Guide to Exercises Hints A in Chapters 3–6 - Wiley: Home to Exercises Hints A in Chapters 3–6 HINT A1 Exercise 3.1 What component casting methods, which give high accu- racy and

Guide to Exercises Hints A in Chapters 3 – 6

HINT A1

Exercise 3.1

What component casting methods, which give high accu-

racy and good surface finish, can be considered? If you

have no suggestion, look for precision casting methods in

Chapter 1.

Hint A14

HINT A2

Exercise 4.1

In this case the solidification is one-dimensional and the

contact between the Cu plate and the solidified metal is

good. Draw a figure of the temperature profile and suggest

a strategy to calculate the solidification time.

Hint A275

HINT A3

Exercise 5.1a

What basic equation do you use to calculate the excess

radiation losses Qcoolrad from the ingot?

Hint A173

HINT A4

Exercise 6.1

What happens in the melt when steel powder of room tem-

perature is poured into the melt?

Hint A127

HINT A5

Exercise 6.4

A solidified ingot has three different crystal zones, surface

crystal zone, columnar crystal zone and central crystal

zone. Divide the curve into a number of relevant time inter-

vals and relate them one by one to the macrostructure of the

alloy.

Hint A138

HINT A6

Exercise 6.10

Draw a sketch of part of the wire.

Hint A22

HINT A7

Exercise 6.7

The condition for grey solidification is that the solidification

rate vgrowth ¼ dyL=dt < 4:0 � 10�4 m/s.

Make a sketch of the temperature distribution at the sur-

face of and inside the rod and set up an analytical expres-

sion of the solidification rate dyL=dt as a function of the

position yL of the solidification front.

Hint A261

Hint A8

(A207)

Qsolrad ¼ AesBðTL

4 � T04Þ t ð3Þ

The liquidus temperature is ¼ 1450 þ 273 ¼ 1723 K. Intro-

duction of known values gives

Qsolrad ¼ 0:4 � 1 � 0:2 � 5:67 � 10�8

� ð17234 � 2934Þ � 64 � 60 ¼ 15:3 � 107 J

The answer is given in Hint A26.

Hint A26Materials Processing during Casting Guide to Exercises H. Fredrikssonand U. Akerlind # 2006 John Wiley & Sons, Ltd.

HINT A9

Exercise 6.2

You want the dendrite arm distance lden as a function of the

thickness of the columnar crystal zone of an ingot. You

know the growth rate vgrowth as a function of the time and

the relation between v and lden.

How do you find a relation between thickness yL of the

columnar crystal zone and time?

Hint A51

HINT A10

Exercise 3.2

Set up an expression for the casting time as a function of the

dimensions of the sprue in case of uphill casting.

Hint A81

HINT A11

Exercise 6.3

The heat transfer coefficient is mentioned in the text. It is

therefore reasonable to set up a heat balance equation,

which might be useful.

Hint A176

HINT A12

Exercise 5.3

Why is the solidification front non-planar?

Hint A178

HINT A13

(A137)

Figure 4.20 on page 78 in Chapter 4 shows the temperature

distribution in the mould and the melt during casting in a

sand mould.

The solidification time can be calculated from Chvori-

nov’s rule [Equation (A4.74a) on page 80 in Chapter 4].

In this case we obtain

ttotal ¼ CVmetal

A

� �2

¼ CAyL

A þ A

� �2

ð1Þ

Find an expression for the constant C.

Hint A293

HINT A14

(A1)

The best choice may be either the Shaw process (Chapter 1,

page 7) or the investment casting method (wax melting

method) (Chapter 1, pages 7–8). Compare them.

Hint A45

HINT A15

Exercise 5.6

The heat transfer coefficient hw depends on the water flow

and the temperature of the cooling water. Only empirical

relations are available. Use one of them.

Hint A114

HINT A16

(A35)

The flow is assumed to be laminar. In this case, Bernoulli’s

equation is valid [Equation (3.2) on page 29 in Chapter 3].

Suggest suitable points as points 1 and 2. Set up Ber-

noulli’s equation, introduce convenient variables and

apply it in the present system.

Hint A88

HINT A17

(A44)

The outlet speed v3 can be calculated if you use the conti-

nuity principle:

A2v2 ¼ A3v3

where A2 and A3 are known. Inserting the v2 function, you

obtain

v3 ¼ A2

A3

v2 ¼ A2

A3

ffiffiffiffiffiffiffiffi2gh

p¼

p0:010

2

� �2

0:140 � 0:140


pð3Þ

2 Guide to Exercises

Answer

The casting rate is v3 ¼ 0:010ffiffiffih

p, where v3 and h are mea-

sured in SI units.

HINT A18

(A133)

The heat lost to the surroundings per unit time is, with nor-

mal designations

� dQ

dt¼ �pR2dy rLcL

p

dT

dt

dT

dt< 0

� �ð1Þ

To continue, you have to set up a heat balance.

Hint A254

HINT A19

(A202)

(A154)

The driving force for heat transport during and after solidi-

fication is proportional to the temperature difference

ðTL � T0Þ.

Answer

(a) The solidification time of the aluminium ingot is about

1.5 h.

(b) The solidification time of the steel ingot is about 1 h.

In spite of its much higher mass, the steel ingot solidifies

more rapidly than the aluminium ingot, because Fe has a

lower heat of fusion and a higher driving force for heat

transport than Al.

HINT A20

Exercise 3.4

Draw a figure of the tundish with molten steel. What equa-

tion is valid for the steel flow? Think of a strategy to solve

the problem.

Hint A95

HINT A21

Exercise 5.12a

Describe the process and make reasonable assumptions.

Hint A109

HINT A22

(A6)

Consider a wire element with radius r0 and length �z. It

solidifies inwards from radius r þ dr to radius r during

the time dt. The solidification heat at the solidification

front is removed by a heat flow in the opposite direction.

Set up a heat balance.

Hint A179

HINT A23

Exercise 3.9a

Set up a heat balance for a volume element of the spiral.

Hint A195

HINT A24

(A305)

Equation (2) gives

Aoutlet

¼ pdoutlet2

4¼ Astrandvcast

voutlet

ð5Þ

V = πR2dy

R

dy

��H TL � T0 Solidification time

Quantity (kJ/kg) (K) (h)

Aluminium 398 635 1.5

Steel 272 1510 1.0

∆ z

r0 r0 y r

r + dr

Materials Processing during Casting 3

or

doutlet¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4

p�Astrandvcast

voutlet

r¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4

p�1:5�0:20�0:0090

3:13

r¼0:033m

Answer

The casting rate is 0.0090 m/s or 0.54 m/min.

The outlet diameter of the tundish has to be 33 mm.

HINT A25

(A126)

The solidification time is obtained if you combine

Equations (1) and (2):

t ¼ 1

4ametal

y

l

� �2

¼rmetalc

metalp

4 kmetal

y

l

� �2

where y ¼ thickness of solidified ‘shell’

t ¼ 7:878 � 103 � 830

4 � 32

0:10

0:90

� �2

� 630 s

Answer

The solidification time of a turbine blade is about 10.5 min

(l ¼ 0:90).

HINT A26

(A165)

(A8)

(A221)

(A38)

Answer

(a) The fraction radiated excess heat is 6 %.

(b) The total radiated heat during the solidification is

1:5 � 102 MJ.

(c) The thickness of the maximum layer is calculated to be

15 cm, which is an unrealistic value. (A pore is often

formed below the solidified layer. It insulates the melt

from the solidified shell.)

(d) Very little radiation heat is emitted provided that the

upper surface is properly insulated. The main advantage

is that the upper surface solidifies later than the centre

of the ingot.

HINT A27

(A208)

You insert these values into Equation (10), which gives

t ¼ 2:7 � 103 � 390 � 103

660 � 25

R2

4 � 220þ R

2 � 1:68 � 103

� �

or

t ¼ 1:66 � 103 R2

0:880þ R

3:36

� �ð11Þ

Answer and plot are given in Hint A79.

Hint A79

HINT A28

(A191)

Composition of the molten alloy (page 49 in Chapter 3).

Small additions of impurities increase the viscosity

strongly and reduce the fluidity. Pure metals and eutectic

alloys have low viscosities and high fluidity lengths. The

larger the liquidus–solidus interval is, the higher will be

the viscosity and the lower the maximum fluidity length.

The best fluidity is obtained if the metal freezes on the

mould wall and the solidification front becomes planar. An

alloying element leads to a solidification interval of the

alloy, which prevents a planar front. Instead, so-called den-

drites (network of crystal arms) are formed. The viscosity

increases and the fluidity decreases.


Hint A164

HINT A29

(A227)

According to the text, you have

yL þ YL ¼ 0:0060 m ð5Þ

Combine equations (4) and (5), introduce known values and

material constants and solve YL. Accurate calculations are

necessary!

Hint A156

HINT A30

Exercise 4.2

How do you classify the type of casting conditions? What

model can be used to the solidification process?

Hint A281


HINT A31

Exercise 6.8a

What is the influence of convection on the nucleation of

equiaxed crystals in a solidifying ingot?

Hint A223

HINT A32

(A199)

Introduce Equation (4) into Equation (3) and replace vingot

by dh=dt:

Asprue

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2gðH � hÞ

p¼ 6Aingot

dh

dt

How do you solve this differential equation?

Hint A75

HINT A33

(A204)

You choose some values of l and use Table 4.4 on page 67

in Chapter 4.

Interpolation gives l � 0:79.

How do you proceed?

Hint A122

HINT A34

(A119)

The heat flow from the superheated melt to the surround-

ings can be written in two ways:

Integrate Equation (5) and solve the cooling time t3.

Hint A180

HINT A35

Exercise 3.3

Make some reasonable assumptions about the steel flow. Is

it turbulent or laminar? What equation is applicable?

Hint A16

HINT A36

(A116)

The heat flux from the melt to the solid phase can be written

as

dq

dt¼ �h2ðTmelt � TLÞ ð1Þ

The heat flux through the solid shell is

dq

dt¼ �ks

TL � Ti metal

yð2Þ

The heat flux from the outer surface to the surroundings is

dq

dt¼ �h1ðTi metal � T0Þ ð3Þ

You want y as a function of the excess temperature,

�T ¼ Tmelt � TL.

How do you proceed?

Hint A184

HINT A37

(A232)

According to geometry and Figure 4 in Hint A232, the

radius r is two-thirds of the height in the equilateral triangle

T1T2T3:

r ¼ Rffiffiffi2

p

2

ffiffiffi3

p� 2

3¼ R

ffiffiffi6

p

3

or

R ¼ rffiffiffi6

p

2ð5Þ

Now you are close to the final answer!

Hint A183

@Q

@t¼ �rVcp

dT

dtCooling heat

per unit time

¼ A

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffikmouldrmouldcmould

p

pt

sðTi � T0Þ

Heat flow across the interface mould/

melt Compare Equation ð4:70Þ on

page 79 in Chapter 4

ð5Þ

l erf(l) 0.325 þ erf(l)ffiffiffip

plel

2 ffiffiffip

plel

2 ½0:325 þ erfðlÞ�0.10 0.1125 0.4375 0.1790 0.078

0.70 0.6778 1.0028 2.0252 2.03

0.75 0.7112 1.0362 2.3331 2.42

0.80 0.7421 1.0671 2.6891 2.87

1.00 0.8427 1.1677 4.8180 5.6


HINT A38

Exercise 5.1d

As you will see later (Chapter 10), it is impossible to accept

that the upper surface of the melt solidifies before the centre

of the ingot. The thickness you got in Exercise 5.1c is also

unreasonably large and unrealistic.

If the upper surface were to solidify, then a pore would

be formed below the solid layer owing to shrinkage. This

pore would stop the further growth of the solid layer.


Hint A26

HINT A39

(A178)

Heat flux across the solidification front:

ks

TL � Ti metal

yL

Heat flux through

the solidified shell

¼ kL

Tmelt � TL

dHeat flux

through the

boundary

layer

þ rð��HÞ dyL

dtSolidification

heat per

unit time

Discuss the convection flow within the boundary

layer and its influence on the shape of the solidification

front.

Hint A246

HINT A40

Exercise 4.6

The solidification conditions are different in the two cases

and different equations are valid. What equation is valid for

the sand mould?

Hint A162

HINT A41

(A344)

lden ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi10�10

ffiffit

p

1:5 � 10�2

s

lden ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

10�10yL

1:5 � 10�2 � 3:0 � 10�2

r¼ 4:71 � 10�4 ffiffiffiffiffi

yLp

Answer

lden ¼ 4:7 � 10�4 ffiffiffiffiffiyL

p

HINT A42

Exercise 6.8c

Compare the values of vfront and vcrystal. You can find them

in Hints A309 and A171.

Hint A267

HINT A43

(A258)

rLLðsin 5�ÞcLp

�dT

dt

� �¼


p

ptcool

sðTE �T0Þ ð8Þ

Calculate the cooling rate as a function of L.

Hint A291

T

T meltT L

T i metal

T 0

y

y L


HINT A44

(A95)

p1 þ rgh1 þrv1

2

2¼ p2 þ rgh2 þ

rv22

2ð1Þ

Point 1 is chosen at the upper surface of the steel melt

and point 2 at the outlet from the tundish. This level is cho-

sen as zero level. As A1 is very large, v1 � 0.

Inserting p1 ¼ p2 ¼ patm, h2 ¼ 0 and h1 ¼ h into Equa-

tion (1), you obtain

patm þ rgh þ 0 ¼ patm þ 0 þ rv22

2) v2 ¼


pð2Þ

How do you proceed when the v2 function is known?

Hint A17

HINT A45

(A14)

What would you choose on the basis of this knowledge?

Hint A72

HINT A46

Exercise 6.9a

How do you get an equation for calculation of the cooling

rate in the centre of the wedge-shaped sample? Start with a

figure and make reasonable assumptions in order to set up

an expression for the heat flow to the surroundings.

Hint A247

HINT A47

(A75)

Inserting h ¼ hfill gives

tfill ¼6Aingot � 2

Asprue

ffiffiffiffiffi2g

p ðffiffiffiffiH

p�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiH � hfill

pÞ

Inserting the values given in the figure in the text in SI

units, you obtain

tfill ¼6 � 0:5 � 2

ð0:15Þ2ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 � 9:81

p ðffiffiffiffiffiffiffi1:7

p�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:7 � 1:5

pÞ ¼ 51

Answer

The filling time is 50 s.

HINT A48

Exercise 4.10b

How do you estimate the value of the heat transfer

coefficient?

Hint A155

HINT A49

Exercise 4.9a

Can you find a relation between the surface temperature and

the thickness of the solidified shell?

Hint A157

HINT A50

Exercise 3.6

Set up a mass balance for sulfur.

Hint A112

HINT A51

(A9)

Integrate the relation vgrowthðtÞ. Then you get the thickness

yL of the columnar zone as a function of t.

Hint A282

Shaw Investment casting

process method (wax melting)

Model built of wood or

gypsum

Model of wax

Mould of refractory material.

Silicon acid as binding agent.

The mould built by dipping

the model in a mixture of

Mould dried and heated in an

oven (1000 �C). Parted mould

ceramic material with silicon

acid as binding agent. The

mould is dried and the wax

is melted away. The mould

is burnt

Good accuracy Very good accuracy

Large components. Small compounds, 0.1– 0.5 kg

Small series or single

components possible from

Large series necessary for

economic reasons

economic point of view


HINT A52

Exercise 6.5a

The coarseness of the structure is determined by the growth

rate. Describe this fact mathematically.

