8. Forced Convection Heat Transfer

8. Forced Convection Heat Transfer 8.1 Introdction The general definition for convection ma be smmarized to this definition "energ transfer between the srface and flid de to temperatre difference" and this energ transfer b either forced eternal, internal flow or natral convection. Heat transfer b forced convection generall makes se of a fan, blower, or pmp to provide highvelocit flid gas or liqid. The high-velocit flid reslts in a decreased thermal resistance across the bondar laer from the flid to the heated srface. This, in trn, increases the amont of heat that is carried awa b the flid To nderstand the convection heat transfer we mst know some of the simple relations in flid dnamics and bondar laer analsis. Firstl we std bondar laer with forced convection flow sstems. 8. Bondar Laer over Flat Plate We consider the direction along the wall with direction normal to the wall as in Figre8.1. Where the laminar bondar laer begins at leading edge 0, and followed b transition region and finall to the trblent region to the trailing edge L. The velocit and temperatre of the flid far awa from the srface ot side the bondar laer thickness δ are the free-stream velocit and free-stream temperatre T. U T Bondar laer edge δ Laminar Transition Trblent Figre 8.1 Flid flow over flat plate with laminar, transition, and trblent bondar laers 8..1 Laminar Bondar Laer Eqations over Flat Plate Re 510 5 The assmptions made to give the simplicit on the analsis are: 1- Stead flow - Two-dimensional incompressible viscos flow 3- No pressre variation in the direction 4- No shear force in the direction 5- Neglect bod force de to gravit 54

All the basic differential eqations can be derived b considering an element control volme inside the laminar region as shown in Figre 8.. ρ v ρ ρ ρ v Figre 8. Element control volme on laminar region Continit eqation: {Rate of mass accmlation within control volme} {Net rate of mass fl ot of control volme} 0 Rate of mass accmlation within control volme ρ 0 stead state assmption t ρ Net rate of mass fl in - direction per nit depth ρ ] ρ ] Net rate of mass fl in - direction per nit depth ρ v] ρ v] ρυ Sbstitte in continit eqation epression it prodce: ρ ρυ 0 ρ ρυ 0 υ 0 8.1 Momentm eqation: B appling Newton's nd low on the same element control volme as in Figre 8. Net rate of linear Time rate of change of Net rate of linear momentm fl ot of linear momentm within momentm fl ot of the control volme For the -direction: control volme the control volme smmation of eternal force acting in the control Time rate of change of linear momentm within the control volme volme ρ 0 t 55

56 Net rate of linear momentm ot of the control nit ρ] ρ] ρv] ρv] υ ρ ρ Eternal forces divided into: - Pressre force P - Viscos force µ Sbstitting in the Newton's nd low eqation it eilds υ ρ ρ P µ Or, ρυ ρυ ρ ρ P µ From continit eqation we have: 0 υ ρ ρ Then the momentm eqation for laminar bondar laer is ρυ ρ µ P Energ eqation: For the shown element control volme as in Figre 8.3.with neglected heat condction in -direction and appling energ balance, the energ eqation ma be written as follows: Energ convected in left face Energ convected in bottom face heat condction in bottom face net viscos work done on element energ convected ot right face energ convected ot top face heat condction ot top face 8.

57 ρ vc p T -k T/ ρ c p T ρ c p [ / ][T T/ ] -k [ T/ / T/ ] ρ c p [v v/ ][T T/ ] Figre 8.3 Element control volme for energ balance The viscos shear force is the prodct of the shear stress and the area per nit depth µ And the distance throgh which it moves per nit time in respect to the element control volme is The net viscos energ delivered to the element control volme µ B appling energ balance on the element control volme shown in Figre8.3 neglecting the second order differentials ields to T k T T T c p µ υ υ ρ From continit Eqation 8.1 the energ eqation can be written as follow

T T T ρ c p υ k µ Dividing b ρc p T T T µ υ α ρc p 8.3 The viscos work term is important onl at high velocities since its magnitde will be small compared with other terms when low velocit flow is stdied. This ma be shown with an order-of-magnitde analsis of the two terms on the right side of energ eqation. For this order-of-magnitde analsis we might consider the velocit as having order of the free stream velocit and the dimension of the order of velocit bondar laer thickness δ. So that And and δ T α α T δ µ ρc p µ ρc p δ 8.4 8.5 Now if the ratio between Eqations 8.5 and 8.4 is µ ρc T pα Pr c T p 1 Then we can neglect this term compared to other terms and we can write the energ eqation in this simple form. T T T 8.6 υ α The soltion of these eqations continit, momentm and energ is simplified b the fact that, for conditions in the velocit hdrodnamic bondar laer flid properties are independent of temperatre. We ma begin b solving the Eqations 8.1 and 8. continit, momentm to get and v. Then the energ eqation can be solved which depending on calclated reslts. The soltion of Eqations 8.1 and 8. can be solved b Blasis eact analtic soltion for: P - 0 - Laminar flow. In Blasis eact soltion, the velocit components are defined in terms of a stream fnction Ψ, where 58

