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Part G-3: Solved ProblemsPart G-3: Solved ProblemsMPE 635: Electronics Cooling1

Part G-3: Solved Problems1. A square silicon chip (k 150 W/m. K) is of width w 5 mm on a side and of thickness t 1mm. The chip is mounted in a substrate such that its side and back surfaces are insulated, while thefront surface is exposed to a coolant. If 4 W are being dissipated in circuits mounted to the backsurface of the chip, what is the steady-state temperature difference between back and frontsurfaces?Data given: Chip dimensions, its thermal conductivity, and 4 W input power to the chip fromthe back surface of the chip.Require: Temperature difference across the chip.Assumptions:(a) Steady-state conductions.(b) Constant properties.(c) One-dimensional conduction in the chip.(d) Neglect heat loss from back and sides.Solution:From Fourier's law,q kAdTdxOr,P q kAThen, T Tt0.001 Χ 4tP 1.07 oC2kA 150 Χ (0.005)2. A square isothermal chip is of width w 5 mm on a side and is mounted in a substrate such thatits side and back surfaces are well insulated, while the front surface is exposed to the flow of acoolant at T 15 C. From reliability considerations, the chip temperature must not exceed T 85 C.If the coolant is air and the corresponding convection coefficient is h 200 W/m2 K. What is themaximum allowable chip power? If the coolant is a dielectric liquid for which h 3000 W/m2.K.What is the maximum allowable power?MPE 635: Electronics Cooling2

Part G-3: Solved ProblemsData given: Chip width, coolant conditions, and maximum allowable chip temperature.Require: maximum allowable chip power at air and dielectric liquid.Assumptions:(a) Steady-state conditions.(b) Neglect heat loss from back surface and sides.(c) Neglect the heat transferred by radiation.(d) Chip is at uniform temperature (isothermal).Solution:According to Newton's law,q hA(Ts T ) PFor air cooling,Pmax hA(Ts ,max T ) 200 Χ (0.005) 2 Χ (85 15) 0.35 WFor dielectric liquid cooling,Pmax hA(Ts ,max T ) 3000 Χ (0.005) 2 Χ (85 15) 5.25 WComment: at comparison between both air and liquid cooling. It appears the air heat transfer ispoorer than the liquid heat transfer but cooling with liquid is higher cost.3. The case of a power transistor, which is of length L 10mm and diameter D 12 mm, iscooled by an air stream of temperature T 25 C. Under conditions for which the air maintainsan average convection coefficient of h 100 W/m2.K on the surface of the case, what is themaximum allowable power dissipation if the surface temperature is not to exceed 85 C?Data given: transistor dimensions, air coolant conditions, and maximum allowable chiptemperature.MPE 635: Electronics Cooling3

Part G-3: Solved ProblemsRequire: maximum allowable transistor power.Assumptions:(a) Steady-state conditions.(b) Neglect heat loss from base, and top surfaces.(c) Neglect the heat transferred by radiation.(d) Transistor is at uniform temperature (isothermal).Solution:According to Newton's law,q hA(Ts T ) PAccording to the maximum surface transistor temperature, the maximum allowable transistorpower is,Pmax hA(Ts ,max T ) 100 Χ (π Χ 0.012 Χ0.01) Χ (85 25) 2.262 W4. The use of impinging air jets is proposed as a means of effectively cooling high-power logicchips in a computer. However, before the technique can be implemented, the convectioncoefficient associated with jet impingement on a chip surface must be known. Design anexperiment that could be used to determine convection coefficients associated with air jetimpingement on a chip measuring approximately 10 mm by 10 mm on a side.Given data: chip dimensions.Required: determine the convection heat transfer coefficients experimentally.Solution:We will give the experiment in steps as follow,1) Construct the system including its components as shown in figure below.2) Bring voltmeter to measure the electric potential volt.3) Bring ammeter to measure the electric current.4) Bring thermometer to measure the surface temperature.5) Close the electric circuit key.6) Let constant power supply (IV const.), plate surface area (A const.), and freestream air temperature (T const.).7) The heat transfer coefficient depends on Reynolds number, and Prandtl number. Thenby changing the jet air velocities according to its flow rates it will gives different heattransfer coefficients, which obtained according to the following relation, by known eachmeasured plate surface temperature Ts (varied with each jet air velocity).q IV hA(Ts T )8) Plot the relation between the jet air velocities and heat transfer coefficients.MPE 635: Electronics Cooling4