Hint A185

HINT A53

Exercise 3.5

The casting rate vcast is the velocity of the strand when it

leaves the chill-mould.

How do you calculate vcast?

Hint A226

HINT A54

(A270)

vmax ¼ pD

2t¼ pD

2

4havðTL � T0Þarð��HÞ

or

vmax ¼ 2pDhavðTL � T0Þarð��HÞ ð7Þ

The average heat transfer coefficient has been calculated

as 925 W/m2 K in Hint A321. The temperature of the cool-

ing water is assumed to be 100 �C. D ¼ 2:0 m and

a ¼ 60 � 10�3 m. These values and material constants,

given in the text, are inserted into Equation (7):

vmax ¼ 2p� 2:0 � 925ð1083 � 100Þ0:060 � 8940 � 206 � 103

¼ 0:10 m=s


Hint A310

HINT A55

Exercise 5.2a

Draw a sketch of the temperature profile of a cross-section

of the inlet.

Hint A116

HINT A56

(A300)

Other factors are temperature, structure of the solidified

melt, thermal conductivity, heat of fusion and magnitude

of the flow.

How do they affect the fluidity?

Hint A164

HINT A57

(A229)

The expression of �Tmelt in Equation (5) is introduced into

Equation (4):

dr

dt¼

Ahcon vfront �dr

dt

� �r� 4pr2ð��HÞm

1

Nð8Þ

Introduction of known values gives

dr

dt¼

1:76 � 40 � 103 4:0 � 10�4 � dr

dt

� �7:0 � 103 � 4pð10 � 10�6Þ2 � 272 � 103 � 0:010

1

N

Solve dr=dt.

Hint A171

HINT A58

(A173)

It is reasonable to use an average value of T because the

temperatures are high.

T ¼ 1500 þ 273 ¼ 1773 K

If you insert the given values into Equation (1) you

obtain the radiated excess heat:

Qcoolrad ¼ AesBðT4 � T4

0 Þt ¼ 0:40 � 1:00 � 0:2 � 5:67

�10�8ð17734 � 2934Þ � 10 � 60 ¼ 2:68 � 107 J

How do you calculate the total excess heat?

Hint A296

HINT A59

Exercise 6.6

Consider the temperature profile of the casting and illustrate

it graphically. Set up the heat balance of the casting when

convection is taken into consideration.

Hint A259


HINT A60

Exercise 5.7

Discuss the heat transport in the strip, illustrate its tempera-

ture profile during cooling and solidification processes and

discuss reasonable approximations.

Hint A152

HINT A61

(A149)

Answer

Region VI.

When all the melt has solidified the cooling rate

becomes constant. dT=dt < 0. The solid phase cools at a

constant cooling rate. The cooling process is controlled

by the heat flow to the surroundings. It is described by

the heat capacity csp of Cu.

dQ

dt¼ Vrcs

p � dT

dt

� �

HINT A62

Exercise 4.5

What equation can be used to calculate the solidification

time in case (a)?

Hint A159

HINT A63

(A102)

Assume that the melt has the volume V. Set up a heat

balance.

Hint A211

HINT A64

(A256)

�2prLrð��HÞ dr

dt¼ � k � 2pLðTmelt � TiÞ

lnr

R

� � ð4Þ

The temperature Ti is not a constant, but is a function of

the position of the solidification front. To find an expression

of Ti, you need another equation. Which one?

Hint A163

HINT A65

(A339)

Inserting given data and material constants, you obtain

t ¼ rð��HÞ2sBeðT4

L � T40 Þ

R

t ¼ 7:8 � 103 � 276 � 103

2 � 5:67 � 10�8ð17534 � 3004ÞR ¼ 2:0 � 103 R

Answer

Provided that the wire is so thin that the temperature gradi-

ent can be neglected, the solidification time of the wire is

proportional to its radius:

t ¼ rð��HÞ2sBeðT4

L � T40 Þ

R

The given data represents the function

t ¼ 2:0 � 103 R ðSI unitsÞ

which is illustrated in the figure.

HINT A66

(A121)

Ti metal can be solved from Equation (5):

Ti metal ¼TL � T0

1 þ h

kyL

þ T0 ð6Þ

which is identical with Equation (4.45) on page 74 in

Chapter 4.

It can be seen from Equation (6) that if yL ¼ 0 then

Ti metal ¼ TL. Hence Ti metal can be read at the T-axis

where yL ¼ 0. What is yL?

How do you obtain corresponding Ti metal and yL values?

Hint A312

t (s)

1.00

0.80

0.60

0.40

0.20

0

0 100 200 300 400 500

R (µm)


HINT A67

Exercise 6.8b

The natural thing to do is to set up a heat balance. Assume

that the equiaxed crystals can be regarded as spheres.

Hint A160

HINT A68

Exercise 3.10a

Consider the forces that act on the curved metal surface in

the corner.

Hint A115

HINT A69

(A276)

Remember that Ti ¼ TE

ffiffiffiffiffiffiffiffitcool

p¼

rLLðsin 5�ÞcLp

ffiffiffip

p

2ðTE � T0Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffikmouldrmouldcmould

p

q � 100

Introduction of numerical values gives

ffiffiffiffiffiffiffiffitcool

p¼ 7:0 � 103 � 0:0872 � 420

ffiffiffip

p� 100

2ð1153 � 20Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:63 � 1:61 � 103 � 1:05 � 103

p L

¼ 19:43 L

How do you proceed to find the cooling rate as a func-

tion of L?

Hint A258

HINT A70

(A161)

The distance d is given by

d ¼ ut ¼ 8 � 0:10 ¼ 0:80 m


Hint A189

HINT A71

(A160)

vcrystal ¼dr

dt

How do you proceed?

Hint A140

HINT A72

(A45)

Answer

Two methods fulfil the technical demands. In this situation,

the economic aspects become important. The text says

nothing about the number of components.

In the case of small series, choose the Shaw process.

In the case of large series, choose investment casting.

HINT A73

Exercise 5.1b

If you can calculate the heat radiated from the upper steel

surface during the solidification time, you have solved the

problem.

For this purpose you must calculate the solidification

time. How?

Hint A141

HINT A74

Exercise 5.10b

Maybe is not reasonable to use an average of the h

values? More careful calculations will answer this ques-

tion. In what way do you have to revise your

calculations above?

Hint A123

HINT A75

(A32)

Separate the variables and integrate the differential equa-

tion:

dt ¼ 6Aingot

Asprue

ffiffiffiffiffi2g

p dhffiffiffiffiffiffiffiffiffiffiffiffiH � h

p )ðt

0

dt ¼ 6Aingot

Asprue

ffiffiffiffiffi2g

pðh

0

dhffiffiffiffiffiffiffiffiffiffiffiffiH � h

p

which gives

t ¼ 6Aingot

Asprue

ffiffiffiffiffi2g

p ð�2Þ½ffiffiffiffiffiffiffiffiffiffiffiffiH � h

p�h0


or

t ¼ 6Aingot � 2

Asprue

ffiffiffiffiffi2g

p ðffiffiffiffiH

p�

ffiffiffiffiffiffiffiffiffiffiffiffiH � h

pÞ

How do you obtain the filling time?

Hint A47

HINT A76

Exercise 4.8b

How do you obtain the solidification rate?

Hint A124

HINT A77

(A278)

Answer

At the distance that corresponds to the maximum of the

curve, the film of evaporated oil or molten casting powder

is gone and there is good, direct contact between the steel

and the chill-mould.

HINT A78

(A301)

Because ngrowth ¼ dr=dt!Derivatization of Equation (6) with respect to time

gives

dr

dt¼ 1

6C

t�5=6

N1=3ð7Þ

Use the relation (1) which you set up in Hint A185. How do

you proceed?

Hint A251

HINT A79

(A27)

(A188)

Answer

(a)

t ¼ rð��HÞðTmelt � T0Þ

R2

4kþ R

2h

� �

(b)

dyL

dt

�� ¼ Tmelt � T0

rð��HÞ1

R0 � yL

2kþ 1

2h

HINT A80

(A216)

(A341)

On inoculation, the length of the columnar zone decreases.

This reduces the anisotropy of the mechanical properties,

which is an advantage during rolling and forging. This is

especially the case in the production of aluminium sheets.

Answer

With the aid of basic equations the relation

dr

dt¼ 1

6C

t�5=6

N1=3

can be derived.


(a) From this equation, in combination with the equation

mgrowthl2 ¼ constant, it can be shown that when N is

increased, owing to inoculation, the lamella distance lalso increases, i.e. the structure becomes coarser, which

was to be proved.

(b) From this equation, in combination with the equation

ngrowth ¼ mðTE � TÞn, it can be shown that the under-

cooling decreases when N is increased. This reduces

the risk of white solidification and is the main reason

for inoculation of a cast iron melt.

Another reason, valid for all metals, is better mechan-

ical properties of the cast metal when the number of

crystals in the melt is increased. Inoculation improves

the quality of the cast metal.

HINT A81

(A10)

The desired expression is Equation (3.13) on page 34 in

Chapter 3:

tfill ¼2Acasting

Asprue

ffiffiffiffiffi2g

p ðffiffiffiffiffiffiffiffiffihtotal

p�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffihtotal � hcasting

pÞ ð1Þ

Which quantities in Equation (1) are known or can be

calculated?

Hint A166

HINT A82

(A138)

Answer

Region I.

The slope of the curve is determined by the heat trans-

port to the surroundings, which is controlled by the natural

convection in the melt. The cooling process is described by

the heat capacity cLp of Cu.

dQ

dt¼ VrcL

p � dT

dt

� �

The melt cools and region I lasts until the excess tem-

perature of the melt is gone. Equiaxed crystals are formed

at the outer surface of the ingot, close to the mould, as soon

as T � T� (critical temperature for nucleation) and the sur-

face crystal zone is formed during the casting operation.

The transition from the equiaxed crystals in the surface

zone to dendrites in the columnar zone cannot be observed

in this cooling curve as it shows the temperature of the

centre of the ingot. Compare Figure 6.38 at page 163 in

Chapter 6. The columnar crystals start to grow inwards.

Characterize region II.

Hint A299

HINT A83

(A215)

Introduce known values and the measured valued of yL and

t used above into Equation (1) and proceed as above. Trans-

formation of Equation (1) gives

thIIðTL � T0Þ ¼ rmetalð��HÞyL þ hIIrmetalð��HÞ2k

yL2 ð4Þ

or

hII ¼yL

tðTL � T0Þrmetalð��HÞ �

yL2

2k

¼ 14 � 10�3

140ð1500 � 25Þ7:9 � 103 � 270 � 103

� ð14 � 10�3Þ2

2 � 30

¼ 144 W=m2

K

If you use another pair of values to check the value, for

example yL ¼ 216 mm and t ¼ 100 min ¼ 6000 s, and

introduce these values into Equation (1), you obtain

hII¼yL

tsolðTL�T0Þrmetalð��HÞ�

yL2

2k

¼ 0:216

6000�ð1500�25Þ7:9�103�270�103

�ð0:216Þ2

2�30

¼64

This value does not agree with that obtained with the aid

of the ‘knee’ values. The reason is that the mould is heated

by the metal at high values of t. In this case, Equation (1) is

not valid as Ti is no longer constant and equal to TL.

A upper h total

A lower

A casting

h casting

A runner


Answer

(b)

The heat transfer coefficient h is of magnitude 1:5�102 W=m

2K in region I and at the beginning of region II.

HINT A84

(A210)

According to Equation (2), dr=dt ¼ 0, i.e. the nuclei cannot

grow.

Answer

Nuclei can probably be formed ahead of the solidification

front but they cannot grow because the melt is not under-

cooled. Neither the free crystals nor the solidification

front can grow.

HINT A85

Exercise 3.9b

To get a numerical value of Lf you need a value of the velo-

city of the melt. How do you get v?

Hint A148

HINT A86

(A155)

Region I.

Equation (2) can be written as

hI ¼rmetalð��HÞ

TL � T0

yL

tð3Þ

The corresponding values of yL and t at the ‘knee’ of the

curve are 14 mm andffiffit

p¼ 1:5 min½, respectively, the latter

corresponding to t ¼ 1:52 min � 2:3 min ¼ 140 s.

Calculate hI and try to find hII in a similar way.

Hint A215

HINT A87

Exercise 5.1c

You want to calculate the maximum thickness of the solidi-

fied upper layer. Set up a heat balance.

Hint A248

HINT A88

(A16)

The natural choice of points 1 and 2 can be seen in the fig-

ure. Bernoulli’s equation can be written as

p1 þ rgh1 þrv1

2

2¼ p2 þ rgh2 þ

rv22

2¼ constant

If you choose the inlet level as the zero level you obtain

patm þrg� 0þrvsprue2

2¼ patm þrgðH � hÞþrvingot

2

2ð1Þ

What additional equations can you find?

Hint A103

HINT A89

(A271)

The two expressions must be equal.

rLLðsin 5�ÞcLp � dT

dt

� �¼


p

pt

sðTi � T0Þ ð5Þ

Equation (5) is a differential equation. Solve it and cal-

culate the time required to remove the excess temperature

of the melt by choice of convenient limits of the integrals.

Hint A276

HINT A90

(A324)

pdupper2

4¼ Aupper ) dupper

¼ 2

ffiffiffiffiffiffiffiffiffiffiffiAupper

p

r¼ 2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:21� 10�4

p

r¼ 1:24� 10�2 m

pdlower2

4¼ Alower ) dlower

¼ 2

ffiffiffiffiffiffiffiffiffiffiffiAlower

p

r¼ 2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1:10� 10�4

p

r¼ 1:19� 10�2 m

1

A sprue = 0.15 × 0.15 m2

Zero level

Aingot = 0.5 m2

H = 1.7 m Sprue Ingot h fill = 1.5 m

2 h

•

•

↓


Answer

The calculated difference between the upper and lower dia-

meters is small and of the order of 0.5 mm. This value is

probably lower than the uncertainty limits. The difference

can be neglected. A straight sprue can be used.

HINT A91

(A187)

1:5 � 10�2ffiffit

p lden2 ¼ 10�10

which gives

lden ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi10�10

ffiffit

p

1:5 � 10�2

sð4Þ

How do you proceed?

Hint A344

HINT A92

(A150)

The heat transfer coefficient h between the rolls and the

metal is high but the thermal conductivity k of Al (and

most metals) is high and the distance between the solidifi-

cation front and the melt is small. We have reasons to

assume that

Nu ¼ hs

k¼ hyL

k� 1

In this case the temperature profile is the one seen in the

figure.

Set up a balance of the heat flux (energy per unit time

and unit cross-sectional area).

Hint A174

HINT A93

(A176)

dyL=dt represents the solidification rate or growth rate

vgrowth. By means of the relation vgrowth lden2 ¼ 1:0�

10�12 m3=s in the text you obtain

dyL

dt¼ vgrowth ¼ 10�12

lden2

ð2Þ

What use can you make of the other relation in the text?

Hint A298

HINT A94

Exercise 5.12d

Discuss the optimal process parameters.