Ψ And Ψ υ So that the continit eqation ma be intrinsicall satisfied 8.7 8.8 Considering the partial differential eqation describing the momentm eqation Eqation 8., we ma se the similarit method in order to convert it into an ordinar differential eqation. Defining the dependent and independent dimensionless variables ƒ and η, will help s in this analtical approach. Ψ f η 8.9 v / 8.10 η / v The Blasis eact soltion is termed a similarit soltion, and η is the similarit variable. This terminolog is sed becase, despite growth of the bondar laer with distance from the leading edge, the velocit profile /, remains geometricall similar as shown in Figre 8.4. fn. δ Where δ is the bondar laer thickness and sall difficlt to measre Assming this thickness to var as ν / 1/, its follows that fn.η Hence the velocit profile is assmed to be niqel determined b the similarit variable η which depends on and directions. δ Figre 8.4 the profile / geometricall similar From Eqations 8.7 and 8.8 Ψ Ψ η η v df dη v df dη 8.11 59

And Ψ υ v f v f υ 1 v df η dη f 8.1 B differentiating the velocit components, it ma also be shown that d f η 8.13 d η d f 8.14 v dη 3 d f 8.15 3 v dη Sbstitting these eqations in the momentm eqation, then we obtain 3 d f d f f 0 3 dη dη 8.16 This is non linear, third-order differential eqation, so that we need three bondar conditions to get a soltion, these bondar conditions are: At η 0 ƒ / η ƒ η 0 At η ƒ / η / 1 The soltion of Eqation 8.16 b nmerical integration and the reslts are given in Table 8.1. At / 0.99 for η 5, sbstitte in Eqation 8.10 δ 5 Re The shear stress can be epressed as τ s µ µ d f v dη 0 η 0 8.17 From the Table 8.1 τ 0. 33 s ρµ And the wall local shear stress coefficient C ƒ,is given b 60

C f τ s 1 ρ 0.664 Re 8.18 Table 8.1: The fnction ƒ η for laminar bondar laer over flat plate η df ƒη d f v dη dη 0 0 0 0.33 0.4 0.07 0.133 0.331 0.8 0.106 0.65 0.37 1. 0.38 0.394 0.317 1.6 0.4 0.517 0.97 0.65 0.63 0.67.4 0.9 0.79 0.8.8 1.31 0.81 0.184 3. 1.569 0.876 0.139 3.6 1.93 0.93 0.098 4.306 0.956 0.064 4.4.69 0.976 0.039 4.8 3.085 0.988 0.0 5. 3.48 0.994 0.011 5.6 3.88 0.997 0.005 6 4.8 0.999 0.00 6.4 4.679 1 0.001 6.8 5.079 1 0 Eample 8.1: Consider flid flow at 0.3 m/s past a flat plate 0.3 m long. Compte the bondar laer thickness at the trailing edge for a air and b water at 0 o C. Soltion: Part a From air properties table at 0 o C ν 15.6710-6 m /s. The trailing edge Renolds nmber is 0.30.3 Re L 5895 6 15.6710 L v The flow is laminar, from Eqation 8.17 the predicted laminar thickness is δ 5 0.065 5895 At 0.3 m δ 0.0195 m 61

Part b From satrated water properties table at 0 o C: ν water 1.043710-6 m /s. The trailing edge Renolds nmber is Part B: Heat Transfer Principals in Electronics Cooling L 0.30.3 Re 8631.67 6 1.0437Χ10 L v This again satisfies the laminar condition the laminar thickness is δ 5 0.017 8631.67 At 0.3 m δ 0.0051 m Solving the energ eqation Eqation 8.6: Let the dimensionless temperatre T T Ts T T T T η, and sbstitte in energ eqation to give this form. s and assme similarit soltion of the form d T Pr dt f 0 dη dη 8.19 Where the variable ƒ depend on the Prandtl nmber vales The appropriate bondar conditions are T 0 0 And T 1 The soltion ma be achieved b nmerical integration method for 0.6 Pr < 50 dt It will prodce the srface temperatre gradient as the following relation. dη η0 If T s >T dη dt 0.33 Pr 1/3 Epression for the local heat convection coefficient determined as follow. // T q h Ts T k k T T Ts T h k Ts T Ts T h k v 1/ η0 T η η 0 0 6

