Temperature Change for Uniform Illumination

Transcription

1 Temperature Change for Uniform Illumination Assume that the surface is uniformly illuminated by the laser Energy absorbed at the surface in a ery small depth H=I(1-R) where R = reflectiity I = light intensity The heat DE has been soled for depth z and time t (by Carslaw & Jaeger, 1959) H z T( z,t ) = α t ierfc α t The ierfc is the integral of the complementary error function

2 Error Function Related Equations Heat flow equations are related to the Error Function erf erf ( x ) = π x e s ds This is the integral of a Gaussian between 0 and x The Complementary Error Function erfc ( x) = 1 erf ( x) erfc erfc is the integral from x to infinity The ierfc is related to the error function by x 1 ierfc x = erfc s ds = exp x π ierfc(1) = 0.05 and is falling rapidly ( ) ( ) ( ) x 1 erf ( x) 0 [ ]

3 Useful Error Function Approximations Error function erf(x), Complementry Error Function erfc(x) are erf ( x ) = π x e s ds s erfc( x ) = 1 erf ( x ) = e ds π erf(x) hard to find but easy to approximate with erf ( x) = ( a t + a t + a t ) 1 t = where 1+ px p 1 x e x = a 1 = , a = , a 3 = See Abramowitz & Segun (Handboo of Mathematical Functions) Error on this is <.5x10-5 for all x We are using complementary error function erfc(x) = 1 - erf(x) erfc(0) = 1 erfc( ) = 0 Approximation has <% error for x << 5.5 For x > 5.5 use asymptotic approximation e 1 erfc( x ) 1 as x x π x Excel & Quatropro spreadsheet hae erf() and erfc() built in. Must actiate analysis toolpac & soler first but become inaccurate for x>5.4 then use asymptotic For x > 5.4 then ierfc(x) becomes x e ierfc( x ) x x π as x

4 Temperature Rise for Uniform Illumination From DE solution since ierfc is small for x>1 Thus find that T rise is small when z α t > 1 Hence small rise when z > 4α t Heat will diffuse a depth L in time of order L t = 4α Change in surface temperature with time substitute z = 0 and note ( 0) ierfc = 1 π Thus surface temperature change is: H T (0,t ) = α t ierfc[] 0 = Thus temp increases with α t H α t π

5 Temperature Change with Finite Time Laser Pulse If hae a square pulse of duration t p The for t < t p follow the preious formula For Time greater than the pulse ( z,t) = δ T ( z,t) T ( z,t t ) T > δ t t p eg. Consider Cooper with H = W/m for t p = 10-6 sec From table α = 1.16x10-4 m /s T rises highest at surface (z=0) and changes fastest At pulse end heat has diffused about L p L 4α t 4(1.16 x10 )(10 ) =.15x At depth pea T occurs much later, and lower alues m

6 Laser Focused into a Spot If laser focused into uniform spot radius a then formula changes to (by Carslaw & Jaeger, 1959) T( z,t ) = H α t ierfc z α t Term on right caused by sideways diffusion At the centre of the spot (z = 0 ) T( z,t ) = H α t This gies same as uniform heating if z + a ierfc α t 1 ierfc π a α t a ierfc << 1 t α This is true for ierfc(>1) thus a t < 4α eg for Copper with a = 1 mm and α = 1.16x10-4 m /s, t < 4 3 ( 10 ) 4 ( 1.16 x10 ) or t <.16 x10 3 s

7 Laser Focused into a Spot As t goes to infinity (ery long times) it can be shown T H ( ) [ ] z,t = z + a z Thus for finite spot temperature reaches a limit Highest surface temp ( 0, ) T = Ha Effects of beam Gaussian distribution is not that different In practice as thermal conductiity, reflectance R, thermal diffusiity α all ary with temperature Thus tend to use numerical simulations for real details

8 Example Focused Laser Spot Calculation

9 Phase Changes and Energy Balance Energy Balance: Energy in = Energy to raise temp + heat flow Note: a rough rule of thumb if near steady state half the energy goes into heat flow so energy required is twice that to raise temperature As heating increase will get melting of the surface Eentually also get aporization point All requires energy to heat In general the specific heat of the material changes C s = specific heat of solid C l = specific heat of liquid phase L f = Latent Heat of Fusion: energy for melting L = Latent Heat of Vaporization: energy to aporize Energy required to melt a unite olume of material m [ C ( T T ) L ] E = ρ + where T m is the melting point, T the starting temp. ρ = density of material Note: this does not include energy lost to heat flow s m f