Part G-3: Solved ProblemsAir jetdU , T Ts (Plate surface temperature)IVh (W/m2.K)V (m/s)The suggested out put chart: Shows the effect of jet air velocities on heat transfer coefficients.5. An instrumentation package has a spherical outer surface of diameter D 100 mm andemissivity ε 0.25. The package is placed in a large space simulation chamber whose walls aremaintained at 77 K. If operation of the electronic components is restricted to the temperaturerange 40 T 85 C, what is the range of acceptable power dissipation for the package? Displayyour results graphically, showing also the effect of variations in the emissivity by consideringvalues of 0.20 and 0.30.Given data: Instrumentation emissivities (ε) and its surface temperature range 40 T 85 CRequire: The range of acceptable power dissipation for the packageAssumptions:(a) Steady-state conditions.MPE 635: Electronics Cooling5

Part G-3: Solved Problems(b) The chamber is very large compared to package size.(c) Constant chamber wall temperature is maintained at 77 K.(d) The chamber is evacuated.Solution:TsqradChamber wallSpherical electronic componentTBecause the electronic component in a large enclosure then the geometrical factor ƒ 1.Then the radiation heat transfer from the electronic component to the chamber wall is:q σε fA(T 4 Ts4 ) 5.67 Χ 10 8 Χ ε Χ 1 Χ (π Χ 0.12 )(T 4 77 4 ) 1.7813 Χ 10 8 ε (T 4 77 4 )1) The range of acceptable power dissipation for the package at ε 0.25MPE 635: Electronics Cooling6

Part G-3: Solved Problems2) The effect of variations in the emissivity6. Consider the conditions of Problem 2. With heat transfer by convection to air, the maximumallowable chip power is found to be 0.35 W. If consideration is also given to net heat transfer byradiation from the chip surface to large surroundings at 15 C, what is the percentage increase inthe maximum allowable chip power afforded by this consideration?The chip surface has an emissivity of 0.9.qconv 0.35 WqradTsurrData given: Chip width, coolant conditions, and maximum allowable chip power due toconvective air cooling, maximum allowable chip temperature, and large surroundingstemperature.Require: percentage increase in the maximum allowable chip power due to radiation effect.Assumptions:(a) Steady-state conditions.(b) Radiation exchange between small surface and large enclosure.(c) Chip is at uniform temperature (isothermal).MPE 635: Electronics Cooling7

Part G-3: Solved ProblemsSolution:The radiation heat transfer is4q σε fA(T 4 Tsurr) 5.67 Χ 10 8 Χ 0.9 Χ 1 Χ (0.005) 2 (358 4 288 4 ) 0.0122 WPercentage increase in chip power due to radiation effect is% Pmax ((0.35 0.0122) / 0.35) 1) *100 3.49 %7. A square chips of width L 15 mm on a side are mounted to a substrate that is installed in anenclosure whose walls and air are maintained at a temperature of T Tsurr 25 C. The chipshave an emissivity of ε 0.60 and a maximum allowable temperature of Ts 85 C.(a) If heat is rejected from the chips by radiation and natural convection, what is the maximumoperating power of each chip? The convection coefficient depends on the chip-to-airtemperature difference and may be approximated as h C (Ts - T ) 0.25, Where C 4.2W/m2.K5/4.(b) If a fan is used to maintain air flow through the enclosure and heat transfer is by forcedconvection, with h 250 W/m2.K, what is the maximum operating power?Given data: Chip width, walls and air temperatures, the chip emissivity, and maximum allowablechip temperature.Require:(a) Maximum operating power of each chip.(b) Maximum operating power if a fan is used and heat transfer is by forced convection, withh 250 W/m2.Assumptions:(a) Steady-state conditions.(b) Chip is at uniform temperature (isothermal).MPE 635: Electronics Cooling8

