7/31/2019 Laser Processing Of Materials - Teil 6 Von 8 - Temperature Distributions - Skript - Prof Dr Frank Mcklich - Dr Andr
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Functional MaterialsFunctional Materials Saarland UniversitySaarland University
Laser processing of materials
Prof. Dr. Frank Mcklich
Dr. Andrs Lasagni
Lehrstuhl fr FunktionswerkstoffeSommersemester 2007
Temperature distributions
Functional MaterialsFunctional Materials Saarland UniversitySaarland University
Contents:
1. Definitions
2. Diffusion length
3. Temperature distribution in Bulk materials
i. Heat diffusion equation
ii. Constrain conditions
iii. The error functioniv. General solution
v. Formulae for different conditions
vi. Examples of laser heating
4. Temperature distribution in Thin Films
i. Heat diffusion equation
ii. Constrain conditions
iii. Formulae for different conditions
iv. Cooling of thin-filmsv. Heat transfer to Substrate
vi. Lateral Heat-transfer in the film
Temperature distributions
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Temperature distributions in bulk materials
LASER BEAM
txhp heat
heat
: light penetration depth
xhp: heat affected zone
t (diffusion length)
k: thermal diffusivity
t: interaction time
The temperature rise is basically
controlled by , xhp and r0.
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Temperature distributions in bulk materials
EXAMPLE
light penetration depth ()for metals:
10-6 - 10-7 mts
Thermal diffusivity (k)
for metals:10-4 - 10-5 m/s
For t = 10 ns
Xhp ~ 0.1-1m
For t = 1ms
Xhp ~ 100-300m
LASER TYPE:
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Temperature distributions in bulk materials
How to calculate the temperature
as function of the time and depth?
t
T
cqx
T
Kx p
=+
.
q
T (,y,z,t) = T0
Temporal condition:
T (x,y,z,0) = T0
0),,,0( =
tzyx
TK
(no heat lost (no radiation))
x
q=0 q=0
xeRqxq
= )1()( 0.
=1/: absorption coefficient
R: reflectivity;
K: thermal conductivity
q0: Power density (laser-light
flux density [W/cm])
cp: specific heat
Functional MaterialsFunctional Materials Saarland UniversitySaarland University
Temperature distributions in bulk materials
In one-dimensional case, for uniform surface irradiation, and
constant K, cp, :
++
+
+
+=
kt
xkterfcxkt
K
qR
kt
xkterfcxktKqR
eK
qR
kt
xierfckt
K
qRTtxT
x
2)exp(
2
)1(
2)exp(
2)1(
)1(
2
)1(2),(
20
20
000
Where: erfc(u) is the complementary error function and ierfc(u)
its integral
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Temperature distributions in bulk materials
Complementary error function:
Integral of the complementary
error function:
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Temperature distributions in bulk materials
However, different simplified equations can be used in different
situations (, xhp and r0): (q = q0 (1-R))
metalspolymers
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Temperature distributions in bulk materials
Functional MaterialsFunctional Materials Saarland UniversitySaarland University
Temperature distributions in bulk materials
Example: Laser heating of stainless steel AISI 304:
t = tp/2
t = tp
Melting point
Melting point
Temperature at different intensities before laser is turned off
tp = 10-3 s
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Temperature distributions in bulk materials
Example: Laser heating of stainless steel AISI 304:
t = 2 tp
t = 10 tp
Melting point
Melting point
Temperature at different intensities after laser is turned off
tp = 10-3 s
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Temperature distributions in bulk materials
Example: Laser heating of stainless steel AISI 304:
tp = 10-3 s
Melting point
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Temperature distributions in thin films
q0
0),,,0(/ = tzyxTK
x
=1/: absorption coefficient
for h>=> we can neglect the wavereflected from the film substrate
In case of thicker films it is
necessary to consider that heat
release is not uniform in depth
h
However, heat release does not
follow light-absorption law given that
the diffusion length >> (up to h ~ 5m)
(1)
(2)
xeAqxq 1)()( 10
.=
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Temperature distributions in thin films
t
Tctxq
x
TK p
=+
2,1
2,12,1
.
2,12
2,12
),(
h
Aqq 10
.
1 =
)](exp[)1( 21102
.
2 hxARqq =
(this means that
T1=C along x)
Considering that r0>>(1)1/2 =>
q0
0),,,0(/ = tzyxTK
x
h (1)
(2)(1-R1-A1) = D1 (x = h)
xeAqxq 1)()( 10
.=
Observation: R1+A1+D1 = 1
T1(x)=Cte
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Temperature distributions in thin films
Considering:
00/)(1 == xatxtT
0)()( 021 === tatTxTxT
hxattTtT == )()( 21
hxatx
tTk
x
tTk =
=
)()( 22
11
== xatTtT 02 )(
q0
0),,,0(/ = tzyxTK
x
h (1)
(2)
xeAqxq 1)()( 10
.=
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Temperature distributions in thin films
D1 = 0 (Transmission)
D1
> 0 ; A1
>0
(A1+D1 = 1 R1)
D2 = 0
R1+A1+D1 = 1
D1 > 0 ; A1>0
(A1+D1 = 1 R1)
D2 > 0
D1 > 0 ; A1~0 (D1 = 1 R1)
D2 = 0
D1 > 0 ; A1~0 (D1 = 1 R1)
D2 > 0
= laser pulse duration Thin-film temperature-time dependence
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Temperature distributions in thin films
q0 = 1012 W/m
Film thickness = 5m
Substrate: Glass
Numerical calculations
Films temperature Films temperature
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Temperature distributions in thin films
Cooling of thin-films
Cooling of the film depends on the heating regime
For the case of an opaque film on any substrate, the temperature at the film
after the laser interaction time () is given by:
nstk
herfTtT 100
)()()(
2
1 =
hc
AqT
p11
101 )(
=
2
2
102
2)(
k
K
AqT =
Example:
considering: = 10ns, h=100nm, k2 = 6E-3 cm/s (glass),
the time tto cool the film up to 0.1T1 is 100 ns!
