2 2 δQ p p αV αV = u2 − u1 + 2 − 1 + gz2 − gz1 + ( 2 2 − 1 1 ) dm ρ ρ 2 2 ( THERMAL FUNDAMENTALS 2 2 p1 α1V1 p αV δQ + + gz1 ) − ( 2 + 2 2 + gz2 ) = (u2 − u1 ) − ρ 2 ρ 2 dm 2 p αV + + gz ρ 2 Hydraulic Resistance and its (u −Role u) h in Electronics Cooling δQ 2 lT In both thermal and fluid sciences, predicting the air and surface temperatures requires an approximate value for the air flow rate. A reasonable estimate for flow rate in any type of enclosure is obtained by accounting for all pressure losses in that system. These pressure drops result from friction generated by fluid flow over the enclosure walls and surfaces, as well as fluid dynamic momentum changes due to sudden expansion or contraction of the enclosure. To get a sense of these losses, we can apply the first law of thermodynamics, conservation of energy, to a viscous fluid flowing through a duct from point 1 to point 2. Assuming a steady state and incompressible flow, the basic energy equation per unit of mass of the fluid reduces to Equation 1: (1) 1 (− δQ ) dm 2 2 2 1 2 δ δ Q p α V 2p=2 u2 p−1u1 +pp2 −αpV αQ α−+1Vu(1α+2Vp2 2 − αp11V gz + 22V− 2=gz u 1 =(u2fluid −gz ) +− gz ( + gzloss, −2 ,1 )2 1 − udm 1+1 22 − gz ρ21 +2 head Had + 1been + total dm2 2 ) = h2 ρ ρ2 lT dm the ρ+frictionless, ρ1 ( ρρthe ρ 2 2 would have been zero and the equation would2have2 2 2 2 p p1 α1Vp1 αp21V2 α1V1 2 p1 reduced αρg α been equation incompressible + ) =δ(Q u − up2) 1V1) ( to Bernoulli’s 2 + gz 21V 2− ( for ( ( + +ρ gz1 ) − 2 ( + +ρgz+2 () =2(+u2 +−gz u1 2)+)−gz 1 )2− ( 1 ρδ Q 2energydm inviscid a2change in internal ρ ρ flow. 2 In this situation, ρ p2 pρ1 u u = − + − + 2 could only2occur through of heat transfer.22No 1 2 pδ Q α V the effect dm α V α V ρα V ρ2 p 2 2 p α V p p gz 2 to1internal +energy would + +energy conversion mechanical 2 1 1 +2 gz t gz + [ L ∗ of + ρdm ] = [=Lu]22 − u1 + ρ − ρ + gzρ2 −pgz1α2+V( 2 2 − 2 p ) take ρ place. 2 2 t L ( 1 + 1 1 + gz1 ) − ( 2 2 2 δQ ρδQ (up21 − uα11)V1 δQ p−2 α()2u V22ρ− u1 ) 2 (the (The u2 −reader uh1 ) may ()−− δ ( + + gz ) − ( + + gz ) = ( u − u ( − ) now wonder why term “head loss” 1 2 2 1 2 lf dm ρ an2energy ρ dmDuring 2p the d dm quantity. αV is used to describe gz + + h hlTρ lT 2 hlT hlm 2 nineteenth century, p αwhen V most of the empirical science 2 + gz 2 + 2 2 2 p of hydraulics was it was common p1 developed, α21Vp1 αp(2u1V2practice pδQ ρ 2 2 −αu )1 +)togz p1 α V α1V α 11V 2 gz gz ( + + ) − ( + + 1 2 21 2 V ( + ) −((− 2 2 = h gz1 ) −balance (express + the energy +ρ (2 + in terms +ρgz 1 lT of energy 2 ) =2hlTper unit 2 2 ρdm δQ ρ ρ (u2 − u1fluid. ) ρ So, 2 2 2the weight of2 have (− ) 3 would h δ Q flowing p2 pEquation α V α V lT 2 2 dm u1 + specific gz = uthe − 1+ − 1 1 ) ( ρg ) − fluid ρg ) (rather2 than 2 2 − (gz 1+ weight, (been ρg ) divided ρ ρ 2 p1dmpby h2lT p α1V1net 2 fluid density, ( − ρ, as ) =ishlfthe routine practice today. ρ p1 +The + gz1 ) − ( 2 ρ 2 2 ( ρ ρ 2 2 2 p α V p α V δ Q ρ 2 dimension of1 