10 Phase Changes and Energy Balance When aporization occurs [ C ( T T ) + C ( T T ) + L L ] E = ρ + s m Generally true that heat capacity does not change much with T C m C C Generally Latent heat of apourization >heat of fusion L < f L Vapourization temperature is much < base or melting T<T m <T Energy input required is approximately ( CT L ) E ρ + m f

11 Melting Depths Consider light pulse on surface Will get melting to some depth Eentually also get surface at aporization point This is the laser welding situation Can estimate the depth of melt front after some time t Recall temperature distribution T( z,t ) = H α t ierfc z α t Ratio of the Temperatures changes with depth are t T( z,t ) z = π ierfc T(0,t ) α t Eentually surface raises to aporization point It cannot rise higher without apourization thus stays at T Hence can calculate the melt pool depth with this.

12 Melt Depth Estimate Estimate the depth of melt front for that after some time t Bottom of melt is at melting point, Top at apourization point and assume base temperature is near 0 o C (ie ~room temp) ( z,t) = Tm T ( 0,t) T T = T T Recall that at the surface m = π ierfc T (0,t ) = H z α t α t π Thus time can be eliminate by soling for Depth of melt is gien by T π = H α t zmh T ierfc = T π T Note: for a gien material Hz m is fixed Thus large welding depths gien by low heat intensities applied for long time proided that there is sufficient energy in the beam m π

13 Example of Melting Calculations What is the heat flow required for weld depth of 0.1mm in copper From the table for copper T m = 1060 o C T = 570 o C K = 400 W/m o C Thus z H ierfc T π From the graph or calculation Thus T π T = T π π m m = = ierfc(x=0.44) = ( 570) π (0.44 ) 9 H = = = zm x10 W / m

14 Vaporization of Material When material remoed by aporization Get a melt front and a heated front liquid front moes with elocity s From the heat balance, assume that all power goes into heating Then the melt front should be s ( CT L ) H ν ρ + where H is power density per square area Note this is the minimum power alue Good rule of thumb is actual power required twice this loss about the same by heat flow to substrate Can calculate depth d of holes by nowing laser pulse duration t p and front elocity d = ν t ν s p

15 Example Depth of Hole with Vaporization Heat pulse of H = W/m and t = 500 microsec hits copper. What will be the resulting max hole depth From the tables T = 570 o C ρ= 8960 g/m 3 C = 385 J/g o C L = 4.75x10 6 J/g d ν = ν t s p = ρ = 0.95x10 3 Ht ( CT + L ) p = m = 0.95mm ( 5x10 ) 6 ( 385[ 570] x10 )

16 Keyholes and Increased Welding/Cutting Depths When the laser forms hole in material Beam penetrates to much greater depth Creates a large deep melt pool behind moing beam Melt fills in hole behind moing beam If not true welding limited 1 mm in steel

17 Keyholes Formulas Modeled by Swift Hoo and Gic, 1973 Assume linear heat source power P (W) Note P is total power while H is unreflected power because eyhole absorbs all the power (reflections do no escape) Extends into metal depth a Moing forward with elocity (weld speed) in direction y direction across weld is x (centred on the heating point) Temperature distribution becomes P x α x + y T = exp K 0 π a α where K 0 is the Bessel function nd ind order 0 Width w of the weld is gien by the point where T=melting α P w at m

18 Laser Machining Processes Laser heat processing diided into 3 regions Heating Melting Vaporization

19 Laser Surface Treatment Annealing or Transformation Hardening surface hardness Surface Melting homogenization, recrystallization Alloying changing surface composition improes corrosion, wear or cosmetic properties Cladding Applying a different material to surface improes corrosion, wear or cosmetic properties Texturing changing surface appearance Plating By Chemical Vapor Deposition

20 Laser Annealing Uses the rapid, local, high temperature, heating and cooling (quenching) Materials where heating with quenching changes characteristics Best examples: Iron/Steel With laser can mae local changes in material parameters Increase hardness, strength Temper (mae more ductile)

21 Comparison of Material Heating Processes