In this Unit we start by taking the electrical analogy for heat transfer which we met in Unit 8 and extending the concept of thermal resistance, first to deal with more complex structures, where the heat flow is primarily by conduction, and then to estimate the temperatures in convection-cooled situations.
Our aim is to give a flavour of the ways in which analytical methods can be used to solve thermal management problems, and give some ideas that you can experiment with in the first part of Assignment 1. However, we must emphasise that our discussion merely “scratches the surface”, because non-simulation methods have been used in electronics thermal management for over 40 years, and there is a great deal of material to be explored.
Whilst we can’t possibly cover the whole field, we have attempted to illustrate it, concentrating on concepts that do not require sophisticated mathematical techniques. Many of the results of the computations might be regarded as “quick and dirty”, but we believe that there is real benefit in obtaining estimates that are independent of simulation techniques. Also, we have found that the process of comparing estimates with simulation is really helpful, both for uncovering errors and for estimating the likely inaccuracies embedded in answers generated by simulation software.
As we saw in Unit 8, simple models can be created where the primary parameter is a ‘thermal resistance’, which relates the heat of flow between two points to the temperature difference between them. For conduction, the thermal resistance is defined by the equation:
. . . . . . . Eqn. 1
where TA, TB are the temperatures of isothermal surfaces (surfaces at constant temperature) and PAB is the total thermal energy flowing between the surfaces in unit time.
Figure 1 shows simple cases of heat flow by conduction that can be modelled in one dimension, as a block, a cylinder, or a hollow cylinder. Where heat is flowing across a material with significant conductivity, lateral spreading takes place, but this can also be modelled as one dimension, by using the correction for spreading resistance described later.
For a composite structure, we can calculate the combined thermal resistance from the dimensions of the elements of the structure and the properties of the materials from which they are made. We will be doing this later when we consider how a package is built up from its components, and we will be returning to the topic both in Unit 12 and Assignment 1.
Fortunately, conduction is relatively “well behaved”, in that the heat flow is a linear function of temperature difference, and conductivity is a property of the material that is both measurable and relatively constant over the normal operating temperature range. Admittedly, there will be some variation within materials that are nominally the same, depending on factors such as impurity level, and some uncertainty as to the thermal resistance of boundaries, where the performance of an interface will depend on factors such as surface roughness, pressure, and the presence of any intermediate thermal interface material.
However, these variations are minute when compared with the variability in the corresponding equations for conduction. We saw in Unit 8 that the representative thermal resistance is given by:
. . . . . . . Eqn. 2
Here the area can be estimated, but the value of h will vary widely, as you will have seen during the web research activity at the end of the Unit. The situation is similar with radiation, though here typical values of the main unknown, emissivity, are both better-documented and less variable.
Overall, there are two problems:
Whatever estimates are made, prudent practitioners will:
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We are starting by tackling the complex model problem in relation to conduction, and considering first the modelling activities that are involved in a typical situation, where a number of power devices are mounted on a board. For each of these devices, we need to estimate the temperature difference that will exist between the die and its surrounding board, and for the whole assembly we will need to identify where the assembly is hottest, and estimate the temperature rise across the board.
In this and the next main section, we are looking at this modelling challenge, at the simplifications that are used, and at broader aspects of the thermal resistance analogy.
In Unit 8, in the final exercise of the section The electrical analogy for heat transfer, we made the simplifying assumption that heat goes straight through the board, but in real life there will be some sideways heat spreading, even with a homogenous material. More so if the structure contains power and ground planes, as copper has a much higher thermal conductivity than the laminate. Although an approximation, based on experience and the result of more exact simulation, much use is made in simple calculations of the concept of a ‘spread angle’, where the heat is visualised as spreading outwards in a cone shape, rather than travelling vertically.
The spreading angle α is defined in Figure 2. If the thermal conductivity K of the underlying layers is significantly higher than the current layer, spreading will be at a minimum, although there will always be some lateral heat spreading due to the randomness of thermal energy, so the spreading angle will actually be greater than 0°. However, where the conductivity of the underlying materials is very low compared with the current layer, the spreading angle will be large, approaching but never reaching 90°. Nguyen gives an approximation of the spreading angle based upon the ratio of thermal conductivities of the layers.