the would be 1α 2 pp2 2αp p21 1head Q1loss consequently αV ρ 2δ1V 2 2V2 ( + + gz=+1 )ugz − (−)u− (++ 2+− +12gz = (u 2 2 2 1) − +2t)2gz =1−h+u + gz − 2 )gz lT( L( ρg 2 1 ρ1 2 − 2 [ρ2 L2 t 2 ρ [(Lρ ∗2+t dm ] = [ L ] ) 2 ρ ρ 2dm ∗ ] = [L ] µ L V [ 2 ∗ ] = [L]t, 2or millimeters 2 flowing fluid. tρ Even LV 22 fL pV2 2L= 32 2 of t ∆Lp =p p1(−α 2 ρg ) p α V p α D h h h = + = though “head+ loss” is+still 1 1 2lmmeaning 2V2 ( in + ΣK ) lf ( gz + in h1use, +lits gzphysical +Dgz2 ) = (u2 − 1 ) − (h + h 2 2 ρ 2 2 2 ρ lf h h Where α and V represent energy, kinetic Equation 3 is a loss inρmechanical energy per L t22 2 ρlfexpressed lf p1 αu, p2 αinternal δ Q 2 2 [ 1V1 2V2 ∗ ] = [L ] 2(u − u 2) − 64 L V L V (energy + + gz ) − ( + + gz ) = 2 velocity, unit mass of flowing 2 Vrespectively. 2 α V1 δρQ coefficient, p1 andp1mean fluid α h−lm fla(2−minδQ hlmt L Lu21)) t 2pfluid.α= V ar ) = u2 + 2 − 1ρ + gz2 2− gz1 + ( 2 2 − 1 1 ) dm hlm hlf (=u2( Re 2 − u1Equation [ ∗ ] = [ L ] D 2 Rearranging yields: gzDh 2 V 2 fL + +dm h 2 2 2 dm ρ ρ 2 2 t 2 Lρ 2 h hlf + hlm = hlf 2( major + ΣK(or) Rt 2l =subcomponents: δQ p2 p1 αTotal α1hVloss 2V2 2head 1 (2) consists of two p αV 2 lT D 2 α V = u − u2 + − + gz2 − gz1 + ( α V −e ) αV h V p + α V + gz p 2α V 1 ρ δQ loss, minor (or local) loss, ρ(u − u ) − δ Q friction) 2 2 hlf , and 2(u 2 hlm . Major − u ) hl =losshlf + hlm (ρ 1 + 21 1 + gz1 )dm − ( 2 + 2 2 + gz ) = 2 1 2 = ( − ) 2 2 2 1 p V p V α α 2 h = R ∗ G 1 1 1 2 2 2 es2are D caused ρ 2 ρ 22 2 dm gz2 ) = hl lTflowt ( +hby frictional + gz1 )effects − ( +in fully developed +dm 2 p1 δQ α1V1 p2 α 2V2 δplmQ2 phenclosures. ρ ρ 2 Rt within constant-area Minor losses (u2 − u1 ) 2 α p1 V are p2 caused ( ( −+ ) + gz1 ) − ( + + gz2 ) =p(1u2 −pu ) −( 1 − 2lT) = h p αV 21 ( − ) = h 4 × A lf 2 2 ρ 2 ρ 2 dm ( − ) = h lf dm In Equation the terms in2 the parenthesis by fluid flow to flow2 accelera2 + +2,gz ( ρg )lfρ dynamic ρp αchanges 2 δQ 2 pρthepD V1V2 12 αdue ρ ρV22 h = αV 1 2 1V1 p α α ρ 2 2 p2 hρlTp1 δ Q2 αp2V2 p2α1V1 1 2 2 − gz1( +in(+an enclosure −+ gz1 )with deceleration −) ( varying +hl = Rcross α 2= V2u2 −αu11V+1 tion−and + ∗+Ggz2 2 ) = hlT ρgzP2wetted = u2 − u1 + − + gz2= −ugz−1u+p(+ 2α−V− 1 + gz) − gz +dm 2 t0.828 p p 2 ρ ρ 2 2 ( − ) 1 2 ρ 2 ρ 2 V fL 2 2 1 In + the flow’s through ( 1 +ρ2 ρ ρ −dm − effects )V = h2 lf fLLµV δQ ρ +2gz 2 ) h2l∆=p h2= h−V = = 32 ( LµV+htubes, ΣK )h( ρthe (u uα1 )V 2 2 2 sectional areas. µ L lf lm 2 ( − ) ρ p p V α h = + = ( + ΣK 2 ρ p p 2 2 0.3164 t32 p1lm−these p2A=2 32 1 2 2 2 2 p l and 1p the losses, 2 D ∆ofp friction = pp1f2 −=[predominate = pLα 2 ([L )2 hlocal 1 2ρg p1 α1V1 V δ∆Qplf =thus gz ( p1 + α1 V + gz +δ Q h 2 2 2∗ 2)] = ] = D 2 2 ( − 20.25 D 2 1 ) − ( dm+ 2 2 ) = hlT α1V1 lf 2 ( +at + gzare −major ( +losses. + gz ) = (u2enclosures, − uh1 ) − 2 V ρ 2 α p α V InD electronic however, the h Dh energy per unit mass 1δ)Q t L h2 Re ρ ρ δQ + gz1 ) −represent +1 gz ) = ( u − u ) − h(ρlT 2 (+p1 22+the 1 mechanical 2 2 2 1 LµV ρ 2 ρ 2 dm 0.828 +(2ugz −12() + 2 1−) u (+− gzanalyzed. 