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## Electronics Cooling MPE 635

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**Electronics Cooling MPE 635**Mechanical Power Engineering Dept.**Course Goals**1. To establish fundamental understanding of heat transfer in electronic equipment. 2. To select a suitable cooling processes for electronic components and systems. 3. To increase the capabilities of post-graduate students in design and analysis of cooling of electronic packages. 4. To analysis the thermal failure for electronic components and define the solution.**Fourier Equation for Conduction Heat Transfer**• Conduction is one of the heat transfer modes. Concerning thermal design of electronic packages conduction is a very important factor in electronics cooling specially conduction in PCB’s and chip packages. • The basic law governing the heat transfer by conduction is Fourier’s law**Energy Equation**As a system, energy balance may be applied on any electronic component. A typical energy balance on a control volume can be described as shown in the following equation. And the amount of energy flowing into or out of the system can be described by the Fourier’s law.**Energy Equation in Cartesian Coordinates**• For a constant thermal conductivity the heat equation could be rewritten as Where α = k/ ρ Cp is the thermal diffusivity.**Special cases of one dimensional conduction**• Boundary conditions - Some of the boundary conditions usually met in heat transfer problems for one dimensional system are described below. These conditions are set at the surface x = 0, assuming transfer process in the positive direction of x- axis with temperature distribution which may be time dependent, designated as T(x, t) • One dimensional steady state conduction without heat generation • One dimensional steady state conduction with uniform heat generation**Boundary conditions**• Constant surface temperature - The surface is maintained at a fixed temperature Ts. It is commonly called a Dirichlet condition, which is the boundary condition of the first case. It can be approximated as a surface in contact with a solid or a liquid in a changing phase state (boiling, evaporating, melting or freezing) therefore the temperature, accompanied with heat transfer process, is constant.**Boundary conditions**• Constant surface heat flux In this case fixed or constant heat flux q" at the surface is described. At which the heat flux is a function of the temperature gradient at the surface by Fourier's law. This type of boundary condition is called Neumann condition. Examples of constant surface heat flux are:**One dimensional steady state conduction without heat**generation • The assumptions made for this kind of analysis are: - One dimensional - Steady state - No heat generation - Constant material properties**One dimensional steady state conduction in Cartesian**coordinates without heat generation**Example 4.1**• Calculate the maximum temperature the transistor base attains if it dissipates 7.5 W through the bracket shown in the Figure below. All dimensions are in mm and the bracket is made of duralumin.**Solution:**Given Dimensions on the figure. q = 7.5 W tw = 50 ºC k = 164 W/m.k As the dimensions of the bracket is very small we can consider that heat transfer is a one dimensional conduction through the sides of the bracket. For the dimensions shown on the figure L = 15 +15 + 15 = 45 mm = 0.045 m w = 20 mm = 0.02 m δ = 5 mm = 0.005 m A = w x δ = 1 x 10 -4 m2 From the above values the only unknown in equation 4.23 is the temperature of the transistor base. q = k A (tb - tw)/L tb = tw + (qL/kA) tb = 50 + (7.5 x 0.045/164 x 10-4) tb = 70.58 ºC**One dimensional steady state conduction in cylindrical**coordinates without heat generation Considering the assumptions stated before, dT/dr is constant and heat flow only in one spatial coordinates the general form becomes, Where subscripts 1 and 2 refer to the inner and the outer surfaces respectively.**Example**A hollow stainless (25 % Cr , 20 % Ni ) steel cylinder 35 mm long has an inner diameter of 50 mm and outer diameter of 105 mm. a group of resistors that generate 10 W is to be mounted on the inside surface of the cylinder as shown in figure. If the resistors temperature is not to exceed 100 º C find the maximum allowed temperature on the outer surface of the cylinder.**Solution**• the thermal conductivity of 25 % Cr , 20 % Ni stainless steel is 12.8 W/m.K considering the one-D conduction equation, The only unknown is the temperature of the outside surface of the cylinder. To= 100 – 10 × (ln (52.5/25)/ (2π × 12.8 × 0.035) To = 100 – 2.64 = 97.36 º C**One dimensional steady state conduction in spherical**coordinates without heat generation • The same assumptions are applied to the spherical system giving the following solution • Where the subscripts i and o refer to the inside and the outside surfaces respectively.