1. CONDUCTION WITH PHASE CHANGE: MOVING BOUNDARY PROBLEMS
CHAPTER 6
CONDUCTION WITH PHASE CHANGE:
MOVING BOUNDARY PROBLEMS
6.1 Introduction L k a
liquid ps c s ) , ( t x T pf
moving interface
Fig. 6.1 i dt dx
solid
• Applications:
Melting and freezing
Casting
Ablation
Cryosurgery
Soldering
Permafrost
Food processing

2. • Features: 6.2 The Heat Equations • Assumptions: Moving interface
Properties change with phase changes
Density change
liquid phase motion
Temperature discontinuity at interface
Transient conduction
Non-linear problems
6.2 The Heat Equations
Two heat equations are needed
• Assumptions:
One-dimensional

3. 6.3 Moving Interface Boundary Conditions
One-dimensional
Uniform properties of each phase
Negligible liquid phase motion
No energy generation
(6.1)
(6.2)
6.3 Moving Interface Boundary Conditions
(1) Continuity of temperature

4. (2) Energy equation q (6.3) fusion (freezing or melting) temperature
interface location at time t
(2) Energy equation L q s i x dx = at t
t+dt
liquid
Fig. 6.2
element
interface
Liquid element
Mass
During dt element changes to
solid
Density change
volume change

5. Conservation of energy:
(b)
(c)
dV = change in volume
p = pressure
(d)

6. energy per unit mass (
b), (c) and (d) into (a)
(e)
Fourier’s law
(f)
(g)
(h)

7. (
f), (g) and (h) into (e) and assume p
= constant
(i)
(j) L

8. where = h ˆ
enthalpy per unit mass
L = latent heat of fusion
(j)
into ( i)
(6.4) L
Eq. (6.4) is the interface energy equation for
solidification.
For melting

9. Equation (3) Convection at the interface
(6.5) L
+ solidification, - melting
6.5 Non-linearity of Interface Energy
Equation
(6.4) L
But dt dx i /
depends on temperature gradient:
Total derivative of s T in
eq. (6.3)

10. (6.6)
(6.6) into (6.5)
(6.7) L
(6.7)
has two non-linear terms

11. Equations: Governing Parameters
6.5 Non-dimensional Form of the Governing
Equations: Governing Parameters
(6.8)
Ste = Stefan number
(6.9) L
(6.1), (6.2) and (6.4)
(6.10)

12. (6.11)
(6.12)
Two governing parameters:
and Ste
• NOTE:
(1) Parameter
is eliminated due to the definition of
(2)
Biot number appears in convection BC
(3)
The Stefan number: Ratio of the sensible heat to the
latent heat

13. Approximation 6.6 Simplified Model: Quasi-Steady Sensible heat
Latent heat = L
Stefan number for liquid phase:
(6.13) L
6.6 Simplified Model: Quasi-Steady
Approximation
• Significance of small Stefan number: L
sensible heat << latent heat

14. Limiting case: specific heat = zero,
Ste
= 0
Temperature can be changed instantaneously
by transferring an infinitesimal amount of heat
Alternate limiting case: L
Ste
= 0
stationary interface
Small Stefan number: Interface moves slowly
• Quasi-steady state model: neglect sensible heat for
set
in (6.10) and (6.11)

15. (6.14)
(6.15)
• NOTE:
(1) Interface energy equation (6.12) is unchanged
(2) Temperature distribution and
are time
dependent

16. Fusion Temperature Example 6.1: Solidification of a Slab at the
Thickness = L T
Fig. 6.3 L f o i x
liquid
solid
Initially liquid
is suddenly maintained
is kept at
Find: Time to solidify
Solution
(1) Observations
• Liquid phase remains at

17. (2) Origin and Coordinates
• Solidification starts at time
• Solidification time
(2) Origin and Coordinates
(3) Formulation
(i)
Assumptions:
(1)
One-dimensional conduction
(2)
Constant properties of liquid and solid
(3)
No changes in fusion temperature
(4)
Quasi-steady state,
(ii) Governing Equations

18. (a)
(b) BC T
Fig. 6.3 L f o i x
liquid
solid
(1)
(2)
(3)
(4)