Hint A189

HINT A95

(A20)

The outlet velocity v2 depends on the height h of the upper

steel surface above the bottom of the tundish.

To find this relation, you may use Bernoulli’s equation

[Equation (3.2) on page 29 in Chapter 3].

Hint A44

HINT A96

(A213)

According to Equation (4.48) on page 74 in Chapter 4, you

obtain

t ¼ rð��HÞTL � Tw

yL

h1 þ h

2kyL

� �ð3Þ

How do you get the desired distance?

Hint A227

T TL

Mould Solid Liquid

T0

0 y L

y

1 A 1

v 1

h Tundish

A 2 A 2

v 2

A 3 v 3

·

·

↓

↓


HINT A97

(A284)

Answer

The discontinuity of the curve after about 45 s is due to the

shrinkage of the cooling shell. An air gap is formed which

reduces the heat transfer instantly.

The solidification rate is the derivative of the curve. It

increases at the end of the solidification process owing to

the decreasing area of the solidification front. The heat

flow is reduced inside the ingot while the cooling at the

solid/mould interface is unchanged.

HINT A98

(A261)

Nu � 1 means that ðh=kÞyL is small compared with 1

and can be neglected in Equation (1). Hence you simply

obtain

h ¼ rð��HÞTL � T0

vgrowth ð2Þ

Calculate h.

Hint A134

HINT A99

(A128)

1. Choose an arbitrary value of l, use Table 4.4 on page 67

in Chapter 4 to find erf(l) and use Figure 4.15 on page 72

in Chapter 4 to read an approximate value offfiffiffip

plel

2

.

2. Form the product (4). If the product is 4.33 you have

found the right value of l. If not, try other l values

to come close to 4.33. When you are close, use

Table 4.4 and calculateffiffiffip

plel

2

. List some values of l

and calculate the corresponding values offfiffiffip

plel

2�ð0:410 þ erfðlÞÞ. The final choice of l can be made

by interpolation.

Start for example with l ¼ 0:5.

Hint A308

HINT A100

Exercise 4.10a

There are two alternatives treated in Chapter 4 when heat

transport across a mould/metal interface is concerned.

Which ones?

Hint A153

HINT A101

Exercise 4.7

If you rotate the wedge 90� and replace it with a plate with

two vertical surfaces at a distance yL from each other, it will

be easier to associate the present problem with the theory of

solidification in Section 4.3.3 in Chapter 4. Draw a figure

and try to find an expression for the solidification time.

Hint A169

HINT A102

(A295)

The cooling curve consists of three intervals, two cooling

periods and one solidification period. Consider each time

interval separately.

How do you calculate the cooling time of the melt?

Hint A63

T

Cooling of the melt

T excessT L Cooling of the solid down

to room temperature

t

T0

0 t 1 t 2 t 3

Solidification process at constant temperature.Phase transformation


HINT A103

(A88)

Consider the ingot. The inlet velocity is vsprue. The total

ingot area corresponds to six moulds.

The velocity of the ingot area is vingot. It can also be written

vingot ¼dh

dtð2Þ

The principle of continuity is valid for an incompressible

liquid [Equation (3.1) on page 29 in Chapter 3].

Aspruevsprue ¼ Atotalvingot ¼ 6Aingot

dh

dtð3Þ

How many variables and equations do you have? Is the

number of equations sufficient to find the filling time? If the

answers are satisfactory, derive the time required to fill the

ingots to the level h.

Hint A142

HINT A104

Exercise 5.5a

According to the advice in the text, it might be a good idea

to calculate the average heat transfer coefficient hav.

Hint A198

HINT A105

Exercise 5.4a

It might be helpful to consider Figure 5.18 on page 106 in

Chapter 5. The chill-mould is strongly water cooled. The

figure in the text shows that the heat flux increases with

the distance from the top of the mould in region 1. Why?

Hint A194

HINT A106

(A238)

Hence you obtain

Vmetal

A¼ pr2L

2prL¼ rL

2Lþ 2r¼ 1160� 20

7:2� 103 � 162� 103

� 2ffiffiffip

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:63� 1:5� 103 � 1:05� 103

p ffiffiffiffiffiffiffiffittotal

pþ 1� 0:63ttotal

2r

� �

or

0:15� 0:60

ð2� 0:60Þþ ð2� 0:15Þ ¼1160� 20

7:2� 103 � 162� 103

� 2ffiffiffip

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:63� 1:5� 103 � 1:05� 103

p ffiffiffiffiffiffiffiffittotal

pþ 1� 0:63ttotal

2� 0:15

� �

�6:0� 10�2 ¼ 0:977� 10�6ð1:124� 103 �ffiffiffiffiffiffiffiffittotal

pþ 2:1ttotalÞ

or

2:1ttotal þ 1:124 � 103ffiffiffiffiffiffiffiffittotal

p¼ 6:0 � 10�2

0:977 � 10�6

or

ttotal þ 535ffiffiffiffiffiffiffiffittotal

p¼ 286 � 102

which gives

ffiffit

p¼ �267:5 �

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi267:52 þ 28600

p¼ �267 þ 316 ¼ 49

or

ttotal ¼ 492 s ¼ 40 min

Answer

The solidification time of the cylinder is about 40 min.

HINT A107

(A282)

Now you know yL as a function of time. If you can find lden

as a function of time, you will be close to the solution of the

problem. How?

Hint A187

A ingot A sprue


HINT A108

(A211)

�rVcp

ðTL

Tstart

dT ¼ A

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffikmould rmould cmould

p

p

sðTi � T0Þ

ðt1

0

dtffiffit

p

or

2ffiffiffiffit1

p¼ rVcpðTstart � TLÞ

AðTi � T0Þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip

kmouldrmouldcmouldp

s

or

t1 ¼ rVcpðTstart � TLÞ2AðTi � T0Þ

� 2 pkmould rmouldcmould

p

ð2Þ

Insert material constants and other given values and

calculate t1.

Hint A332

HINT A109

(A21)

The wire solidifies rapidly in air. After complete solidifica-

tion, it starts to cool to the temperature of the air. The wire

is so thin that you may neglect the radial temperature gra-

dient in it. The consequence is that each wire element has a

uniform temperature, i.e. the surface and the centre have the

same temperature.

Discuss possible ways of heat removal.

Hint A277

HINT A110

Exercise 5.8

Consider the heat flow in the strip.

Hint A192

HINT A111

(A281)

Equation (4.26) on page 70 in Chapter 4 describes the

position of the solidification front yL as a function of

time t:

yLðtÞ ¼ lffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4ametalt

pð1Þ

where

ametal ¼kmetal

rmetalcmetalp

ð2Þ

[Equation (4.11) on page 62 in Chapter 4].

l is a constant. How can you determine its value?

Hint A313

HINT A112

(A50)

The sulfur balance in the molten metal when a drop with

volume dV is added to the melt and a volume of equal

size solidifies:

c0dVContribution of S

to the melt due to

addition of the

volume dV from

the electrode

� cmeltdVLoss of S

in the melt

due to soli-

dification of

a volume dV

¼ Vmeltdcmelt

Change of S

content in the

melt

ð1Þ

V and cmelt are variables while Vmelt is kept constant.

Equation (1) is separable. Solve it!

Hint A201

Electrode

Slag bath

V melt Molten metal

V Water-cooled

Solid metal Cu mould

Water-cooled base plate


HINT A113

(A332)

Integrate Equation (3) and solve the solidification

time t2.

Hint A224

HINT A114

(A15)

According to the empirical relation (5.30) on page 112 in

Chapter 5, the heat transfer coefficient will be

h ¼ 1:57 w0:55ð1 � 0:0075TwÞa

where a¼ a machine parameter � 4, w ¼ water flux and

Tw ¼ water temperature (�C).

To calculate the h values for the different zones you must

know the water flux w for the different zones.

Hint A193

HINT A115

(A68)

For symmetry reasons, the forces that act on the curved sur-

face have a resultant force in the direction of the symmetry

line OA, which is marked in the figure.

The pressure pi inside the sphere is the pressure of the

metal melt. It is the sum of the atmospheric pressure p0

and the static pressure rgh of the melt, where h is the height

up to the free metal surface. It is reasonable to use an aver-

age value of h.

The pressure p outside the sphere is the external atmo-

spheric pressure p0. Pressures are perpendicular to the sur-

face, independent of its direction.

In addition, there are surface tension forces acting in

the tangent plane along the boundary line, i.e. along the

circle with radius r. Find the expression for the resulting

surface tension forces in terms of r and y by use of a

force balance.

Hint A144

HINT A116

(A55)

In the figure shown, T0 ¼ temperature of the surround-

ings, Ti metal ¼ temperature of the outer metal surface,

TL ¼ liquidus temperature of the melt and Tmelt ¼ tempera-

ture of the melt, including the excess temperature.

At steady state, the heat flux through the system must be

the same at all interfaces.

No freezing occurs at the inlet. The shell thickness is

constant.

Use conventional designations and set up the heat

transport equations. Assume that the interface areas are

equal.

Hint A36

HINT A117

(A294)

The solidification is complete when the solidification fronts

of two opposite sides meet. Which pair, 2– 4 or 1–3, will

meet first?

@Q

@t¼ r

dV

dtð��HÞ

Heat flow

generated

by solidification

of the melt

¼ A


p

pt

sðTi � T0Þ

Heat flow across the

mould=metal interface:Compare Equation ð4:70Þon page 79 in Chapter 4

ð3Þ

z

A

y

R r

Rx

O

q

∑

∑

∑

T

T melt

T i metal S LT 0

y

y L

T L


As h4 < h, you can conclude from Equation (12) that

y4 < y. Hence side 4 will grow more slowly than the

other three equivalent sides. The conclusion is that sides

1–3 meet before 2–4, which is illustrated in the figure.

When the solidification fronts of the sides 1 and 3 meet,

the casting becomes completely solid. For comparison with

the result in Exercise 5.10a, you have to calculate the soli-

dification time and solidification rate for sides 1–3.

Set up an analogous heat balance for sides 1 and 3.

Hint A329

HINT A118

(A302)

Answer

The solidification time is about 13 s during sand mould

casting and about 5 s during metal mould casting. The pro-

duction capacity will be more than doubled in the latter

case.

HINT A119

(A224)

ðffiffiffiffit2

p�

ffiffiffiffit1

pÞ2 ¼ 2:7 � 103 � 0:253 � 398 � 103

2 � 6 � 0:252 � ð660 � 20Þ

� 2

� p0:63 � 1:61 � 103 � 1:05 � 103

¼ 3610 s

which gives

ffiffiffiffit2

p�

ffiffiffiffit1

p¼

ffiffiffiffiffiffiffiffiffiffi3610

p¼ 60:08

or ffiffiffiffit2

p¼

ffiffiffiffiffi79

pþ 60:08 ¼ 68:97

which gives

t2 � 4757 s ¼ 79:3 min

Solidification time ¼ 79.3 min � 1.3 min ¼ 78 min

Set up a heat balance for the final cooling process.

Hint A34

HINT A120

Exercise 4.11b

List corresponding values of time and thickness yL of the

solidified shell. Plot yL as a function of time.

Hint A284

HINT A121

(A264)

The heat flows are equal, which gives

kAdT

dy¼ hAðTi metal � T0Þ

which gives

h ¼ k

Ti metal � T0

dT

dyð3Þ

The temperature gradient is approximately constant in

the shell and can be replaced by

dT

dy¼ TL � Ti metal

yL � 0ð4Þ

This expression is introduced into Equation (3):

h ¼ k

yL

TL � Ti metal

Ti metal � T0

ð5Þ

If you use the figure to find corresponding Ti metal and yL

values, you can calculate h for each curve. How do you read

Ti metal from the curve?

Hint A66

HINT A122

(A33)

You will use Equation (1), hence you need a value of the

thermal diffusion coefficient ametal [Equation (4.10) on

page 62 in Chapter 4]:

ametal ¼kmetal

rmetalcmetalp

¼ 30

7500 � 650¼ 6:15 � 10�6 m2=s

4

3 1

2


Then you solve the solidification time by squaring

Equation (1):


p) t ¼ 1

4ametal

yL

l

� �2

ð5Þ

What value of yL has to be inserted into Equation (5) to

find the solidification time?

Hint A244

HINT A123

(A74)

Equation (2) will not be introduced. You have to calculate

separate solidification times for the different side pairs of

the casting.

The thicknesses of the solidified shells at time t are

shown in the figure. Owing to a different heat transfer coef-

ficient h4, the thickness of shell 4 is less than that of the

other shells. Assume that

y2 � y1 ¼ y3 ¼ y and y 6¼ y4

You also have

z1 ¼ z3 ¼ a � ðy þ y4Þ ð8Þz2 ¼ z4 ¼ a � 2y ð9Þ

Set up the material balances for sides 2 and 4.

Hint A294

HINT A124

(A76)

Replace R by R0 � yL in Equation (10) in Hint A208 and

derive it with respect to t.

Hint A297

HINT A125

Exercise 5.11

Check Nusselt’s number!

Hint A340

HINT A126

(A308)

Interpolation gives l ¼ 0:90.

How do you calculate the value of the solidification time

when l is known?

Hint A25

HINT A127

(A4)

Assume that you add the m kg of steel powder per kg of

melt. It will be heated by the melt, which cools. Apply

the energy law!

Hint A266

HINT A128

(A275)

The material constants of Fe and Cu are introduced into

Equation (3) in order to solve l. The temperature values

shown in the figure are reasonable assumptions.

y4

z4

y3y1

a z3 z1 a

z2

y2 a

a

R 0

y L R

l erf(l) 0.410 þ erf(l)ffiffiffip

plel

2 ffiffiffip

plel

2 ½0:410 þ erfðlÞ�0.80 0.7421 1.1521 2.6891 3.10

0.85 0.7707 1.1807 3.1029 3.66

0.90 0.7969 1.2069 3.5859 4.33

0.95 0.8209 1.2309 4.1520 5.11


The highest possible value of the temperature at the

external side of the Cu plate is 100 �C.

With the given values inserted into Equation (3), you

obtain

830�ð1808�400Þ272�103

¼ffiffiffip

plel

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi32�7880�830

350�8940�397

rþerf ðlÞ

!

or

4:33 ¼ffiffiffip

plel

2

ð0:410 þ erf ðlÞÞ ð4Þ

You have available Table 4.4 with the error function

erf(z) on page 67 and Figures 4.14 and 4.15 on pages

71–72 in Chapter 4. Use them to find a numerical solution

of l by trial and error.

Hint A99

HINT A129

Exercise 5.2b

Use Equation (8) or rather Equation (7) in Hint A237 to cal-

culate the excess temperature with the aid of the given

values.

Hint A200

HINT A130

Exercise 5.12b

Consider the solidification process and set up a heat

balance.

Hint A288

HINT A131

Exercise 5.10a

The square casting solidifies laterally from all four sides.

Check Nusselt’s number to find the tentative temperature

profile.

Hint A197

HINT A132

(A206)

The maximum casting rate is

vmax ¼ l

t¼ pD

2tð17Þ

where D is the diameter of the wheel and l is half the per-

iphery.

You introduce the expression for t into Equation (17):

vmax ¼ pD

2

hðTL � T0Þ

arð��HÞ 1

2� 1

81 þ h4

h

� �� ð18Þ

Insert values of material data, calculate vmax and com-

pare the result with that in 5.10a.

