Lecture 12, January 30, 2003 Reminder Quiz 1 in the tutorial session Tuesday – up to and including fins (ch1-ch3). Last Day,

• Determined the fin heat rate • Defined fin efficiency and effectiveness • Looked at temperature distributions in fins • Introduced the Biot number

Today:

• Look at CFD simulations of fins (sneak intro to convection) • Look at numerical solution (control volume) technique • Finned arrays

CFD (Computional Fluid Dynamics) Results The following results are obtained using a commercial CFD code, Fluent. This code solves the Navier-Stokes equations, coupled with conservation of energy to obtained detailed solutions of pressure, temperature and velocity fields. From these fields, we can calculate whatever is of interest for our particular problem. Note that these simulations were carried out very quickly without performing the detailed checks and refinements which are required when using CFD for real design/analysis work. Accordingly, the results will certainly be representative of the physics, but will not be quantitatively perfect (especially when turbulence is modelled). The results are however perfect to illustrate the points that are made in this section. We will use this code to solve laminar and turbulent flow over a single fin. We will explore:

• Fin effectiveness • How convection correlations are determined • How the temperature field is coupled to the velocity field in convection

problems.

Let’s begin with a uniform air stream impinging on a hot vertical wall. The air is at 293K, while the wall is at 353 K. Initially, let the uniform air velocity be 0.98 m/s, which will correspond to a laminar flow. The following figure depicts the resulting velocity vectors computed by Fluent. Note that the flow is from right to left, and that the hot wall is the left wall. Since the air cannot pass through the wall, it is deflected

upwards, and exits the domain vertically at the top wall. Note the interesting velocity pattern in the lower left corner. Here the velocities are very low, and the flow is recirculating. This will have implications on the energy field.

The velocity vectors look nice, but a much better way to visualize a two dimensional flow field is to compute the stream function, as discussed in Mech 341. The next figure depicts contours of the stream function. Each contour line is a line upon which the stream function is constant – this is a streamline. The significant feature of a streamlines, which make them a fantastic tool for visualizing flow fields, is that they are everywhere parallel to the velocity vector. This means that no flow can pass through a streamline. Pick any two streamlines, and follow them from the inlet to the outlet. What you are looking at the true path that the fluid takes to traverse our domain. Since the flow rate is constant, and this is an incompressible flow, we can glean even more from this plot. When the streamlines are close together, the air is moving faster than when the streamlines are further apart. Therefore, we can clearly see in the figure that the air is decelerated as it approaches the hot wall and turns the corner, and it accelerates again as it exits the domain. Again, I will defer detailed discussion to both Mech 341, and the convection section of the course, but it is precisely where the flow is decelerating that flow separation and regions of recirculation are likely to form. These regions will have an enormous impact on heat transfer, as we shall see.

Let’s look closely at the velocity vectors in the lower left corner. It is clear from this figure that the air is moving in roughly a circular pattern in the corner. The details of this “flow structure” will change considerably with the addition of a fin along the bottom section of the hot wall, and especially when the flow becomes turbulent.

Now, let’s look at the resulting temperature field. The next figure depicts temperature contours over the entire domain. Most of the domain contains air at the free stream temperature, 293 K. Adjacent to the hot wall though the air is clearly heated by the hot wall. Notice that the temperature contours are very different where the air is recirculating compared to where the flow is “attached” to the hot wall. The recirculating air is heated at the hot wall, and carries this energy away from the wall. Accordingly, the temperature contours are wildly distorted along the paths taken by the air.

This fact is even more apparent in the next two figures which are simply increasingly magnified views of the recirculation zone.