2 ) = (u2 − u1 ) − effect of local losses is as significant 2 2 ρ sections dm ∆Lp V = 2losses. p1 − p2 = 32 cross being 2 2 64 2 as friction 2 ρ 22 2 1 and 2 of theρsystem 2 dm L V dm 2 64 L V 2 2 ( ρg ) h 64 lf LhlfV= ( L ) t L LV 2 2 =µVfla minR p1 difference p2 theαtwo α1V1 ) 510.79=Dhfla min 2 The between art hlf = ( A 2V2 is what is irreversibly p α V Rt L × h ( ) f = = 1 e / D 2.51 [ ∗ ] = [ L ] h 2 ∆ = − = 32 p p p lf la ar min gz gz h ( + + ) − ( + + ) = αV Re Dh 2Re Dh 22 1 2 2Dh h 2 1 lT 2 lT + 2 + gz Equation 2.0log( ) = − + Re D 2 D 2 h 2 t L 3 can be used to evaluate friction losses. For increase in internal energy + gz ρconverted ρ p + 2 αtoVan undesired ρ 2 h h Dh L V ρ 2 + gz 3.7 Re f 0.5 lm f 0.5 2 h = ( 64 ) A 2 2 2 = f min2 fully developed flow through a constant-area (u12-u2ρ ) 2and the losspof energy in the form of heat lf enclosure, V p V α α 2 e h = R ∗ G 1 1 1 2 2 2 e V 2 Re Dhh =2 R ∗laG t 2 hlf L V 2l (Lρg ) t e gz gz h ( + + ) − ( + + ) = 64 L δQ l t 1 2 lT α V [ 2δQ ∗ ] = [L ] the (term at the sections. L × 510.79 hlfis=the (2 same ) = flatwo − ) D ρ 2 δQ ρ (u2 −2u1 ) u1 ) min ar D L V t (uL)2. This (ρ− D − u1 )conversion Re Dh e2 hlm Dh 2 of energy is known ( −mechanical ) 2 hdm A2 4.2 ∗ L dm lf = f dm ( ρg ) 2the total 2 D 2 h Considering that the only losses are friction losses, and 2 as head loss and is designated by in EquaD 4× A lT hlfL t p1 ehp2 4 × A2 A 4 × A h [tion ∗ ] = [ L ] D = α V ( − ) = h lT assuming the effect of elevation is negligible, 0.828 Dh =Equation 30.828 ρ h 22 2 3: lf Dh = 2 t L D P ρ ρ 2 2 p αV pP2wetted α VV2 Pwetted wetted hlmp α V 4× A 2 2 reduce α1V1 h =toK2 Equation ( 1 + 1 1 + gzwill + gz224: ) = hlT A2 2 V Dh =4.2 ∗ L 2 2 p V α 1 ) − ( lm+ A2 + gz1 ) − h( 2 (+p1 2+ α 1+ 1 gz2L) = h 2 2 2 t 2 ρ 2 2 0.3164 PA +[ gz1∗) −lT(] = [L+] + gz2ρ) = hlT2 (3) wetted lfρ 4 × A µ L V 2 2 p p 2 0.3164 0.3164 2 2 = p(2 =1 32 A ∆p =fD p= − 2 ) 2= hlf αV ρ h− 1 t2 L ρ f = × [ 10.25 = f × (1(4) − 1 )]2 0.46 Pwetted hlm ( ρg ) ρ0.25 ρDh 0.25 Re 0.3164 Re Re A A2 2 ( ρg ) L × 510.79 2 f = Re10.25 L × 510.79 hlf ρ 2 2 0.3164 64 L V L V 2 L µV 1 A ρ p2 αpV hlf =1f( = )∆p0.25 =− fp Aar32 = p12.51 ela2min / =D 2.51 1 × [ × (1 − 1 )]2eA/ 2Dh 2 2 h0.46 21 h 1 e / D ( − ) = h Re Re D 2 D 2 2 0.321 lm L t 2.0log( ) = − + h | | lf PUBLISHED THERMAL SOLUTIONS, INC. 89-27 ACCESS ROAD NORWOOD, MA 020620.5 T: 781.769.2800 WWW.QATS.COM h + A1= −2.0log( A2 PAGE h 0.50.5 ) hD 2 ADVANCED t e3.7 / D14h + R ρ2 BY [ 2 ∗ ] = [L] 0.5 = −2.0log( 0.5 3.7 Re f Lρ t2 2 f f1 ] = [L ] 2 2 δQ αV p p αV = u2 − u1 + 2 − 1 + gz2 − gz1 + ( 2 2 − 1 1 ) dm 2 2 ρ ρ ( ρg ) 2 Dh 2 2 dm L × 510.7 p1 α1V1 h p2 ρ α+2V2 2 + gz δQ ρ ( + +lT gz1 ) − ( + + gz2 ) = (u2 − u1 ) − 4× A 0.828 A2hl Dh = ρ 2 ρ 2 2 dm 2 δQ 22 Pwetted p1 α1V L1 t(u2 − u1 )p2 α 2V2 ( − Rt A2 ) 2 + gz ] 1=) [−L(] + 2 ) = hlT p α V ( + [ 2 ∗+ gz 2 dm 2 2 ρ p2 p δ Q 0.3164 