Hint A310

HINT A133

Exercise 3.8

Consider a small volume element dy and discuss the energy

change during the time dt.

Hint A18

HINT A134

(A98)

The maximum heat transfer coefficient is obtained if you

introduce the maximum solidification rate into Equation (2):

hmax ¼ rð��HÞTL � T0

vmaxgrowth ð3Þ

If you insert the given values and the material constants

you obtain

hmax¼7:0�103�272�103

1150�20�4:0�10�4¼6:7�102 W=m

2K

Cu Solid Liquid

1808 K

400 K

373 K

~

Steel belt

Roller Casting Roller

Roller


Answer

The heat transfer coefficient must not exceed 0.6 kW/m2 K

(0.67 kW/m2 K).

HINT A135

Exercise 4.8a

Consider the heat flow through a cylinder element with

radius r and height L and set up a heat balance.

Hint A315

HINT A136

(A212)

Cooling capacity of the mould:

A large cooling capacity results in a small maximum

fluidity length as heat is transported away quickly and the

metal solidifies quickly. The maximum fluidity length

decreases with increasing cooling capacity.

What about the influence of surface tension?

Hint A191

HINT A137

Exercise 4.3a

Owing to the dimensions of the casting it can be regarded as

a plate with a thickness of 100 mm, which solidifies one-

dimensionally. The Al ingot is cast in a sand mould.

Draw a figure which shows the temperature distribution in

the mould and metal. What equation is valid for casting in a

sand mould?

Hint A13

HINT A138

(A5)

The temperature–time curve for the centre of the ingot

depends on

� cooling rate (rate of heat losses to the surroundings)

� nucleation rate

� growth rate

� heat of solidification and heat capacitivities.

The cooling rates and the heat of solidification are

constant.

Characterize region I.

Hint A82

HINT A139

(A311)

The casting rate equals the ratio of the cooling length to the

solidification time:

vcast ¼l

ttotal

ð5Þ

Expression (4) is introduced into Equation (5):

vcast ¼l

ttotal

¼ 2lhðTL � T0Þrdð��HÞ ð6Þ

Find the distance d between the rollers as a function of

the casting rate and material constants.

Hint A205

HINT A140

(A71)

You derive Equation (2) and introduce the derivative into

Equation (1):

Ahcon�Tmelt ¼ Nr� 4pr2 dr

dtð��HÞ ð3Þ

which gives

dr

dt¼ Ahcon�Tmelt

r� 4pr2ð��HÞ1

Nð4Þ

All quantities except A, �T and N are known. How do

you calculate the area A?

Hint A328


HINT A141

(A73)

A smart way is to use the so-called ‘rule of thumb’. You can

find it on page 97 in Chapter 5.

Hint A207

HINT A142

(A103)

Variables are vsprue, vingot, h and t. The number of indep-

endent equations is three, which is sufficient to eliminate

vsprue and vingot and obtain the time t as a function of the

height h.

Derive the time t as a function of the height h.

Hint A199

HINT A143

(A326)

From the figure in the text, it can be seen that the cooling

length L is 2.5 m. With the aid of Equation (3) you can find

the maximum casting rate:

vmax ¼ L

t¼ 2LhðTL � T0Þ

rð��HÞ1

dð4Þ

Introduce material constants and other values, given in

the text, and calculate the relation between vmax and d.

Hint A342

HINT A144

(A115)

The surface tension forces have tangential directions

along the circle with radius r. Their resultant in the

direction OA is 2prscosy (compare the figures in

Example 3.6 on page 52 in Chapter 3). Three forces act

on the curved metal surface.

ppr2

Resulting force

from the external

pressure directed

towards

the centre

þ 2prscosyResulting force

of the surface

tension forces

directed towards

the centre

¼ pipr2

Resulting force

from the internal

pressure; directed

outwards

from the centre

ð1Þ

The external pressure is

pi ¼ p0 þ rgh ð2Þ

Combine Equations (1) and (2) and solve r.

Hint A245

HINT A145

(A156)

rð��HÞTL �Tw

yL

h1þ h

2kyL

� �1¼ rð��HÞ

sBeðTL4 �T0

4Þð0:0060� yLÞ

1

640

yL

10001þ1000

440yL

� �

¼ 1

5:67�10�8ð9334 �2934Þð0:0060� yLÞ

yL 1þ1000

440yL

� �

¼ 640�103

5:67�10�8ð9334 �2934Þð0:0060� yLÞ

yL þ2:2727yL2

¼ 640�103

5:67�10�8 �7:500�1011ð0:0060� yLÞ

yL þ2:2727yL2 ¼ 15:050ð0:0060� yLÞ

yL2 þ0:44yL ¼ 0:0397323�6:622yL

yL2 þ7:062yL ¼ 0:0397323

yL ¼�3:531�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3:5312 þ0:0397323

p¼�3:531�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12:467961þ0:039732

p

You want the positive root

yL ¼�3:531þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi12:507693

p¼�3:531þ 3:5366 ¼ 0:0056m

Check:

The sum of YL (Hint 156) and YL is 6.0 mm, in agreement

with the text in Exercise 5.7.

Answer

The solidification fronts meet at 0.5 mm from the upper

surface of the strip.


HINT A146

(A333)

(A268)

Answer

(a) Figure 4.17 is valid for all cases. Figure 4.27 is valid for

small values of Nusselt’s number hyL=k.

(b) Diagrams for steel and copper for the two h values are

shown below.

HINT A147

Exercise 5.5b

What condition determines the casting rate?

Hint A228

HINT A148

(A85)

You use the equation

v ¼ffiffiffiffiffiffiffiffi2gh

pð4Þ

and read the height of the melt from the figure:

h ¼ ð45 þ 25 þ 50 þ 8Þmm ¼ 128 mm ¼ 0:128 m

v ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 � 9:81 � 0:128

p¼ 1:58 m=s

Find the required material constants of Al and use Equa-

tions (2) and (3).

Hint A322

HINT A149

(A209)

Answer

Region V.

Region V is characterized by dT=dt < 0. The total heat

of solidification is smaller than the heat losses to the sur-

roundings. The heat of solidification decreases owing to

decreasing area A of the solidification front, when the crys-

tals impinge against each other.

Region V lasts until all melt has solidified, fs ¼ 1.

Characterize region VI.

Hint A61

For steel you have

TL ¼ 1530 �C

kFe ¼ 30 W=m K

h ¼ 2 � 102 W=m2

K h ¼ 2 � 103 W=m2

K

yL Ti Fe yL Ti Fe

0.01 1436 0.01 926

0.02 1352 0.02 667

0.03 1278 0.03 523

0.04 1212 0.04 432

0.05 1152 0.05 368

For copper you have

TL ¼ 1083 �C

kCu ¼ 398 W=m K

h ¼ 2 � 102 W=m2

K h ¼ 2 � 103 W=m2

K

yL Ti Cu yL Ti Cu

0.01 1078 0.01 1031

0.02 1072 0.02 986

0.03 1067 0.03 944

0.04 1062 0.04 905

0.05 1057 0.05 870


HINT A150

Exercise 5.9

Check Nusselt’s number to estimate the type of temperature

distribution in the solid metal.

Hint A92

HINT A151

(A293)

If you insert the given material constants and temperature

values, you obtain

C ¼ p4

rmetalð��HÞTL � T0

� 21

kmouldrmouldcmouldp

C ¼ p4

2:7 � 103 � 398 � 103

933 � 298

� �2

� 1

0:63 � 1:61 � 103 � 1:05 � 103¼ 2:1 � 106 s=m

2

Calculate the solidification time!

Hint A202

HINT A152

(A60)

Radiation and heat transfer to the plate occur simulta-

neously and cool the melt to the melting-point temperature.

The strip is thin and it is reasonable to assume that its

temperature is constant as long as the excess temperature

persists. The temperature profile is sketched in the figure.

How do you proceed?

Hint A306

HINT A153

(A100)

Case 1: Poor contact between mould and metal

Equation (4.48) on page 74 in Chapter 4 gives the soli-

dification time:

t ¼ rmetalð��HÞTL � T0

yL

h1 þ h

2kyL

� �ð1Þ

Case 2: hyL=2k � 1

If h and/or yL are/is small and k is large, then the term

hyL=2k � 1 and can be neglected in Equation (1). In this

case the solidification time can be written as


yL

hð2Þ

Apply this information to explain the appearance of the

curve.

Hint A257

HINT A154

(A286)

Completely analogous calculations as in 4.3a with other

material data give

ttotal ¼ CyL

2

� �2

¼ 1:5 � 106 � 0:052 ¼ 1:04 h

List the values of solidification time, heat of fusion for

aluminium and steel and a measure of the driving force

for heat transport.

Hint A19

HINT A155

(A48)

The straight way is to use reasonable values of material

data, temperatures and measured values of yL and t as a

basis for calculation of the heat transfer coefficient for the

two alternatives.

Hint A86

HINT A156

(A29)

t ¼ 1

TL � Tw

0:0060 � YL

h1 þ h

2kð0:0060 � YLÞ

�

¼ 1

sBeðTL4 � T0

4Þ YL

ð6Þ

T excess T L

T w T 0


Solve Equation (6) after introduction of known values

and the material constants for aluminium. Assume that

e � 1.

1

660 � 20

0:0060 � YL

10001 þ 1000

2 � 220ð0:0060 � YLÞ

�

¼ 1

5:67 � 10�8ð9334 � 2934Þ YL

ð0:0060 � YLÞ 1 þ 1000

2 � 220ð0:0060 � YLÞ

�

¼ 640 � 103

5:67 � 10�8ð9334 � 2934Þ YL

0:0060ð1 þ 2:273 � 0:0060Þ � 0:0060YL

� YLð1 þ 2:273 � 0:0060Þ þ YL2

¼ 640 � 103

5:67 � 10�8 � 7504 � 108YL

YL2 � 0:0060YL � YL � 1:01364

� 15:05YL þ 0:0060 � 1:01364 ¼ 0

YL2 � 16:07YL þ 0:0060818 ¼ 0

YL ¼ þ8:035 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffið8:035Þ2 � 0:006082

q

As YL < 0:0060, you want the smallest root:

YL ¼ 8:035 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi64:561225 � 0:006082

p

¼ 8:035 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi64:555143

p¼ 8:035 � 8:0346

¼ 0:0004 m

The accuracy of the known quantities involved is not

particularly satisfactory. As a check you may eliminate

YL between Equations (5) and (6) and solve yL. If you do

not want to practice solving another equation of second

degree, you can look at the next hint.

Hint A145

HINT A157

(A49)

Equation (4.45) on page 74 in Chapter 4 gives


1 þ h

kyL

þ T0 ð1Þ

Use this relation and the figures mentioned in the text to

discuss the validities of the two figures.

Hint A333

HINT A158

(A223)

Answer

1. Owing to the motion in the melt, dendrite arms are torn

away, which gives crystal multiplication due to inhomo-

geneous nucleation.

2. The hot melt causes partial remelting of already solidi-

fied dendrite arms. The new particles in the melt give

crystal multiplication in the same way as in point 1.

3. The convection gives increased heat transport from the

melt to the solidification front and a faster decrease of

the superheat of the melt. The last effect results in better

possibilities for the nuclei to survive and grow when the

temperature of the melt has decreased to the nucleation

temperature T�.

HINT A159

(A62)

The dimensions of the cylinder are given. It is easy to

calculate its volume and the area. Hence you can use

Chvorinov’s rule. Are there any complications?

Hint A238

HINT A160

(A67)

Two expressions of the heat flow can be obtained:

dQ

dt¼ Ahcon�T ¼ N

d rVð Þdt

ð��HÞ ð1Þ

where

V ¼ 4pr3

3ð2Þ

How do you define the growth rate of the equiaxed crys-

tals?

Hint A71

HINT A161

(A288)

t ¼ 7:8 � 103 � 50 � 10�6 � 276 � 103

2 � 5:67 � 10�8 � 1 � ð17534 � 3004Þ ¼ 0:10 s

Now you know the total solidification time. How do you

obtain the distance?

Hint A70


HINT A162

(A40)

Sand mould:

If you cast a metal in a dry sand mould, Equation (4.72)

on page 79 in Chapter 4 gives the relation between shell

thickness and solidification time:

yLðtÞ ¼2ffiffiffip

p Ti � T0

rmetalð��HÞ


p

q ffiffit

pð1Þ

or

t ¼ p4

rmetalð��HÞyL

Ti � T0

� 21

kmouldrmouldcmouldp

ð2Þ

Use the material data and calculate the solidification

time.

Hint A273

HINT A163

(A64)

The heat transfer between the solid metal and the mould

can be described by the relation

dQ

dt¼ h � 2pRLðTi � T0Þ ð5Þ

Combine Equations (2) and (5) to eliminate dQ=dt.

Hint A316

HINT A164

(A28)

(A56)

Answer

(a) Increased cooling capacity of the mould ) low Lf

Increased surface tension ) low Lf.

Increased viscosity ) low Lf.

Wide solidification interval of the alloy ) low Lf.

(b) Pure metals and eutectic alloys ) high Lf .

Increased temperature ) low viscosity ) high Lf .

Formation of dendrites prevents the flow strongly

) low Lf.

Low thermal conductivity and high heat of fusion

increase Lf .

Laminar flow gives a larger fluidity length than turbu-

lent flow.

HINT A165

(A296)

The major part of the excess heat is removed by convection

in the steel melt to the surrounding cast iron mould.

The fraction f removed with the aid of radiation is

f ¼ Qcoolrad

Qcooltotal

¼ 2:68

42¼ 0:064


Hint A26

HINT A166

(A81)

tfill can be calculated with the aid of the empirical equation

in the text, where Acasting is the cross-sectional area of the

cylinder ¼ pr2 ðr ¼ 5 cmÞ, hcasting ¼ 23 cm and htotal ¼ 28 cm.

Asprue ¼ Arunner ¼ Alower is unknown and can be derived

from Equation (1) as all other quantities are known. Calcu-

late tfill and then Asprue.

Hint A285

HINT A167

(A343)

You derive Equation (6) and change to SI units, which

gives

vfront ¼dy

dt¼ 2:5 � 10�2

2ffiffiffiffiffiffiffi60t

p ð7Þ

To find vfront you must find the value of t. How do you

find t?

Hint A241


HINT A168

(A263)

You know that the cooling length l is equal to half the per-

iphery of the wheel. Try again!

Hint A270

HINT A169

(A101)

According to Equation (4.48) on page 74 in Chapter 4,

the solidification time of a shell with thickness yL is

t ¼ rð��HÞTL � T0

yL

h1 þ h

2kyL

� �ð1Þ

Relate the wedge and this equation.

Hint A252

HINT A170

(A320)

The values of lden from the diagram are listed in the table

shown.

Calculation of h for the first row in the table:

h ¼ rð��HÞ�10�11

r0ðTL �T0Þr

lden2¼ 7:0� 103 � 280�103 �10�11

65� 10�6ð1450� 20Þr

lden2

which gives

h ¼ 0:21� r

lden2¼ 0:21� 55� 10�6

ð2:0� 10�6Þ2¼ 29� 105 W=m

2K

The other values are calculated in the same way and are

listed in the table.

You want h as a function of z. How do you calculate z?

