Hydraulic Resistance and Role in Electronics Cool

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.
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PAGE 18