Castanal1

ME 6222: Manufacturing Processes and Systems

Prof. J.S. Colton © GIT 2011

1

Casting Analysis - 1

Melting and Pouring

ver. 1



2

Overview

• Processes

• Analysis – Melting

– Pouring

– Solidification and cooling

– Surface tension

– Gas solubility and porosity

• Defects

• Design rules

• Economics



3

Casting Steps quick route from raw material to finished

product

• Melt metals

• Pour / force liquid into hollow cavity

(mold)

• Cool / Solidify

• Remove

• Finish



4

Temperature vs. Time

Temperature

time

initial

melting solidification

pouring

removal



5

Melting

• Raw material (charge)

– scrap, alloying materials

• Atmosphere

– 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

Raw materials

Furnace

Atmosphere



6

Heat to melt

• H = heat [J]

• r = density

• V = volume

• c = specific heat (s = solid, l = liquid)

• Hf = heat of fusion

meltpourlfinitialmelts TTcHTTcVH r

Melting Time

• Estimate by

• Take into account oven efficiency



7

Power

Energytime



8

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.



9

Melting – Ex. 1-2

• 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

H = 120 * [0.9 * (660-20)

+ 396 + 1.05 * (710 – 660)]

= 123 MJ = 1.17 x 105 BTU

Melting – Ex. 1-3

• time = 123 MJ / (20 kW * 75%)

• time = 2.3 hours

• Should probably buy a bigger furnace



10



11

Pouring - Fluid Flow

• Reynold’s number

• Bernoulli’s equation

• Continuity



12

Reynold’s number (Re)

ratio of momentum (inertia) to viscosity

viscosity

diametervelocitydensity

Re

rVd



13

Critical Reynold’s number

• Re < 2,000

– viscosity dominated, laminar flow

• Re > 20,000

– inertia dominated, turbulent flow

• Controlled through gate and runner

design



14

Bernoulli’s Equation

• Used to calculate flow velocities

• Assumptions: steady state, incompressible, inviscid flow

P = pressure g = gravity

r = density h = height

v = velocity f = losses due to friction

fghv

Pghv

P oo

o 1

2

11

2

22r

rr

r



15

Continuity

Q0 = Q1

A0v0 = A1v1

where:

• Q = volumetric flow rate

• A = area

• v = velocity

Assumption: incompressible flow



16

Pouring – Ex. 2-1

Determine the geometry of the sprue, so that there is no air aspiration (the pressure never is less than atmospheric).

hc

ht

1

2

3

1 = free surface of metal

2 = spue top

3 = sprue bottom

pouring basin

sprue



17

Casting – Ex. 2-2

• Assuming

– entire mold is at atmospheric pressure (no

point below atmospheric)

– metal in the pouring basin is at zero

velocity (reservoir assumption)

3

23

31

21

122

ghv

Pghv

P rr

rr

tghv 23



18

Casting – Ex. 2-3

• Similarly

2

22

21

21

122

ghv

Pghv

P rr

rr

cghv 22



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Casting – Ex. 2-4

• By continuity

A2v2 = A3v3

• Hence

t

c

h

h

v

v

A

A

3

2

2

3



20

Mold Filling Time Estimate

• Using continuity

– Q = Ag vg =A3 v3

• Assuming Ag = A3

• Hence

t3gate 2ghvv

gategatevA

mold of Volume timefilling Mold



21

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



22

Pouring – Ex. 3-2

Q = vbase of sprue *A3

= 198.1 * 2.5 = 495 cm3/s

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

cm/s 1.19820*981*22v sprue of base tgh



23

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.

h hm ht

Am = mold cross-sectional area

Ag = gate cross-sectional area



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• 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 hhgv tg 2



25


• Equating Qm = Qg,

• If tf = filling time,

dthhgAdhA tgm 2

mtt

g

mf

t

m

g

h

t

hhhgA

At

dtA

A

hh

dh

g

fm

2

2

2

1

00



26

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.

Cope

Drag Casting

Parting Line

Sprue Risers

Gate



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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?



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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



29

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



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Pouring - Ex. 4-5

• Now we need to determine the

velocity at the bottom of the sprue

(v1) using Bernoulli’s equation.



31

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.



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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.



33

Pouring - Ex. 4-8

• Substituting into Bernoulli’s equation:

fghv

Pghv

P oo

o 1

2

11

2

22r

rr

r

008.978602

7860

051.08.978602

07860

21

2

vP

P

atm

atm



34

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



35

Pouring - Ex. 4-10

• Now using continuity:

A1v1 = Agatevgate =

2.03 x 10-5 * 1 = Agatevgate

and Agate = prgate2



36

Pouring - Ex. 4-11

• Now, Reynold’s number < 20,000

• Solving gives:

vgate*dgate = 5.72 x 10-3 m2/s

000,20 10 2.25

7860Re

3-

gategate dvVd

r



37

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



38

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



39

Material Specific heat

(kJ/kg-oC)

Density

(kg/m3)

Thermal

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



40

Material Melting

point (oC)

Density

(kg/m3)

Latent heat

of

solidification

(fusion)

(kJ/kg)

Thermal

conductivity

(W/m-oC)

Specific

heat

(kJ/kg-oC)

Viscosity

(mPa-s)

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



41

Summary

• Analysis

– Melting

– Pouring

– Solidification and cooling

– Surface tension

– Gas solubility and porosity



42

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