It follows that the local Nsselt nmber in this form h N.33 Re k 1/ 0 Pr 1 / 3 8.0 And the average heat transfer coefficient can be obtained b integration h h h N 1 h 0 h k 0.664 Re 1 / Re 1 / 3 8.1 8.3 The Thermal Bondar Laer Analogos to the velocit bondar laer there is a thermal bondar laer adjacent to a heated or cooled plate. The temperatre of the flid changes from the srface temperatre at the srface to the free-stream temperatre at the edge of the thermal bondar laer as shown in Figre 8.5. T T, T S > T Figre 8.5.Flid temperatre variations inside the thermal bondar laer T S The velocit bondar laer thickness depends on the Renolds nmber Re X.. Bt the thermal bondar laer thickness depends both on Re X and Pr as shown in Figre8.6. Temperatre bondar laer Velocit bondar laer Pr <1 T, Pr 1 Pr >1 δ δ T T S 63

Figre 8.6 thermal bondar laer thicknesses relative to velocit bondar laer thickness at different Prandtle nmber For laminar flow Re < Re cr : δ 5 Re At Pr 0.7 δ δ T 1/ 3 Pr 8. At Pr «1 δ δ T 1/ Pr For trblent flow Re > Re cr : δ 0.37 Re 0. X δ δ T 8.3 8.4 Cooling Air Fans for Electronic Eqipment Air is the most commonl sed medim for heat transfer. It is available everwhere on the srface of the planet.air is sall taken directl from the srronding atmosphere and retrned to it. Man different tpes of fans are available for cooling electronic eqipment. These can generall be divided into two major tpes: aial and centrifgal fans. These fans can be driven b varios tpes of electric motors, single phase, three phase, 60 ccles, 400 ccles, 800 ccle ac/dc, constant speed, variable speed. The flow rates also var from 1 cfm to several thosand cfm. When a fan is sed for cooling electronic eqipment, the airflow direction can be qite important. The fan can be sed to draw air throgh a bo or to blow air throgh a bo. A blowing fan sstem will raise the internal air pressre within the bo, which will help to keep dst and dirt ot of a bo that is not well sealed. A blowing sstem will also prodce slightl more trblence, which will improve the heat transfer characteristics within the bo. However, when an aial flow fan is sed in a blowing sstem, the air ma be forced to pass over the hot fan motor, which will tend to heat the air as it enters the electronic bo, as shown in Figre 8.7. An ehast fan sstem, which draws air throgh an electronic bo, will redce the internal air pressre within the bo. If the bo is located in a dst or dirt area, the dst and dirt will be plled into the bo throgh all of the small air gaps if the bo is not sealed. In an ehast sstem, the cooling air passes throgh an aial flow fan as the air eits from the bo, as shown in Figre 8.8. The cooling air entering the electronic bo is therefore cooler. Figre 8.7 Aial flow fan blowing cooling air throgh a bo 64

Figre 8.8 Aial flow fan drawing cooling air throgh a bo 8.5 Static Pressre and Velocit Pressre Airflow throgh an electronic bo is de to a pressre difference between two points in the bo, with the air flowing from the high-pressre side to the low-pressre side. The flow of air will reslt in a static pressre and a velocit pressre. Static pressre is the pressre that is eerted on the walls of the container or electronic bo, even when there is no flow of air; it is independent of the air velocit. Static pressre can be positive or negative, depending pon whether it is greater or less than the otside ambient pressre. Velocit pressre is the pressre that forces the air to move throgh the electronic bo at a certain velocit. The velocit pressre depends pon the velocit of the air and alwas acts in the direction of the airflow. The amont of cooling air flowing throgh an electronic bo will sall determine the amont of heat removed from the bo. The higher the air flow rate throgh the bo the higher heat will be removed. As the airflow throgh the bo increases, however, it reqires an even greater pressre to force the air throgh the bo. Static and velocit pressres can be epressed in lb/in and g/cm. However, these vales are sall ver small, so that it is often more convenient to epress these pressres in terms of the height of a colmn of water. The velocit head H v is a convenient reference that is often sed to determine pressre drops throgh electronic boes. The velocit head can be related to the air flow velocit as follow Where: V V Velocit of the air g gravitational acceleration H v velocit head in centimeter of water gh v 8.4 The Eqation 8.4 can be modified sing standard air with a densit of 0.001 g/cm 3 at 0.5 o C and 1 bar, this is shown in Eqation 8.5. V 979.6 cm / sec 3 1 g / cm water H 3 0.001 g / cm air V cm water V 177 H v cm water cm / sec. 8.5 The total head will be the sm of the velocit head and the static head as follow H H H t v s 8.6 65