Hint A214

HINT A171

(A57)

Ndr

dt¼ 2:94 � 106 4:0 � 10�4 � dr

dt

� �

or

vcrystal ¼dr

dt¼ 2:94 � 106 � 4:0 � 10�4

N þ 2:94 � 106¼ 11:76 � 102

N þ 2:94 � 106

Answer

The growth rate of the free crystals in the melt is

vcrystal ¼12 � 102

N þ 3 � 106m=s

HINT A172

(A226)

Consider one of the furnaces and one of the lines.

Mould Solid metal Metal melt

T

TL

T

T i metal

T0

y

0 y (t) yL

y ¼ r0 � r r lden h z

(mm) (mm) (mm) (W/m2 K) (m)

10 55 2.0 29 � 105

20 45 2.0 24 � 105

30 35 2.2 15 � 105

40 25 2.8 6.7 � 105

50 15 3.5 2.6 � 105

60 5 4.3 0.6 � 105

h1

Tundish

A outlet

v outlet

Chill- mould

A strand

v cast

↓

↓


The principle of continuity gives

Aoutletvoutlet ¼ Astrand vcast ð1Þ

You know Astrand from the text and vcast from your own

calculations (Hint A226). Hence you must calculate voutlet

in order to find Aoutlet.

How do you find voutlet?

Hint A305

HINT A173

(A3)

The radiated excess heat can be written as

Qcoolrad ¼ AesBðT4 � T0

4Þt ð1Þ

where A ¼ area of the upper steel surface, T ¼ absolute

temperature of the cooling upper steel surface,

T0 ¼ absolute temperature of the surroundings and

t ¼ solidification time.

Insert given values into Equation (1) and calculate the

radiated excess heat. You have one difficulty to solve: the

temperature of the upper surface varies during the cooling

process. How do you solve that problem?

Hint A58

HINT A174

(A92)

The temperature distribution is stationary. The temperature

is a function of position but not of time. The solidification

heat, developed at the solidification front when it moves

towards the centre at a rate dyL=dt, is transported away

across the interface between the cooled rollers and the

solid metal.

With the aid of basic heat laws, given in Chapter 4, you

obtain

rð��HÞ dyLðtÞdt

Solidification

heat flux

¼ hðTL � T0ÞHeat flux transferred

from the metal to the

cooled wheels ðNu � 1Þ

ð1Þ

How do you proceed to find the solidification time?

Hint A217

HINT A175

(A323)

(A233)

Corresponding calculations for cast iron give

RFe ¼sffiffiffi2

p

rFegh¼ 0:5 �

ffiffiffi2

p

7:8 � 103 � 9:81 h¼ 0:92 � 10�5

h

The ratio of the radii is

RAl

RFe

¼ 8:00

0:92¼ 8:7

Answer

(a) The corner radius R ¼ sffiffiffi2

p=rgh depends on the surface

tension and on the static pressure of the melt. For alumi-

nium it is RAl ¼ 8:0 � 10�5=h (expressed in metres)

where h is the height from the corner up to the free sur-

face of the metal melt.

(b) The corner radius of cast iron is 8.7 times smaller than

that of aluminium. A cast iron corner will therefore be

much ‘sharper’ than that of aluminium.

HINT A176

(A11)

Consider a surface with area A at the solidification front.

The heat transferred across the interface during the time

dt can be written in two ways:

hAðTL � T0Þdt

heat transferred

across the interface

during time dt

¼ rAdyLð��HÞsolidification heat

during time dt

or

hðTL � T0Þ ¼ rð��HÞ dyL

dtð1Þ

How do you proceed? What is the physical significance

of dyL=dt?

Hint A93

Cooled wheel

d S Liquid

y L

l

Cooling length


HINT A177

(A254)

T04 can be neglected as T0

4 � T4. Separation of the vari-

ables and integration gives

ðT

Ti

dT

T4¼ � 2esB

rLcLp R

ðt

0

dt

Continue!

Hint A314

HINT A178

(A12)

The reason is that the heat is transported with the aid of

convection.

As a consequence of large temperature differences

within the melt, natural convection will appear within a

boundary layer in the melt, close to the solidified metal.

The thickness of the boundary layer is [Equation (5.6) on

page 96 in Chapter 5]

dðzÞ ¼ Bg

zðTmelt � TLÞ

� �1=4

ð1Þ

Draw a sketch of the temperature profile of a cross-

section of the ingot and discuss the heat flow in the

solidifying ingot.

Hint A39

HINT A179

(A22)

dQ

dt¼ h � 2pr0�zðTL � T0Þ

heat flow to the

surroundings

per unit time

¼ �r 2prdr

dt�zð��HÞ

solidification heat

per unit time

ð1Þ

where z is the length coordinate of the wire (z ¼ 0 at the

nozzle). How do you continue?

Hint A287

HINT A180

(A34)

�rVcp

ðTfinal

TL

dT ¼ A


p

p

sðTi � T0Þ

ðt3

t2

dtffiffit

p

or

2ðffiffiffiffit3

p�

ffiffiffiffit2

pÞ ¼ rVcpðTL � TfinalÞ

AðTi � T0Þ


kmouldrmouldcmouldp

s

or

ðffiffiffiffit3

p�

ffiffiffiffit2

pÞ2 ¼ rVcpðTL � TfinalÞ

2AðTi � T0Þ

� 2 pkmouldrmouldcmould

p

ð2Þ

Insert material constants and other given values and cal-

culate t3.

Hint A249

HINT A181

(A262)

Answer

h ¼ rð��HÞ � 10�11

r0ðTL � T0Þr0 � y

lden2

or

h ¼ 0:21 � 65 � 10�6 � y

lden2

The heat transfer coefficient decreases with increasing

thickness y ¼ r0 � r of the solidified shell.

The table above shows that t heat transfer coefficient

decreases with increasing distance from the nozzle. The

same is true during the secondary cooling during continu-

ous casting (compare Exercise 5.5 in Chapter 5). In this

case, the reason is probably formation of a steam layer

around the wire.

y ¼ r0 � r r lden h z

(mm) (mm) (mm) (W/m2 K) (mm)

10 55 2.0 29 � 105

20 45 2.0 24 � 105 40

30 35 2.2 15 � 105 80

40 25 2.8 6.7 � 105 140

50 15 3.5 2.6 � 105 230

60 5 4.3 0.6 � 105 360


HINT A182

(A266)

m ¼1:00cL

p ðTmelt � TLÞcs

pðTL � T0Þ þ ð��HÞ

¼ 1:00 � 0:52 � 103ð1520 � 1470Þ0:65 � 103ð1470 � 20Þ þ 272 � 103

¼ 21:4 � 10�3 kg

Answer

More than 22 g of steel powder has to be added; 22 g melts

and becomes part of the molten alloy. The rest of the steel

powder helps to make the steel fine-grained on solidifica-

tion.

HINT A183

(A37)

Equations (3), (4) and (5) give the desired radius of curva-

ture of the melt:

R ¼ rffiffiffi6

p

2¼ 2scosy

rgh

ffiffiffi6

p

2¼ 2s

rgh

1ffiffiffi3

pffiffiffi6

p

2¼ s

ffiffiffi2

p

rghð6Þ

Introduce the known values into Equation (6).

Hint A323

HINT A184

(A36)

Combine Equations (1) and (2) and solve y:

�h2�T ¼ �ks

TL � Ti metal

yð4Þ

which gives

y ¼ ks

h2

TL � Ti metal

�Tð5Þ

TL is known but you have to find Ti metal. How?

Hint A336

HINT A185

(A52)

vgrowthl2 ¼ constant ð1Þ

where vgrowth ¼ dr=dt ¼ growth rate of the nuclei, r ¼radius of the spherical nuclei at time t and l¼ lamella

distance.

To prove that the structure becomes coarser when the

number of cells increases, you must find a relation between

N, the number of cells in the melt, and the lamella distance.

Try to find such a relation.

Hint A234

HINT A186

(A252)

The air gap between mould and solid metal is small. Sim-

plification of Equation (1) is possible if the last term in the

last factor can be neglected in comparison with 1. We cal-

culate Nusselt’s number (Section 4.4.5 in Chapter 4):

Numax ¼ hs

k¼ hymax

k¼ 2:0 � 103 � 0:05

1:84 � 104¼ 0:005

As Numax � 1, Equation (1) can be simplified to


yL

h

Express the solidification time t as a function of z, derive

the function and calculate the total solidification time.

Hint A335

HINT A187

(A107)

You use the relation in the text:

vgrowthlden2 ¼ 10�10 ð3Þ

and combine it with Equation (1)

Hint A91

HINT A188

(A297)

dyL

dt

��¼ 660� 25

2:7� 103 � 390� 103� 1

0:10� yL

2� 220þ 1

2� 1:68� 103


or

dyL

dt

�� ¼ 0:265

231 � 1000yL

for

0 � yL � 0:10 m: ð14Þ

Plot the functions (14) and (11) in Hint A27.

Hint A79

HINT A189

(A277)

(A70)

(A272)

(A94)

The optimal parameters are discussed in the answer.

Answer

(a) Air is a poor thermal conductor, which excludes thermal

conduction. The absence of a temperature gradient in

the melt excludes convection. The assumption is reason-

able.

(b) The metallurgical depth is 0.80 m.

(c) The cooling rate is 6:1 � 103 K=s.

(d) The flow should be laminar and not turbulent. A smooth

beam, which solidifies rapidly, is needed, otherwise the

surface tension may promote drop formation. It is

favourable if a protecting film of oxide is formed at

the surface of the wire. The wire should be cast in an

atmosphere that contains oxygen. Rapid solidification

is important otherwise the beam of melt may be

unstable and form droplets instead of a wire with a con-

stant radius.

HINT A190

(A247)

With conventional designations you obtain

dQ

dt¼ rL½Lðsin 5�ÞdA�cL

p � dT

dt

� �ð2Þ

and

dq

dt¼ rLLðsin 5�ÞcL

p � dT

dt

� �ð3Þ

Try to find another expression for the heat flux from

the cooling volume element to the surroundings. Consult

the section in Chapter 4 where sand mould casting is

treated.

Hint A271

HINT A191

(A136)

Surface tension (page 51 in Chapter 3).

The surface tension of the melt prevents the fluidity of

the melt. A low surface tension corresponds to a large fluid-

ity length.

What about the influence of composition of the melt?

Hint A28

HINT A192

(A110)

Heat flow from the cooling strip:

dQ

dt¼ �cpm

dT

dt¼ �cprAd

dT

dtð1Þ

Heat flow from the strip to the roller:

dQ

dt¼ hAðTstrip � T0Þ ð2Þ

How do you proceed?

Hint A338

HINT A193

(A114)

When w is known you can calculate the hw values.

Hint A255

Cooling Length Water Periphery area Water flux

zone (m) flow (l/s) ¼ 4al (m2) w (l/m2 s)

Spray zone 0.200 1.33 0.080 16.6

Zone 1 1.280 2.92 0.518 5.64

Zone 2 1.850 2.50 0.740 3.38

Zone 3 1.900 2.92 0.760 3.84


HINT A194

(A105)

Answer

Region 1.

It is customary to add colza oil or casting powder to the

melt in order to reduce the friction between the melt or

solid shell and the chill-mould. The oil is evaporated and

the casting powder melts on contact with the molten

steel. A thin film is formed between the steel and the

chill-mould. The film prevents the heat flux and lowers h.

The film thickness decreases with increasing distance

from the top of the mould and the heat flux and heat transfer

coefficient increase gradually.

HINT A195

(A23)

The volume element has the cross-section of a triangle

with two equal sides. The heat per unit time lost to the

surroundings from the volume element is, with normal

designations,

� a2sina2

dyrLcLp

dT

dt¼ h 2a þ 2a sin

a2

� �dyðT � T0Þ ð1Þ

where dT=dt < 0 and h is the heat transfer coefficient (com-

pare Example 3.5 on page 45 in Chapter 3).

Separate the variables and solve the equation.

Hint A325

HINT A196

(A340)

rð��HÞdyLðtÞdt

Solidification heat flux

¼ hðTL �T0Þ

Heat flux transferred from

the metal to the cooled belt

ð1Þ

How do you proceed?

Hint A326

HINT A197

(A131)

Nusselt’s number is defined on pages 85–86 in Chapter 4.

h ¼ 1:0 � 103 W=m2

K

s ¼ 30 � 10�3 m ðhalf the thickness of the castingÞ

kCu ¼ 398 W=m K

Nu ¼ hs

k� 1:0 � 103 � 30 � 10�3

398¼ 0:075 � 1

Draw a sketch of the temperature distribution in the

casting as a function of distance yL of the solidification

front from the surface. How can the solidification time be

calculated?

Hint A253

HINT A198

(A104)

A ‘weighted’ average value indicates that the area influ-

ences the heat transport.

dQ

dt¼ hAðT � T0Þ

The cross-section of the casting is the same in the four

cooling zones. Hence the area is proportional to the length

of each zone. It is therefore natural to calculate the average

h value with respect to the length of the zones:

hav ¼ h1L1 þ h2L2 þ h3L3 þ h4L4

L1 þ L2 þ L3 þ L4

ð1Þ

α 48° 24° a

a

Envelope area:

yaaA d2

sin22 +=α

Volume:

2

ds2 yinaV =

α


The values of h and L in the table in the text are inserted

into Equation (1):

hav ¼ð1000� 1:0Þþ ð440� 4:0Þþ ð300� 5:0Þþ ð200� 10:0Þ

1:0þ 4:0þ 5:0þ 10:0

¼ 313W=m2

K

Check Nusselt’s number and make a sketch of the

temperature profile, which it is reasonable to use as a

basis for your calculations.

Hint A283

HINT A199

(A142)

As Asprue � Aingot, the inlet velocity vsprue � vingot [Equa-

tion (3)]. The latter can be neglected in Equation (1),

which can be reduced to

vsprue ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2gðH � hÞ

pð4Þ

How do you continue?

Hint A32

HINT A200

(A237)

(A129)

Equation (7) can be transformed into

�T ¼ h1

h2

TL � T0

1 þ h1

ks

y

ð9Þ

The maximum value of the temperature T0 of the cooling

water is 100 �C. The value of y is half the slab thickness.

If you insert the given values into Equation (9), you obtain

�T ¼ 1000

800� 1450 � 100

1 þ 1000

30� 0:10

¼ 389 �C

Answer

(a)

y ¼ ks

h1

h1

h2

TL � T0

�T� 1

� �

(b) The excess temperature has to be about 400 �C. If the

channel width tends to decrease, the velocity of the

melt increases. The heat transfer coefficient h2 increases

with increasing velocity of the melt, which helps to keep

the channel open.

HINT A201

(A112)

Solve V as a function of cmelt:

1

Vmelt

dV ¼ dcmelt

c0 � cmelt

Integration gives

1

Vmelt

ðV

V0

dV ¼ðcmax=2

cmax

�dcmelt

cmelt � c0

or

V � V0 ¼ Vmelt lncmax � c0

cmax

2� c0

0B@

1CA ð2Þ

This is not the result you want! How do you find the dis-

tance y½?