Part G-3: Solved Problems(c) Radiation exchange between small surface and large enclosure.Solution:(a) The maximum operating chip power is the summation of heat transfer due to convectionand radiation isPmax q tot q conv q rad4 hA(Ts T ) σε fA(Ts4 Tsurr) 4.2(0.015) 2 (85 25)1.25 5.67 Χ 10 8 Χ 0.6 Χ 1 Χ(0.015) 2 (358 4 298 4 ) 0.2232 W(b) Maximum operating power if a fan used isPmax qtot q conv q rad4 hA(Ts T ) σε fA(Ts4 Tsurr) 250(0.015) 2 (85 25) 5.67 Χ 10 8 Χ 0.6 Χ 1 Χ(0.015) 2 (358 4 298 4 ) 3.44 W8. A computer consists of an array of five printed circuit boards (PCBs). Each dissipating Pb 20 W of power. Cooling of the electronic components on a board is provided by the forcedflow of air, equally distributed in passages formed by adjoining boards, and the convectioncoefficient associated with heat transfer from the components to the air is approximately h 200 W/m2.K. Air enters the computer console at a temperature of Ti 20 C, and flow isdriven by a fan whose power consumption is Pf 25 W.(a) If the temperature rise of the air flow. (To - Ti), is not to exceed 15 C, what is theminimum allowable volumetric flow rate of the air? The density and specific heat ofthe air may be approximated as ρ 1.161kg/m3 and Cp 1007J/kg.K, respectively.(b) The component that is most susceptible to thermal failure dissipates 1 W/cm2 ofsurface area. To minimize the potential for thermal failure, where should thecomponent be installed on a PCB? What is its surface temperature at this location?MPE 635: Electronics Cooling9

Part G-3: Solved ProblemsGiven data: Five printed circuit boards (PCBs) each dissipating Pb 20 W of power, convection coefficient associated with heat transfer from the components to the air, Air inlettemperature, and fan power consumption.Assumptions:(a) Steady-state conditions.(b) Neglect the heat transferred by radiation.Solution:(a) By overall energy balance on the system including fan power is5Pb Pf ma C p (To Ti )5 Χ 20 25 1.161 Χ Va Χ 1007(15)The total volumetric flow rate of the air isVa 0.00713 m 3 / s(b) To minimize the potential for thermal failure, the component should be installed on aPCB at the coolest air condition which at air entrance.The board air inlet temperature which equals to temperature leaving the fan isTb ,i ( Pf / ma C p ) Ti (25 / 1.161Χ0.00713Χ1007) 20 23 o CThe heat flux occurs at maximum temperature difference.MPE 635: Electronics Cooling10

Part G-3: Solved Problemsq // h Tmax h(Ts Tb ,i )10000 200(Ts 23)The surface temperature at this location Ts equals 73 oC.9. Electronic power devices are mounted to a heat sink having an exposed surface area of0.045 m2 and an emissivity of 0.80. When the devices dissipate a total power of 20 W and theair and surroundings are at 27 C, the average sink temperature is 42 oC. What averagetemperature will the heat sink reach when the devices dissipate 30 W for the sameenvironmental condition?Given data: heat sink surface area, the average sink temperature and its emissivity, total powerdissipation, air and surrounding temperaturesRequire: Sink temperature when dissipation is 30 W.Assumptions:(a) Steady-state conditions.(b) All dissipated power in devices transferred to the sink.(c) Sink is at uniform temperature (isothermal).(d) Radiation exchange between small surface (heat sink) and large enclosure(surrounding) case.(e) Convective coefficient is the same for both power levels.Solution:At a total power device of 20 W.P qtot q conv q rad4 hA(Ts T ) σε fA(Ts4 Tsurr)20 0.045h(42 27) 5.67 Χ 10 8 Χ 0.8 Χ 1 Χ0.045(315 4 300 4 )The convective heat transfer coefficient h ish 24.35 W / m 2 .KMPE 635: Electronics Cooling11

