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Introduction

Thermal Analysis (Overview)

In ANSYS, the Thermal analysis application can be steady-state or transient:

• A steady-state analysis implies that the thermal conditions of the system have reached a state of static equilibrium, i.e., this resultant analysis is not time dependent.
• A transient analysis implies that the system is thermally dynamic, i.e., the thermal properties of the system are changing with time (time dependent).

With Thermal analysis, the properties that are considered are listed here:

Temperatures	Points in the model of known temperature
Heat flow	Points in the model of known heat flow rate
Heat flux	Surfaces of the model of known heat flow rate per unit area.
Convection	Surfaces of the model where heat is transferred to or from surroundings by means of convection
Radiation	Surfaces of the model where heat is transferred to or from the model by means of radiation.
Adiabatic Surfaces	Surfaces of the model where no heat is transferred from or to the model (i.e., perfectly insulated)
Heat generation	Regions within the model of known heat generation

Modelling Process when using ANSYS Steady-State Thermal

In this section, a generic modelling process is presented, the aim of which is to establish general good modelling practices:

Example 8

Example 8 – Step 1: Problem Definition

Before rushing to model the problem using software, the first step is to give some consideration to the problem statement. This stage is essential for the following reasons:

1. By suitably framing the problem statement, we ensure that the system is modelled appropriately.
2. We may be able to make estimations or predictions of the modelled results. This will support the validation of models and provide a degree of confidence that our results are correct.
3. By considering any errors or simplifications that we might introduce during numerical analysis, we have a stronger understanding of our system, which will provide context to our result and assist in troubleshooting should our results contain uncertainty.

For clarity, the following assumptions are implicit in our re-framing of the problem in this manner:

• The heat loss is mainly through the top & bottom surfaces, i.e., heat flux through the tip and sides of the fin is negligible.
• The problem is in steady-state conditions, i.e., the heat flux entering and leaving the fin have settled to constant values.
• The fin is not affected by thermal stresses causing structural deformation, i.e., the physical dimensions are independent of temperature.
• The conductivity (k) is constant through the length of the fin and independent of temperature.

Stating the energy balance equation in terms of heat flux (ɸ):

$\frac{d^2{\Phi }_{\left(x\right)}}{d{x}^2}-m^2{\Phi }_{\left(x\right)}=0$ 

where flux (ɸ) is defined as:

${\Phi }_{\left(x\right)}=(T_{\left(x\right)}-T_{fl})$ 

and m² is a constant defined as:

$m^2=hP/Ak$ 

This differential equation has a Particular Solution in the form:

${\Phi }_{\left(x\right)}={\Phi }_0\frac{\text{cosh }m(L-x)}{\text{cosh }m\left(L\right)}$ 

where (ɸ₀) is:

${\Phi }_0=(T_0-T_{fl})$ 

L is the length of the fin and cosh (x) is the hyperbolic cosine function defined as:

$\text{cosh(x)}=\frac{1}{2}(e^x+e^{-x})$ 

Using the above solution, we can input known conditions from our problem statement and make an estimation of the temperature at the tip of the cooling fin:

T0 = 100°C Tfl = 24°C

kALU = 237.5 W/m K h = 100 W/m2 K

P = 2 x 1000 = 2 000 mm = 2 m A = 1000 x 10 = 10 000 mm2 = 0.01 m2 L = 300 mm = 0.3 m

Using these parameters, we can compute our constant m:

$m^2=hP/Ak$ 

$m=\sqrt{hP/Ak}$ 

$m=\sqrt{\frac{100\times 2}{0.01\times 237.5}}=9.18$ 

And with those, we can calculate the flux through the fin and therefore the Temperature at the tip T_L:

Remembering that: ${\Phi }_{\left(x\right)}=(T_{\left(x\right)}-T_{fl})$ 

${\Phi }_{\left(x\right)}={\Phi }_0\frac{\text{cosh}\left(m\right(L-x\left)\right)}{\text{cosh}\left(m\right(L\left)\right)}$ 

