Transcript Electronics Cooling MEP 635

Electronics Cooling
MPE 635
Mechanical Power Engineering
Dept.
Course Goals
1. To establish fundamental understanding of
heat transfer in electronic equipment.
2. To select a suitable cooling processes for
electronic components and systems.
3. To increase the capabilities of post-graduate
students in design and analysis of cooling of
electronic packages.
4. To analysis the thermal failure for electronic
components and define the solution.
• Part-A
• Main topics
• Introduction to electronics cooling and thermal packaging
• Introduction to basic modes of heat transfer
• Conduction heat transfer and extended surfaces in electronic
devices
• Transient conduction
• Natural convection heat transfer (i.e. PCB cooling)
• Forced convection heat transfer (Internal and External flow )
• Fan performance
• Radiation heat transfer and its applications in electronic devices
• Solving the electronics cooling problems with EES software
• Electronics cooling problems
• Solution of selected electronics cooling problems
3. Basics of Heat Transfer
Modes of heat transfer
Conduction
• Conduction heat
transfer as diffusion
of energy due to
molecular activity.
• Conduction in liquids
and solids ascribed to
molecules vibration
(solids), translational
and rotational (liquids)
Conduction
• Fourier’s law
dT
q x   k
dx
T2  T1
qx   k
L
Thermal convection
• The heat transfer by convection is described by the
Newton's law of cooling:
q  hA(TW  T )
Air movement due to temperature difference
Forced fan
Air
(b)Forced convection on electric components chips
(a)Free convection on electric components chips
Thermal convection
• convection heat transfer ranges
Process
Free convection
- gases
- liquids
Forced convection
- gases
- liquids
Convection with two phase
- boiling or condensation
h(w/m2.k)
2-25
50-1000
25-250
50-20,000
2500-100,000
Thermal convection
•
Example 3.1: An electric current is passed through a wire 1mm
diameter and 10 cm long. This wire is submerged in liquid water at
atmospheric pressure, and the current is increased until the water
boils. For this situation h = 5000 W/m2.oC. And the water will be 100
oC. How much electric power must be supplied to the wire to maintain
the wire surface at 114 oC?
Schematic:
Water
Electric wire
Thermal convection
• Solution:
• The total convection loss from the wire is given
by.
q  hA(TW  T )
• For this problem the surface area of the wire is
A= π d L = π (1 x 10-3) (10 x 10-2) = 3.142 x10-4 m2
• The heat transfer is therefore
4
q  5000  3.142 10  (114  100)  21.99W
• And this is equal to the electric power which
must be applied.
Thermal radiation
• The mechanism of heat transfer by radiation
depends on the transfer of energy between
surfaces by electromagnetic waves in the wave
length interval between 0.1 to 100 μm.
• Radiation heat transfer can travel in vacuum
such as solar energy.
• Radiation heat transfer depends on the surface
properties such as colors, surface orientation
and fourth power of the absolute temperature
(T4) of the surface.
• The basic equation for radiation heat transfer
between two gray surfaces is given by
q   fA(T  T )
4
1
4
2
Thermal radiation
• Example 3.2. A horizontal steel pipe
having a diameter of 10 cm is
maintained at a temperature of 60 oC in
a large room where the air and wall
temperature are at 20 oC with average
heat transfer coefficient 6.5 W/m2K.
The emissivity of the steel is 0.6
calculate the total heat lost from the
pipe per unit length.
Thermal radiation
• Solution:
• The total heat lost from the pipe due to
convection and radiation
qtotal  qconvection  q radiation
 hA(TS  T )   fA(T  T )
4
S
4

• Because the pipe in a large enclosure then
the geometrical factor ƒ = 1
qtotal  6.5 ( x0.1)(60  20)  5.67 x10 8 (0.6)(1)( x0.1)(3334  2934 )
 134.33 W / m
Analogy between Heat Transfer
and Electric Circuits
• There exists an analogy between the
diffusion of heat and electrical charge.
Just as an electrical resistance is
associated with the conduction of
electricity, a thermal resistance may be
associated with the conduction of heat.
Analogy between Heat Transfer
and Electric Circuits
T
 qcond  kA
 qcond  Rt ,cond T
L
L
 Rt ,cond 
kA
 qconv  hAT  qconv  Rt ,convT
 Rt ,conv
1

ha
E2  E 1
L
Re 

I
A
Series Circuits:
E
E1  E2
i 

Re Re,1   Re, 2   Re,3 
• By analogy
q 
Toverall
Rt
T1  T 2
q 
Rt ,conv   Rt ,cond   Rt ,conv 
q 
T1  T 2
 1   L   1 

  

  
 h1 A   kA   h2 A 
Parallel Circuit:
L1, k1, A1
T T
qi  ki Ai

Li Rt ,i
qtot  qi
T1
T2
L3, k3, A3
qtot
L4, k4, A4
qtot
L5, k5, A5
 1

1
1
1
1
1
1

 qtot  T  






R
R
R
R
R
R
R
t
,
1
t
,
2
t
,
3
t
,
4
t
,
5
t
,
6
t
,
7


T
 qtot 
Rt ,tot
1
1

Rt ,tot
Rt ,i
L2, k2, A2
qtot
L6, k6, A6
L7, k7, A7
q1
q2
q2
R2
q3
R3
q3
q4
R4
q4
q5
R5
q5
q6
R6
q6
q7
R7
q7
qtot
Combined Modes of Heat Transfer
•
Combined Convection and Radiation
qnet  qconv  qrad
q rad  hr  A  (Ts  T f )
q rad
(Ts4  Te4 )
hr 
     Fse 
A  (Ts  T f )
(Ts  T f )
(Ts4  T f4 )
hr      Fse 
(Ts  T f )
 hr      Fse 
(Ts2  T f2 )  (Ts  T f )  (Ts  T f )
(Ts  T f )
Combined Modes of Heat Transfer
•
Combined Convection and Radiation
 hr      Fse  (Ts2  T f2 )  (Ts  T f )
 hr      Fse  ((Ts  T f ) 2  2TsT f )  (Ts  T f )
Now if we define the arithmetic mean temperature as:
Tm 
Ts  T f
If further Ts-Te<<Ts then
2
Tm  Ts T f
So we may define the radiation heat transfer coefficient as:
 hr  4      Fse  Tm3
And finally;
htot  hconv  hrad
Where
q net  htot  A  (Ts  T f )
Combined Modes of Heat Transfer
•
Combined Convection and Conduction
•
This combination is likely to occur with the use of
extended surfaces where the primary surface exchanges
heat by convection to the adjacent fluid flow and by
conduction through the extended surfaces. This case
may be considered in a similar manner as the above, but
here the problem doesn't need extra work as the
conduction thermal resistance is predefined.
qnet  qconv  qcond
q net
(Ts  Text) 

  hconv(Ts  Tf )  k
 A
L


Overall Heat Transfer Coefficient
U h Ah  U c Ac 
1
R"f ,c
R"f ,h
1
x
1

 

hc ( Ac, p   f ,c Ac,s ) ( Ac, p   f ,c Ac,s ) kAm ( Ah, p   f ,h Ah,s ) hh ( Ah, p   f ,h Ah,s )
Fluid combination
U, W/m2.ºK.
Water to water
850-1700
Water to oil
110-350
Steam condenser, water in tube
1000-6000
Ammonia condenser, water in tube
800-1400
Finned tube heat exchanger, water
in tubes air in cross flow
25-50