p2 p1 α 2V2 + +ρgz 2t δ QL αV αV 4.2 ∗− u u gz gz = − + − + − + ( hlT= u2 − u1 + 2 − 1 + gz2 − gz1 + ( 2 2 − 1 1 ) 2 ρ 2 2 2 1 f = 2 0.25 1 ρ h ρ= R ∗ G V 2 fL dm ρ ρ 2 2 20 l t ( ρg ) hlf dm Re h h h = + = ( A 2 l lfL × 510.79 lm δQp1 α21V12 p2 α 2V2 2 2 2 Dh (u2 − u1 ) ρ ( − ) p α V p α V gz gz h ( + + ) − ( + + ) = p α V p α V δ Q 2 THERMAL FUNDAMENTALS 2 2 2 lT hlm( dm 1 ( 11 + 1 1 + gz1e)/−D( 2 +2.51 +Agz2 ) = (u2 +ρ 1 1 2+ gz1 ) −1 ( 2 +ρ 2 2 2+ gz 2 2 ) = (u 2 2 − u1 ) − hρ ρ = −22.0log( 2 ) p22 p1 ρ 2α 2V2 α1V1 dm + hlT Lδ2 Q t=2 u −ρ − + gz2 − gz1 + ( − ) 2u [ 2 ∗ α] =V 2 [L 1 +) 3.7 0.828 Re f 0.5V 2 fL1 f 0.5 (] ρg A 2 2 ρ ρ 2 2 2 t dm L 2 0.46 × p α V hl = hlf + h KL−) × ( [ ∗ +L×Σ(1 p α V 2 lm = p1 α1V1 p V α 4.2 ρ gz + + A 2 2 2 gz + + A A 2 Dh 1 22 ( + + gz +pgz2 )α=VhlT2 ρ 2L V 2 p 1) − α1(Vρρ1 +loss22is calculated δQ 2 2 2 ρ 2 hfriction Therefore, from the) static (9) A2 hl = R lf( 1 + head 2gz1 )2− ( + + + gz = ( u − u ) − h f = 2 2 1 lf p p L t 2 ρ two dm δQ at the δQis laminar ( 21(sections. ]lf =ρ[L] the2flow (u2 − u1D )h 2 0.321 u−2[ −2u)∗1=) hWhen h (pressures ρg ) (− ) lm ( − ) Rt ρ ρ t L 2 is given as: L × 510.79 the static pressure drop dm 2 dm p αV A2 ρ V 2 2+ Local, or minor, losseshhlTare the types that occur h + gz A when1 A 2 lm = K αV LµV ρ 2 lT hlf 2 2 × [ toh × (1 − ∗1 G 0.46 )]0.8 2 2 (5) ever the dynamic momentum of the flow alters due ∆ = − = 32 p p p L t 2 2 l = RA t 1 2 2 2 2 A 1 2 p V p V α α [ 2 ∗ ] = [ L2 ] h p αV D δQ 1 2 2= R + R + .... (u2 − u1 )( p1 +lm α1V1 (+ ( 1cross + 1 sectional + gz1 ) −area. ( 2 +InRfluid + 1gz2 )2= hlTA eq hlTabrupt change in the t L − gz1h))− ( 2 + 2 2 + gz2 ) =an 4.2 ∗ L ρ 2 ρ 2 p1 p2 ρ 2 ρ 2 dm flow through tubes, for example, these losses are seen 2L V2 ( − ) = hαlf64 L V2 0.321 1 V) Equation hSubstituting f = A2 are 0.46 lf ρhEquation ρhlf (=ρg 5( into 4 will yield the in fittings, bends, valves, reducers, etc. The losses ×[ × lT la ar min ( ρg ) ) 2 0.828 Re 2 Dh 2 A 2ofDahfriction L ×A51 1 head loss in terms factor2 that, for a the result of energy dissipation due to mixing 2 hfriction ρ 1 of fluid 1 in A 12 lm p1 α1Vρ1 p V α 2 2 2 µ L V ..... = + + gz gz h ( + + ) − ( + + ) = A given length diameter, is only dependant the separated regions. Figure 1 depicts the flow vortices 2 lT ∆p and =ρ p1hydraulic −e p = p32 1 1 p22 Req R1 R2 2( 2− 22LReynolds ρ hlf 2 )number: = 2 0 L2 when t 2 a fluid flow t D 1 A in the separated regions encounters a R = R + R + ..... R αonVthe fluid velocity 2 h D or 1 n [ 2 ∗ ] =0.46 [L] × [ eq [ 2 ∗ρ ] =ρ [L] × (1 − 1 )] 2 L ( ρg ) sudden contraction. t t L A1 A2 2 64 L 4V×2 A L V2 L × 510.79 4.2 (6) µ L V hlfρ = ( D ) = = fla min ar h ∆ = − = 32 p p p h p1 p2 h lf Re D 2 A 2 Dh 2 2 A2 ( − ) = hlf 2 2 lf hPwetted 1 Dh 0.321 R 1 1 1 11 + R2 eq = R L t h ρ ρ h = + + ..... lm 2 e[ 2 ∗ ] =lm[L] 2 2 Req RA R2 Rn 0.3164such 64 L V L V 1 For turbulent tflow,Lfhowever, analytical correlation is = hlf2 0.25 4.2 ∗ L =( ) = fla min ar D 2 LµV 1 αV Re αV of theRe friction for circular Dh factor 2 Dh 2 ∆not = The 32 value p =available. p1 − p2 h 2 0.46 × A[2 1× (1 lf D 1 1 R = R + R + ..... R h tubes can be extracted 2 4 × A 2from Moody’s diagram, given that eq 1 2 n= + A1 + e Dhhlmnumber, =2 1 2.51 2 e/D the Reynolds , the ratio of the pipe inside R R R2 h eq 1 p1 p2 64 L V Pwetted L2 V ) =p −2.0log( + p 0.5 ( − ) = h hlf = ( ) =0.5fla(min1D − ) = h lf ar 3.7 Re f f 0.3 lf ρ ρ Re Dα 22 1 A1 2 ρ Dρh 2 are known. Even h V 0.3164 surface roughness and diameter, 0.46 ×1[ × (1 − )]A 1 1 1 f = 24 × A A1 A2 = + + ..... 2Re0.25 DLh =V though the Moody diagram has been developed for circue µ L V LµV R=eqp − pR1= 32 R2 Rn hlf ∆=pf= p −Ppwetted ∆ p = 32 1 2 lar ducts when 2 2 Dh1 2for2 non-circular D tubes, it can p1 also p2 be used Dh Dh diameter, ( − ) = h Req = R1 + R2 + 0.321 lf the tube diameter is replaced with hydraulic 1ρ ρ e2/0.3164 Dh 2.51 2.0log( ) = − + 2 2 2 f V= 4 × Af 0.5 64 L V L V A 64Re0.25 L Re V 2 f 0.5 L V2 hlm = K 3.7 Dh = h ( ) f = = Vortices (2 ) = fla min ar lf la min ar lf = rectangular Pwetted . The usehof Re Dh 2 Dh 2 ReLµDVh 2cross section Dh ducts 2 ∆ = − = 32 p p p 1 2 2 2 to width is less than are valid only if the ratio of height 1 1 1 L V 1 Dh e / Dh 2.51 Req = R =1 + R2 ++ .....Rn+ .. 0.3164 h4lf [1]. =f e e 2.0log( ) = − + about 3 or 0.5 f = Req R1 R2 Dh 2 f 0.5 2 Re0.25 DSeparated Regions. 64D L V L3.7 V 2 Re f Figure 1. Flow Vortices in h =( 2 ) = fla min ar For turbulentlfflowV in smooth pipes withDReynolds numh 2 hlm = K5 Re Dh 42 × A L V2 4 ×using A 1Equation 1 10, where 1 1 1 less than 10 e/D 2.51 bers , 2Blasius’ correlation is used to estimate Local losses can be estimated f = h D h= Dh = = + + ..... ) 2 = −2.0log( +h lf 0.5 D e 0.5 P K represents the loss coefficient. K values are pubwetted The 3.7 ReP fthe friction factor: f wettedh Req R1 R2 Rn D 2 lished in hydraulic resistance handbooks for a variety of V 0.3164 0.3164 (7) situations that could bef encountered in practical applica= L V2 f =hlm = K 0.25 4 × A Re0.25 2 hlf = f Reexpansion tions. These include sudden and contraction, D = Dh 2 h Pwetted flow over barriers, orifices, etc. In some cases, values 1 e/D 2.51that 1 e / Dh 2.51 V 2 common formula most to estimate of K are provided by manufacturers ) = −2.0log(of theh devices + 2.0log( ) = −used + turbulent hThe 0.3164 0.5 lm = K 0.5 0.5 Re f 0.5 is f the value of the3.7 3.7 Re f f0.25 2 f is= Colebrook’s friction factor Equation: cause the head loss. If loss coefficient Re L V2 hlf = f e / Dh 1 2.51 (8) = −2.0log(Dh 2 + ) 0.5 3.7 Re f 0.5 f V2 hlm 2= K L V 2 and iterations are needed This equation hlf is=transcendental f Dh 2 to evaluate the friction factor. After estimating the friction factor, the friction2head loss in turbulent flow can be V hlm = the K following equation: calculated using 2 not available in literature or provided by the supplier it can L V2 be determined by a