Notes on Ivantsov’s solution
G. Phanikumar
July 18, 2011
1 Thermal field around a dendrite tip
1.1 Assumptions
• Dendrite grows at steady state velocity v
• Surface tension effects are negligible
• Dendrite grows in a shape perserving manner
• Changes of density upon solidification are negligible
1.2 Paraboloid case
Write the governing equations and boundary conditions in a coordinate system (z, r) fixed at the center of a
circle that fits the tip of the paraboloid and moves at a constant velocity v in the z direction. Let R0 be the
radius of the tip.
Shape of the parabola is given by a single parameter R0 as follows:
R0 r2
z= − = P (r; R0 ) (1)
2 2R0
The governing equation for the solid domain defined by : z ≤ P (r; R0 ) is
∂Ts
= αs ∇2 Ts
−v (2)
∂z
The governing equation for the liquid domain defined by : z ≥ P (r; R0 ) is
∂Tl
−v = αl ∇2 Tl (3)
∂z
Boundary condition at the interface at z = P (r; R0 ) is
Ts = Tl = Tm (4)
Stefan’s condition at the interface at z = P (r; R0 ) is
ks ∇Ts · n̂ − kl ∇Tl · n̂ = ρLf v n̂ (5)
Far field temperature at r2 + z 2 → ∞ is given by:
Tl = T∞ (6)
1.3 Non-dimensionalization
Scale the distances using R0 and the temperature using Tm − T∞ to bring dimensionless numbers out of the
equations above. Assume that the dendrite is isothermal such at ∇Ts = 0.
z
z∗ = (7)
R0
r
r∗ = (8)
R0
Tl − T∞
θl = (9)
Tm − T∞
1
We can rewrite the equations from the previous section as follows:
Shape of the parabola:
1
z∗ = 1 − r∗ 2 = P ∗ (r∗ ) (10)
2
With hindsight, define the following two dimensionless numbers:
vR0
Pe ≡ (11)
2αl
Cpl (Tm − T∞ )
Ste ≡ (12)
Lf
Governing equation in the liquid domain where z ∗ < P ∗ (r∗ ) is
∂θl
−2Pe = ∇ 2 θl (13)
∂z ∗
Interface temperature at z ∗ = P ∗ (r∗ ) is
θl = 1 (14)
Stefan condition at z ∗ = P ∗ (r∗ ) is
2Pe
−∇θl · n̂ = n̂ (15)
Ste
1.4 Paraboloidal coordinate system
Define a paraboloidal coordinate system with ξ, η as below.
r∗ = ξη (16)
1 2
z∗ = ξ − η2
(17)
2
p
ξ 2 = z ∗ + r∗ 2 + z ∗ 2 (18)
p
η 2 = −z ∗ + r∗ 2 + z ∗ 2 (19)
In this coordinate system, the solid-liquid interface is at ξ = 1
The vector normal to the solid-liquid interface will then be ξ.ˆ
Thus the expression in Stefan’s condition can be simplified using:
∂θ
∇θl · n̂ = (20)
∂ξ
The solution (in 2D) is given by:
√ √
πPe
θl (ξ) = · exp (Pe) · erfc ξ Pe (21)
Ste
The solution (in 3D) is given by:
E1 Pe ξ 2
θl = (22)
E1 (Pe)
Where, the exponential integral is defined as
∞
e−s
Z
E1 (x) = ds (23)
x s
∂ e−x dx
E1 (x) = − (24)
∂ξ x dξ