. . . . . . . Eqn. 3
where α = spreading angle, K1 is the thermal conductivity of the current layer, and K2 the thermal conductivity of the underlying layer. For many applications one can assume a 45° spreading angle to obtain first-order approximations of temperature rise.
The heat-spreading model is widely used, and holds if there are no thermal mismatches or thermal boundaries. If thermal boundaries exist as shown in Figure 3a then the analysis needs to be broken down into two components in series. The first component would have a trapezoidal heat path due to spreading as shown in Figure 3b, the second, the rectangular section depicted in Figure 3c, because there is no opportunity for heat spreading. The total thermal resistance is the series combination of the thermal resistances of the trapezoidal and rectangular sections.
In his paper Simple formulas for estimating thermal spreading resistance, Robert Simons of IBM tackled the problem of thermal spreading from a chip to the package, looking for a solution that could be incorporated in a spreadsheet. A key simplification that proved unexpectedly precise was to transform the square package into an equivalent circular geometry. His comment is that the simplified formulae provide a useful tool, capable of less than 8% inaccuracy even for the largest component.
Simon’s concerns about the basis for thermal resistance are explained, albeit in a slightly different way, by this article on thermal resistance from CoolingZone. Whilst the removal of heat by methods other than conduction is both important and a variable quantity, the concept of thermal resistance is still agreed to be a “very useful figure of merit if you know how to use it”.
As well as the analogies between electrical and thermal parameters, a number of the equations, concepts and principles that will be discussed in later sections are the same in both electrical and thermal terms:
resistances in series
resistances in parallel
When several materials are stacked in series, such as a die attached with epoxy to a substrate that is soldered to a package base, the equivalent (total) thermal resistance becomes the sum of the individual thermal resistances. This is analogous to having several resistors in series. For N thermal resistors in series the equivalent thermal resistance is
. . . . . . . Eqn. 4
The temperature of any particular interface may be calculated from
. . . . . . . Eqn. 5
where Tj,j-1 is temperature of the interface between layers j and j−1 and σ is the sum of thermal resistances from that interface to the heat sink. Note that the temperature of the interface of any two layers cannot change discontinuously, so that, for example, the temperature at the bottom of the die will be the same as the temperature at the top of the die attach material.
Figure 4 illustrates the stacking of thermal resistances using this electrical circuit analogy.
θ1 = Thermal resistance of die
θ2 = Thermal resistance of die attach
θ3 = Thermal resistance of substrate
θ4 = Thermal resistance of substrate attach
θ5 = Thermal resistance of package
Tj = Junction temperature
Tc = Case temperature
Whilst such analysis is very much a simplification, it is still useful in demonstrating which of the elements within the construction are the biggest contributors to thermal resistance. Use a spreadsheet or similar to determine the thermal resistance for the package shown in Figure 4, using the following typical parameters:
Note that we haven’t supplied conductivity values for the different materials, or information about spreading, as these have already been given in earlier Units.
Our calculation is shown in the attached spreadsheet, which includes a pie chart of the contributions from the different elements. You might like to reflect on:
[Note: Depending on your browser settings, and on which version of Windows you are running, you may experience problems in opening our spreadsheet within your browser window. In which case, right-click on the link above, save the file on your machine by choosing the “Save Target As...” option, and open it within Excel.]
In this exercise we have indicated how one might use basic information on the dimensions and materials of a package to calculate the thermal resistance of a selected path. As we'll see later, this ‘analytical’ approach can be extended into further dimensions, although the calculation rapidly becomes more complex. But don’t knock the simple approach and the obvious conclusions that can be drawn, such as “consistent with intuition, decreasing the thickness or increasing the cross-sectional area or thermal conductivity of an object will decrease its thermal resistance”, as Bruce Guenin puts it in One-dimensional heat flow. It is surprising how many engineers lose sight of this.
In combination with estimates of the way heat spreads, the concept of thermal resistance can be applied to model a more complex structure. Whilst we won’t attempt to look at this level of detail until Unit 12, look at Figure 5 to see one way in which a resistor mesh is used to model a complete semiconductor structure, including a heat sink.
