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Casting Analysis - 1

Melting and Pouring

ver. 1

ME 6222: Manufacturing Processes and Systems 1
Prof. J.S. Colton © GIT 2011

Overview
• Processes
• Analysis
– Melting
– Pouring
– Solidification and cooling
– Surface tension
– Gas solubility and porosity
• Defects
• Design rules
• Economics


Casting Steps
quick route from raw material to finished
product
• Melt metals
• Pour / force liquid into hollow cavity
(mold)
• Cool / Solidify
• Remove
• Finish


Temperature vs. Time

Temperature
pouring

melting solidification

removal
initial
time


Melting
Furnace
Heat Atmosphere
• Raw material (charge)
– scrap, alloying materials
• Atmosphere Raw materials

– Air (oxygen), vacuum, inert gas (argon)
• Heating
– External - electric, gas, oil
– Internal - induction, mix fuel with charge
• steel making in blast furnace -mix coke with steel
• Furnace material - refractory
– high melting point metals, ceramics

Heat to melt

 
H  rV cs Tmelt  Tinitial   H f  cl Tpour  Tmelt 
• H = heat [J]
• r = density
• V = volume
• c = specific heat (s = solid, l = liquid)
• Hf = heat of fusion


Melting Time

• Estimate by

Energy
time 
Power

• Take into account oven efficiency


Melting – Ex. 1-1

Calculate the time required to raise the
temperature of a 120 kg aluminum billet
from 20oC to 50oC above its melting
point using a 20 kW furnace that is 75%
efficient.


Melting – Ex. 1-2
H = 120 * [0.9 * (660-20)
+ 396 + 1.05 * (710 – 660)]
= 123 MJ = 1.17 x 105 BTU
• density = 2700 kg/m3
• melting point = 660oC
• heat of fusion = 396 kJ/kg
• specific heat of liquid = 1.05 kJ/kg-K
• specific heat of solid = 0.9 kJ/kg-K


Melting – Ex. 1-3

• time = 123 MJ / (20 kW * 75%)
• time = 2.3 hours

• Should probably buy a bigger furnace


Pouring - Fluid Flow

• Reynold’s number
• Bernoulli’s equation
• Continuity


Reynold’s number (Re)
ratio of momentum (inertia) to viscosity

rVd
Re 


density  velocity  diameter
viscosity


Critical Reynold’s number

• Re < 2,000
– viscosity dominated, laminar flow
• Re > 20,000
– inertia dominated, turbulent flow
• Controlled through gate and runner
design


Bernoulli’s Equation
• Used to calculate flow velocities
• Assumptions: steady state, incompressible, inviscid
flow

rv 2
rv 2
Po  o
 rgho  P 
1
1
 rgh1  f
2 2

P = pressure g = gravity
r = density h = height
v = velocity f = losses due to friction


Continuity

Q0 = Q1
A0v0 = A1v1

where:
• Q = volumetric flow rate
• A = area
• v = velocity

Assumption: incompressible flow

Pouring – Ex. 2-1
Determine the geometry of the sprue, so that
there is no air aspiration (the pressure never
is less than atmospheric).

1 pouring basin
hc sprue
2
ht

1 = free surface of metal
2 = spue top
3 = sprue bottom
3


Casting – Ex. 2-2
rv1
2
rv3
2
P
1  rgh  P 
1 3  rgh3
2 2
• Assuming
– entire mold is at atmospheric pressure (no
point below atmospheric)
– metal in the pouring basin is at zero
velocity (reservoir assumption)

v3  2 ght

Casting – Ex. 2-3
• Similarly

rv1
2
rv2
2
P
1  rgh  P2 
1  rgh2
2 2

v2  2 ghc


Casting – Ex. 2-4

• By continuity

A2v2 = A3v3
• Hence
A3 v2 hc
 
A2 v3 ht


Mold Filling Time Estimate

• Using continuity
– Q = Ag vg =A3 v3
• Assuming Ag = A3

vgate  v3  2ght

• Hence Mold filling time 
Volume of mold
Agatevgate


Pouring – Ex. 3-1

• Given
– height of sprue (ht) = 20 cm
– area of sprue (A3) = 2.5 cm2
– volume of mold cavity (V) = 1560 cm3
• Find
– vsprue
– Flow rate (Q)
– Mold filling time