Hint A303

HINT A202

(A151)

ttotal ¼ CyL

2

� �2

¼ 2:1 � 106 0:10

2

� �2

¼ 2:1 � 106 � 0:052 s ¼ 1:46 h


Hint A19

HINT A203

(A314)

1

T3¼ 6esB

rLcLp R

t þ 1

Ti3

or

T3 ¼ Ct þ 1

Ti3

� ��1

ð3Þ


where

C ¼ 6esB

rLcLp R

The excess temperature will be

�T ¼ T � TL ¼ Ct þ 1

Ti3

� ��1=3

�TL

Answer

The excess temperature is

�T ¼ T � TL ¼ ðCt þ T�3i Þ�1=3 � TL

where

C ¼ 6esB

rLcLp R

The equation is valid until �T ¼ 0.

HINT A204

(A313)

The material constants of Fe and Cu and the tempera-

ture values are introduced into Equation (3) in order to

solve l.

When the material data are introduced into Equation (3),

you obtain

2:82 ¼ffiffiffip

plel

2

½0:325 þ erf ðlÞ� ð4Þ

Solve this equation numerically.

Hint A33

HINT A205

(A139)

The temperature of the cooling water is 100 �C. This value

and the given values in the text are inserted into Equa-

tion (6):

vcast ¼2lhðTL � T0Þrdð��HÞ

¼ 2 � 0:050 � 3:0 � 103ð933 � 373Þ2:7 � 103 � 398 � 103 d

¼ 0:156 � 10�3

d

ð7Þ

or

vcast ¼1:56 � 10�4 m2=s

d¼ 60 � 1:56 � 10�4 m2=min

d

¼ 9:38 � 10�3 m2=min

d

ð8Þ

where d is measured in metres.

Discussion of the Result

Equation (8) is valid if Nu ¼ hs=k � 1; s is half the thick-

ness of the casting at bilateral cooling.

If you allow an upper limit for Nu ¼ 0:10 and h and k are

known, the maximum thickness of the casting can be calcu-

lated. The thermal conductivity kAl ¼ 220 W/m K. The

maximum thickness dmax of the casting can be calculated

as follows:

Nu ¼ hs

k¼ hyL

k¼ 0:10

which gives

yL ¼ 0:10 k

h¼ 0:10 � 220

3:0 � 103¼ 7:33 � 10�3 m

dmax ¼ 2yL 15 � 10�3 m

Plot the function (8) in a diagram.

Hint A292

HINT A206

(A329)

ða=2

0

a � y 1 þ h4

h

� �� dy ¼ ahðTL � T0Þ

rð��HÞ

ðt

0

dt

ay � y2

21 þ h4

h

� �� a=2

0

¼ ahðTL � T0Þrð��HÞ t

a

2� a

81 þ h4

h

� �¼ hðTL � T0Þ

rð��HÞ t

which can be written as

t ¼ arð��HÞhavðTL � T0Þ

1

2� 1

81 þ h4

h

� �� ð16Þ

Find the maximum casting rate in the same way as in

Hint A270 and Hint A54 in exercise 5.10a.

Hint A132


HINT A207

(A141)

The rule of thumb is the relation

yL ¼ 2:5ffiffit

pð1Þ

where yL is measured in centimetres and t in minutes. In

this case yL is half the smallest dimension of the ingot,

i.e. yL ¼ 20 cm. Hence the solidification time will be

t ¼ yL2

6:25¼ 202

6:25¼ 64 min ð2Þ

Now you can calculate the total radiation losses through

the upper unshielded ingot surface during the solidification

time.

Hint A8

HINT A208

(A265)

Inserting r ¼ 0 gives, as the limit of r2 lnðr=RÞ is zero when

r approaches 0,

t ¼ rð��HÞðTmelt � T0Þ

R2

4kþ R

2h

� �ð10Þ

Equation (10) is the desired function. Insert material

constants and other known values.

Hint A27

HINT A209

(A345)

Answer

Region IV.

The derivative of the curve is zero. The solidification

heat balances the heat losses. Region IV can be character-

ized as the ‘steady state’ of formation and growth of

equiaxed crystals in the central part of the ingot.

Characterize region V.

Hint A149

HINT A210

(A289)

The general condition for growth of the free nuclei, close to

the solidification front, is

vcrystal ¼dr

dt¼ mðTL � TcrystalÞn ð2Þ

If Tfront ¼ TL and dyL=dt ¼ 0, the consequence is that

the temperature of the melt in front of the solidification

front must be equal to TL. As the nuclei are formed in the

melt ahead of the solidification front, the temperature Tcrystal

equals Tfront ¼ TL.

What is the conclusion from this statement?

Hint A84

HINT A211

(A63)

The heat flow from the superheated melt to the surround-

ings can be written in two ways:

@Q

@t¼ �rVcp

dT

dtCooling heat

per unit time

¼ A


p

pt

sðTi � T0Þ

Heat flow across the

interface mould=melt:

Compare Equation ð4:70Þon page 79 in Chapter 4

ð1Þ

Integrate Equation (1) and solve the cooling time t1.

Hint A108

HINT A212

Exercise 3.7a

Use your general knowledge or, if necessary, have a look on

the text in Chapter 3.

Hint A136

HINT A213

(A306)

If you neglect the thermal conduction in the solid layer at

the upper surface of the strip and neglect the cooling heat

during the solidification process, you obtain the heat flow at

the upper strip surface:

AsBeðTL4 � T0

4Þ ¼ rAð��HÞ dYL

dtð1Þ

Integration givesðt

0

dt ¼ rð��HÞsBeðTL

4 � T04Þ

ðYL

0

dYL

or

t ¼ rð��HÞsBeðTL

4 � T04Þ YL ð2Þ


Find the solidification time for the solid layer yLat the

lower part of the strip.

Hint A96

HINT A214

(A170)

You know the casting rate vcast ¼ 10 m=s.

z ¼ vcastt

How do you calculate the solidification time t?

Hint A290

HINT A215

(A86)

hI ¼rmetalð��HÞ

TL � T0

yL

t

¼ 7:9 � 103 � 270 � 103

1500 � 25� 14 � 10�3

140¼ 144 W=m

2K

Obviously you have to use Equation (1) to find hII. How?

Hint A83

HINT A216

(A251)

On inoculation, the number of crystals is increased; l2 is

proportional to N1=3. Hence a large number of cells (large

N) corresponds to a coarse structure (large l).


Hint A80

HINT A217

(A174)

Separate the variables and integrate Equation (1). Integra-

tion limits?

Hint A243

HINT A218

(A318)

Use the rule of thumb. See page 97 in Chapter 5.

Hint A343

HINT A219

(A342)

Choose a number of d values, calculate the corresponding

values of vmax with the aid of Equation (4) and plot them

in a diagram.

Answer

The maximum casting rate as a function of the slab thick-

ness is

vmax ¼ 64 � 10�3 m2=min

dðd � 0:20 mÞ:

HINT A220

(A312)

Answer

(a)

d vmax

(m) ðm=minÞ0.010 6.4

0.020 3.2

0.050 1.3

0.080 0.8

0.10 0.6

0.20 0.3

Ti metal yL hcalc

ð�CÞ (m) ðW=m2

KÞ635 0.012 1390

615 0.018 1708

600 0.032 1344

610 0.040 868

600 0.048 896

595 0.057 831

580 0.090 681


The first value of h is too low, owing to the excess tem-

perature of the melt.

The heat transfer coefficient is not constant. The reason

is that the surface temperature Ti metal of the solidified metal

is not constant.

The heat transfer coefficient hcalc decreases with increas-

ing distance from the mould/metal interface.

HINT A221

(A248)

y ¼ esBðTL4 � T0

4Þtrð��HÞ

The solidification time was found to be 64 min in Hint

A207.

y ¼ 0:2 � 5:67 � 10�8ð17234 � 2934Þ � 64 � 60

7:88 � 103 � 272 � 103

� 2 � 10�14ð1723 � 293Þ � 17233 ¼ 0:146 m


Hint A26

HINT A222

(A325)

According to the text, the velocity v of the melt is assumed

to be constant and you obtain

Lf ¼ vtf

which gives

Lf ¼ vtf ¼asina

4 1 þ sina2

� � vrLcLp

hln

TL þ ð�TÞi � T0

TL � T0

� ð2Þ

The first factor is determined by the design of the equip-

ment:

asina

4 1 þ sina2

� � ¼ 0:0080 � sin 48�

4ð1 þ sin 24�Þ

¼ 0:0020 � 0:7431

1:4067¼ 0:001056

ð3Þ


Hint A322

HINT A223

(A31)

The convection increases the number of small nuclei. In

what ways?

Hint A158

HINT A224

(A113)

rð��HÞðV

0

dV ¼ A


p

p

sðTi � T0Þ

ðt2

t1

dtffiffit

p

or

2ðffiffiffiffit2

p�

ffiffiffiffit1

pÞ ¼ rVð��HÞ

AðTi � T0Þ


kmouldrmouldcmouldp

s

or

ðffiffiffiffit2

p�

ffiffiffiffit1

pÞ2 ¼ rVð��HÞ

2AðTi � T0Þ

� 2 pkmouldrmouldcmould

p

ð4Þ

Insert material constants and other given values and

calculate t2.

Hint A119

HINT A225

(A328)

�Tmelt ¼ Tcrystal � Tfront

where Tcrystal is the temperature of the melt close to the sur-

face of the free crystals and Tfront the temperature of the

melt close to the solidification front.

To advance you need to couple the temperatures to

growth rates. How?

Hint A242

HINT A226

(A53)

You set up a mass balance.

The mass m ¼ 70 tons of steel leaves the tundish per

hour and enters the chill-mould.

m ¼ Astrandvcastrt ð1Þ


If you insert m ¼ 70 tons and t ¼ 1 h ¼ 3600 s into

Equation (1), you obtain

vcast ¼m

Astrandrt¼ 70 � 103

0:20 � 1:5 � 7:2 � 103 � 3600

¼ 0:0090 m=s ¼ 0:54 m=min

Draw a sketch of the tundish and the chill-mould and

introduce vcast and voutlet (the velocity of the melt on its

way out of the tundish) into the figure.

You want the outlet diameter of the tundish. If you can

calculate Aoutlet it will be easy to derive the diameter. How

do you get Aoutlet?

Hint A172

HINT A227

(A96)

Provided that the solidification processes starts simulta-

neously at the upper and lower surfaces of the strip, the

solidification times are equal:

t ¼ rð��HÞTL � Tw

yL

h1 þ h

2kyL

� �¼ rð��HÞ

sBeðTL4 � T0

4Þ YL ð4Þ

You have one equation but two unknown quantities, YL

and yL. How do you get a second equation?

Hint A29

HINT A228

(A317)

(A147)

The casting rate is determined by the condition that the

casting must have solidified completely before it leaves

the last cooling section. Hence the maximum casting rate

must be

vmax ¼ L

t¼ 20

21:2¼ 0:943 m=min

Answer

(a) The solidification time is 21 min.

(b) The maximum casting rate is 0.94 m/min.

HINT A229

(A309)

Take Equation (4) in Hint A140 again and combine it with

Equation (5) in Hint A318. Introduce the values of A in Hint

A328 and vfront in Hint A309 and all the other known values

into a combined equation.

Hint A57

HINT A230

(A274)

Material constants for steel are given in the text. Assume

that T0 ¼ 100 �C.

If you insert these values and material constants, you

obtain

� dT

dt¼ 2:0 � 103

650 � 7:8 � 103 � 100 � 10�6ðTstrip � 100Þ

¼ 3:94ðTstrip � 100Þð5Þ

If you insert Tstrip ¼ 1400 �C into Equation (5), you obtain

� dT

dt¼ 3:94ðTstrip � 100Þ ¼ 3:94ð1400 � 100Þ ¼ 5120 K=s

Answer

The desired relation is

� dT

dt¼ h

cprdðTstrip � T0Þ

The cooling rate is 5:1 � 103 K=s.

HINT A231

(A334)

lden ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffirð��HÞ � 10�12

400ðTL � T0Þ

s1ffiffiffip

p ð5Þ

Introduce the following temperatures and material con-

stants into Equation (5): TL ¼ 660 �C, T0 ¼ 25 �C, rAl ¼2:7 � 103 kg=m

3and ð��HÞAl ¼ 398 � 103 J=kg. This

gives you the answer.

Answer

lden ¼ 6:5 � 10�5ffiffiffip

p m

where p is measured in atmospheres.


HINT A232

(A304)

Figures 1 and 2:

The xy-, yz- and xz-planes are tangent planes to the sphere.

The distance from the corner point O to each tangent point

(T3 and T2 are seen in the figures) is Rffiffiffi2

p. The distance

between the tangent points, e.g. T3 and T2, is also Rffiffiffi2

p.

Figures 3 and 4:

The three distances OT1, OT2 and OT3 define a tetrahedron

with the side Rffiffiffi2

pfor symmetry reasons. The plane,

defined by the three tangent points, is perpendicular to

the line OA. This plane is the one drawn in the figure

in the text and Hint A115. The circle in Figure 4 represents

the boundary line along which the surface tension forces

act. Its radius is equal to r, which was introduced in Hint

A115 and calculated in Hint A245.

How do you get the relation between r and R?

Hint A37

HINT A233

Exercise 3.10b

Calculate the corresponding radius for cast iron.

Hint A175

HINT A234

(A185)

Examine the heat flow from the ingot. The heat flow is to a

great extent controlled by the slow heat flow through the

sand mould. Equation (4.70) on page 79 in Chapter 4 can

be used to find the heat flow through the interface solid

metal/sand mould:

dQ

dt¼ A


p

pt


where A ¼ area of the interface mould/ingot, Ti ¼ temperature

of the metal at the metal/mould interface and t ¼ time.

Set up an expression for the heat flow from the ingot.

The melt has been inoculated with N nuclei, which grow.

Hint A279

HINT A235

Exercise 5.4c

The heat flux decreases continuously in region 2. There are

two reasons. Which ones?

Hint A319

HINT A236

(A330)

Integrate Equation (4):

Nrð��HÞðr

0

4pr2dr ¼ A


p

pt

sðTi �T0Þ

ðt

0

dtffiffit

p

which gives

Nrð��HÞ4pr3

3¼A


p

pt

sðTi �T0Þ�2

ffiffit

pð5Þ

Solve r as a function of N and t.

Hint A301

HINT A237

(A336)

The desired function is obtained if you introduce Equation

(6) into Equation (5):

y ¼ ks

h2

TL � TL � T0

1 þ h1

ks

y

þ T0

0BB@

1CCA

�T¼ ks

h2

ðTL � T0Þh1

ks

y

�T 1 þ h1

ks

y

� �

yz-p

lane

T3

R

O

xz-plane

R •

z

R T3

yR

x R

O

T2

••

1 2

T3 r A

T2

T1

O

P

·

··

·

T3

r

T1 T2

2 R

·

··

3 4


After reduction you obtain

1 þ h1

ks

y ¼ h1

h2

TL � T0

�Tð7Þ

or

y ¼ ks

h1

h1

h2

TL � T0

�T� 1

� �ð8Þ

Equation (8) is the desired function. The answer is given

in Hint A200.