The thermal resistance simplification has long been used as a way of describing the parameters of a semiconductor package. Often this has been in terms of a pair of thermal resistances, one from the die to the case (by conduction), and the other representing the thermal impedance from the package to the environment (by convection). Whether this simplification is sufficient is the subject of an extended discussion in Unit 12. The value of the thermal resistance junction-to-case is also well understood and useful in projecting the likely die temperature from the known temperature of a heat sink; we will be doing this in Unit 14.
We will be considering the limitations of case-to-ambient thermal resistance in more detail in following Units but, if you wish to get ahead of the game, you might like to look now at the JESD51 series specifications that are available on (free) registration from the JEDEC website (http://www.jedec.org/).
Printed circuit boards are composite materials, with enormous differences in conductivity between copper and the base laminate, in addition to having features that are far from uniformly-distributed. So, when trying to model conduction in the board, we need to allow for the fact that the board is neither uniform nor isotropic.
One way of doing this, whilst still making a very useful simplification, is to define an “effective thermal conductivity” for the overall board. To see the way in which this is derived for conduction in the plane of the board, take the simplest case, of the single-sided board shown in Figure 6.
Here, for simplicity, we are considering only heat conduction along the length of the board. Heat flows in parallel along the laminate and copper layer, and the total thermal conductivity (the reciprocal of the thermal resistance) is the sum of the two conductivities, as shown in Equation 6.
. . . . . . . Eqn. 6
where A is the cross-sectional area of the board, and t is the thickness. We can define the effective thermal conductivity as:
. . . . . . Eqn. 7
The parameters that affect the conductivity value are the cross-sectional area, and the percentage of the board volume that is copper. So this equation be reduced to:
or . . . . . . . Eqn. 8
where n is the percentage of copper in the stack up. Although variations in copper coverage will result in localised temperature variations, we can use our knowledge of the thickness of the foils and of the copper coverage on each layer to calculate an adequately accurate effective percentage for any board.
As far as thermal flow in the plane of the laminate is concerned, the conductivities through laminate and conductor are in parallel. However, they are in series for conduction through the board. In this case, we need to add the thermal resistances of the foil (very low) and the laminate (regrettably high).
Derive the simplified equations equivalent to Equation 8 for the case of thermal flow through the board.
In an electrical resistor-capacitor network, the voltage across the capacitor charges exponentially from zero to its final value of V0 according to
. . . . . . . Eqn. 9
The rate at which the capacitor charges to the equilibrium voltage is the time constant τ:
The thermal analogue to the circuit is shown in Figure 6. The temperature ΔT across the material can be calculated as a function of time by:
. . . . . . . Eqn. 10
where ΔT is the temperature difference, and TE the equilibrium temperature. Again the rate at which the material approaches equilibrium is the thermal time constant:
The rate at which system warm up and cool down can be quite important in practice, as this has an effect upon the mechanical stresses within the system, and hence on the potential for failure. The worked example at this link is an unedited extract from Chris Hill’s submission for a previous version of Assignment 1. Look at the way in which he approaches the problem, and examine the calculations in the associated spreadsheet.
Reflect on the alternative way in which this calculation might have been attempted using the RC product approach, and note the way in which the spreadsheet is used for an iterative calculation. Such calculations are quite common in thermal engineering, with iterations continued until the result has stabilised within the required limits.
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Whilst the analysis of a thermal performance of a power device is of value, in most applications we will find multiple heat sources. On some occasions we will be able to simplify calculations by treating the less significant thermal contributions from a large number of circuit elements by “smearing” their dissipation over the whole board. You will come across this valuable simplification during Assignment 1.
But sometimes we cannot avoid considering individual circuit elements as sources of heat. The first of the techniques we use for this is “superposition”, which you may remember seeing in the earlier list of analogies between electrical and thermal parameters. This looks at a problem containing multiple sources by calculating the potential temperature rise due to each independently.
In an electrical circuit, superposition means that the voltage at any point is equal to the sum of the various voltages when each of the current sources acts on its own. This principle can be extended to thermal circuits using the analogy that the temperature at any point in a thermal circuit is equal to the sum of the temperatures that would arise if each of the heat sources acted on its own.