${\Phi }_{\left(L\right)}={\Phi }_0\frac{\text{cosh}\left(m\right(L-L\left)\right)}{\text{cosh}\left(m\right(L\left)\right)}$ $=$ ${\Phi }_0\frac{\text{cosh}0}{\text{cosh}\left(m\right(L\left)\right)}$ $={\Phi }_0\frac{1}{\text{cosh}\left(m\right(L\left)\right)}$  

$(T_L-T_{fl})=(T_0-T_{fl})\frac{1}{\text{cosh}\left(m\right(L\left)\right)}$ 

$T_L=(T_0-T_{fl})\frac{1}{\text{cosh}\left(m\right(L\left)\right)}+T_{fl}$ 

$T_L=\frac{76}{\text{cosh}(9.18\times 0.3)}+24$ 

$T_L=33.6\text{°}C$ 

We can also make an estimation of the total cooling capacity of the fin by estimating the heat flux at the base of the fin, i.e., the rate of heat flow into the fin from the hot object.:

Above we derived the equation for temperature:

${\Phi }_{\left(x\right)}={\Phi }_0\frac{\text{cosh }m(L-x)}{\text{cosh }m\left(L\right)}$ 

where

${\Phi }_{\left(x\right)}=(T_{\left(x\right)}-T_{fl})$ 

We may then state:

$\frac{dT}{dx}=\frac{d}{dx}\left({\Phi }_0\frac{\text{cosh }m(L-x)}{\text{cosh }m\left(L\right)}\right)={\Phi }_0\frac{(-m)\text{ sinh }m(L-x)}{\text{cosh }m\left(L\right)}$ 

At the base of the fin; the derivative of Temperature with respect to displacement is therefore:

$\frac{dT}{dx}{|}_{x=0}=(-m{\Phi }_0)\frac{\text{sinh }m\left(L\right)}{\text{cosh }m\left(L\right)}=(-m{\Phi }_0)\frac{e^{mL}-e^{-mL}}{e^{mL}+e^{-mL}}$

where:

$mL=9.17\times 0.3=2.75$ 

$e^{mL}=15.689$ 

$e^{-mL}=0.064$ 

so

$\frac{dT}{dx}{|}_{x=0}=-9.17\times (100-24)\times \mathrm{0,992}=-691.8$ 

Remembering, the equation for rate of heat transfer is:

$q=-kAdT/dx=-237.5\times 0.01\times (-691.8)$ 

$q=1643W$ 

We should only expect these results to approximate that produced by ANSYS, as several of our assumptions inherent in these hand calculations will not be applicable within a modelling environment.

Example 8 – Step 2: Pre-processing

Prior to generating the solution, the model requires a degree of setup and configuration that is captured in the pre-processing stage. In this stage we shall work through the following steps:

1. Model Setup:
   1. Create the ANSYS model, set a Title for the project Save the project.
   2. Set preferences.
2. Specify material properties.
3. Geometric design.
4. Generating the mesh and element type

Example 8 – Step 3: Generating the Numerical Solution

The model has now been prepared and is ready to generate numerical solutions. In this step we shall configure the initial conditions and solve the model. In summary we shall complete the following steps:

1. Setup the initial conditions:
   1. Initial thermal conditions
   2. Thermal loads.
2. Solve the model.

Example 8 – Step 4: Post-processing

In step 3, we generated a numerical solution to the problem statement. Using this solution, we may now solve the model and then apply methods of visualising the data or extract specifically relevant data from the solution. In stage we shall:

1. Introduce the ‘Results’ tool to visualise the data to provide a better understanding of the behaviour of the system.
2. Introduce the ‘Scope’ tool to directly extract relevant data from the solution.
3. Compare our results to our estimated predictions to provide a degree of results verification.

