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

• 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

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

(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

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

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

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

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

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)

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

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)

(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

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

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)

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

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

(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

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

(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)

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)

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

(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

(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

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

(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

(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)

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

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

(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:

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)

(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

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

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

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

(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)

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

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

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.

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)

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

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

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

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

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

• 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)

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.

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)

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:

(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)

• (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)

• 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

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

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

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

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

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):

• 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

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

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

(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

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