I. Analytical Solutions in Conduction Heat Transfer I. Analytical ...

I. Analytical Solutions in Conduction Heat Transfer
G. Instantaneous and Continuous Sources
1. Application to heat treatment of surfaces with lasers and to
welding
power
input q (W)
melt pool
fusion zone
y
z
x
What is size and shape of
melt pool How is it affected
by power input and velocity
plate velocity u
Assume continuous point source
fixed at origin in a semi-infinite solid
that moves with steady velocity u in
x-direction.
lesson 7
I. Analytical Solutions in Conduction Heat Transfer
1. Application to heat treatment of surfaces with laser and to
welding (cont.)
power
input q (W)
y
fusion zone
melt pool
x
z
plate velocity u
Additional assumptions
•point heat source
•constant thermal properties
•no melting or solidification
•no heat loss from surface of plate
Approach
•instantaneous planar source
•instantaneous point source
•surface treatment of moving
solid
lesson 7
1

I. Analytical Solutions in Conduction Heat Transfer
2. The fundamental solution - one-dimensional, instantaneous,
planar source
a. pulse of thermal energy released uniformly over plane at
x = 0 in infinite solid with initial temperature T = 0.
Let Q = strength of source, K⋅m, where Qρc (=) J/m 2
(I.G.1)
(I.G.2)
We want to find T(x,t) where x ranges from -∞ to +∞.
By conservation of energy,
Q ρc
=
∫ ∞ −∞
ρcT(x,t)
dx
(I.G.3)
lesson 7
I. Analytical Solutions in Conduction Heat Transfer
b. governing equations
2
∂T
∂ T
= α
∂t
2
∂x
at t = 0, T(x,0) = Qδ(x)
at x = ±∞,
T( ±∞,t)
= 0
(I.G.4)
(I.G.5)
(I.G.6)
∞
δ(x)dx
= 1
−∞
∞
f(x) δ(x)dx
= f(x)
−∞
∫
∫
(I.G.7)
(I.G.8)
c. solution (see Lesson 12 of Farlow)
Also known as
Green’s function,
the fundamental
solution, and the
impulse response
function.
lesson 7
2
Q ⎛ x ⎞
T( x,t ) = exp
1 / 2
2( παt
)
⎜ −
4αt
⎟ (I.G.9)
⎝ ⎠
2
x = 2αt
(I.G.10)
r
r
2
2
=
=
x
x
2
2
+
+
y
y
2
2
= 4αt
2
+ z = 6αt
(I.G.11)
(I.G.12)
2

I. Analytical Solutions in Conduction Heat Transfer
3. Three-dimensional, instantaneous, point source
a. statement of problem for point source in infinite solid
Let S = strength of source, K⋅m 3 , where Sρc (=) J
(I.G.13)
We want to find T(x,y,z,t) where x, y, and z range from -∞ to +∞.
By conservation of energy,
S ρc
=
∫ ∞ −∞
ρcT(x,y,z,t)
dxdydz
(I.G.14)
lesson 7
I. Analytical Solutions in Conduction Heat Transfer
3. Three-dimensional, instantaneous, point source (cont.)
b. governing equations
2 2 2
∂T
⎛ T T T ⎞
⎜ ∂ ∂ ∂
= α + + ⎟
∂t
⎜ 2 2 2
x y z ⎟
⎝ ∂ ∂ ∂ ⎠
at t = 0, T(x,y,z,0) = Sδ(x)
δ(y)
δ(z)
at x,y,z = ±∞,
T( ±∞ , ±∞ , ±∞,t)
= 0
(I.G.15)
Then
lesson 7
c. solution
Let T(x,y,z,t)=T x (x,t)T y (y,t)T z (z,t)
2
∂Tx
∂ T
= α x
∂t
2
∂x
1/ 3
and
at t = 0, Tx
(x,0) = S δ(x)
at x = ±∞,
Tx
( ±∞,t)
= 0
(I.G.17)
T(r ,t )
where
(I.G.16)
2
=
πα
exp
3 / 2
⎜
α
r
8(
2
= x
S
t )
2
+ y
2
(I.G.18)
+ z
⎛ r ⎞ −
4 t
⎟
⎝ ⎠
2
3

