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.
4.Conduction Heat Transfer
Fourier Equation for Conduction
Heat Transfer
• Conduction is one of the heat transfer
modes. Concerning thermal design of
electronic packages conduction is a very
important factor in electronics cooling
specially conduction in PCB’s and chip
packages.
• The basic law governing the heat
transfer by conduction is Fourier’s law
dT
q x   k
dx
Energy Equation
As a system, energy balance may be
applied on any electronic component. A
typical energy balance on a control
volume can be described as shown in the
following equation. And the amount of
energy flowing into or out of the system
can be described by the Fourier’s law.




E in  E g  E out  E stored
Energy Equation in Cartesian
Coordinates
  T
k
x  x
   T
  k
 y  y
   T
   k
 z  z
T
 
  q   CP
t

Energy Equation in Cartesian
Coordinates
• For a constant thermal conductivity the
heat equation could be rewritten as

 T  T  T q 1 T


 
2
2
2
k  t
x
y
z
2
2
2
Where α = k/ ρ Cp is the thermal diffusivity.
Energy Equation in Cylindrical
Coordinates
1   T
kr

r r  r
 1   T    T
   k
k
 2

 r     z  z
T
 

q


C

P
t

Energy Equation in Spherical
Coordinates
1   2 T
kr
2

r r 
r
1
  T 
1
 
T  
T



k

k
sin


q


C




P
2
2
 r 2 sin   

r
sin






z
t





Special cases of one dimensional
conduction
• Boundary conditions
- Some of the boundary conditions usually met in
heat transfer problems for one dimensional system
are described below. These conditions are set at
the surface x = 0, assuming transfer process in
the positive direction of x- axis with temperature
distribution which may be time dependent,
designated as T(x, t)
• One dimensional steady state
conduction without heat generation
• One dimensional steady state
conduction with uniform heat
generation
Boundary conditions
• Constant surface
temperature
- The surface is maintained at a
fixed temperature Ts. It is
commonly called a Dirichlet
condition, which is the boundary
condition of the first case. It can
be approximated as a surface in
contact with a solid or a liquid in a
changing phase state (boiling,
evaporating, melting or freezing)
therefore the temperature,
accompanied with heat transfer
process, is constant.
T (0, t )  Ts
Boundary conditions
• Constant surface heat flux
Finite heat flux
k
T
x
In this case fixed or constant heat flux q" at the surface
is described. At which the heat flux is a function of the
temperature gradient at the surface by Fourier's law.
This type of boundary condition is called Neumann
condition. Examples of constant surface heat flux are:
Adiabatic heat flux
 q s
x 0
T
x
 0s
x 0
Convection surface
condition
k
T
x
 h[T  T (0, t )]
x 0
One dimensional steady state
conduction without heat generation
• The assumptions made for this kind of
analysis are:
-
One dimensional
Steady state
No heat generation
Constant material properties
One dimensional steady state conduction
in Cartesian coordinates without heat
generation
Ts , 2  Ts ,1
q x  k A(
)
L
x
T ( x)  (Ts , 2  Ts ,1 )  Ts ,1
L
Example 4.1
• Calculate the maximum temperature the transistor base
attains if it dissipates 7.5 W through the bracket shown
in the Figure below. All dimensions are in mm and the
bracket is made of duralumin.
Solution:
Given
Dimensions on the figure.
q = 7.5 W
tw = 50 ºC
k = 164 W/m.k
As the dimensions of the bracket is very small we can consider that heat transfer
is a one dimensional conduction through the sides of the bracket. For the
dimensions shown on the figure
L = 15 +15 + 15 = 45 mm = 0.045 m
w = 20 mm = 0.02 m
δ = 5 mm = 0.005 m
A = w x δ = 1 x 10 -4 m2
From the above values the only unknown in equation 4.23 is the temperature of the
transistor base.
q = k A (tb - tw)/L
tb = tw + (qL/kA)
tb = 50 + (7.5 x 0.045/164 x 10-4)
tb = 70.58 ºC
One dimensional steady state conduction in
cylindrical coordinates without heat generation
Considering the assumptions stated before,
dT/dr is constant and heat flow only in one
spatial coordinates the general form becomes,
2kL (T1  T2 )
q
ln( r2 / r1 )
Where subscripts 1 and 2 refer to the inner and
the outer surfaces respectively.
Example
A hollow stainless (25 % Cr , 20 % Ni ) steel cylinder 35 mm long has
an inner diameter of 50 mm and outer diameter of 105 mm. a group
of resistors that generate 10 W is to be mounted on the inside
surface of the cylinder as shown in figure.
If the resistors temperature is not to exceed 100 º C find the
maximum allowed temperature on the outer surface of the cylinder.
Solution
• the thermal conductivity of 25 % Cr , 20 % Ni stainless steel is 12.8
W/m.K considering the one-D conduction equation, The only
unknown is the temperature of the outside surface of the cylinder.
q (ln( ro / ri ))
To  Ti 
2kL
To= 100 – 10 × (ln (52.5/25)/ (2π × 12.8 × 0.035)
To = 100 – 2.64 = 97.36 º C
One dimensional steady state
conduction in spherical coordinates
without heat generation
• The same assumptions are applied to
the spherical system giving the
following solution
(Ti  To )
qr 
(1 / ro  1 / ri )
4k
• Where the subscripts i and o refer to
the inside and the outside surfaces
respectively.
One dimensional steady state conduction
with uniform heat generation
• The assumptions made for this kind of
analysis are:
-
One dimensional
Steady state
Uniform heat generation
Constant material properties
One dimensional steady state
conduction in Cartesian coordinates
with uniform heat generation
k
d 2T
dx
T 
q 2
T   ( L  x 2 )  Ts
2k
2
 q   0
q  2
x  C1 x  C 2
2k
LL c q 2 2
T  Tc  q
 (L  x )
kc
2k
q =q''' ×A × L = Ts – Tc/(Lc/kc Ac)
One dimensional steady state
conduction in cylindrical coordinates
with uniform heat generation
1   T
 k r
r r  r
T = Ts + q''' / 4k( ro2 – r2 )
For a convective boundaries
T = T∞ + q''' / 4k( ro2 – r2 ) + q''' ro/ 2h