Many thermal problems can be solved easily by combining several thermal resistances using series/parallel combinations and applying the principle of superposition. An example is the multiple heat sources shown in Figure 7a, where the electrical analogue is shown in Figures 7b and 7c. Assuming a substrate base at a constant temperature, we need to solve for the temperature rises ΔT1 and ΔT2 in heat sources Q1 and Q2.
Using the electrical analogue:
. . . . . . . Eqn. 11
where θT is the equivalent thermal resistance of θ11, θ22 and θ11-22 in parallel.
With heat source Q2 open circuit, the temperature T12 at node a is:
. . . . . . . Eqn. 12
and with heat source Q1 open circuit, the temperature at node a is:
. . . . . . . Eqn. 13
By superposition, with both heat sources on, the temperature rise is the sum of the individual temperature rises:
. . . . . . . Eqn. 14
so the temperature rise from Q1 to the heat sink is:
. . . . . . . Eqn. 15
As well as superposition, another general concept that can sometimes be useful is reciprocity. The principle of reciprocity states that the effect at position A due to an event at position B is the same as the effect at position B should the event have occurred at position A.
In the thermal case, this can be re stated as “the temperature at point A due to a heat dissipation Q at point B is equal to the temperature at point B due to a heat dissipation Q at point A”.
Typically components are distributed across a circuit board, all contributing to the temperature rise. Intuitively, we can appreciate that the highest temperatures will be reached in the centre of the array, but what does the temperature distribution look like? Here the answers will differ slightly, depending on the physical arrangements. However, if we take the type of board that might be found in a conduction-cooled military assembly, where the heat is extracted from the board edges, we can concentrate on the conduction aspects.
If we ignore convected heat, our analysis of the array of parts shown in Figure 7 can be simplified in three ways:
These simplifications are indicated in Figure 8, which also shows a temperature distribution with a parabolic shape, which is given by Equation 16.
. . . . . . . Eqn. 16
where q is the heat input per unit length, K is the thermal conductivity, and A the cross-sectional area. When we evaluate this for the central position (x=0), we can see that the maximum temperature rise will be:
.. . . . . . . Eqn. 17
where Q is the overall heat input. The derivation of this equation is shown at this link.
Such equations, the result of “exact solutions”, are fairly rare, but we can get a very similar result using a spreadsheet, and dividing the length into a number of discrete lengths, rather than the infinitesimal lengths used by the integration method. One can even argue that the result may be more representative of the real world, in that heat inputs are generally not uniform, but the result of discrete heat-generating elements.
Take the example above, with a strip of length L = 100mm and 10mm wide, a heat input of 800mW evenly distributed over that area, a board thickness of 1.6mm, including a total of 105µm of copper. First, calculate the temperature rise using Equation 17. Then use Excel to calculate the heat rise based on different numbers of cells. Finally, work out the temperature of the hottest device, given the situation where 8 SON packages are spread evenly along the length, each dissipating 100mW. Compare your answer with the attached spreadsheet.
Although this example is relatively trivial, you will have discovered the way in which a solution to a complex problem can be sought, even if the outcome is not known initially, either using the principle of superposition or by repeated calculation. This approach can readily be extended to two dimensions.
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Again, for simplicity, we will consider the situation of an assembly with a uniformly-distributed heat load. As you were reminded earlier, Equation 2 gives a simple way of estimating the thermal resistance between the surface of the assembly and the environment:
. . . . . . . Eqn. 2
Given this value, and also the overall dissipation in the assembly, we can calculate the mean temperature of the surface. However, we need to decide what value of heat transfer coefficient to use, and this depends on a number of factors, including board orientation.
Intuitively, we can appreciate that the results of cooling will be different in the case of a horizontal board, compared with an assembly mounted vertically (Figure 9):
Fortunately, there is quite a lot to help us in the literature. For example, Steinberg’s “rule of thumb” that the convection coefficient for the lower surface will be about half that of the upper surface.
The degree of cooling will depend on whether the fluid flow is laminar or turbulent. Typically natural convection starts out laminar, but becomes turbulent when the temperature difference is high, or the body is long, as measured in the direction of airflow. For typical conditions in electronic equipment, where temperature differences are under 100°C, and the “characteristic length” of surfaces are below 500mm, laminar flow is generally stated to be a fairly safe assumption, at least for natural convection.