Question: why are such cooling-times extremely short?
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Temperature distributions in thin films
Heat transfer to Substrate
1 - During laser irradiation, a more or less significant part of the thermal
energy is drained through the substrate2 - Only a part of the substrate with depth lp is heated up:
lp ~ (2 )1/2
3 - The energy efficiency of the treatment can be described as:
= 1 Qd/QaQa = energy absorbed
Qd = energy dissipated
4- The energy efficiency can be written in terms of thermal properties of
the substrate and the film:
with:2
1+= tkc
hc
p
p
222
11
=
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Temperature distributions in thin films
Heat transfer to Substrate
Qd = 0 Qd = Qa
With the rise of (lower), thefilm-heating efficiency drops
quickly!
If:2
22
22
11
4 k
h
c
c
p
p
then >
Consequently: laser thin-film
treatment should be carried out in a
pulsed regime at short times. This
provides lower energy loss and lower
risk of substrate damage!
E.g.: 100 nm Cu-film on quartz
substrate : < 36 ns
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Lateral Heat-transfer in the film
Temperature distributions in thin films
2
0r
hr02h
Generally, the lateral heat flow weakly
affects the film heating due to:
2
0r hr02>
q
q
From 3D analysis it can be proved that
if:
Then the lateral heating is negligible =>
10 2 kr >
Strong lateral
heat flow
T0c: temperature at the center
without considering lateral heat flow
Trc: temperature at the centerconsidering lateral heat flow
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Temperature distributions in thin films
Lateral Heat-transfer in the film
Example:
Cu-Film
0.00
0.50
1.00
1.50
2.00
2.50
3.00
3.50
4.00
4.50
5.00
0.00 10.00 20.00 30.00 40.00 50.00
][2 1 mk
Laser interaction time [ns]
10 2 kr >
10 2 kr liquid expulsion, vaporization
Liquid expulsion
Vapor/plasma
plume
Vapor consist on: clusters, molecules,
atoms, ions, and electrons
The higher the laser-light intensity the
higher the density of species
The energy required to remove an atom
from a solid can be estimated from:
Ha [J/atom] = HV[J/g] / Ns
HV: enthalpy of vaporization
Ns=L/M: atom number density(L=Avogadro Number; M=atomic weight)
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Plume
P
PS
PL
Laser ablation - vaporization
However the plume will absorb and scatter the incident laser radiation!
Considering only evaporation:
( )( )[ ]Lsinput PRPPE = 1Energy input:P: laser power
Ps: power absorbed by the vapor plume
PL: energy looses (heat conduction, radiation, convection, reaction
enthalpies, etc.)
R: reflectivity
: dwell time of the laser beam
( ) ( )mvplmpsmvvap TTcTTcLLhAE +++= 0Energy to required to vaporize a volume A.h:
Lv: latent heat of vaporization; Lm: latent heat of melting
Cps, Cpl: specific heat of solid and liquid, respectively
Tm: melting point; Tv: vaporization point; T0: initial temperature
: density
h: ablated depth; A: ablated area
h
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Plume
P
PS
PL
When plasma is produced, an important part of the energy is absorbed
by the plume and the
calculation is more
complicated
Laser ablation - vaporization
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( )( )( ) ( )[ ]
mvplmpsmv
ls
TTcTTcLLA
PRPPh
+++
0
1
Laser ablation - vaporization
Combining both equations:
( )( )
v
ls
H
Rh
1
: laser fluence (J/cm)
This equation is only valid if: l(optical penetration depth) bulk heating is minimized => PL ~ 0
( )( )
vH
Rh
1 Ablation rate is proportional to Laser fluence(relative low energy densities, without considering
liquid expulsion)
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Laser ablation - vaporization
One-dimensional model (surface temperature ):
Considerations:
We ignore any liquid layer
Tvs is the temperature at the solid-vapor interface and vvs is its velocity Any attenuation of incident laser light by the plume is ignored
( )
+
+= mv
vsp
s LLv
PR
cTT
)1(
2
1)1( 0
=
s
vvs
Tvv
exp)2( 0
we need to solve equations (1) and (2) simultaneously
B
a
v
B
vv
k
H
k
E
=
vo is in the order of the sound velocity within the solid
is the activation energy for vaporization and can be
replaced by the enthalpy of vaporization per atom/moleculev
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Laser ablation
Influence of liquid layer:
Vapor
Recoil force
Liquid
expulsion
( )ssatrec Tpp ~
The recoil pressure (prec) is originated
because of the momentum conservation
of the evaporated species. This is in the
order of the saturated vapor pressure at
Ts (psat) and increases nonlinearly with
P (W/cm)
The melt-ejection flux (J)
is given by:
(w = laser-beam radius)
4/11~ recm p
wJ