Part G-3: Solved ProblemsWhen the devices dissipate 30 W. Using the same value of heat transfer coefficient.30 0.045 Χ 24.35(Ts 300) 5.67 Χ 10 8 Χ 0.8 Χ 1 Χ0.045(Ts4 300 4 )By trial-and-error, the temperature of the heat sink isTs 322 K 49 oC10. Consider a surface-mount type transistor on a circuit board whose temperature ismaintained at 35 C. Air at 20 C flows over the upper surface of dimensions 4 mm by 8 mmwith a convection coefficient of 50 W/m2.K.Three wire leads, each of cross section 1 mm by0.25 mm and length 4 mm, conduct heat from the case to the circuit board. The gap betweenthe case and the board is 0.2 mm.(a) Assuming the case is isothermal and neglecting radiation; estimate the casetemperature when 150 mW are dissipated by the transistor and (i) stagnant air or (ii) aconductive paste fills the gap. The thermal conductivities of the wire leads, air. Andconductive pastes are 25, 0.0263, and 0.12 W/m.K. respectively.(b) Using the conductive paste to fill the gap, we wish to determine the extent to whichincreased heat dissipation may be accommodated, subject to the constraint that thecase temperature not exceeds 40 C. Options include increasing the air speed toachieve a larger convection coefficient h and/or changing the lead wire material toone of larger thermal conductivity. Independently considering leads fabricated frommaterials with thermal conductivities of 200 and 400 W/m.K, compute and plot themaximum allowable heat dissipation for variations in h over the range 50 h 250W/m2.K.Given data: Surface-mount transistor, power dissipation by conduction and convectionRequired:(a) The case temperature with (i) air-gap and (ii) conductive paste fills the gap.Assumptions:(a) Steady-state conductions.(b) Constant properties.(c) Transistor case is isothermal.MPE 635: Electronics Cooling12

Part G-3: Solved Problems(d) One-dimensional conduction.(e) Neglect heat loss from edges.Solution:By energy balance across the transistor caseP 3qlead q conv q cond , gapqlead k l Al (Tc Tb ) / LWhere Tc is the case temperature, and Tb is the board temperatureq conv hAc (Tc T )q cond , gap k g Ag (Tc Tb ) / δ gapSubstitute in the energy equation:P 3k l Al (Tc Tb ) / L hAc (Tc T ) k g Ag (Tc Tb ) / δ gap(i) By substitute in numerical values for air-gap condition0.15 [3(25)(0.001Χ0.00025) / 0.004 0.0263(0.008Χ0.004) / 0.0002](Tc 35) 50(0.008Χ0.004)(Tc 20)The case temperature with air-gap isTc 47 oC(ii) By substitute in numerical values for conductive paste fills the gap condition0.15 [3( 25)(0.001Χ 0.00025) / 0.004 0.12(0.008Χ 0.004) / 0.0002 ](Tc 35) 50(0.008 Χ 0.004)(Tc 20)The case temperature with conductive paste fills the gap isTc 40 oC11. A transistor that dissipates10W is mounted on a duralumin heat sink at 50 C by aduralumin bracket 20 mm wide as shown in the opposite figure. The bracket is attached tothe heat sink by a rivet. With convective cooling negligibly small, estimate the surfacetemperature of the transistor.MPE 635: Electronics Cooling13