As we zoom in really close, note that that the temperature contour lines are very nearly parallel to the hot wall very close the wall, where the velocities are very nearly zero (no-slip condition at the wall). Remember from Fourier’s law, that heat energy will be transferred from regions of higher temperature to regions of lower, and that this will occur in a direction perpendicular to the temperature contour lines (i.e. in the opposite direction of the temperature gradient). It is clear that energy is being transferred from the wall to the air. Notice too, that the contour lines are perpendicular to the bottom boundary. I have chosen this to be a symmetry boundary, which physically means that the geometry and flow is mirrored about this line. If that is the case, then there can be no heat transfer across this line since there can be no temperature gradient there (since the temperature is the same at the same distance from the line). Temperature contour lines intersecting a boundary at 90O like this indicate that there is no heat transfer through this boundary.

Now Lets add a fin, made of a metallic material with a conductivity of 202 W/mK, and use the same free stream velocity. Notice that the velocity vectors look similar, but there is definitely more distortion near the fin,

Looking closely at the fin tip, we see that the velocity vectors closest to the fin are actually directed towards the inlet – the fluid nearest the fin surface is moving against the main flow! This means that the flow has “separated” from the fin surface very near the fin tip, and this the recirulation zone extends along the entire length of the fin.

Looking now near the lower left corner, we see a fairly similar plot compared to when there was no fin, though the flow certainly has been perturbed by the fin. Note that structure is evident within the recirculation region. Flow comes down from the vertical

sidewall, and turns to flow along the fin in the reverse direction from the bulk flow. Some of this flow is directed upwards fairly quickly, to form the circular recirculation, while some of it continues along the fin. Again, this will have implications on the heat transfer.

Again, the streamlines are a usefull tool for describing the general flow.

Let’s now look at the temperature contours. Again, the temperature contours follow the general features of the frecirculating flow. The prime difference between this case, and the case with no fin, is that there are many temperature contour lines (nearly) parallel to the fin surface, indicating that there is significant heat transfer from the fin surface to the air. Note too, that the high temperature region extends further out into the flow in the middle of the fin compared to the base and the tip sections. This should not be surprising since the air velocities in this region were the lowest.

This feature is increasingly evident as we magnify the contours. It is very clear that the moving air is carrying heat energy with it – clearly demonstrating how convection enhances heat transfer compared to conduction alone. Finally, not that the lone temperature contour in the fin is practically a vertical line. This suggests that heat transfer is practically 1-D along the direction of the fin – as we would expect from calculating the Biot Number.

We will now calculate the heat transferred from the hot wall to the air, and break this out into two sections. The first section comprising the fin only (remember qfin is calculated by determing how much heat energy is conducted in at the base as this must be dissipated at the surface of the fin) and the second comprising the rest of the wall. The fin is 0.5 cm high, while the rest of the wall shown here is 20 cm high. This means that the fin cross section is 1/40th the cross section of the remaining wall. Further, we will vary the air velocity, to see the effect that this has on heat transfer. In order to describe the different air velocities then, we will calculate a Reynolds number at the end of the fin (which is 15 cm long).

µρVL

L =Re

where L= 0.15 m, ρ=1.225 kg/m3 is the air density, and µ = 1.8 x 10-5 kg/(ms) is the air viscosity. Our 0.98 m/s velocity, presented above corresponds to ReL=10,000. A summary of the heat rates for various cases are given in the following table. Notice that when there is no fin present there is very little heat transfer through the area corresponding to where the fin will go (0.855 W), adding the fin (and keeping the air velocity constant) increases this to 43.5 W, for a fin effectiveness over 50. Note too under these conditions that the fin is dissipating almost half as much energy as the remaining wall, which has 40 times the cross sectional area! This clearly demonstrates why fins are used. In fact, as the velocity increases, this gets even better

ReL Fin heat rate [W] Wall heat rate [W]

Fin effeciveness [-]