Hint A200

HINT A238

(A159)

Use Equation (4.73) on page 80 in Chapter 4.

yL ¼ 2ffiffiffip

p Ti � T0

rmetalð��HÞ


p

q ffiffit

pð1Þ

It is valid for a planar solidification front (one dimen-

sion). In the case of a cylinder (three dimensions) you

have to add a correction term to Equation (1). According

to Equation (4.76) on page 82 in Chapter 4, you have

Vmetal

A¼ Ti � T0

rmetalð��HÞ

� 2ffiffiffip

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffikmould rmould cmould

p

q ffiffiffiffiffiffiffiffittotal

pþ nkmould ttotal

2r

� �ð2Þ

where n ¼ 1 for a cylinder.

Hint A106

HINT A239

(A280)

If Equation (7) is combined with the expression [Equation

(6.14) on page 147 in Chapter 6]

ngrowth ¼ mðTE � TÞn ð9Þ

you obtain

ngrowth ¼ dr

dt¼ 1

6C

t�5=6

N1=3¼ mðTE � TÞn ð10Þ

What is your conclusion from Equation (10)?

Hint A341

HINT A240

Exercise 5.12e

Set up a heat balance for the solidification process.

Hint A339

HINT A241

(A167)

The text tells you to perform the calculations at a shell

thickness of 10 cm. Introduce this value of y into Equation

(6) and solve t.

Hint A309

HINT A242

(A225)

Use the common trick to involve the liquidus temperature

and apply the growth law, given in the text.

Hint A318

HINT A243

(A217) ðyL

0

dyL ¼ hðTL � T0Þrð��HÞ

ðt

0

dt ð2Þ

yL ¼ hðTL � T0Þrð��HÞ t ð3Þ

What is the value of yL when the casting has solidified

completely?

Hint A311

HINT A244

(A122)

10 cm


yL will be equal to the thickness of the casting as it is

cooled only from one side. In this case it is 0.10 m.

Hence you obtain

t¼ 1

4ametal

yL

l¼ 1

4�6:15�10�6

0:10

0:79¼ 651s¼10:8 min

Answer

The solidification time is 11 min (l ¼ 0:79).

HINT A245

(A144)

p0pr2 þ 2prscos y ¼ ðp0 þ rghÞpr2

or

r ¼ 2scos yrgh

ð3Þ

The desired radius of curvature is the radius R of the

sphere. R must be derived as a function of r and cosymust be calculated. How do you get cosy?

Hint A304

HINT A246

(A39)

Answer

The convection flow is illustrated in the figure. The direc-

tion of the convection flow is opposite to the growth of the

solidification front at the top of the mould. Comparatively

hot melt moves from the right- to the left-hand side and

delays the solidification. The shell will therefore be thinner

at the top than at the middle. The boundary layer is thinnest

at the top.

At the bottom of the mould the convection flow has the

same direction as the growth of the solidification front.

Comparatively cold melt, moving in the direction of the

shell growth, promotes the solidification. The shell will

therefore be thicker at the bottom than at the middle. The

boundary layer is largest at the bottom.

HINT A247

(A46)

Consider a volume element dV. The length of the volume

element (not seen in the figure) is b:

dV ¼ ydA ¼ Lðsin 5�ÞdA ð1Þ

where y is the perpendicular distance from dA to the central

axis and L the distance from the wedge edge to the volume

element.

Find expressions for the heat flow and the heat flux from

the cooling volume element.

Hint A190

HINT A248

(A87)

Assume that all the solidification heat, which disappears

with the aid of radiation, emanates from the solidification

heat of the frozen surface layer:

AesBðTL4 � T0

4Þt ¼ ryAð��HÞ

dL

dA

y

L

5°


Solve y and insert material constants and other known

values.

Hint A221

HINT A249

(A180)

ðffiffiffiffit3

p�

ffiffiffiffit2

pÞ2 ¼ 2:7 � 103 � 0:253 � 1:25 � 103 � 50

2 � 6 � 0:252ð660 � 20Þ

� 2

� p0:63 � 1:61 � 103 � 1:05 � 103

¼ 89 s

which gives (t2 � 4757 s from Hint 119)

ffiffiffiffit3

p�

ffiffiffiffit2

p¼

ffiffiffiffiffi89

p)

ffiffiffiffit3

p

¼ffiffiffiffiffiffiffiffiffiffi4757

pþ

ffiffiffiffiffi89

p¼ 68:97 þ 9:43 ¼ 78:40

t3 � 6146:6 s ¼ 102:4 min

Cooling time ¼ t3 � t2 ¼ 102:4 � 79:3 ¼ 23 min.

Answer

The cooling time of the melt is 1 min. The solidifica-

tion time is 78 min. The cooling time of the solid metal is

23 min. The cooling rate after solidification is much

slower than cooling rate of the melt because the mould

becomes heated during the solidification process.

The cooling of the cube to room temperature requires a

longer time than the solidification process.

HINT A250

Exercise 6.9b

If the cooling rate ð�dT=dtÞ � 60 K=s, white solidification

occurs. The change from white to grey solidification will

therefore occur at the height that is obtained at a cooling

rate of 60 K/s. Insert the cooling rate ¼ 60 K/s into Equa-

tion (10) in Hint A269 and solve L.

Hint A260

HINT A251

(A78)

You insert the expression (7) into Equation (1):

vgrowthl2 ¼ constant ð1Þ

which gives

1

6C

t�5=6

N1=3l2 ¼ constant ð8Þ

What is your conclusion?

Hint A216

HINT A252

(A169)

The wedge is cooled bilaterally. The heat flow is perpendi-

cular to the surface. The solidification is complete at the

distance z from the edge when the thickness of the ‘shell’

is yL

yL

z¼ tan5� � 0:0875

yL � 0:0875z ð2Þ

As sin5� � tan5�, we obtain

ymax ¼ 0:0875 � 0:10 ¼ 0:00875 m

Use the material constants for the Al–Si alloy, given in

the text, and check if it is possible to simplify Equation (1).

If the answer is yes, do it!

Hint A186

T

Coolingof the melt

T excess

T L

T 0

Cooling of the solid

t

0 t 1 t 2 t 31 min 78 min 23 min

Solidification process atconstant temperature.Phase transformation

5°O O′

0 z 10 cm

yL y max

dtdQ


HINT A253

(A197)

Equation (4.48) on page 74 in Chapter 4 gives the solidifi-

cation time. If Nu � 1, the last term in the last factor can be

neglected in comparison with 1, which gives


yL

hð1Þ

Apply this equation on the present case and set up a heat

balance.

Hint A321

HINT A254

(A18)

�pR2dyrLcLp

dT

dtEmitted heat per unit

time from the element

¼ 2pRdyesBðT4 � T04Þ

Radiated heat per unit

time to the surroundings

from the element

ð2Þ

Solve T as a function of t.

Hint A177

HINT A255

(A193)

hw ¼ 1:57ð1 � 0:0075 � 40Þ4

w0:55 ¼ 0:275 w0:55

Answer

The heat transfer coefficient hw is 1.3 kW/m2 K for the

spray zone and 0.71, 0.54 and 0.57 kW/m2 K for zones 1,

2 and 3, respectively.

HINT A256

(A315)

dQ=dt is the heat of solidification per unit time, developed

at the solidification front:

dQ

dt¼ � 2prdrLrð��HÞ

dt¼ �2prLrð��HÞ dr

dtð3Þ

Eliminate dQ=dt with the aid of Equations (2) and (3).

Hint A64

HINT A257

(A153)

On the axes of the curve you have the thickness of the soli-

dified layer yL versus the square root of time.

� Equation (1) in Hint 153 corresponds to a parabolic

growth law of the type yL ¼ Affiffit

pþ B. This equation cor-

responds to a straight line in the type of diagram you have

here.

� Equation (2) in Hint 153 corresponds to a linear growth

law of the type yL ¼ Cðffiffit

pÞ2

. This equation corresponds

to a curved line in the type of diagram you have here

(withffiffit

pinstead of t on the horizontal axis).

Equation (1) represents the parabolic growth law and

corresponds to region II.

Equation (2) represents the linear growth law and corre-

sponds to region I.

Answer

(a)

Region I corresponds to a linear growth law, which cor-

responds to a curved line in the type of diagram you have

here. Region II corresponds to a parabolic growth law,

which corresponds to a straight line in the type of diagram

you have here.

HINT A258

(A69)

You want the cooling rate at distance L from the wedge

edge when the excess temperature is gone, i.e. when the

temperature is equal to the eutectic temperature. At that

time t ¼ tcool.

Go back to Equation (5) in Hint A89 and replace t by

tcool and Ti by TE.

Hint A43

T

TL

Mould Solid Liquid

T0

0 yL y


HINT A259

(A59)

Call the temperature at the solidification front Tfront. The

heat balance can be written as

ks

Tfront � Ti

yL

Heat removed

by conduction

¼ rð��HÞ dyL

dtSolidification heat

developed at the

solidification front

þ hconðTmelt � TfrontÞConvection heat

from the interior

of the melt

Discuss the solidification heat developed at the solidifi-

cation front.

Hint A289

HINT A260

(A269)

(A250)

White solidification occurs if

L �ffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:1324

� dT

dt

vuut ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:1324

60

r¼ 0:0470 m

Answer

(a) The cooling rate as a function of the distance L from the

wedge edge is � dT

dt¼ 0:13

L2

(b)

The change from white to grey cast iron will occur at

L ¼ 4:7 cm.

HINT A261

(A7)

During continuous casting, the contact with the wall of

the chill-mould is poor. This leads to the temperature

profile illustrated in the figure (Figure 4.17 on page 73 in

Chapter 4).

We have a cylindrical geometry but will use a one-

dimensional model as an approximation to calculate the

solidification rate at the melt/metal interface.

Equation (4.46) on page 74 in Chapter 4 gives the soli-

dification rate:

dyL

dt¼ vgrowth ¼ TL � T0

rð��HÞh

1 þ h

kyL

ð1Þ

Solve the heat transfer number h from Equation (1).

Hint A98

HINT A262

(A337)

The casting rate is constant and equal to 10 m/s.

z1 ¼ vcast t1 ¼ 40 � 10�6 m

z2 ¼ vcastðt1 þ t2Þ ¼ 83 � 10�6 m

z3 ¼ vcastðt1 þ t2 þ t3Þ ¼ 140 � 10�6 m

z4 ¼ vcastðt1 þ t2 þ t3 þ t4Þ ¼ 230 � 10�6 m

z5 ¼ vcastðt1 þ t2 þ t3 þ t4 þ t5Þ ¼ 360 � 10�6 m

List the values of z in the last column in the table in Hint

A170. Look at the table and answer the following questions:

How does h change with the thickness ðr0 � rÞ of the

shell?

How does h change with the distance from the nozzle?

Hint A181

T

Tfront

Ti

yyL

Tmelt

Grey cast iron

White4.7 cm cast iron


T

T L

T

T i metal

T0

y 0 y (t) yL


HINT A263

(A321)

ða=2

0

ða � 2yÞdy ¼ ahavðTL � T0Þrð��HÞ

ðt

0

dt ð4Þ

½ay � y2�a=20 ¼ ahavðTL � T0Þ

rð��HÞ t ) a

4¼ havðTL � T0Þ

rð��HÞ t

which gives

t ¼ arð��HÞ4havðTL � T0Þ

ð5Þ

Now you know the solidification time. How do you get

the maximum casting rate?

Hint A168

HINT A264

(A331)

The heat flow in the solid shell:

dQ

dt¼ �kA

dT

dyð1Þ

The heat flow at the interface between the mould and

metal is

dQ

dt¼ �hAðTi metal � T0Þ ð2Þ

Find an expression for the heat transfer coefficient.

Hint A121

HINT A265

(A316)

Introduce Ti [Equation (6)] into Equation (4):

�2prLrð��HÞ dr

dt¼

�k2pL Tmelt �T0hR ln

r

R� kTmelt

hR lnr

R� k

264

375

lnr

R

ð7Þ

The variables are separable and Equation (7) can be inte-

grated. After reduction you obtain

rrð��HÞ dr

dt¼ khRðTmelt � T0Þ

hR � lnr

R� k

ðt

0

dt ¼ðr

R

rð��HÞkhRðTmelt � T0Þ

hrR lnr

R� kr

� �dr ð8Þ

or

t ¼ rð��HÞkhRðTmelt � T0Þ

ðr

R

hrR lnr

R� kr

� �dr

or

t ¼ rð��HÞkhRðTmelt � T0Þ

hRr2

2ln

r

R� r2

4

� �r

R

� kr2 � R2

2

�

or

t¼ rð��HÞkhRðTmelt�T0Þ

hRr2

2ln

r

R�r2

4þR2

4

� �þk

R2�r2

2

� ð9Þ

The solidification time is obtained for r ¼ 0. Find it!

Hint A208

HINT A266

(A127)

If you add too little steel powder, it will be heated to the

temperature of the melt, which still is an excess tempera-

ture. You must at least add so much steel powder that it

removes all the excess temperature of the melt. If you

add more steel powder then there is no ‘excess energy’

left to melt the additional powder. It will be left in the

melt and gives the desired fine-grain effect.


T

TL

T

T i metal

T0

y0 y (t) yL


The heat required to heat the steel powder and melt it is

taken from the ‘excess heat’ of the melt.

m½cspðTL � T0Þ þ ð��HÞ� ¼ 1:00cL

p ðTmelt � TLÞ ð1Þ

Solve the equation and insert numerical values.

Hint A182

HINT A267

(A42)

In Hint A309 you found that vfront ¼ 4:0 � 10�4 m=s at a

shell thickness of 10 cm.

You obtain the maximum growth rate of the free crystals

for small values of N:

vcrystal �dr

dt

� �max

¼ 12 � 102

3 � 106m=s ¼ 4 � 10�4 m=s

Answer

ðvcrystalÞmax ¼ 4 � 10�4 m=s.

The growth rate of the free crystals is always smaller than

the growth rate of the dendrites at the solidification front.

If a nucleating agent is added to the melt, N increases

strongly and the freely floating crystals will stop the den-

drite growth. The transition from the columnar to the cen-

tral zone occurs and vfront becomes zero.

HINT A268

Exercise 4.9b

List some values of yL and calculate the corresponding

values of Ti metal for the two cases h ¼ 2 � 102 and

2 � 103 W=m2

K for steel and copper, respectively.

Hint A146

HINT A269

(A291)

� dT

dt¼


p

qffiffiffip

prLLðsin 5�ÞcL

p

TE � T0ffiffiffiffiffiffiffiffitcool

p ð9Þ

The expression you found forffiffiffiffiffiffiffiffitcool

pin Hint A69 is intro-

duced into Equation (9) together with the temperatures and

material constants:

�dT

dt¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:63� 1:61� 103 � 1:05� 103

pffiffiffip

p� 7:0� 103 � 0:0872� 420L

� 1153� 20

19:43L¼ 0:1324

L2

ð10Þ

This is the desired function. The answer is given in Hint

A260.

Hint A260

HINT A270

(A168)

vmax ¼ l

t¼ pD

2tð6Þ

Combine Equations (5) and (6) and list material data.

Hint A54

HINT A271

(A190)

Equation (4.70) on page 79 in Chapter 4 is what you are

looking for. Note that the temperature Ti of the solid

metal at the metal/mould interface at sand mould casting

is constant and in this case equal to the eutectic temperature

TE [Equation (4.60) on page 78 in Chapter 4).

@q

@t¼


p

pt


Compare the two expressions (3) and (4).