19. (4) Solution Interface energy condition (5) L Initial conditions (6)
Fig. 6.3 L f o i x
liquid
solid
(6)
(7)
(4) Solution
Integrate (a) and (b)
(c)

20. A, B, C and D are constants. They can be functions of time.
BC (1) to (4):
(e)
(f)
• NOTE:
Liquid remains at
Interface location: (e) and (f) into BC (5)

21. L L
Integrate L
IC (6) gives 1 = C
(6.16a) L

22. (5 ) Checking Dimensionless form: Use (6.8) (6.16b) Set
Time to solidify
in (6.16a)
(6.17a)
s L L2
Or at
(6.16b) gives
(6.17b) (5
) Checking
Dimensional check

23. (6) Comments Limiting check: no solidification, thus Setting then
(i) If
no solidification, thus
Setting
in (6.17a) gives
then
Setting
in (6.17a) gives
(ii) If
(6) Comments
(i) Initial condition (6) is not used. However, it is
satisfied
(ii) Fluid motion plays no role in the solution

24. Example 6.2: Melting of Slab with Time Dependent Surface Temperature
Initial solid temperature i x
liquid
Fig. 6.4 L
solid f T t o b
exp
Thickness = L
is at
is at
Find : ) ( t x i
and
Melting time

25. (2) Origin and Coordinates
Solution
(1) Observations
• Solid phase remains at
• Melting starts at time
• Melting time
(2) Origin and Coordinates
(3) Formulation (
i) Assumptions: (1
) One-dimensional
(2)
Constant properties (liquid and solid)
(3)
No changes in fusion temperature

26. (ii) Governing Equation(4
) Quasi-steady state, 1 .
< te S
(5)
Neglect motion of the liquid phase
(ii) Governing Equation
(a)
No heat transfer to the solid:
(b)
BC:
(1)
(2)

27. (4) Solution Interface energy condition L (3) IC: (4) Integrate (a)
BC (1) and (2)
(c)

28. (c) into condition (3)
(d)
Integrate, use IC (4)
Solving for ) ( t x i

29. (5) Checking (6.18) L L Melt time Set in (6.18) (6.19) L L Solve for
by trial and error.
(5) Checking
Dimensional check
Limiting check
: Special case:

30. Constant temperature at
Can not set
in (6.18). Expanded
for small values of
L L
Set
(6.20)
L L
This result agrees with
eq. (6.16a)

31. (6) Comments 6.7 Exact Solutions 6.7.1 Stefan’s Solution (
i) No liquid exists at time t
= 0, no initial condition is
needed.
(ii) Quasi-steady state model is suitable for time
dependent boundary conditions.
6.7 Exact Solutions
6.7.1 Stefan’s Solution
• Semi-infinite liquid region
• Initially at
• Boundary at
is suddenly maintained at
• Liquid remains at

32. Determine: Temperature distribution and interface
location
Liquid solution:
(a)
(b)
Solid phase:
BC:
(1)
(2)
(a)
into (6.4)
(3) L

33. Solution IC: (4) Use similarity method (6.21) Assume (c)
(b) transforms to
(d)

34. Solution to (d)
(6.22)
BC (2) and (6.21)
(e)
This requires that
Let
(f)
is a constant
• NOTE:

35. (f) satisfies initial condition (4)
BC (1):
(g)
BC (2):
(h)
(g) and (h) into (6.22)
(6.23)
Interface energy condition (3) determines
(f) and
(6.23) into BC (3)

36. L or
(6.24) L
• NOTE:
(1) (6.24) does not give
explicitly
depends on the material as well as temperature at
(2)
• Special Case: Small

37. Small
corresponds to small
Evaluate eq. (6.24) for
small
First expand
and
and
Substituting into
eq. (6.24) gives
for small
(6.25) L

38. Examine the definition of Stefan number
(6.13) L
Therefore a small
corresponds to a small Stefan
number.
Substitute (6.25) into (f)
for small
(6.26)
(small Ste)
s L
Eq. (6.26) is identical to eq. (6.16a) of the quasi-steady
state model.