 

Example 9

Example 9 – Step 1: Problem Definition

The rate of heat transfer through the pipe by conduction is:

$q_{COND}=kA(T_{r1}-T_{r2})/\left(\Delta r\right)$ 

The rate of heat transfer from the surface of the pipe by convection is:

$q_{CONV}=hA(T_{r2}-T_{fl})$ 

Rearranging these equations in terms of temperatures, remembering  $q_{COND}=q_{CONV}=q$  we get:

$(T_{r1}-T_{r2})=\left(\Delta r\right)q/kA$ 

$(T_{r2}-T_{fl})=q/hA$ 

Adding these equations, we can eliminate the unknown outer surface temperature term (T_r2):

$(T_{r1}-T_{fl})=\frac{q}{A}\left(\frac{\Delta r}{k}+\frac{1}{h}\right)$ 

Solving for the rate of heat transfer ‘q’ gives:

$q=A(T_{r1}-T_{fl}){\left(\frac{\Delta r}{k}+\frac{1}{h}\right)}^{-1}$ 

For clarity, the following assumptions / simplifications are implicit in our ‘hand calc’ estimated solutions:

Using the above solution, we can input the known system parameters stated in our problem statement and make an estimation of the rate of heat transfer from the pipe and the temperature on the outer surface of the pipe:

Tr1 = 180°C Tfl = 0°C

kLEAD = 35 W/m K h = 100 W/m2 K

Δr = 0.03 – 0.02 = 0.01 m A = l x πD2 = 1 x 3.14 x 0.06 = 0.19 m2

Using our solution, we can estimate the rate of heat transfer q through the pipe wall and into the environment:

$q=A(T_{r1}-T_{fl}){\left(\frac{\Delta r}{k}+\frac{1}{h}\right)}^{-1}$ 

$q=0.19\times 180\times {\left(\frac{0.01}{35}+\frac{1}{100}\right)}^{-1}=3299W$ 

We can state this estimation of the rate of heat transfer q in terms of heat flux f by dividing it by the outer surface area A of the pipe.

$f=\frac{q}{A}=\frac{3299}{0.19}=17.5kW/m^2$ 

We can also use the estimation of rate of heat transfer q in either of our independent statements of heat transfer to estimate the temperature on the outer surface of the pipe:

$q_{COND}=kA(T_{r1}-T_{r2})/\left(\Delta r\right)$ 

$T_{r2}=T_{r1}-\left(\frac{q\Delta r}{kA}\right)=180-\text{(5)}=175\text{°}C$ 

$q_{CONV}=hA(T_{r2}-T_{fl})$ 

$T_{r2}=T_{fl}+\left(\frac{q}{hA}\right)=0+\text{(175)}=175\text{°}C$ 

Example 9 – Step 2: Pre-processing

As in Example 8, prior to generating the solution, we require to setup and configure the model in this pre-processing stage. As a reminder the following steps should be completed:

1. Model Setup:
   1. Create the ANSYS model using a Steady-State Thermal analysis toolset, set a Title for the project Save the project.
   2. Set preferences, remembering to use the Metric Unit System, and setting the solver to 3D Geometry.
2. Specify material properties:
   1. Select Lead as the new default material within the ‘Engineering Data Sources’ / ‘Thermal Material’ as per the problem statement.
   2. Remembering to check that the Isotropic Thermal Conductivity is 35W/m K.
3. Geometric design.
   1. Create a new DesignModeler geometry and sketch the inner and outer circumferences of the pipe in the XYPlane, setting the dimensions as per the problem statement.
   2. Select the sketch from within the XYPlane tree and extrude the object 1 meter into the z-plane.
4. Generating the mesh and element type
   1. In this example, we shall allow DesignModeler to improve upon the default quadratic mesh used in Example 8. Under Default Mesh settings select Element Order: Program Controlled, and Element Size: Default. Under Sizing Mesh settings select Use Adaptive Sizing: Yes, and Mesh Defeaturing: Yes.

Example 9 – Step 3: Generating the Numerical Solution

The model has now been prepared and is ready to generate numerical solutions. In this step we shall configure the initial conditions and solve the model. In summary we shall complete the following steps:

1. Setup the initial conditions:
   1. Initial thermal conditions
   2. Thermal loads.
2. Solve the model.

Example 9 – Step 4: Post-processing

In step 3, we generated a numerical solution to the problem statement. Using this solution, we may now solve the model and then apply methods of visualising the data or extract specifically relevant data from the solution. In stage we shall:

1. Introduce the ‘Results’ tool to visualise the data to provide a better understanding of the behaviour of the system.
2. Introduce the ‘Scope’ tool to directly extract relevant data from the solution.
3. Compare our results to our estimated predictions to provide a degree of results verification.