**One dimensional steady state conduction with uniform heat**generation • The assumptions made for this kind of analysis are: - One dimensional - Steady state - Uniform heat generation - Constant material properties**One dimensional steady state conduction in Cartesian**coordinates with uniform heat generation q =q''' ×A × L = Ts – Tc/(Lc/kc Ac)**One dimensional steady state conduction in cylindrical**coordinates with uniform heat generation T = Ts + q''' / 4k( ro2 – r2 ) For a convective boundaries T = T∞ + q''' / 4k( ro2 – r2 ) + q''' ro/ 2h**Extended surfaces (Fins)**From the rate equations, it is clear that enhancing heat transfer could be done by several methods: - Increasing the temperature difference - Increasing the heat transfer coefficient - Increasing the surface area A , Fins are used to add a secondary surface to the primary surface and thus increasing the heat transfer area. In electronic equipment cooling straight rectangular fins are mostly used and are done of good conducting material to attain the root temperature through the fins in order to increase the heat transferred. Fins used in electronics cooling are usually used of aluminum and are quite thin about 1.3 to 1.5 mm thick.**Extended surfaces (Fins)**In electronics cooling fins are usually considered to have an insulated tip. The heat transferred by fins are expressed in its effectiveness which is defined as ηf = qf / qmax where qf is the heat actually transferred by the fin and qmax is the maximum heat that could be dissipated by the fin (i.e. when the fin has a uniform temperature equals to the root temperature tr.**Extended surfaces (Fins)**ηf = (tanh ml )/ml Where m = and l and b are defined on the fin sketch. Also the fin effectiveness could be considered as the fraction of the total surface area of the fin Af that is effective for the heat transfer by convection maintained at root temperature tr. ηf = Af, eff/Af,tot q = h [( Atot – Af) + ηf Af] (tr- ta) = ηo Atot h (tr- ta) Where ηo = 1 – (Af/Atot) (1- ηf ) If the fin is of convective tip a correction could be done as lc= l + (t/2)**Fin Geometries**Fin descriptions**Factors affected on the fin selection**Heat transfer/pressure drop**Factors affected on the fin selection**Size figure of merit**Factors affected on the fin selection**Weight figure of merit.**Factors affected on the fin selection**A relative comparison of the fin configurations, based on all the factors discussed is critical in determining the proper design. Comparison of All Parameters**Example**• In order to enhance the heat transfer in logic chips a heat sink is attached to the chip surface in order to increase the surface area available for convection, the heat sink consists of an array of square fins of width w and the sum of the fin spacing and its width is the fin pitch S, the fins are attached to the chip causing a contact resistance R''t,c . • Consider the chip width = 16 mm, cooling is provided by a dielectric liquid with T∞= 25 °C and h =1500 W/m2. k and the heat sink is fabricated from copper (k= 400 W/m .k) and its characteristic dimension are w = 0.25 mm and S = 0.50 mm and Lf = 6 mm , and Lb = 3 mm**Two-Dimensional, Steady-State Conduction**Under the assumptions of two dimensional steady state conduction the general heat equation is reduced to • Now we have two goals for solving the above equation, the first is to determine the temperature distribution across the flied which became a function in the two coordinates x and y T(x, y), then to determine the heat fluxes qx” and qy”in the two direction x and y respectively. • There are many techniques for solving this equation including: • Analytical, • Graphical, and • Numerical solution (finite element and finite difference approaches).**The Method of separation of Variables**Consider the rectangular plate shown in figure, with three boundaries maintained at T1, while the fourth side is maintained at T2, where T2 ≠ T1, the solution of this problem should give the temperature distribution T(x, y) at any point in the solution domain.**The Method of separation of Variables**For solution purpose, the following transformation is done: And thus the heat equation yields: Since the equation is second order in both x and y, two boundary equation are required for each coordinate which are: The separation of variables technique is applied by assuming that the required function is the product of the two functions X (x) and Y (y):**The Method of separation of Variables**Substituting in heat equation and dividing by XY: The above equation is separable as the left-hand-side depends only on x, and the right-hand-side depends only on y. Therefore, the equality can only apply if both sides are equal to the same constant, λ2, called the separation constant. Thus the partial differential equation is converted to two second order ordinary differential equations. The value of λ2must not be negative nor zero in order that the solution satisfies the prescribed boundary equation.