39. 6.7.2 Neumann’s Solution: Solidification of Semi- Infinite Region
• Stefan’s problem:
• Neumann’s problem:
• Assumptions:
One-dimensional
Constant properties
Stationary
• Differential equations:
(6.1)

40. (6.2)
BC:
(1)
(2)
(3)
(4)
Interface energy equation:
(5) L

41. IC:
(6)
(7)
Solution
Similarity method:
(6.21)
Assume
(a)
and
(b)

42. Equations (6.1) and (6.2) transform to
(c)
(d)
where
(e)
Solutions to (c) and (d) are
(f)

43. Transformation of BC and IC:
(g)
BC (2):
Thus or
(h)
Transformation of BC and IC:

44. liquid
Fig. 6.7
solid o T h l i f
(1)
(2)
(3)
(4)
(5) L
(6)
(7)

45. • NOTE:
• (h) satisfies IC (7)
• Conditions (4) and (6) are identical
• Interface appears stationary at
• BC (1) - (4) give A, B, C and D. Solutions (f) and (g)
become
(6.27)
and
(6.28)

46. Determine
substitute (6.27) and (6.28) into the energy
condition (5)
• NOTE:
in Neumann’s equations (6.27), (6.28) and
Set
(6.29) to obtain Stefan’s solution.

47. Solution 6.7.3 Neumann’s Solution: Melting of Semi-infinite Region
• Follow the same procedure used in solving the
solidification problem
• Modify energy condition (5):
Solution
(6.30)
(6.31)
(6.32)

48. Phase Change 6.8 Effect of Density Change on the Liquid where
is given by
6.8 Effect of Density Change on the Liquid
Phase Change i x
Fig. 6.8 dt dx ) ( t u
solid
liquid
Changes in density cause the
liquid phase to move.
Heat equation becomes:

49. (6.34) Determine u first dx = liquid element dx =
interface at t
t+dt
liquid
element L 9 . 6
Fig
Determine u first L dx =
liquid element s dx =
element in solid phase i dx =
displacement of interface during
solidification
(a) L dx i - =
displacement of liquid
(b)

50. • (c) (6.35) dt dx / is time dependent. Thus Eq. (6.35) into
Conservation of mass for the element or
(c)
(c) into (b)
(6.35) dt dx i /
is time dependent. Thus •
Eq. (6.35) into
eq. (6.34)
(6.36)

51. • Limitations: Infinite domain
Solution:
By similarity method
6.9 Radial Conduction
with Phase Change
• Limitations: Infinite domain
Example: Solidification due to a
line heat sink
Liquid is initially at
Heat is removed at a rate o Q
per
unit length

52. Assume: (1) Constant properties in each phase (2)
Neglect effect of liquid motion
(6.37)
(6.38)
BC:
(1)
(2)

53. (3)
(4)
Interface energy equation
(5) L
IC:
(6)
(7)

54. Solution Similarity method: (6.39)
Equations (6.37) and (6.38) transform to
(6.40)
(6.41)
Separate variables and integrate eq. (6.40)

55. Rearrange, separate variables and integrate
A and B are constants of integration.

56. Similarly, solution to eq. (6.41) is
(b)
is value of
at the interface. It is determined by
setting
in eq. (6.39)
(6.42)
BC (2):

57. • NOTE: constant, therefore is independent of time. Thus (6.43) (6.43)
satisfies IC (7).
Eq. (6.43) into (6.42):
(c)
where
constant.
Eq. (a) and BC (1) give A

58. Solving for A
(d) (
a), (c) and
B.C. (2) give B
(e)

59. (b), (c) and BC (3) give D
(f)
Similarly, BC (4) gives
(g)
Interface energy equation (5) gives

60. (6.44) L
• Examine integrals in the solution:
The
exponential integral function Ei ( x
) is defined as
(6.45)
Values of this function at
and
are
(6.46)
Use the definition of ) ( x Ei
in (6.45) and the constants
in (d)-(e), equations (a), (b) and (6.44), become

61. 6.10 Phase Change in Finite Regions
(6.47)
(6.48)
(6.49) L
6.10 Phase Change in Finite Regions