Calculating the relationships that produce a value of h is far from straightforward, and involves a number of the dimensionless ratios that we touched upon in Unit 5, where we suggested you read our short commentary on Dimensionless numbers and Dimensionless Numbers in Heat Transfer in the CoolingZone tutorial section.
Typical relationships embody constants that reflect the geometry, size and orientation of the surface, and these parameters are experimentally determined, rather than the result of analysis. For example, Çengel’s Introduction to Thermal Dynamics and Heat Transfer suggests that, for natural convection and laminar flow, the heat transfer coefficient for air can be expressed in the form
. . . . . . . Eqn. 18
where ΔT is the temperature difference between the surface and the fluid, L is the length of the body along the heat flow path (the so-called “characteristic length”), and K is a constant whose value depends on the geometry and orientation of the part being cooled. The value of K (expressed in W/m2·°C) is of the order of 1.3 for the top surface of a horizontal plate, but 1.4 for a vertical plate, and 2.4 for a typical case for a vertical board with components on one surface. These figures have to be regarded as for indication only!
For more information on estimating the natural convection heat transfer coefficient, see these papers by Robert Simons:
When we choose an analytical solution for modelling thermal flow, we often make an assumption that the whole extended area of the heat sink is at the same temperature. That is, a “isothermal boundary condition” applies. Whether or not this is valid, and whether there are temperature gradients within structures can be estimated by calculating the Biot number calculated as . Where Bi<0.1, this indicates that the heat flow is essentially one-dimensional. Fortunately, typical heat sinks fall within this range, so the hypothesis that the base plate of a heat sink is isothermal is generally valid.
More information on the modelling of heat sinks, together with some web tools, can be found in the paper by Culham, Teertstra and Yovanovich, Natural convection modeling of heat sinks using web-based tools. The tools are particularly useful in allowing estimates to be made for simple heat sinks in a natural convection environment.
With natural convection, the velocity of air movement is quite low, so this type of cooling is limited to low power applications. However, the more air that goes past the surface, the larger the heat flow, and the higher will be the heat transfer rate. Changing from natural convection to forced convection (more accurately, adding forced convection to the natural convection that will occur anyway) has a very significant impact on performance. Whilst we should always design products so as to get the best possibility cooling without a fan (to cover the eventuality of fan failure), we are potentially looking at an order of magnitude improvement in heat transfer coefficient.
Estimating the heat transfer coefficient becomes quite complex, depending primarily on whether there is any thermal boundary layer between the body and the fluid, and whether the flow is laminar or turbulent. Both considerations are affected by the nature and dynamic viscosity of the fluid, so the cooling available will vary according to temperature and pressure, and will be totally different if a liquid is used rather than a gas. The difficulties in estimating h are compounded with surfaces that are not flat, for example, the cylinder shown in Figure 10.
Here the fluid wraps around the leading edge, but the boundary layer detaches from the rear surface, forming a wake behind. This type of flow separation typically needs to be studied experimentally, rather than attempted analytically.
Of course, most real situations combine convection and conduction. The way in which this is tackled conceptually is shown in Figure 11. Here the assembly has been modelled as a series of smaller plates, each of which is associated with thermal resistances to the surrounding plates and a convection thermal resistance to the environment.
Some kind of spreadsheet analysis is helpful for tasks of this kind, and we have included as a supplementary sheet a second example from Chris Hills’ work, where he uses ThermXL, a thermal analysis add-in to Excel, which automates the process to some extent, although one has to be extremely careful to ensure that the detail of the model is absolutely correct. In this case, for reasons, that will be obvious, we have removed some sections of the text. But we have included a link to the spreadsheet that was produced. This is interesting in that it shows the type of result that can be obtained with relatively straightforward tools.
Unfortunately, for internal administrative reasons, we have been unable to arrange for ThermXL to be made available to you on the AMI site. However, if you want to explore this route, you may be able to negotiate direct with GEC Alsthom a time-limited (five days only!) demonstration.