simple hlf = fdifferential pressure measureDh 2 ment across the device: hlm = K V2 2 (10) Adding the friction and local losses, Equations 9 and 10, will give the expression for the total pressure loss. Please PUBLISHED BY ADVANCED THERMAL SOLUTIONS, INC. | 89-27 ACCESS ROAD NORWOOD, MA 02062 | T: 781.769.2800 WWW.QATS.COM PAGE 15 THERMAL FUNDAMENTALS hl = hlf + hlm = V 2 fL ( + ΣK ) 2 Dh (11) The concept of hydraulic losses explained in the preceding sections is now extended to electronic Rt enclosures. A simple forced air cooled cabinet for an electronic system, Vortices shown in Figure 2, is one in which hl = air Rt enters ∗ G 2 an inlet panel at the bottom, is drawn through the chassis to the fan tray on the rear, and is exhausted out into the room. 0.828 Power Supply Enclosure A2 Exhaust Fan L × 510.79 A2 Perforated Plates at Inlet & Output of PS the fan manufacturer, the point where these curves intersect will be considered the operating point from which the system airflow is determined. Such a graph is shown in Figure 3. A system should be designed so that its operating point falls below the stall range of the fan curve. In Figure 3, the stall range of the fan is the intermediate section of the fan curve where it becomes fairly flat before its sharp final descent. Static Pressure note that the summation in Equation 11 indicates that all discrete local head loss coefficients must be evaluated and accounted for in determining the total head loss. 10 9 8 7 6 5 4 3 2 1 0 0 5 10 15 20 Volumetric Airflow 4.2 ∗ L A2 Fan Curve Low resistance System High resistance System Circuit Cards 0.46 × [ Filter Inlet Grill 1 A × (1 − 1 )]2 A1 A2 0.321 A2 Req = REnclosure. Figure 2. A Simple Electronic 1 + R2 + .....Rn The air inlet and outlet panels and the internal mechanical structure of1 the chassis filters, perforated plates, 1 2 (air V12 + fL..... 1 = + V fL h h K = + = ( + Σ ) honeycomb, hcircuit cards, component blockade, fan tray h = h Rl+eqhlmlf = Rlm ( 2 R+2 ΣDKh ) Rn plenum,l etc.)lf resist the21airflow and cause the system Dh pressure loss. The total pressure drop of the system can be calculated from the net resistance, Rt including all the Rt air volumetric flow aforementioned resistances and the rate, G. h = R ∗ G2 hl = Rt l∗ G 2 t (12) In a graphical presentation of Equation 12, the flow rate versus the pressure drop0.828 would0.828 produce a system curve for the electronic chassis. When the A2 system curve is A2 drawn with the same scale on the fan curve provided by Figure 3. System and Fan Characteristic Curves. Multiple fans in a fan tray can be represented by a single fan curve. Theoretically, the volumetric flow rate of multiple fans coupled in parallel is the sum of the individual flow rates, while the static pressure will not differ much compared to that of a single fan. However, in practice the available rate will be about 20 to 30% less. In contrast, fans configured in series (push-pull) would have