Pouring – Ex. 3-2

vbase of sprue  2 ght  2 * 981 * 20  198.1 cm/s

Q = vbase of sprue *A3
= 198.1 * 2.5 = 495 cm3/s

Mold filling time = V/Q =1560 / 495 = 3.2 s


Bottom Gated Systems
• Air forced up and out of mold
• Reduced splashing and oxidation
• As metal is poured into the system,
effective head is reduced.
Ag = gate cross-sectional area
Am = mold cross-sectional area

ht h hm



• In time dt,
– height of metal in mold increases dh
– flow rate in mold, Qm = Amdh/dt
– flow rate of metal delivered by gate to mold,
Qg = Agvg, where
vg  2 g ht  h


• Equating Qm = Qg,
Amdh  Ag 2 g ht  hdt
• If tf = filling time,
hm tf
1 dh Ag
2g 
0

ht  h Am 0 dt

t f 
2 Am
Ag 2 g
 ht  ht  hm 

Pouring - Ex. 4-1
• You are pouring liquid iron into a mold.
• The mold has a sprue height of 2 inches.
• The bottom of the sprue has a diameter
of 0.2 in.
Risers
Sprue

Gate
Cope
Parting Line
Drag
Casting


Pouring - Ex. 4-2
• You wish to pour the metal so that you do
not entrain air.

• What should the diameter of the gate
(dgate) be?


Pouring - Ex. 4-3

• Here we need to use:
– Reynold’s number
• Values below 20,000 are OK in casting
• To prevent entrainment of air resulting from
turbulence
– Bernoulli’s equation
– Continuity


Pouring - Ex. 4-4
• Iron data:
– density (r = 7860 kg/m3
– viscosity at pouring temp ( = 2.25 cp
= 2.25 x 10-3 N*s/m2
• h0 = 2 in. = 0.051 m
• h1 = 0 m
• g = 9.8 m/s2
• A1 = pr2 = 3.14 * 0.002542 =
2.03 x 10-5 m2


Pouring - Ex. 4-5

• Now we need to determine the
velocity at the bottom of the sprue
(v1) using Bernoulli’s equation.


Pouring - Ex. 4-6
• We can assume that the velocity
at the top of the mold (vo) is zero,
if there is a pouring basin, which is
typical.


Pouring - Ex. 4-7

• Ignore friction effects (f=0).
• Assume the mold is open to
atmospheric pressure
(P0=P1=Patm).
• Ignore the effect of the height of
the metal in the mold.


Pouring - Ex. 4-8
• Substituting into Bernoulli’s equation:

rvo
2
rv12
Po   rgho  P 
1  rgh1  f
2 2

7860  02
Patm   7860  9.8  0.051 
2
7860  v1
2
Patm   7860  9.8  0  0
2


Pouring - Ex. 4-9
• And solving:

v1 = 1 m/s

• Checking Reynolds number

Re = 7860*1*0.00508/2.25x10-3
=17,746 < 20,000


Pouring - Ex. 4-10
• Now using continuity:

A1v1 = Agatevgate =

2.03 x 10-5 * 1 = Agatevgate

and Agate = prgate2


Pouring - Ex. 4-11
• Now, Reynold’s number < 20,000

rVd 7860  vgate  d gate
Re    20,000
 2.25  10 -3

• Solving gives:
vgate*dgate = 5.72 x 10-3 m2/s


Pouring - Ex. 4-12
• Combining the following equations:

• 2.03 x 10-5 * 1 = Agatevgate

• Agate = prgate2

• vgate*dgate = 5.72 x 10-3 m2/s


Pouring - Ex. 4-13

• And solving gives:

dgate = 4.5 mm = 0.18 in.

A not unreasonable answer, given
the sprue is 5 mm in diameter.


Data for Solid Materials
Room Temperature
Material Specific heat Density Thermal
(kJ/kg-oC) (kg/m3) conductivity
(W/m-oC)
Sand 1.16 1500 0.60
Aluminum 0.90 2700 222
Nickel 0.44 8910 92.1
Magnesium 1.03 1740 154
Copper 0.38 8960 394
Iron 0.46 7860 75.4
Steel 0.434 7832 59


Data for Liquid Metals
Material Melting Density Latent heat Thermal Specific Viscosity
point (oC) (kg/m3) of conductivity heat (mPa-s)
solidification (W/m-oC) (kJ/kg-oC)
(fusion)
(kJ/kg)

Aluminum 660 2390 396 94 1.05 4.5

Nickel 1453 7900 297 0.73 4.1

Magnesium 650 1585 384 139 1.38 1.24

Copper 1083 7960 220 49.4 0.52 3.36

Iron 1537 7150 211 0.34 2.2


Summary

• Analysis
– Melting
– Pouring
– Solidification and cooling
– Surface tension
– Gas solubility and porosity


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