No Fin (u=0.98m/s) 0.855 102.16

2,500 23.17 46.53 27.1

5,000 30.78 70.52 36.0 10,000 43.5 96.92 50.9 20,000 68.42 118.61 80.0

100,000 358.9 1422.79 419.8 The first 5 cases were assumed to be laminar, while the final case, at ReL=100,000, was assumed to be turbulent. Notice that for the laminar cases, when we double the flow rate, we can about a 50% increase in heat dissipated. However, when we jumped from ReL=20,000 to ReL=100,000, a fivefold increase, we got a very similar increase in heat dissipated. Turbulence changes the flow field enormously and provides for much more heat transfer. This occurs in two ways. First, the turbulent flow creates far more mixing, and hence is more effective at carrying hot fluid away from the surface of the fin to the cooler surrounding fluid. Second, the increased energy in the flow is able to radically alter the flow structure, resulting in much smaller recirculation regions. The following figure shows the velocity vectors for the turbulent flow case (ReL=100,000). Notice that the recirclations regin is clearly smaller.

If we again focus on the tip region of the fin, we see that now the flow is attached to the fin – the velocity vectors close to the fin surface point in the same direction as the main flow. The recirculation zone is limited to a much smaller area near the base of the fin.

The temperature field is also radically altered by the turbulent flow. Now, the high temperature region is confined very close to the fin virtually everywhere. Since the fin is now dissipating considerable heat energy, there is also a mich greater temperature drop within the fin itself. Again, we can see that temperature contours inside the fin are vertical lines, indicating that heat transfer is 1-D in this case. If we tried this again with a fin having a much higher Biot number (lower conductivity) we would see a very different story.

If we keep everything else constant, and reduce the fin conductivity to 0.1 W/mK (as in the next two figures), the temperature field again changes radically, and the heat dissipated through the fin drops to 4.6 W from 358.9 W.

Notice that the temperature of the fin has dropped to 293 K (the ambient) about halfway along its length, meaning that the rest of the material is simply wasted. Also note that the temperature contours in the fin are very two dimensional. Clearly, a 1-D analysis would not work in this case. Note the value of the Biot number in indicating this before we have done any analysis.

Finally, lets calculate a local value of a convection coefficient at each point along the fin. Note that this is easy to do since we know the entire temperature field, and we can calculate the heat rate at each and every point by calculating the gradient of the temperature distribution. At the fin surface, the fluid velocity is zero (no-slip condition), and heat transfer is purely by conduction at this point. Therefore, the heat rate is Calculated using Fourier’s law. We then compare this with Newton’s law of cooling, q = hA(Ts - T∞). We know the area of each control volume, and we know the local surface temperature and our reference ambient temperature. It is therefore a simple calculation to determine a local h value. The next figure plots the variation of the local convection coefficient over the fin for the various cases. Note that the assumption of a constant h, that we made when considering the fin only, is clearly a big approximation. In general, the convection coefficients are high near the fin tip, and drop as the base is approached. In the laminar flow cases, though it the effect of the recirculating air is clearly apparent as h drops to a minimum and then increases again.

How do we actually solve the 1-D fin equation numerically? Stepping back to what we did last day with MATLAB, how do the provided programs actually solve for the temperature distribution? Examine the programs available on the web site (unifin.m for example). The script ‘finfun2004.m’ is a script that produces all the outout given in the powerpoint slides last day. The following description should help you understand exactly how the functions work.

Now, applying conservation of energy for a steady 1D system with no generation,

Express this in a convenient form, identifying the unknowns,

And, considering each control volume, we have exactly as many unknowns as equations and we can solve for the temperature distribution and hence the fin heat rate.

Fin Arrays (section 3.6.5 in text) How do we extend our analysis to include multiple fins? We shall develop resistance networks to describe the geometry of interest. First define a thermal resistance for a single fin based on what we have already done with fins (numerically or analytically)

And reform this equation to define a thermal resistance for a fin.

Both the fins, and the area between them will dissipate heat, and so we have to consider the spaces as well. If there are N fins, then the total area dissipating heat is

The total heat rate is then,

These equations, and the concept of thermal resistance networks allow us to analyze any array of fins, and even the effects of contact resistance. Be very careful though about the approximation of constant convection coefficient.