a higher static pressure while the volumetric flow rate nearly stays unchanged. Fan tray characteristics must be determined experimentally if accurate results are needed. In estimating the total system resistance of an electronic chassis, all individual resistances must be considered. Unfortunately, very limited resistance data is available for the devices typically used in electronic enclosures. Some of the published data for the common chassis elements is presented in Table 1. L × 510.79 L × 510.79 2 A INC. | 89-27 ACCESS ROAD NORWOOD, MA 02062 | T: 781.769.2800 WWW.QATS.COM PUBLISHED BY ADVANCED THERMAL SOLUTIONS, A2 PAGE 16 δQ ==((uu2 −−uu1 ))−− δQ 2 1 dm δQ = (u2 − u1 ) − dm dm = hlT z2 ) = hlT ==hhlT lT = hlT hl = hlf + hlm = ( h + ΣK )R t 2 Dh hl = Rt ∗RR Gt 2 2 hl = Rt ∗t G Rt hhl ==RRt ∗∗GG22 l t hl 0.828 = Rt ∗ G 2 THERMAL FUNDAMENTALS 0.828 A2 2 A 0.828 Chassis Element Resistance Comment 0.828 2 2 fL 2V A 2 0.828 hl = hlf + hlmV= fL( + ΣK ) A Perforatedhplate A - open area L × 510.79 (2 D + hΣK ) 2 l = hlf + hlm = 2 Dh LA×2510.79 A 2 A Filter Rt LL××510.79 A-filterexposedarea 510.79 Rt L × A510.79 22 L-filterlosscoefficientprovidedby 4.2 A ∗L the manufacturer ∗L 2 AA24.2 hl = Rt ∗2 G 2 hl = Rt ∗ G A2 Circuit board L - circuit board length 4.2 ∗ L 4.22∗ L A - effective free area of the channel AA2∗ L 4.2 between two circuit boards 1 2 A1 2 A 0.828 FanTray-SuddenExpansion 0.828 2 0.46 × [ A×1[ ×1(1×−(1A−2 )]A1 )]2 A1 - small area 0.46 2 A A - large area A A A 2 11 1 AA1 2 2 0.46 (1 0.46××[[A1××0.321 (1−−AA1)])]2 Fan Tray- Sudden Contraction A - small area 1 A A2 1 2 0.46 × [ 1 × (10.321 2− 2 )] L × 510.79 A1 A A22 L × 510.79 A 2 0.321 2 Ain Electronic Chassis0.321 left chamber, there are 5 evenly spaced circuit cards. The Table 1. Resistance of Common Elements A A+2chamber 2.....R to the right contains the power supply enclosure [2]. Req = R1 + R0.321 2A n 2 is entirely sealed except for its perforated inlet and Req = R1 + R 2 + .....Rn Athat 4.2and ∗ L the system outlet. When obtaining the total pressure drop 4.2 ∗ L 2 R = R + R + .....R Air is exhausted to a room by a fan in the back. A Reqeq =resiscurve of an electronic enclosure, the2equivalent R11 + R22 + .....Rnn A1 air 1 1n flow in the enclosure can be modeled according = R +1 R +The .....R tance must be calculated from the individualRresistances. = eq 1+ 1 12+ ..... 1 1 the electrical circuit shown in Figure 4. In this network, The procedure is analogous to finding Reqequivalent Rn =R1 resis+R2 to + ..... pressure losses due to expansions and contractions are tance in an electrical network. However,Rthe simple linear 11 Rn 1 11 eq A1 112 R1 11 R2 not considered. 