Whilst we have made much of the analogy between electrical resistance and thermal resistance, and will continue to make use of this, it is important to remember that too much simplification is dangerous. This is emphasized by Clement Lasance in his paper Thermal Resistance: an oxymoron?
Another of the points he makes is that the difference between insulation and conduction is 20 orders of magnitude for electrical engineers, but only about three orders of magnitude in thermal terms.
Note also his concerns that thermal parameters should be seen as mathematical constructs rather than representations of the physical reality. Whilst the results may be dependable when applied to a transistor on a heat sink, the same cannot be said for thermal resistance used in modelling heat exchange with the surroundings. Or are his fears just a different way of expressing the fact that heat transfer due to conduction varies enormously according to the convection heat transfer coefficient, because this depends on the physical conditions?
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We saw in Unit 8 a simple way of quantifying the air flow needed, relating this to the total dissipation and to good practice values for temperature rise. However, although we may have decided we need to use forced convection to provide the necessary cooling, we will only have an idea of the specification required for the fan once we have been able to strike a balance between the volume/pressure relationship for the fan and the impedance to air flow provided by the equipment. [You will find more about the practical aspects of selecting fans in Unit 15]
There are a number of ways of tackling the problem of resistance to flow of the overall structure. Experimental pragmatists would use a fan to force air through the system and measure both the flow and pressure drop for varying values of flow. Engineers with a simulation programme available would use this tool to model the components of the system. Designers wishing to estimate the resistance to flow, but without access to simulation tools, have a significant challenge, but one that can be approached by severe simplification of the assembly and by applying general principles of fluid mechanics.
To gain an appreciation of this approach, skim read this sequence of articles by Cathy Biber, and then our comments:
Even with such simple structures, the principal problem is that the resistance of flow experienced depends on the flow rate, whilst the flow rate depends on the resistance to flow presented to the fan. As a result, in order to estimate the volume flow rate, and thus be able to perform a thermal analysis, we need to approach the problem through successive approximations.
We warmed to Dr Biber’s comment that analytical calculations are not perfect, but are “better than rules of thumb and . . . better than waiting for a test to tell you that you are in big trouble”!
One key concept that occurs in the first article is that of the difference between “developing flow” and “developed flow”, shown in Figure 12. This attempts to show the distortion in the airflow pattern that exists at the input end of the channel being considered, creating a transition region.
You will also notice that the literature suggests a number of different correlations. In other words, there is no single set of equations that yields accurate results regardless of the situation. There may be an exact solution for fully-developed laminar flow between parallel plates, but that is about all.
However, for calculating the volume flow rate from a fan to a system whose main resistance consists of rectangular channels in series with changes of area, it is an approach which gives a reasonable fit, thought the prudent will always verify an estimate by comparing theory and measurement.
Flow across a plate, or in a tube, represents a condition very different from a typical application, where the surfaces are rough, and airflow is frequently constrained in some directions. This is where simulation becomes particularly useful. However, a number of comparatively simple flow circuit elements have been subjected to analysis. These include expansions and contractions, both abrupt changes and gradual transitions. Not only must the loss be considered, but transitions typically introduce turbulence in the flow.
Similar analyses have been applied to smooth and sharp-radius bends, and to the ‘manifold’, where the coolant flow is divided between a number of branches, shown schematically in Figure 13. Such analysis has obvious application to chassis constructions with multiple printed circuit assemblies.
Circuit boards are only flat as fabricated, and the populated board contains components that protrude into the fluid stream. If the flow is laminar, the protrusion will only have a significant effect if its height exceeds the thickness of the boundary layer. However, if the flow is turbulent, even minor roughness, such as that on a sand casting, can affect the pressure drop.
A typical simplification used in analysis is the uniform, in-line array of modules assembled on a board that is shown in side view in Figure 14. Note that this ignores the diversity of real packages, and the presence of items such as cables and card separators than may partially block the flow of air.
For a uniform array of equally sized elements, we can define its packaging density as:
where Lx and Ly are the dimensions of each element, and Sx and Sy are the pitch of the packages in x and y directions respectively. For typical packages and applications array densities will lie in the range 0.25–0.9 (25–90%).
