Cooling of Electronics Lecture 2

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1 Cooling of Electronics Lecture 2 Hans Jonsson Agenda Lecture 2 Introduction to Cooling of Electronics Cooling at different levels Cooling demand calculations

2 Introduction to Cooling of Electronics Both Power and Heat Flux increasing! 2

3 Is this a new trend? Forecast from 992 Is this a new trend? Forecast from 998 3

4 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 8th month Moores law Are the forecasts reliable? Forecast from 990 4

5 Are the forecasts reliable? 50 Forecast from Electronics have a maximum termperature! 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. 00 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! 5

6 Heat fluxes and temperature differences for different cooling applications = h A Δt q = = h Δt A Där: q h A Δt Heat transfer basics Newton s law of cooling : Heat dissipation : Heat Flux : Heat transfer coefficient : Heat transfer area : Temperature difference 6

7 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? Component 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! q = A = h Δt Still, the components are getting smaller and smaller! 7

8 Cooling gets more and more important Electronics 80% Cooling 20% Electronics 60% Cooling 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. 8

9 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 Air FC media Water Air FC media Water Natural Convection Single-phase Forced Convection FC media Water Boiling Heat Transfer Coefficient, h (W/(m 2 K)) 9

10 Cooling at different levels Cooling at different levels Room and Cabinet level Board level (PCB level) Component level 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. 0

11 Cooling at Room and Cabinet Level 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. 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.

12 Raised floor EIA Rack housing computers with given air temperature rise Hot aisle Raised floor Modular AC unit Cooling Coil Air mover Plenum with cold air return Cool fluid Hot fluid Hatching signifies blocking from cables, piping, etc. Vent for cold air return AC-units beneath the ceiling 2

13 Air conditioning units underneath ceiling EIA Rack housing computers with given air temperature rise Heat exchanger (Evaporator) Cool fluid Hot fluid Hot aisle Cold aisle Cooling demand calculations Traditionally, only the cooling load calculation was considered. Nowadays also the energy consumption achieving low temperatures is considered. 3

14 Cooling demand of a room Cooling El.appl. Solar Transmission People Ventilation Infiltration + Cooling Transm. Cooling demand = + People Inf. + + El.appl. Vent. + Solar + 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 t indoor < t outdoor! 4

15 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. Solar = + = 0.9 I A + q Solar,windows v U A q v = I a = I ε I: Solar intensity perpendicular to surface (W/m²) a: absorptivity, ε: emissivity α Solar,walls = 5

16 Solar irradiation Solar irradiation 6

17 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. Transm. = U A envelope ( t t ) t o is the outdoor and t i is the indoor temperature. o i 7

18 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. Inf. = m& Inf. c p ( t t ) o i 8

19 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. Vent. = m& Vent. c p ( t t ) sup ply The supply temperature is dependent on the ventilation system! i How do we remove the heat? Cooling El.appl. Solar Transmission People Ventilation Infiltration System : 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 9

20 System : AC-unit in the room Condenser t supply = t o Cooling, t i Vent. = m& Vent. c p ( t t ) o i System : AC-unit in supply air Condenser t o t i + t Vent., sup ply Cooling, Transm. = t t supply Cooling, = m& = + o Vent., Inf. c People p + + Cooling, (t o Vent., (m& t i El.appl. ) + = Vent., Solar Const. c p ) + + Vent., 20

21 System 2: AC-unit with heat exchanger Heat exchanger t i t o Condenser t m t t m = t Vent.,2 Cooling,2 sup ply o η = m& = t ( t t ) = m o Vent. i c Const. p t supply Cooling,2 ( t t ) = m& c ( η) ( t t ) + Cooling,2 m Vent.,2 (m& i Vent. c Vent. p ) p o i System 3: AC-unit without ventilation t i Condenser t supply Cooling,3 t Vent.,3 Cooling,3 sup ply = 0 = t = m Const. + Cooling,3 Vent.,3 (m& = Vent. c Const. p ) 2

22 System 4: AC-unit with recirculation t i m& Vent t o Condenser ( x) & t t mix sup ply m Vent Vent.,4 = t x m& Vent t mix = ( x) m& = x ti + ( x) to = + Cooling,4 mix Const. Cooling,4 t supply Cooling,4 Vent. c p Vent.,4 (m& ( t t ) o Vent. i c p ) System 5: Hybrid between 2, 3, and 4 Heat exchanger t i m& Vent t o Condenser ( x) & t t m mix t m m Vent =? Vent.,5 =? x & =? m Vent t t mix Cooling,5 sup ply t supply Cooling,5 = t = mix Const. + Cooling,5 Vent.,5 (m& Vent. c p ) 22

23 Heat Exchanger Problem A heat exchanger has a heat transfer area of 0 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 00 C shall heat cold water with a mass flow rate of 0.6 kg/s and inlet temperature of 0 C. At what temperature is the heated water leaving the heat exchanger is the heat exchanger is connected a) Counter-flow b) Parallel-flow Ans: a) 43,7 C b) 38, C.

24 H. Jonsson: Applied Thermodynamics - Collection of Formulas.4 Heat Exchangers Three types of heat exchangers are distinguished Recuperative Regenerative (Ljungström preheater) Evaporative (Cooling towers) ϑ 2 Δ t 2 ϑ Δ 2 2 t t Counterflow 2 t θ = Δ ϑ t 2 2 t Parallel-flow ϑ Δ 2 θ 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 = ( m& cp Δ) = ( m& cp Δ) 2 [.20] &Q= k A ϑ m [.2] ϑ ϑ2 ϑ m = [.22] ln ( ϑ ϑ ) 2 The temperature efficiencies are defined Δ = [.23] θ η θ Cross flow Δ Δ 2 Δ 2 = [.24] θ η 2 Introduce: W& = ( m& c p ) [.25] hence W& 2 = ( m& c p ) 2 [.26] η W& 2 = η = Y W& 2 η [.27] For counterflow heat exchangers it can be shown that where X ( Y) e η = [.28] X ( Y) Y e k A X = & [.29] W 35

25 H. Jonsson: Applied Thermodynamics - Collection of Formulas W& Y = W& [.30] 2 For parallel-flow heat exchangers it can be shown that X (+ Y) e η = + Y To be able to use the diagrams on page 64, set W & < W & 2. [.3] 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, Heat Transfer Through Walls α i λ δ α o &Q ϑ The heat flow is given by &Q = k A ϑ [.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 k δ = + + [.33] α n λ α i n o where α i : heat transfer coefficient on the inside (convection + radiation) α o : 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 (A i and A o are the inside and outside surface areas respectively) k A = α A i i δ + n λ A m n + α o A o [.34] where, for cylindrical walls (r i : inside radius and r o : outside radius) A m 2 π L (ro ri ) = [.35] ln ( r r ) o i for spherical walls (r i : inside radius and r o : outside radius) A = 4 π r r [.36] m o i 36

26 64

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