cooling of electronics – lecture 2
TRANSCRIPT
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Cooling of Electronics –Lecture 2
Hans Jonsson
• Introduction to Cooling of Electronics • Cooling at different levels• Cooling demand calculations
Agenda – Lecture 2
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Introduction to Cooling of Electronics
Both Power and Heat Flux increasing!
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Is this a new trend?Forecast from 1992
Is this a new trend?Forecast from 1998
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Is this a new trend?
The increase in heat dissipation has been so even, it has been formulated as a ”law” and given a name:
The heat dissipation from an electronic component/product/system is doubled every 18th month
Moores law
Are the forecasts reliable?Forecast from 1990
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Are the forecasts reliable?
1980 1995 2010 2025
150
100
50
Forecast from 2004
• Most electronic components consists (or contains) Silicon.
• Silicon is ageing faster if it is exposed to high temperatures.
• In the industry there are temperature limits to prevent premature ageing of the silicon (e.g. 100 °C). Different limits exists depending on type of component etc.
• Electronic products are usually designed to operate in an ambient temperature of 40 °C (other ambient conditions can of course be used).
• The temperature difference is hence constant!
Electronics have a maximum termperature!
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Heat fluxes and temperature differences for different ”cooling applications”
Heat transfer basicsNewton’s ”law” of coolingtAhQ Δ⋅⋅=&
thAQq Δ⋅==′′&
: Heat dissipation: Heat Flux: Heat transfer coefficient: Heat transfer area: Temperature difference
q ′′
tΔAh
Q&Där:
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This is the problem!
• Heat dissipation from electronics increases• Electronics is decreasing in size• Temperature difference constant
• This gives rise to more and ”harder” cooling problems as more heat has to be dissipated from a smaller heat transfer area!
What does this mean?
If the heat dissipation is increasing and the temperature difference is constant, the heat transfer area and/or the heat transfer coefficient have to increase!
Still, the components are getting smaller and smaller!th
AQq Δ⋅==′′&
Component
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Cooling gets more and more important
Electronics80%
Cool-ing
20%Electronics
60%
Cool-ing
40%
”Conventional” cooling• Dominated by air cooling – both natural and
forced convection• Fan driven forced convection more and more
dominant• Vapor compression evaporation cooling is
used to cool rooms (i.e. computer centers, server rooms, radio base stations, etc).
• New techniques already implemented in some products.
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Future -Possibilities and limitations
• An increase in heat transfer coefficient is needed to be able to handle future cooling needs
• One way to achieve this is by changing cooling media (i.e. use water instead of air).
• Another way is to use a more efficient method of cooling (impinging jets, heat pipes, thermosyphons, etc).
Achievable Heat Transfer Coefficients
AirFC mediaWater
1 10 100 1000 10.000 100.000 1.000.000Heat Transfer Coefficient, h (W/(m2·K))
Natural Convection
Single-phase Forced Convection
Boiling
AirFC mediaWater
FC mediaWater
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Cooling at different levels
• Room and Cabinet level• Board level (PCB level)• Component level
Cooling at different levels
Different cooling problems at the different levels forces us to use different solutions at the different levels, i.e. usually more than one technique needs to be used.
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• Typical applications: Telecommunication Switch boards, Radio base stations
• Heat dissipation at the level of several kW/m³• Cooling of the room by using conventional air
conditioners that circulate air inside the room.
Cooling at Room and Cabinet Level
Air Conditioning• Data centers, server rooms is nowadays always
air conditioned.• The cool air can be supplied in different ways:• Through the floor using a so-called ”Raised
floor”• Putting the AC-units beneath the ceiling.
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Raised floor
Cool fluidHot fluidHatching signifies blocking from cables, piping, etc.Vent for cold air return
Plenum with cold air return
EIA Rack housing computerswith given air temperature rise
Hot aisle
ModularAC unit
Cooling CoilAir mover
Raisedfloor
AC-units beneath the ceiling
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Air conditioning units underneath ceiling
Cool fluidHot fluid
EIA Rack housing computerswith given air temperature rise
Hot aisle
Heat exchanger (Evaporator)
Cold aisle
Cooling demand calculations
• Traditionally, only the cooling load calculation was considered.
• Nowadays also the energy consumption achieving low temperatures is considered.