In most practical applications the assembly will not be in free air, but the coolant air will flow in a channel formed between the board and the system enclosure or between two adjacent boards. Coolant entering the channel will be divided, some going into the space above the modules (‘bypass flow’) and some being channelled into the spaces between array elements (‘array flow’).
With tall components, the fluid velocity in the array is significantly less than that in the bypass region, so that little cooling takes place. However, for the lower-profile packages typical of surface mount parts, convection plays a more significant role.
Wirtz suggests the following equation for bypass flow, based on the cross-sectional area of the channel:
where a is the element height and H the height of the channel. Since both a/H and Ly/Sy are positive and less than unity, this equation shows that the bypass flow will be faster than the array flow, caused by the extra blockage that the array elements present. Wirtz estimated that, for a 60% packing density and 4mm high packages spaced 20mm apart, the average bypass velocity would be of the order of 20% greater than the average inlet velocity.
Using the values for velocity in both the array and bypass flow, analytical techniques can be used to demonstrate that the array flow will be turbulent, and estimate how much cooling power is actually “wasted”. Whilst the detail of this analysis is not needed by this module, the idea of bypass flow being generated as a result of a high impedance channel not only equates well with common-sense expectations, but also points out the need for balancing between different channels within a complex assembly, so as to get sufficient flow where it is needed.
Much of Wirtz’s argument is fairly impenetrable, though you may wish to consult the original paper. But the concept of using an analytical approach to make decisions on such aspects as heat sink design still has some application.
Another concept that comes from the Wirtz paper that is more generally useful is what is sometimes referred to as the “thermal wake” effect. In an array, the temperature rise that is experienced by a package will be the combined effect of any self-heating within the package, and the increase in fluid temperature produced by components that are “upstream” of the package. This heat will affect all downstream packages, irrespective of whether or not they generate heat themselves.
The principle of superposition that we introduced above, allows these two effects to be added, so that the temperature of any element can be calculated as:
The first term on the right of this equation gives the self-heating temperature rise of element k in terms of the heat released qk, the surface area over which heat is transferred Ak and the adiabatic5 heat transfer coefficient hk. The second element expresses the temperature rise of element k due to the thermal wakes from all upstream elements.
Real systems are inevitably difficult to model, though less so to conceptualise. Figure 15 shows a typical traditional, high-specification, high-density cabinet, where the available air flow is split between a number of paths. Two aspects of this design are worth noting:
This second aspect means that the inlet air temperature will be raised because of the fan motor dissipation. In other words, when considering the overall dissipation, we should include that of the fan motor. This is probably a prudent design philosophy in any case, since fan motor dissipation will always affect the surrounding air temperature, even when fan and airflow are not so directly coupled as in this example.
We can use the inlet-outlet temperature difference, either based on best practice temperature rise, or on allowable maximum outlet temperature, combined with the overall power dissipation, to calculate the total airflow required, as we did in Unit 8. Then, to calculate the hot-spot temperatures we will need to examine the flow in each path, estimate the applicable value for h, and calculate the temperature rise based on the local dissipation.
But this still doesn’t tell the designer what pressure needs to be applied by the fan in order to give the designed airflow. For this we will have to build a model of the complete system. This can again use a resistance model: referring to Figure 15, the resistance to flow of the inter-board spaces will be in parallel, whereas most of the other resistances to flow will be additive.
Use your knowledge of enclosure practice and your intuitive feeling for how air flow may be constrained, to identify in Figure 15 what are the main contributors to static pressure that the designer will need to calculate. Our answer may be found here.
Using hand calculations or spreadsheets to estimate airflow is inevitably going to be difficult, but software has been developed that analyses the flow distribution in commonly-encountered situations, modelling the physical system in terms of a network of ducts, screens and tee-junctions. How this is done is illustrated by Innovative Research in their paper Flow in a fan-cooled cabinet containing a card array. Such an alternative to a full simulation can be particularly helpful during the early part of the design cycle.
Visit the Innovative Research MacroFlow web site, and look for other illustrations of what is referred to as Flow Network modelling. In their Case Studies section, you will come across papers by Belady and Minchiello that are referred to elsewhere in this module.
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Each of these lists is in the order in which the material is referenced in the Unit text. However, note that links to non-documentary items such as spreadsheets are not included, nor are links to SAQ answers!
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