Part G-3: Solved ProblemsGiven data: Transistor power dissipation, and heat sink temperature and bracket dimensionsRequire: Surface temperature of the transistor.Assumptions:(a) Neglect the convection cooling.(b) Neglect the contact resistances(c) One-dimension conduction.(d) Steady state condition.Solution:The heat source is the base of the transistor and the rivets connecting the bracket to the heatsink,The heat flow path length isL 20 20 20 60 mmThe base transistor area equals the heat flow area isA width x thickness 20 x 5 100 mm2The duralumin thermal conductivity could be obtained from appendix isk 164 W/m.KFrom Fourier's law,P q kA TLThen the surface temperature of the transistor isTs 50 (10 Χ 0.06) /(164 Χ 100 Χ 10 6 ) 86.58 oCMPE 635: Electronics Cooling14

Part G-3: Solved ProblemsComment: Essentially in this case, conduction is not one-dimensional conduction. But thesolution based on one-dimensional conduction only for simplicity with loss of accuracy.12. A cable 10 mm diameter is to be insulated to maximize its current carrying capacity. Forcertain reasons, the outside diameter of the insulation should be 30 mm. The heat transfercoefficient for the outer surface is estimated to be 10 W /m2 .K.What should be the thermal conductivity of the chosen insulation? By what percentage wouldthe insulation increase the energy carrying capacity of the bare cable?Given data: Cable diameter, insulation diameter and heat transfer coefficient.Require: thermal conductivity of insulation required to maximize the current carryingcapacity, and the percentage increase in the energy carrying capacity due to insulation.Assumption:(a) One-dimensional conduction.(b) Steady-state conditionsSolution:Insulation (k)Tidins (d2)d1qT The heat transfer from the cable isTi T Ti T T q Rth Rcond Rconv ln r2 / (2πkL ) 1 r h(2πr L) 2 1 For increasing the radius of insulation the conduction resistance increases, on another handthe convection resistance decreases. So that there's a minimum total thermal resistance causesmaximum heat transfer (or current carrying capacity) as shown.MPE 635: Electronics Cooling15

Part G-3: Solved ProblemsqRtotalRcondRconvr2r2c(critical radius)To find the maximum heat transfer: differentiate the heat transfer to the radius of insulation(r2). Then equal it to zero. It givesr2 c khAt h 10 W /m2 .K. and r2c 0.015 mThe thermal conductivity of insulation required to maximize the current carrying capacity isk 0.015 x 10 0.15 W /m .K./ q max T rln 2c r1 1 / (2πk ) h(2πr2c ) T 1 ln(15 / 5) /( 2π Χ 0.15) 10(2π Χ 0.015) T / 2.2267 W/mFor bare cable the heat transfer is T T/ q bare T / 3.183 W/m 1 1 Χπ10(20.005)πh(2r) 1 The percentage increase in the energy carrying capacity due to insulation is{}//% qincrease (q max/ qbare) 1 100 43 %MPE 635: Electronics Cooling16

Part G-3: Solved Problems13. The vertical side of an electronics box is 40 x 30 cm with the 40 cm side vertical. What isthe maximum radiation energy that could be dissipated by this side if its temperature is not toexceed 60 C in an environment of 40 C, and its emissivity is 0.8?Given data: electronics box vertical side area, maximum temperature of the electronics boxvertical side, its emissivity, and the environment temperature.Require: maximum radiation energy dissipates from this side of the electronic boxAssumptions:(a) Steady-state conditions.(b) Radiation exchange between small surface and large enclosure.(c) The side is at uniform temperature (isothermal).(d) The temperature of the enclosure equals the temperature of the environment.Solution:wT LTsTeThe maximum radiation energy dissipates from this side of the electronic box isq max σε fA(Ts4 Te4 ) 5.67 Χ 10 8 Χ 0.8 Χ 1 Χ(0.4 Χ 0.3)(333 4 313 4 ) 14.7 W14. In a manufacturing process, a transparent film is being bonded to a substrate as shown inthe sketch. To cure the bond at a temperature To, a radiant source is used to provide a heat fluxq o'' (W/m2), All of which is absorbed at the bonded surface. The back of the substrate ismaintained at T1 while the free surface of the film is exposed to air at T and a convection heattransfer coefficient h.(a) Show the thermal circuit representing the steady-state heat transfer situation. Be sure tolabel all elements, nodes, and heat rates. Leave in symbolic form.MPE 635: Electronics Cooling17