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Cooling demand of a room
SolarQ&
PeopleQ&
onInfiltratiQ&
nVentilatioQ&
.appl.ElQ&
onTransmissiQ&
CoolingQ&
Cooling demand
• Solar irradiation, heat dissipated from people, and heat losses from electric appliances are always heat gains!
• Heat gains due to transmission, ventilation, and infiltration are dependent on temperature gradient and are heat gains if tindoor < toutdoor!
.Vent.Inf.Transm
Solar.appl.ElPeopleCooling
QQQ
QQQQ&&&
&&&&
+++
+++=
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Heat gains• Heat gains from people depends on the number
of people, and their activity.• Heat gains from electrical appliances are the
sum of their heat losses at a given point in time.
Heat gains• Heat gains from solar irradiation consists of
two parts; irradiation transmitted through windows, and increased heat transmission through walls.
qv = I·a = I·εI: Solar intensity perpendicular to surface (W/m²)a: absorptivity, ε: emissivity
1v
walls,Solarwindows,SolarSolar
AUqAI9.0
QQQ
α⋅⋅+⋅⋅=
=+= &&&
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Solar irradiation
Solar irradiation
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Solar irradiation
Heat gains• Transmission heat gain through building
envelope (walls, windows, roof, floor) is dependent on overall heat transfer coefficient (U), surface area (A), and temperature difference.
to is the outdoor and ti is the indoor temperature.
( )∑ −⋅⋅=envelope
io.Transm ttAUQ&
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Design outdoor temperature
Heat gains• Infiltration heat gains through building envelope
(walls, windows, roof, floor) is dependent on total infiltration flow rate, specific heat of air, and temperature difference.
( )iop.Inf.Inf ttcmQ −⋅⋅= &&
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Heat gains• With ventilation we mean: exhaust air that is
replaced by fresh air!• Ventilation heat gains is dependent on total
ventilation flow rate, specific heat of air, and temperature difference.
• The supply temperature is dependent on the ventilation system!
( )iplysupp.Vent.Vent ttcmQ −⋅⋅= &&
How do we remove the heat?
• System 1: AC-unit in the room• System 2: AC-unit with heat exchanger• System 3: AC-unit without ventilation• System 4: AC-unit with recirculation• System 5: Hybrid between 2, 3, and 4
SolarQ&
PeopleQ&
onInfiltratiQ&
nVentilatioQ&
.appl.ElQ&
onTransmissiQ&
CoolingQ&
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System 1: AC-unit in the room
( )iop.Vent1.Vent ttcmQ −⋅⋅= &&
CondenserQ&
Cooling,1Q&tsupply = to
ti
System 1: AC-unit in supply air
to ti
CondenserQ&
Cooling,1Q&tsupply
)cm(Qtt
QQQQQ
QQQQ
)tt(cmQ
p1.,Vent1,Coolingoplysup
1.,Vent.Const1.,Vent.Inf.Transm
Solar.appl.ElPeople1,Cooling
iop1.,Vent1.,Vent
⋅−=
+=+++
+++=
−⋅⋅=
&&
&&&&&
&&&&
&&
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System 2: AC-unit with heat exchangerti
to CondenserQ&
Cooling,2Q&tsupply
Heat exchanger
tm
( )( ) ( )
)cm(Qtt
QQQ
tt)1(cmttcmQ
tttt
p.Vent2,Coolingmplysup
2.,Vent.Const2,Cooling
iop.Ventimp.Vent2.,Vent
ioom
⋅−=
+=
−⋅η−⋅⋅=−⋅⋅=
−⋅η−=
&&
&&&
&&&
System 3: AC-unit without ventilationti
CondenserQ&
Cooling,3Q&tsupply
)cm(Qtt
QQQQ
0Q
p.Vent3,Coolingmplysup
.Const3.,Vent.Const3,Cooling
3.,Vent
⋅−=
=+=
=
&&
&&&&
&
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System 4: AC-unit with recirculationti
to CondenserQ&
Cooling,4Q&tsupply
tmix
Ventmx &⋅
Ventm)x1( &⋅−
Ventm&
( )
)cm(Qtt
QQQ
t)x1(txt
ttcm)x1(Q
p.Vent4,Coolingmixplysup
4.,Vent.Const4,Cooling
oimix
iop.Vent4.,Vent
⋅−=
+=
⋅−+⋅=
−⋅⋅⋅−=
&&
&&&
&&
System 5: Hybrid between 2, 3, and 4ti
to CondenserQ&
Cooling,5Q&tsupply
Heat exchanger
tm
tmix
Ventmx &⋅
Ventm)x1( &⋅−
Ventm&
?t
?Q
?t
mix
5.,Vent
m
=
=
=& )cm(Qtt
QQQ
p.Vent5,Coolingmixplysup
5.,Vent.Const5,Cooling
⋅−=
+=
&&
&&&
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Heat Exchanger ProblemA heat exchanger has a heat transfer area of 10 m² and an overall heat transfer coefficient of 230 w/(m²·K). Hot water with a mass flow rate of 0.3 kg/s and inlet temperature of 100 °C shall heat cold water with a mass flow rate of 0.6 kg/s and inlet temperature of 10 °C. At what temperature is the heated water leaving the heat exchanger is the heat exchanger is connected
a) Counter-flowb) Parallel-flowAns: a) 43,7 °Cb) 38,1 °C.
