value of Y Plus is maintained between the 0 to 5 region which is the correct range for the K-Omega turbulence model.
- To get the accurate value of heat transfer coefficient we must select the proper turbulence model and the correct Y plus value to get the correct first layer thickness value.