Part G-3: Solved Problems(b) Assume the following conditions: T 20 oC, h 50 W/m2.K and T1 30 oC.Calculate the heat flux qo'' that is required to maintain the boded surface at To 60 oC.(c) Compute and plot the required heat flux as a function of the film thickness for 0 Lf 1mm.Given data: A radiant source is used to provide a heat flux at the bond to cure the bond at atemperature To.Assumptions:(a) Steady-state conditions.(b) Constant properties.(c) One-dimensional conduction heat transfer.(d) Neglect the contact resistance.Solution:(a) Thermal circuit based on heat flux distribution represented below.q1//q 2//ToT T1R1 1/hR2 Lf / KfR3 Ls / Ksq o//(b) Using this thermal circuit and performing energy balance on film-substrate interface,q o// q1// q 2// (To T1 )(To T ) ( Ls / k s ) ( L f / k f ) (1 / h) 60 3060 20 (0.001 / 0.05) (0.00025 / 0.025) (1 / 50) 2833 W/m 2MPE 635: Electronics Cooling18

Part G-3: Solved Problems(c) The heat flux as a function of the film thickness,q o// q1// q 2// (To T1 )(To T ) ( Ls / k s ) ( L f / k f ) (1 / h) 1500 4040 L f 0.02W/m 215. A silicon chip is encapsulated such that, under steady-state conditions, all of the power itdissipates is transferred by convection to a fluid stream for which h 1000 W/m2.K and T 25 C. The chip is separated from the fluid by a 2-mm-thick aluminum cover plate, and thecontact resistance of the chip/aluminum interface is 0.5 x 10-4 m2.K/W. If the chip surfacearea is 100 mm2 and its maximum allowable temperature is 85 oC, what is the maximumallowable power dissipation in the chip?Given data: chip surface area is 100 mm2 and its maximum allowable temperature, contactresistance of the chip/aluminum interface and fluid stream conditions.Require: maximum allowable power dissipation in the chip.Assumption:(a) Steady-state conditionsMPE 635: Electronics Cooling19

Part G-3: Solved Problems(b)(c)(d)(e)Neglect the radiation heat transfer.One-dimensional conduction.Neglect the heat loss from the back and side surfaces.Chip at uniform temperature (isothermal).Solution:The thermal circuit of the system represented as shown in the following figure, Conductionheat transfer from the chip equals the convection heat transfer to the fluid stream,RconvRcond RcontactT Tc,maxq P c,maxq P c,maxAccording to the thermal circuit, the maximum allowable power dissipation in the chip isq Pc , max Tc ,max T Rcontact Rcond Rconv Tc ,max T (1 / hA) contact ( L / KA) Alu min um (1 / hA) conv.(85 25) Χ 10 4 5.667 W(0.5 Χ 10 4 ) (0.002 / 238) (1 / 1000)16. Approximately 106 discrete electrical components can be placed on a single integratedcircuit (chip), with electrical heat dissipation as high as 30,000 W/m2. The chip, which isvery thin, is exposed to a dielectric liquid at its outer surface, with ho 1000 W/m2. K andT ,o 20 C, and is joined to a circuit board at its inner surface. The thermal contactresistance between the chip and the board is 10-4 m2.K/W. and the board thickness andthermal conductivity are Lb 5 mm and kb 1 W/m.K, respectively. The other surface of theboard is exposed to ambient air for which hi 40 W/m2.K and T ,i 20 C.(a) Sketch the equivalent thermal circuit corresponding to steady-state conditions. Invariable form, label appropriate resistances, temperatures, and heat fluxes.(b) Under steady-state conditions for which the chip heat dissipation is qc'' 30,000W/m2. What is the chip temperature?(c) The maximum allowable heat flux q c'',m , is determined by the constraint that the chiptemperature must not exceed 85 C. Determine q c'',m for the foregoing conditions. If airis used in lieu of the dielectric liquid, the convection coefficient is reduced byapproximately an order of magnitude. What is the value of q c'',m for ho 100 W/m2.K?With air cooling, can significant improvements be realized by using an aluminumoxide circuit board and/or by using a conductive paste at the chip/board interface forwhich Rt'',c 10-5 m2.K/W?MPE 635: Electronics Cooling20