I. Analytical Solutions in Conduction Heat Transfer
4. Solution to heat treatment or welding problem
a. governing equations for infinite solid
Recall
⎛ ∂T
c V ~ ⎞
ρ ⎜ + ⋅∇T⎟
= ∇ ⋅ +
⎝ ∂t
⎠
( k∇T) u'' '
(I.B.18)
For our problem, (I.B.18) becomes
2 2 2
∂T
⎛ T T T ⎞
⎜ ∂ ∂ ∂ ∂T
= α + + ⎟ − u
∂t
⎜ 2 2 2
x y z ⎟
⎝ ∂ ∂ ∂ ⎠
∂x
at t = 0, T(x,y,z,0) = Sδ(x)
δ(y)
δ(z)
at x,y,z = ±∞,
T( ±∞ , ±∞ , ±∞,t)
= 0
(I.G.19)
(I.G.19) is the diffusion-advection or diffusion-convection equation.
lesson 7
I. Analytical Solutions in Conduction Heat Transfer
x’ = x - ut
y’ = y
z’ = z
t’ = t
lesson 7
4. Solution to heat treatment or welding problem (cont.)
b. solution - define coordinates that move with the solid (x’,
y’, z’, t’) so that the source is always at the origin. Solve
the problem in these coordinates (making use of (I.G.18))
and then transform back to x, y, z, t coordinates. See
Lesson 15 in Farlow.
Hot spot
at t = 0
S
z
x
ut
x’ = x - ut
The solution for instantaneous pulse of strength S in
primed coordinates is (I.G.18):
T(x',y',z',t' )
u
Hot spot
at time t
⎛ 2 2 2
S
⎞
⎜ (x')
+ (y')
+ (z' )
=
exp −
⎟
3 / 2
8( παt'
) ⎜
⎟ (I.G.20)
⎝
4αt'
⎠
4

I. Analytical Solutions in Conduction Heat Transfer
b. solution (cont.)
In stationary coordinates (I.G.20) becomes
⎛ 2 2 2
S
⎞
⎜ (x − ut) + (y) + (z)
T(x,y,z,t) = exp −
⎟
3 / 2
8( παt)
⎜
⎟
⎝
4αt
⎠
c. solution with continuous source
Suppose S is zero until time t > w, then it has value S.
The new temperature field becomes
⎧0,
t ≤ w
φ(x,y,z,t)
= ⎨
⎩T(x,y,z,t
− w), t > w
(I.G.21)
∴if ρcS = qdt, q = constant (W), and q is switched on at t = w,
and left on, temperature at time t is (see Lesson 10)
(I.G.22)
t
2 2 2
q ⎛ [x u(t w)] (y) (z) ⎞ dw
T(x,y,z,t)
exp⎜
− − + +
= ⎟
3 / 2 ∫ −
0
4 (t w)
3 / 2
lesson 7 8ρc(
πα)
⎜
⎟
⎝
α −
⎠ (t − w)
I. Analytical Solutions in Conduction Heat Transfer
c. solution with continuous source (cont.)
Now let t→∞ in (I.G.22) to give
T (x,y,z)
= q ⎡ u
exp⎢−
4πkr
⎣ 2α
x
⎤
( r − ) ⎥⎦
(I.G.23)
d. now consider heat treatment on surface of semi-infinite
solid and neglect all heat losses from surface. (I.G.23)
becomes
T (x,y,z)
= q ⎡ u
exp⎢−
2πkr
⎣ 2α
x
⎤
( r − ) ⎥⎦
(I.G.24)
lesson 7
5

I. Analytical Solutions in Conduction Heat Transfer
d. heat treatment on surface of semi-infinite solid (cont.)
= q ⎡ u
T(x,y,z) exp⎢−
2πkr
⎣ 2α
x
(I.G.24)
⎤
( r − ) ⎥⎦
Pe = ua/α (Peclet no., a
dimensionless velocity)
Pe = 0, 2,10, 50
a = q/(2 k π T M )
T M = melting point of solid
lesson 7
I. Analytical Solutions in Conduction Heat Transfer
d. heat treatment on surface of semi-infinite solid (cont.)
T (x,y,z)
= q ⎡ u
exp⎢−
2πkr
⎣ 2α
x
⎤
( r − ) ⎥⎦
(I.G.24)
Typical laser treatment conditions:
12-by-12-mm uniform square beam
5.7 kW
u = 38 mm/s
Consider T(x,0,0) = 850 C (Actual T = 850 + T 0 )
lesson 7
q
T(x,0,0) =
2πkr
q 5700 W
x = =
= 21 mm
2πkT
2π(50
W /(m K)850 C
6