Thermal Analysis Correlations of Pressure Drop K=0.2065+0.1549xCv in Electronics Packaging Calculating pressure drop is one of the most basic and important aspects of any thermal analysis. Knowing the pressure drop in a system is key to determining how much fluid is available for the cooling process. This is true for heat sinks, heat exchangers, telecom chassis or any other device using some form of liquid or gas. -0.4224 In 1839, experiments by Hagen showed that pressure drop in laminar flow was proportional to flow rate, but for turbulent flow it is roughly proportional to-0.4224 the square 2 of the flow velocity. K=0.2065+0.1549xCv L V ∆P hf = f = The pressureD drop 2 for a laminar flow in a tube was 1 ρU12 2 g found by Hagen to be: K=0.2065+0.1549xCv-0.4224 ∆P 8µLG L V2 ∆P h f == f D4 2 ∆P 1 2 K ρU1 =1 g R -0.4224 2 K=0.2065+0.1549xCv Pressure drop can be divided into two fundamental 2 L3/4 V1/4 Where G is the volumetric flow rate, and μ is the categories: frictional and dynamic. The pressure drop ∆P 8µLG µ D-4.75G1.75 ∆P=0.241Lρ hf =viscosity. f dynamic from frictional effects is the dominant part of the overall 1 D 2g ρU ∆P = 4 pressure drop. Dynamic pressure drop owing to K2 = The Darcy-Weisbach equation for turbulent flow R momentum changes is typically small, unless the fluid pressure drop in a tube can L V2 be shown as: velocity reaches Mach 0.2. Dynamical changes can ∆P 4A 8µLG 1/4 -4.75 1.75 h be attributed to entrance effects, sudden expansion 1 Dh∆P=0.241Lρ =f = f D 3/4 2gµ D G ρU1 or contraction, elbows, valves, etc. 2 ∆P P = 4 K R A fundamental formula for pressure drop is the This relation shows that pressure drop in turbulent flow Darcy-Weisbach equation. This correlates head loss varies by∆P the power of 1.75 of velocity. L is the length 8µLG =the 4A 3/4µ 1/4 -4.75 1.75 due to friction in a pipe of any cross section, factoring K of pipe.2 ∆P=0.241Lρ D G ρV /2g ∆P = D = 4 in velocity and pipe dimensions for either laminar or h K= Ran equivalent diameter, called P pipes For noncircular turbulent flow. Head loss is simply the pressure drop A1 -0.4224 the hydraulic2diameter, Dh, can be defined as: K=0.2065+0.1549xC divided by the density, which is equivalentv to a column V∆P f1 of that fluid exerting the same pressure as the 4A ∆h = +3/4ΣK µ1/4D-4.75G1.75 U1 ∆P=0.241Lρ K loss pressure drop. 2 D=h = 2g D ρVP/2g ( This equation is stated as: ( ) Where A is the cross section and P is the wetted 2 perimeter. ( ) (( ) ) A1 L1 ∆P 4A V f1 L V 2 2 1 ∆P A1 f =VA2 ∆P A DKhdynamical ∆h + aΣK 2 hf = f == = -1+ 1-drop, + coefficient, K, U1 comp =1Forloss pressure loss ρV /2g P 1 1 2 2 2g D D 2g C2vρU1 = , A1 A2 D ρU1 A 2 is defined as: Vchannel A1 A1 ∆P Where f is the Darcy friction factor, L is the length of the L U1 1 V2 f1 K = 8µLGof the pipe and V is the velocity. ∆h V= 2 pipe, D is the diameter + ΣK U1 /2g D lossρV 2g comp The∆P friction =factor can4be taken from the Moody Chart, C = , which plots the friction R coefficient as a function of the Kv= 1.4 A1 V A1 2 density and g is the gravitational Reynolds number for both laminar and turbulent flow, Where ρ is channel theV fluid f1 and for smooth and rough pipes. constant. = 1 ∆h + ΣKU1 U1 loss V 2g ∆P=0.241Lρ3/4µ1/4D-4.75G1.75 comp D Cv = , V A1 2 ( ) ( ) Copyright© Advanced Thermal Solutions, inc. | 89-27 Access Road Norwood, MA 02062 channel usa | T: 781.769.2800 www.qats.com ( Page 8 g 3/4µ 1/4 -4.75 -0.4224 2 ∆P=0.241Lρ D vG1.75 K=0.2065+0.1549xC 2 LG 4A K = 1.4 L V2 P h = f THermal Analysis 4 Dh = f ρ3/4µ1/4D-4.75G1.75 2g 1 2 ∆P ρV2/2g 8µLG K= ρU12 = ( ) ( ) A2 2 -1+ 1- A1 A1 ( ) V f1 ∆hloss = + ΣK 3/4µ 1/4 -4.75 1.75 2g D ∆P=0.241Lρ D G -0.4224 + U1 ( ) f1 ∆P A2 =11 2 ρU1 A1 2 D U2 A2 1 2 Where V is the fluid velocity. L1 A1 we show the pressure drop for some simple geometries: In the following, U2 V4A f1 comp Sudden expansion [2] D = + ΣK C = 1 h U1 , D v V P ) ( ρV) /2g( ) channel ∆P = K= 1 ρU12 2 , G 2 A2 The total∆P head= loss of a complicated system can be calculated as: A1 K = 1.4 