  q   0

q 2 2
(ro  r )
4k
 ln(r / r )
1  2
 q  c o 
 ro L
2
k
2
hr
L
c
c 

T  T 
Extended surfaces (Fins)
From the rate equations, it is clear that enhancing
heat transfer could be done by several methods:
- Increasing the temperature difference
- Increasing the heat transfer coefficient
- Increasing the surface area A ,
Fins are used to add a secondary surface to the
primary surface and thus increasing the heat transfer
area. In electronic equipment cooling straight
rectangular fins are mostly used and are done of good
conducting material to attain the root temperature
through the fins in order to increase the heat
transferred. Fins used in electronics cooling are usually
used of aluminum and are quite thin about 1.3 to 1.5
mm thick.
Extended surfaces (Fins)
In electronics cooling fins are
usually considered to have an
insulated tip.
The heat transferred by fins are
expressed in its effectiveness
which is defined as
ηf = qf / qmax
where qf is the heat actually
transferred by the fin
and qmax is the maximum heat
that could be dissipated by the
fin (i.e. when the fin has a
uniform temperature equals to
the root temperature tr .
Extended surfaces (Fins)
ηf = (tanh ml )/ml
Where m = 2h( L  b) / k (b L)
and l and b are defined on the fin sketch.
Also the fin effectiveness could be considered as the fraction of
the total surface area of the fin Af that is effective for the heat
transfer by convection maintained at root temperature tr.
ηf = Af, eff/Af,tot
q = h [( Atot – Af) + ηf Af] (tr- ta)
= ηo Atot h (tr- ta)
Where ηo = 1 – (Af/Atot) (1- ηf )
If the fin is of convective tip a correction could be done as
lc= l + (t/2)
Fin Geometries
Fin descriptions
Factors affected on the fin selection
Heat transfer/pressure drop
Factors affected on the fin selection
Size figure of merit
Factors affected on the fin selection
Weight figure of merit.
Factors affected on the fin selection
A relative comparison of the fin configurations, based on all the
factors discussed is critical in determining the proper design.
Comparison of All Parameters
Fin Configuration
P
Size
Weight
Cost
Average
Straight
1
5
4
2
3
Offset
4
3
3
4
3.5
Pin
5
1
5
1
3
Wavy
3
4
2
3
3
Louver
2
2
1
5
2.5
Example
• In order to enhance the heat transfer in logic chips a
heat sink is attached to the chip surface in order to
increase the surface area available for convection, the
heat sink consists of an array of square fins of width w
and the sum of the fin spacing and its width is the fin
pitch S, the fins are attached to the chip causing a
contact resistance R''t,c .
• Consider the chip width = 16 mm, cooling is provided by
a dielectric liquid with T∞= 25 °C and h =1500 W/m2. k
and the heat sink is fabricated from copper (k= 400
W/m .k) and its characteristic dimension are w = 0.25
mm and S = 0.50 mm and Lf = 6 mm , and Lb = 3 mm