Solidification Dr.Peerapong Triyacharoen 75 Department of - - PowerPoint PPT Presentation

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Solidification Dr.Peerapong Triyacharoen 75 Department of - - PowerPoint PPT Presentation

Kasetsart University 213211: Imperfection Solidification Dr.Peerapong Triyacharoen 75 Department of Materials Engineering Kasetsart University 213211: Imperfection Schematic Solidification Stages Formation of Nuclei Growth of nuclei into


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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

75

Solidification

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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

76

Schematic Solidification Stages

Formation of Nuclei Growth of nuclei into crystals Completely solidified

Nucleation Growth

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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

77

Nucleation

v

G r ∆ =

3

π 3 4

Embryo Nucleus

γ π π

2 3

4 r G r G

v T

+ ∆ = ∆ 3 4

r = radius of embryo or nucleus ∆Gv = volume free energy γ = specific surface free energy

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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

78

Homogeneous Nucleation

  • Occurs when the undercooling (∆T) becomes large

enough to cause formation of a stable nucleus.

T H T G r

f m v

∆ ∆ − = ∆ − = γ γ 2 2

*

r* = critical radius Tm = melting temperature ∆T = undercooling (Tm – T) ∆Hf = latent heat of fusion represents the heat given up during the liquid-to-solid transformation

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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

79

Heterogeneous Nucleation

  • Occurs in a liquid on the surfaces of its container,

insoluble impurities, or other structural material which lower the critical free energy required to form a stable nucleus.

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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

80

Glasses

  • Amorphous (glassy) solid
  • Insufficient time for nuclei to form and grow by

rapid cooling (rapid solidification processing)

Tg Tm Temperature Density

C r y s t a l l i n e S

  • l

i d G l a s s Undercooled liquid Liquid

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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

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Growth Mechanisms

Need to remove two types of heat:

1. Specific heat: required to change temperature of a unit weight of the material by one degree. 2. Latent heat of fusion

Planar Growth Dendritic Growth

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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

82

Casting or Ingot Structure

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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

83

Solidification Time

Time required to completely solidify for a simple casting (ts) :

n s

A V B t ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =

Chvorinov’s rule B = mold constant, depending on the properties and initial temperature of both the metal and the mold V = volume of the casting A = surface area of the casting in contact with the mold n = constant (usually about 2)

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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

84

Effect on Structure and Properties

m s

kt SDAS =

k and m = constants depending on the metal composition Secondary Dendrite Arm Spacing:

Aluminum alloy

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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

85

Effect on Structure and Properties (con.)

Aluminum casting alloy

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Kasetsart University

Dr.Peerapong Triyacharoen Department of Materials Engineering

213211: Imperfection

86

Example: Design of an Aluminum Casting

Design the thickness of an aluminum casting whose length is 12 in. and width is 8 in., in order to produce a tensile strength of 40,000 psi. The mold constant in Chvorinov’s rule for aluminum alloys cast in a sand mold is 45 min/in2.

n s

A V B t ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =

Solution

  • Required SDAS = 0.007 cm to obtain a T.S. = 42,000 psi
  • Required solidification time = 300 s or 5 min

B = 45 min/in2 V = (8)(12)(x) = 96x A = (2)(8)(12) + (2)(x)(8) + (2)(x)(12) = 40x + 192

2 2

192 40 96 ) min/ 45 ( min 5 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = x x in

333 . ) 45 / 5 ( 192 40 96 = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + x x

⇒ x = 0.77 in.