R4 2 gv 5 D ∆P 2 A2 ∆P A1 -1+ 1L1 A1 2 A2 V U1 ∆h loss = 1 K = 1.4 2g 2 A1 1.75 Cv = ( 2 2 Sudden Contraction [2] A1 f1 ( ) ( ) ∆P A2 2 1 A1 f1L1 f2L2 1 A2 =1+ 2 1- 2 + + 1 2 ρU1 A1 A2 D1 D2 2 + U1 D L2 A1 ) f1 + ΣK2 A2 D L U1 Vchannel U2 1 , A1 U1 A2 2 L1 L Vcomp L2 A2 L2 Introducing THE BWT-104 1 A2 2 Benchtop Wind Tunnel L A1 U2 U1 1 Overall dimensions (L x W x H) 91 cm x 44 cm x 44 cm A2 x 17.25”) (36 x 17.25 2Dimensions Test Section 50.8 cm x 44 cm x 10 cm (20 x 17.25 x 4”) L1 L2 Materials Aluminum, Plexiglas Flow Range 0 to 6 m/s (1200 ft/min) A1 1 A2 2 Weight 14.5 kg (32 lbs.) The BWT-104™ is a research quality, open loop, benchtop wind tunnel for thermal characterization of components, circuit boards and cooling devices such as heat sinks, heat exchangers and cold plates. It provides homogeneous flow, up to 6 m/s (1200 ft/min) within its Plexiglas® test section, has 12 ports for probes and sensors and can be operated on any axis, making it ideal for laboratory environments. For further technical information, please contact Advanced Thermal Solutions, Inc. at 1-781-769-2800 or www.qats.com L Copyright© Advanced Thermal Solutions, inc. | 89-27 Access Road Norwood, MA 02062 usa | T: 781.769.2800 www.qats.com Page 9 2 + v ∆P=0.241Lρ3/4µ1/4D-4.75G1.75 Dh = 4A P K= ∆P ρV2/2g Thermal Analysis -0.4224 ( A1 ) V2 boardf1channels [3] K value in circuit ∆hloss = + ΣK U1 2g D -0.4224 K=0.2065+0.1549xCv U2 ( () )( () ) ( ) ( ) 8µLG 1. White, F., Fluid Mechanics, McGraw-Hill, 1979. K = 1.4 2. Blevins, R., Applied Fluid Dynamics Handbook, Van Nostrand R K = 1.4 Reinhold, 1984. 3. Azar, K., Electronic Cooling Theory and Application, Lucent Technologies, 1994. 4. Kays, M. and London, A., Compact Heat Exchangers, 3/4 1/4 -4.75 1.75 ∆P=0.241Lρ G Pressure drop for a heatµsinkD [4] S α= Third Edition, McGraw-Hill, 1984. H 4A ∆P friction P =f Dh = ( L −3 10 Dh (S) ) ( )( 2 ( 1.7012 − (( ) (1 ) ) 2 0.5 ρU ∆P Kf == 96 1 −21.3553 +1.9467 ReρV /2g α α α 3 + 0.9564 α 4 − 0.2537 α 5 ) ATS-619 A1 ) V2L 10 f1 0.5 (ρU2) D (S) + ΣK ∆hloss = U1 2g D 2 Uapp 2 A22 (U2 ∆P expansion = ρ 1- σ - 1.0257σ +2.029σ - 1.0058) ∆P friction = f A1 U1 −3 h Vcomp ( 2 2 ) U2 1 L1 2 L1 total L2 L2 U1 1 A2 = ∆P friction +∆P contraction +∆P expansion A1 L σ = Open channel area / Total frontage area 1 A2 2 Where S is channel spacing, H is the fin height, U is velocity between fins, Uapp is the approach velocity, and Dh(s) isLthe channel hydraulic diameter. A2 2 Uapp C∆P = , ( 1- σ2 - 0.4405σ2 +0.039σ - 0.4011) = ρ v contraction Vchannel 2 A1 ∆P ( ) ( L References: K value for a sharp4corner turn ∆P = G1.75 A2 Readers should note that the literature contains many 1 correlations for pressure2drop that should be used with caution. While these correlations can provide good first order approximations for quick analyses, they may break Where the volume fraction coefficient is defined by: L1 L2 down in cases that do not fall into their range of Vcomp applicability. For example, some correlations will Cv = ,2 reasonably predict the pressure drop across a heat Vchannel L2V A1 2 ∆P A2 2 a bypass f1 the fin density 2 1 ∆Pf2L2 A2 2 1 A1 21- A1 flow +when is=1moderate. A2 A1 f h∆P = f = sink inA2 -1+ ∆P A1 f1L1 + 2 11 f = D -1+ 1+ U1 1 A2 2 1 1 2 2 =1+ 1+ + 2 ρU1 A1 A2 D ρU1 A1 A2 may fail 2 if the heat sink has2 3-4 mm tall, 1 Here,2Vcomp is the total g volume ρU1 A1 A2 occupied D by the 2 12 ρU12 But they A1 A2 D1 D2 2 components and V closely packed, thick fins. channel is the total volume of the circuit board forming the channel. 2 The ATS-619 is made from black anodized, extruded aluminum, and employs ATS’ patent pending maxiGRIPTM attachment solution. The ATS-619 is just 16.25mm high with a footprint of 22mm by 24mm (LxW) and has a thermal resistance of 6.11◦C/W. For further technical information, please contact Advanced Thermal Solutions, Inc. at: 1-781-769-2800 or www.qats.com Copyright© Advanced Thermal Solutions, inc. | 89-27 Access Road Norwood, MA 02062 usa | T: 781.769.2800 www.qats.com Page 10