Part G-3: Solved ProblemsGiven data: chip joined to a circuit board and its electrical heat dissipation, board properties,and coolant fluids conditions.Assumptions:(a) Steady-state conditions(b) Neglect the radiation heat transfer.(c) One-dimensional conduction.(d) Neglect chip thermal resistance.Solution:(a) The equivalent thermal circuit as shownT ,i q i// (1/hi)(L/k)bT ,oTc R''c(1/ho )q o//q c//(b) According to the thermal circuit, the chip heat dissipation is qi// q o//q c//Tc T ,i 30000 Rc// ( L / k ) b (1 / hi ) Tc T ,o(1 / ho )Tc 20T 20 c( 10 ) (0.005 / 1) (1 / 40) (1 / 1000) 4Then the chip temperature isTc 49 oC(c) At chip temperature 85 C, the maximum allowable heat flux q c'',m isq c//,m qi// q o// q c//,m Tc T ,iRc// ( L / k ) b (1 / hi ) Tc T ,o(1 / ho )85 2085 20 67160 W / m 2( 10 ) (0.005 / 1) (1 / 40) (1 / 1000) 4MPE 635: Electronics Cooling21

Part G-3: Solved ProblemsThe maximum allowable heat flux q c'',m at ho 100 W/m2.k, q i// q o//q c//,mTc T ,i q c//,m Rc// ( L / k ) b (1 / hi ) Tc T ,o(1 / ho )85 2085 20 8660 W / m 2( 10 ) (0.005 / 1) (1 / 40) (1 / 100) 4At an aluminum oxide circuit board (k 38 W/m.K), the maximum allowable heat flux q c'',m isq c//,m q i// q o//Tc T ,i q c//,m R ( L / k ) b (1 / hi )//c Tc T ,o(1 / ho )85 2085 20 9076 W / m 2( 10 ) (0.005 / 38) (1 / 40) (1 / 100) 4By using a conductive paste at the chip/board interface for which Rt'',c 10-5 m2.K/W, themaximum allowable heat flux q c'',m isq c//,m q i// q o// q c//,m Tc T ,iR ( L / k ) b (1 / hi )//c Tc T ,o(1 / ho )85 2085 20 8666 W / m 2( 10 ) (0.005 / 1) (1 / 40) (1 / 100) 5Using an aluminum oxide circuit board gives higher maximum allowable heat flux q c'',m thanusing conductive paste at the chip/board interface.17. Consider a power transistor encapsulated in an aluminum case that is attached at its baseto a square aluminum plate of thermal conductivity k 240 W/m.K, thickness L 6 mm, andwidth W 20 mm. The case is joined to the plate by screws that maintain a contact pressureof 1 bar, and the back surface of the plate transfers heat by natural convection and radiationto ambient air and large surroundings at T Tsur 25 C. The surface has an emissivity of ε 0.9, and the convection coefficient is h 4 W/m2.K. The case is completely enclosed suchthat heat transfer may be assumed to occur exclusively through the base plate.(a) If the air-filled aluminum-to-aluminum interface is characterized by an area of Ac 2x 10-4 m2 and a roughness of 10 µ.m. what is the maximum allowable power dissipationif the surface temperature of the case, Ts,c, is not to exceed 85 C?MPE 635: Electronics Cooling22