**The Method of separation of Variables**The solutions equation of the above ODE gives: The general solution of the heat equation is: Then applying the boundary condition that θ (0, y) = 0, we get that C1 = 0, Then the condition that θ (x, 0) = 0, we get The above equation is satisfied by either C3 = - C4 or C2 = 0, but the later will eliminate the solution dependency on x coordinate, which is a refused solution, thus the first solution is chosen C3 = - C4, applying the condition that θ (L, y) = 0, we get**The Method of separation of Variables**The only acceptable solution is that sin (λ L) = 0, this is satisfied for the values of Rearranging Where Cn is a combined constant. The above equation have an infinite number of solutions depending on n, however it is a linear problem. Thus a more general solution may be obtained by superposing all the solutions as:**The Method of separation of Variables**Now in order to determine Cn the remaining boundary condition should be applied: An analogous infinite series expansion is used in order to evaluate the value of Cn resulting that Substituting in previous equation we get: Then we obtain the solution of the rectangular shape in terms of θ as represented in the following figure in the form of Isotherms for the schematic of the rectangular plate.**The graphical Method**The graphical approach is applied for two dimensional conduction problems with adiabatic and isothermal boundaries The idea of the graphical solution comes from the fact that the constant temperature lines must be perpendicular on the direction of heat flow, the objective in this method is to form a symmetrical network of isotherms and heat flow lines (adiabatic) which is called plot flux. Consider a square, two-dimensional channel whose inner and outer surfaces are maintained at T1 and T2 respectively as shown in figure (a). The plot flux and isothermal lines are shown in figure (b).**The graphical Method**• The procedure for constructing the flux plot is enumerated as follows: • Identify all relevant lines of symmetry. Such lines are determined by thermal, as well as geometrical, conditions. • (For the square channel of Figure (a), such lines include the designated vertical, horizontal, and diagonal lines. For this system it is therefore possible to consider only one-eighth of the configuration, as shown in Figure (b)). • Lines of symmetry are adiabatic in the sense that there can be no heat transfer in a direction perpendicular to the lines. They are therefore heat flow lines and should be treated as such. • Sketch lines of constant temperature within the system. Note that isotherms should always be perpendicular to adiabatic lines. • The heat flow lines should then be drawn with an eye toward creating a network of curvilinear squares. This is done by having the heat flow lines and isotherms intersect at right angles and by requiring that all sides of each square be of approximately the same length. It is often impossible satisfy this second requirement exactly, and it is more realistic to strive for equivalence between the sums of the opposite sides of each square.**The graphical Method**Assigning the x coordinate to the direction of the flow and the y coordinate to the direction normal to this flow, the requirement may be expressed as The rate at which energy is conducted through a heat flow path, which is the region between adjoining adiabatic lines, is designated as qi. If the flux plot is properly constructed, the value of qi will be the same for all heat flow paths and the total heat transfer rate may be expressed as Where M is the number of heat flow paths associated with the plot. qi may be expressed as Where ΔTj is the temperature difference between successive isotherms, A, is the conduction heat transfer area for the heat flow path, and l is the length of the channel normal to the page.**The graphical Method**However, if the flux plot is properly constructed, the temperature increment is the same for all adjoining isotherms, and the overall temperature difference between boundaries, ΔT1 – 2 may be expressed as Where N is the total number of temperature increments. Combining these equations and recognizing that Δx ≈ Δy for curvilinear squares, we obtain**The graphical Method**Now we may be used to define the shape factor, S, of a two-dimensional system as being the ratio (Ml / N). Hence, the heat transfer rate may be expressed as From the previous equation, it also follows that a two-dimensional conduction resistance may he expressed as Shape factors for numerous two-dimensional systems and results are summarized in Table 5.1 for some common configurations. For each case, two-dimensional conduction is supposed to occur between boundaries that are maintained at uniform temperatures.**Example**A hole of diameter D = 0.25 m is drilled through the centre of a solid block of square cross section with w = 1 m on a side. The hole is drilled along the length l = 2 m of the block, which has a thermal conductivity of k = 150 W/m K. A warm fluid passing through the hole maintains an inner surface temperature of T1 75°C, while the outer surface of the block is kept at T2 = 25°C. 1. Using the flux plot method, determine the shape factor for the system. 2. What is the rate of heat transfer through the block? Solution Assumptions 1. Steady-state Conditions. 2. Two-dimensional conduction. 3. Constant properties. 4. Ends of block are well insulated.**Solution**Analysis 1. The flux plot may be simplified by identifying lines of symmetry and reducing the system to the one-eighth section shown in the schematic. Using a fairly coarse grid involving N = 6 temperature increments, the flux plot was generated. And the resulting network of curvilinear squares is as follows.