H. Jonsson: Applied Thermodynamics - Collection of Formulas
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11.4 Heat Exchangers Three types of heat exchangers are distinguished • Recuperative • Regenerative (Ljungström preheater) • Evaporative (Cooling towers)
The recuperative heat exchangers can be divided into: • Counterflow heat exchangers • Parallel-flow heat exchangers • Cross flow heat exchangers For calculations, the following equations apply & ( & ) ( & )Q m c m cp p= ⋅ ⋅ = ⋅ ⋅Δ Δ1 2 [11.20]
&Q k A m= ⋅ ⋅ϑ [11.21]
( )21
21m ln ϑϑ
ϑ−ϑ=ϑ [11.22]
The temperature efficiencies are defined
θΔ
=η 11 [11.23]
θΔ
=η 22 [11.24]
Introduce:
& ( & )W m cp1 1= ⋅ [11.25] & ( & )W m cp2 2= ⋅ [11.26] hence
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112 Y
WW
η⋅=⋅η=η &
& [11.27]
For counterflow heat exchangers it can be shown that
)Y1(X
)Y1(X
1eY1
e1−⋅−
−⋅−
⋅−
−=η [11.28]
where
1WAkX &
⋅= [11.29]
ϑ2
ϑ1
ϑ1
2
1
t1t2
tΔ1
Δ2
θ
Counterflow
2
1
tθ = Δ1
Δ2
t1
t2
Parallel-flow
θΔ2
Δ1
Cross flow
ϑ2
H. Jonsson: Applied Thermodynamics - Collection of Formulas
36
2
1WWY &
&= [11.30]
For parallel-flow heat exchangers it can be shown that
Y1
e1 )Y1(X
1 +−
=η+⋅−
[11.31]
To be able to use the diagrams on page 64, set 21 WW && < . For cross flow heat exchangers matters are a bit more complicated. For solving these kind of problems, the reader is referred to Compact Heat Exchangers by W. M. Kays and A. L. London, 1964. 11.5 Heat Transfer Through Walls
The heat flow is given by
&Q k A= ⋅ ⋅ϑ [11.32]
where k : overall heat transfer coefficient ϑ : temperature difference The overall heat transfer coefficient for a plane wall consisting of multiple layers is calculated as
on ni
11k1
α+∑ ⎟
⎠⎞
⎜⎝⎛
λδ
+α
= [11.33]
where αi : heat transfer coefficient on the inside (convection + radiation) αo : heat transfer coefficient on the outside (convection + radiation). For cylindrical and spherical walls consisting of multiple layers, the overall heat transfer coefficient can be found by (Ai and Ao are the inside and outside surface areas respectively)
oon nmii A
1AA
1Ak
1⋅α
+∑ ⎟⎟⎠
⎞⎜⎜⎝
⎛⋅λδ
+⋅α
=⋅
[11.34]
where, for cylindrical walls (ri: inside radius and ro: outside radius)
( )io
iom rrln
)rr(L2A −⋅⋅π⋅= [11.35]
for spherical walls (ri: inside radius and ro: outside radius) A r rm o i= ⋅ ⋅ ⋅4 π [11.36]
αi
αo
ϑ
δ
λ
&Q
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