We have man cases for measrement of the total heads inside the electronic bo as shown in the Figres 8.9, 8.10, and 8.11. In case of a pressrized electronic bo with no air flows the total pressre eqal to the static pressre as shown in Figre 8.9. While in case of a fan blows air throgh the electronic bo, the pressre within the bo will be slightl higher than the otside air pressre. A velocit head will now be developed, as shown in Figre 8.10. Bt, in case of a fan draws the air throgh the bo, the pressre within the bo will be slightl lower than the otside air pressre, and the pressre head characteristics will appear as shown in Figre 8.11. And the total head is still constant as shown b Eqation. 8.6. Figre 8.9 a pressrized electronic bo with no air flow Figre 8.10 pressre head characteristics when the fan blows air throgh an electronic bo Figre 8.11 pressre head characteristics when the fan draws air throgh an electronic bo 66

8.6 Fan Performance Crve Electronic boes that are cooled with the se of fans mst be carefll evalated to make sre the fan will provide the proper cooling. If the fan is too small for the bo, the electronic sstem ma over heat and fail. If the fan is too big for the bo, the cooling will be adeqate, bt the larger fan will be more epensive, heavier, and will draw more power. Air flowing throgh the electronic bo will enconter resistance as it enters the different chambers and is forced to make man trns. The flow resistance is approimatel proportional to the sqare of the velocit, so that it is approimatel proportional to the sqare of the flow rate in cbic feet per minte cfm. When the static pressre of the air flow throgh a bo is plotted against the air flow rate, the reslt will be a parabolic crve. This crve can be generated b considering the varios flow resistances the airflow will enconter as it flows throgh the bo. The method of analsis is to assme several different flow rates throgh the bo and then to calclate a static pressre drop thogh the bo for each flow rate. This crve is called impedance crve for the electronic bo as shown in Figre 8.1. Once the bo flow impedance crve has been developed, it is necessar to eamine different fan performance crves to see how well the fans will match the bo. A tpical fan performance crve is shown in Figre 8.13. Figre 8.1 air flow impedance crve for the electronic bo 67

Figre 8.13 fan impedance crve If the impedance crve for the bo is sperimposed on the impedance crve for the fan, the will intersect. The point of intersection represents the actal operating point for the sstem, as shown in Figre 8.14. Figre 8.14 intersection of fan and bo impedance crves 8.7 Cabinet Cooling Hints In addition to selecting a fan, there ma be some choice in the location of the fan or fans, and in this regard, the illstration in Figre 8.15 ma prove sefl. The following comments shold also be kept in mind with regard to fan location: 1 Locate components with highest heat dissipation near the enclosre air eits Size the enclosre air inlet and eit vents at least as large as the ventri opening of the fan sed 3 Allow enogh free area for air to pass with velocit less than 7 meters/sec 4 Avoid hot spots b spot cooling with a small fan 5 Locate components with the most critical temperatre sensitivit nearest to inlet air to provide the coolest air flow 6 Blow air into cabinet to keep dst ot, i.e. pressrize the cabinet 7 Use the largest filter possible, in order to redce pressre drop and keep the sstem from the dst 8.8 Design Steps for Fan-Cooled Electronic Bo Sstems The sstem as shown in Figre 8.16 mst be capable of continos operation in a 55 C 131 F ambient at sea level condition. The maimm allowable hot spot component srface temperatre is limited to 100 C 1 F. The sstem contains seven PCBs, each dissipating 0 watts, for a total power dissipation of 140 watts. This does not inclde fan power dissipation. 68

Figre 8.15 Cabinet cooling hints Figre 8.16 plan view of fan-cooled electronic bo 69

The flow area at the partitions designed on the drawing are as follows Inlet to fan is an annals dimensions are in mm 48 8 90 o trn and transition from a rond section at the fan to an oval section 74 100 13 Contraction and transition to rectanglar section a rectangle 8 15 mm Plenm entrance to a PCB dct rectangle each 1.5 155 mm PCB channel dct as shown in the following drawing Rectangle each.5 30 mm 0.1 9 in Design procedres: Two fans are available for cooling the bo: Fan A is three phase 15500 rpm that has a 5 Watts motor. Fan B is single phase, with an 18 Watts motor that operates at 11000 rpm at sea level. The bo mst be eamined in two phases to ensre the integrit of the complete design. In phase 1, the thermal design of the bo is eamined, with the proposed fan, to make sre the component hot spot temperatre of 100 C 1 F is not eceeded. In phase, the electronic chassis airflow impedance crve is developed and matched with several fans, to make sre there is sfficient cooling air available for the sstem. 70