1 A ..... = + + relation of the Ohm’s Law does not apply to the fluid flow 0.46 × [ × (1 −1 = 2 )] + 0.46 × [ ×A(1 )]A2 R1 12 + ..... RR 1n R− system. Instead, the flow pressure R1eqeq R11 + RR A1 1 drop A2is=approximately 2 + ..... n R-CC1 proportional to the square of the flowRrate. This R1nonlinearR2 Rn R-CC2 eq Circuit ity makes iterative solutions necessary, and the resulting Cards R-CC3 0.321 0.321 R-CC4 calculations can be quite lengthy and tedious. Table 2 2 R-CC5 2 A R-in shows how equivalent resistanceA is calculated when the R-filter Air-out enclosure elements are in series and parallel. Series R = R1 + R2 + .....Rn Req =eqR1 + R 2 + .....Rn 1 1 1 1 =1 +1 + .....1 Parallel 1 ..... = + + R R R R Req eq R1 1 R2 2 Rn n Table 2. Equivalent Resistance Calculation. To illustrate how the equivalent resistance is calculated from individual resistances, refer again to the electronic enclosure depicted in Figure 2. Air enters the enclosure through the inlet grill at the bottom, passes through a filter and reaches a section with two adjacent chambers. In the Air-in R-in R-internal R-out POWER SUPPLY Figure 4. Forced Air Flow Circuit for the Enclosure Shown in Figure 2. As shown in Figure 4, the flow is in series going through the inlet grill and filter. However, it would have to divide when it reaches the card cage and power supply chambers. In the card cage chamber, the flow is distributed among the cards before reaching the exhaust fan. In the power supply chamber, the equivalent resistance will include the inlet and outlet perforated sections and the internal resistance in the power supply chamber. To evaluate total flow resistance, the equations for the PUBLISHED BY ADVANCED THERMAL SOLUTIONS, INC. | 89-27 ACCESS ROAD NORWOOD, MA 02062 | T: 781.769.2800 WWW.QATS.COM PAGE 17 THERMAL FUNDAMENTALS appropriate enclosure elements given in Table 1 must be used in calculating the individual resistances. In analyzing the modeled electrical circuit, the equations in Table 2 also must be implemented to find equivalent resistance in series and parallel connections. Once the total equivalent resistance of the system is obtained, it must be substituted in Equation 12, and a curve of the system pressure versus the volumetric flow rate must be constructed. Given that this curve is drawn with the same scale on the fan curve, the point where the system and fan curves intersect will be the operating point from which the amount of available air flow for the system is found. References: 1.Fox,R.,McDonald,A,IntroductiontoFluidMechanics, Wiley,1985. 2.Ellison,G.,ThermalComputationsforElectronicEquipment,VanNostrandReinhold,1984. MynameisFrancisco. IdesignGPSdevicesforthedefenseindustry. Yesterday,ittookmeahalf-hour tofindmycaratthemall. IAMHOT ( ) ANDI’MNOTAFRAIDTOADMITIT They say the first step to solving your problems is admitting you have one. The second is calling us. Now that ADVANCED THERMAL SOLUTIONS, INC. has teamed up with DIGI-KEY, your electronics cooling solutions are only a click away. Visit www.Digikey.com to place an order or www.qats.com for further technical information. PUBLISHED BY ADVANCED THERMAL SOLUTIONS, INC. | 89-27 ACCESS ROAD NORWOOD, MA 02062 | T: 781.769.2800 WWW.QATS.COM PAGE 18