Part G-3: Solved Problems(b) The convection coefficient may be increased by subjecting the plate surface to a forcedflow of air. Explore the effect of increasing the coefficient over the range 4 h 200W/ m2.K.Given data: A power transistor attached at its base to a square aluminum plate with a contactpressure of 1 bar, the plate whose emissivity of ε transfers heat by natural convection andradiation to ambient air and large surroundings.Assumptions:(a) Steady-state conditions.(b) One-dimensional conduction.(c) Heat transfers occur exclusively through the base plate only.Solution:(a) For air interfacial fluid between the aluminum case and the aluminum plate with aroughness 10µ.m, and 1 bar contact pressure. Then the contact resistance Rc// 2.75 x 10-4m2.K/W.The thermal circuit represented as shown(1/hAp )(R''c/Ac) (L/Ak)pqcqc Tp,oTs,cT T surr(1/ε Ap )According to the thermal circuit, the maximum allowable power dissipation at Ts,c 85 C.Pmax q c Ts ,c T p ,oRc// / Ac ( L / Ak ) p4σ (T p4,o Tsurr) (1 / hA p )(1 / εA p )T p ,o T To get the plate out side temperature Tp ,o ,358 T p ,o(2.75 Χ 10 -4 ) /( 2 Χ 10 -4 ) (0.006 / 240(0.02) 2 ) T p ,o 2981 / 4(0.02) 2MPE 635: Electronics Cooling 5.67 Χ 10 8 (T p4,o 298 4 )1 / 0.9(0.02) 223

Part G-3: Solved ProblemsBy trial and error,T p ,o 357 oCThe maximum allowable power dissipation isTs ,c T p ,o1 0 .5 WPmax q c //Rc / Ac ( L / Ak ) p 1.4375(b) The effect of increasing the out side convection coefficient18. A transistor, which may be approximated as a hemispherical heat source of radius ro 0.1mm, is embedded in a large silicon substrate (k 125 W/m.K) and dissipates heat at a rate q.All boundaries of the silicon are maintained at an ambient temperature of T 27 C, exceptfor a plane surface that is well insulated. Obtain a general expression for the substrate temperature distribution and evaluate the surface temperature of the heat source for q 4 W.MPE 635: Electronics Cooling24

Part G-3: Solved ProblemsGiven data: A heat source embedded in a large silicon substrate, source and substrate boundaryconditions.Require: substrate temperature distribution and surface temperature of heat source for q 4W.Assumption:(a) Steady-state conditions.(b) One-dimensional conduction.Solution:From energy equation reduced to1 ddT(kr 2) 02drr drAt constant silicon thermal conductivityd 2 dT(r) 0drdrBy integration to the substrate radiusdT C1drCT (r ) 1 C 2rr2Boundary conditionsT ( ) T , and T (ro ) TsThen the constantsC 2 T C1 ro (T Ts )The substrate temperature distributionT (r ) (Ts T )ro / r T The heat rate isq kA()dT k (2π r 2 ) (Ts T )ro / r 2 2π ro (Ts T )drThe surface temperature of heat source for q 4 W isMPE 635: Electronics Cooling25

Part G-3: Solved ProblemsTs 4 / 125(2π Χ10 4 ) 27 78 oC21. An isothermal silicon chip of width W 20 mm on a side is soldered to an aluminum heatsink (k 180W/m.K) of equivalent width. The heat sink has a base thickness of Lb 3 mm andan array of rectangular fins, each of length Lf 15 mm. Air flow at T 20 C is maintainedthrough channels formed by the fins and a cover plate, and for a convection coefficient of h 100 W/m2.K, a minimum fin spacing of 1.8 mm is dictated b

MPE 635: Electronics Cooling 9 (c) Radiation exchange between small surface and large enclosure. Solution: (a) The maximum operating chip power is the summa tion of heat transfer due to convection and radiation is W hA T T fA T T P q q q s s surr tot conv rad 0.2232 4.2(0.015) (85 25) 5.67 10 0.6 1 (0.015) (358 298 ) ( ) ( ) 2 1.25 8 2 4 4 4 4 .