Phase 1: Electronic bo thermal design: To be on the safe side, base the calclation on the se of larger, 5 W motor, Fan A The total energ to be dissipated wold then be q 140 5 165 W Air reqired for cooling is q m c t t p a, o a, i Past eperience with air-cooled electronic sstems has shown that satisfactor thermal performance can be obtained if the cooling air eit temperatre from the electronic chassis does not eceed 70 C 160 F, so that assme eit air temperatre t a,o 70 C From the air propert table at mean temperatre t m 7055/ 6.5 C ρ 1.05 kg/m 3 ν 19.3 10-6 m /s Pr 0.7 C p 1008 J/kg.K k 0.089 W/m.. o C Air reqired for cooling is 165 m 0.01091kg/s 100870 55 B calclating the heat transfer coefficient between the air and PCB ' s and, hence, the temperatre rise of the PCB ' s above the ambient air. For this prpose we calclate the Renolds' nmber VDH Re v 4A 4 9 0.1 Q DH 0.197 in 0.005 m P 9 0.1 and V m 0.01091 ρ A 1.05 790.1 0.054.55 0.005 Re 663-6 19.3 10.55 m / s The heat transfer coefficient for laminar flow throgh dcts is shown in the following relation h D k H N D h 4. W/m Re Pr 1.86 / L DH 663 0.7 0.005 1.86 8 0.054.K 1/ 3 1/ 3 4.19 The total heat transfer is q h S eff t h 71

Actall, the back srface of a PCB is not available for heat transfer; the practice is to assme 30 percent onl available for this prpose S 7 1.3 8 9 0.054 0.43 m eff t h 165/ 4. 0.43 16.1 Therefore, maimm component srface temperatre t ma t a,o t h 70 16.1 86.1 o C This is acceptable srface temperatre since it is less than 100 C o C Phase : Electronic chassis air flow impedance crve: The air flow conditions are eamined at si different points in the chaises, where the maimm static pressre losses are epected to occr, as shown in Figre 8.16. These static pressre losses are itemized as follow: 1- Air inlet to fan - 90 o trn and transition to an oval section 3- Concentration and transition to a rectanglar section 4- Plenm entrance to PCB dct 5- Flow throgh PCB channel dct 6- Ehast from PCB dct and chaises The following table gives the ratio between static head loss to velocit head at the different positions Position nmber H s / H v 1 1 0.9 3 0.4 4 5 1 6 1 The flow areas at each position are: π π A1 d o d i 0.048 0.08 1.194 10 4 4 Position : A 0.06 0.074 π 0.013. 455 m 3 Position 3: A3 0.008 0.15 1 10 m 3 Position 4: A4 0.0015 0.155 7slots 1.68 10 m 3 Position 5: A5 0.005 0.3 7 4.05 10 m 3 Position 1: Position 6: A 3 5 A6 4.05 10 m m The following table gives velocit, velocit heads at each position at 10 cfm 0.83 m 3 /min 7

Position V cm/s. H v cm H O 1 400 0.098 376 0.087 3 488 0.146 4 90 0.0516 5 116 0.008 6 116 0.008 Performing the test nder different cfm air flow: let the flow rate also at 0 cfm, and 30 cfm The losses for each another flow calclated from V H H s / H v H v, 10 10 The following table gives the static pressre loss in cm H O at 10 cfm, 0 cfm, and 30 cfm Position 10 cfm 0 cfm 30 cfm 1 0.098 0.39 0.88 0.086 0.345 0.777 3 0.079 0.9 0.657 4 0.103 0.411 0.97 5 0.008 0.035 0.0731 6 0.008 0.035 0.0731 Total 0.3761 1.5 3.389 Then drawing the chassis air flow impedance crve at different fan crves as shown below The minimm flow rate reqired for this sstem is m 0.01091 3 3 V 0.01 m / s 10370.7 cm / s ρ 1.05 73

From the impedance crve it shows: The flow rate spplied b fans A is 3 V 30cfm 14157 cm / s The flow rate spplied b fans B is 3 V 3cfm 10854 cm / s So that both fans A and B can sppl more than the minimm reqired flow rate, either fan will be acceptable. 74