Hint A89

HINT A272

(A307)

The whole cylinder is solid and starts to cool. The heat flow

of the cooling wire can be written in two ways:

2pLR0sBeðT4 � T40 Þdt

Cooling heat radiated

during the time dt

¼ cprpR02Lð�dTÞ

Cooling heat released when the

temperature of the cylinder

decreases by an amount dT:

ð3Þ

The cooling rate will be

� dT

dt¼ 2sBeðT4 � T0

4ÞcprR0

ð4Þ

Inserting the known values into Equation (4) gives

�dTmelt

dt¼2�5:67�10�8�1�ð17534�3004Þ

450�7:8�103�50�10�6¼6:1�103K=s


Hint A189


HINT A273

(A162)

Inserting the values into Equation (1), you find that

t ¼ p4

2:7 � 103 � 398 � 103 � 2:5 � 10�3

660 � 25

� �2

� 1

0:63 � 1:61 � 103 � 1:05 � 103¼ 13:2 s

Next you consider the solidification process in the metal

(Cu mould). Discuss the temperature profile in the casting

before you start.

Hint A302

HINT A274

(A338)

You just solve the cooling rate from Equation (3):

� dT

dt¼ h

cprdðTstrip � T0Þ ð4Þ

It is now time to introduce known values and constants

into Equation (4).

Hint A230

HINT A275

(A2)

The temperature distribution and solidification time at ideal

cooling have been treated in Section 4.3.2 in Chapter 4.

The general heat equation is solved and five constants in

the solution are determined with the aid of five boundary

conditions (pages 69–71 in Chapter 4).

Equation (4.26) on page 70 in Chapter 4 describes the

position of the solidification front yL as a function of time t:


pð1Þ

where

ametal ¼kmetal

rmetalcmetalp

ð2Þ

(Equation (4.11), page 62 in Chapter 4); ametal is the thermal

diffusion constant of the metal (compare Example 4.2

on page 67 in Chapter 4) and l is a constant, which

has to be determined from Equation (4.36) on page 71 in

Chapter 4:

cmetalp ðTL �T0Þ

��H¼

ffiffiffip

plel

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffikmetalrmetalc

metalp

kmouldrmouldcmouldp

sþ erfðlÞ

!ð3Þ

How can you solve l from Equation (3)?

Hint A128

HINT A276

(A89)

ðtcool

0

dtffiffit

p ¼rLLðsin 5�ÞcL

p

ffiffiffip

p

ðTi � T0Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffikmouldrmouldcmould

p

q ¼ðTE

TEþ100

�dT ð6Þ

or

2ffiffiffiffiffiffiffiffitcool

p¼

rLLðsin 5�ÞcLp

ffiffiffip

p

ðTi � T0Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffikmouldrmouldcmould

p

q � 100 ð7Þ

Solveffiffiffiffiffiffiffiffitcool

pand introduce the given numerical values of

temperatures and material constants.

Hint A69

HINT A277

(A109)

There are three possible ways to remove heat, by radiation,

conduction and convection.

Air is a very poor thermal conductor, which results in a

low value of the heat transfer coefficient between the wire

and the air. This alternative can be abandoned.

Convection in the air around the wire occurs but gives no

major contribution to the heat transport to the surroundings.

Obviously the main alternative for the heat transport is

radiation. The answer is given in Hint A189.

Hint A189

T

Mould Solid phase MeltTL

Ti

y 0 yL (t)


HINT A278

Exercise 5.4b

What is the situation in the chill-mould at the maximum of

the curve?

Hint A77

HINT A279

(A234)

The developed heat flow comes from cooling of the solid

and liquid parts of the ingot and from solidification heat

of the growing cells in the melt. The cooling rate is low

and the cooling heat can be neglected in comparison with

the solidification heat from the growing cells.

dQ

dt¼ 4pr2 dr

dtNrð��HÞ ð3Þ

where r ¼ radius of the cells, N ¼ number of cells in the

melt and r¼ density of the equiaxed crystals.

Compare the two heat flow expressions.

Hint A330

HINT A280

Exercise 6.5b

It seems to be more than doubtful to inoculate an ingot melt

before casting as the mechanical properties of the cast iron

will be deteriorated. Mention two advantages that some-

times dominate over the disadvantage.

Hint A239

HINT A281

(A30)

Centrifugal casting provides good contact between the Cu

chill-mould and the metal. The casting process can be

regarded as a one-dimensional solidification.

In this case, the temperature distribution can be assumed

to be as illustrated in the figure.

What equation is valid for the solidification process?

Hint A111

HINT A282

(A51)

vgrowth ¼ dy

dt¼ 1:5 � 10�2ffiffi

tp m=s ð1Þ

Integration: ðyL

0

dy ¼ 1:5 � 10�2

ðt

0

dtffiffit

p

which gives

yL ¼ 1:5 � 10�2 � 2ffiffit

p

or

yL ¼ 3:0 � 10�2ffiffit

pð2Þ

How do you proceed?

Hint A107

HINT A283

(A198)

We have unidirectional solidification from four sides. From

the first figure we realize that total solidification occurs

when s in Nusselt’s equation equals half of the shortest

side or 0:5 � 0:290 ¼ 0:145 m.

Nu ¼ hs

k¼ 313 � 0:145

30� 1:5

As Nusselt’s number is not small, the thermal conduc-

tion through the solid metal layer can not be neglected in

the present case.

What equation is valid for the solidification time in this

case?

Hint A317

1500 mm

290 mm

T

TLT

Ti metal

T0

y 0 y (t) yL


HINT A284

(A120)

No calculations are required to obtain the new table. You

use the figure in the text and the table in Hint 220. The

curve is shown in Hint A97.

How do you explain the discontinuity of the curve?

Hint A97

Can you get the solidification rate from the curve in the

text?

Hint A97

Why does the solidification rate increase at the end of the

solidification process?

Hint A97

HINT A285

(A166)

Mass of the cylinder mcasting ¼ pr2hcastingr

¼ p� 0:052 � 0:23 � 6:9 � 103 kg ¼ 12:46 kg

tfill ¼ 3:4ðmcastingÞ0;42 ¼ 3:4ð12:46Þ0:42 ¼ 9:81 s ð2Þ

Asprue is solved from Equation (1):

Asprue ¼2Acasting

tfill

ffiffiffiffiffi2g

p ðffiffiffiffiffiffiffiffiffihtotal

p�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffihtotal � hcasting

pÞ

¼ 2p� 0:052

9:81ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 � 9:81

p � ðffiffiffiffiffiffiffiffiffi0:28

p�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi0:28 � 0:23

pÞ

¼ 1:10 � 10�4 m2

How do you obtain the upper cross-sectional area of the

sprue?

Hint A324

HINT A286

(A327)

Calculate the value of the constant C for steel in the same

way.

C ¼ p4

rmetalð��HÞTL � T0

� 2

� 1

kmouldrmouldcmouldp

C ¼ p4

7:88 � 103 � 272 � 103

1808 � 298

� �2

� 1

0:63 � 1:61 � 103 � 1:05 � 103¼ 1:5 � 106 s=m

2

Calculate the solidification time for steel.

Hint A154

HINT A287

(A179)

You realize that dr=dt ¼ vgrowth and use the relation in the

text between lden and vgrowth.

Hint A320

HINT A288

(A130)

The heat flow of the solidifying volume element of the wire

can be written in two ways:

2pR0LsBeðTL4 � T0

4Þdt

Heat radiated during the

time dt from the surface

of the outer cylinder

surface with the

constant radius R0

¼ r� 2prð�drÞLð��HÞReleased solidification

heat when the cylindrical

shell with radius r;thickness dr

and height L solidifies

ð1Þ

Equation (1) is integrated:

R0sBeðT4L � T0

4Þðt

0

dt ¼ �rð��HÞð0

R0

rdr

yL(exp) t (exp)

(m) (min)

0.012 0.25

0.018 0.50

0.032 1.0

0.040 1.5

0.048 2.0

0.057 2.5

0.090 3.5

r

L

R0


which gives

t ¼ rð��HÞr2

2sBeR0ðTL4 � T0

4Þ

� R0

0

¼ rR0ð��HÞ2sBeðTL

4 � T04Þ

ð2Þ

Insert given values and material constants into Equa-

tion (2).

Hint A161

HINT A289

(A259)

The growth of the columnar crystals stops only if the

condition dyL=dt ¼ 0 is fulfilled. A consequence of this

condition and the equation

vfront ¼dyL

dt¼ mðTL � TfrontÞn ð1Þ

is that Tfront ¼ TL.

Provided that the convection in the melt close to the soli-

dification front is strong, plenty of dendrite fragments will

be formed, which serve as heterogeneities and small crys-

tals are nucleated in the melt ahead of the solidification

front.

What is the condition for growth of these nuclei?

Hint A210

HINT A290

(A214)

You use Equation (1):

h2pr0�zðTL � T0Þ ¼ �r� 2prdr

dt�zð��HÞ ð4Þ

which can be transformed into

dt ¼ rð��HÞhr0ðTL � T0Þ

rdr

This equation can be integrated:

ðt

0

dt ¼ rð��HÞhr0ðTL � T0Þ

ðr2

r1

r dr ¼ rð��HÞ2r0ðTL � T0Þ

r22 � r1

2

hð5Þ

You have to be careful with integration because h varies

as a function of r. This difficulty can be overcome by inte-

gration in steps. Within each interval you use an average

value of h.

Hint A337

HINT A291

(A43)

� dT

dt¼


p

qffiffiffip

prLLðsin 5�ÞcL

p

TE � T0ffiffiffiffiffiffiffiffitcool

p ð9Þ

Introduce numerical values into Equation (9).

Hint A269

HINT A292

(A205)

Choose a number of d values, calculate the corresponding

values of v with the aid of Equation (8) and plot them in

a diagram.

Answer

The maximum casting rate is

vmax ¼ 9:4 � 10�3 m2=min

dðd < 15 � 10�3 mÞ

d v

(m) (m=min)

1 � 10�3 9.4

2 � 10�3 4.7

5 � 10�3 1.9

8 � 10�3 1.2

10 � 10�3 0.9

15 � 10�3 0.6


HINT A293

(A13)

The expression for the constant C is obtained by combining

Equations (4.73) and (4.74) on pages 80–81 in Chapter 4 or

with the aid of Example 4.4 on page 81 in Chapter 4:

C ¼ p4

rmetalð��HÞTi � T0

� 21

kmouldrmouldcmouldp

ð2Þ

where Ti ¼ TL in this case.

Calculate C with the aid of material data.

Hint A151

HINT A294

(A123)

In analogy with Equation (3) in Exercise 5.10a (Hint 321),

you obtain for side 4

z4dlrð��HÞ dy4

dt¼ adlh4ðTL � T0Þ ð10Þ

Similarly you obtain for side 2



From Equation (9) and the figure in Hint A123, you

know that z2 ¼ z4. If you divide Equations (10) and (11)

you obtain after reduction

dy4

h4

¼ dy2

h2

or, after integration and introduction of the designations

h2 ¼ h and y2 ¼ y,

y4 ¼ h4

hy ð12Þ

Before you go on with the calculations, consider the sig-

nificance of Equation (12).

Hint A117

HINT A295

Exercise 4.4

The cooling curve of a central thermocouple consists of

three time intervals. Draw a schematic curve of the tem-

perature of the thermocouple as a function of time. Explain

the physical process during each time interval.

Hint A102

HINT A296

(A58)

Qcooltotal ¼ cpm�T ð4Þ

or

Qcooltotal ¼ 420 � 104 � 100 ¼ 42 � 107 J

Calculate the fraction of the total excess heat removed

with the aid of radiation. The major part of the excess

heat is removed by means of some other mechanism.

Which one?

Hint A165

HINT A297

(A124)

t ¼ rð��HÞTmelt � T0

ðR0 � yLÞ2

4kþ R0 � yL

2h

" #ð12Þ

Derivatization with respect to t gives

1 ¼ rð��HÞTmelt � T0

2ðR0 � yLÞ4k

þ 1

2h

� �� dyL

dt

� �

or

dyL

dt

�� ¼ Tmelt � T0

rð��HÞ1

R0 � yL

2kþ 1

2h

ð13Þ

Insert material data and other constants.

Hint A188

HINT A298

(A93)

Introduce the relation

h ¼ 400 p ð3Þ

and Equation (2) into Equation (1).

Hint A334


HINT A299

(A82)

Answer

Region II.

Nucleation of new crystals in the melt, ahead of the

growing columnar crystals, occurs when the temperature of

the cooling melt has declined to the critical temperature T�.

The total heat of solidification is smaller than the heat

losses to the surroundings because the crystals initially

are small. For this reason, the slope of the curve is still

negative, dT=dt < 0, but it becomes less negative when

the crystals grow.

Characterize region III.

Hint A345

HINT A300

Exercise 3.7b

Other factors of influence?

Hint A56

HINT A301

(A236)

r ¼ Ct1=6

N1=3ð6Þ

where C is a constant [a summary of the constant quantities

in Equation (5)].

Equation (6) can be used to derive dr=dt. Why do you

need this relation?

Hint A78

HINT A302

(A273)

Metal mould:

First of all you check Nusselt’s number:

Nu ¼ hs

kmetal

¼ hyL

kmetal

¼ 900 � 2:5 � 10�3

0:23 � 103� 1

In such cases, the temperature is that given in the figure

and Equation (4.85) on page 86 in Chapter 4 is valid:


yL

hð3Þ

where yL is half the thickness of the casting.

t ¼ 2:7 � 103 � 398 � 103

660 � 25� 2:5 � 10�3

900¼ 4:7 s

It is time for comparison and answer.

Hint A118

HINT A303

(A201)

V � V0 ¼ Aingot y½ ¼ pD2

4y½ ð3Þ

where Aingot is the cross-sectional area of the ingot.

If you combine Equations (2) and (3) you obtain

y½ ¼ Vmelt

Aingot

lncmelt � c0

cmelt

2� c0

0B@

1CA

¼ 130 � 10�6

p� 0:1002

4

ln0:37 � 0:03

0:185 � 0:03

� �¼ 1:30 � 10�2 m

Answer

The sulfur concentration has decreased to half its original

value after 13 mm.

HINT A304

(A245)

Let the sphere be inscribed in a cube with the side 2R. The

angle y is the angle between the line OA and a coordinate

axis. For symmetry reasons, the angles between OA and the

x-, y- and z-axes are equal.

T

TL

Mould Solid Liquid

T0

0 yL

y

x

z

A

B

2R 3 2R

D

O

2 2R

q


You use the triangle OBA and obtain directly

cosy ¼ 2R

2Rffiffiffi3

p ¼ 1ffiffiffi3

p ð4Þ

which gives y ¼ 55 �.With the aid of three-dimensional geometry you can find

a relation between R and r.

Hint A232

HINT A305

(A172)

Apply Bernoulli’s equation on points 1 and 2 in the tundish.

p2 þ rgh2 þrv2

2

2¼ p1 þ rgh1 þ

rv21

2ð2Þ

At both points, the pressure equals the atmospheric pres-

sure patm.

v2 ¼ voutlet and h2 ¼ 0 gives

p þ rg � 0 þ rvoutlet2

2¼ p þ rgh1 þ

rv12

2ð3Þ

or

voutlet2 ¼ 2gh1 þ v1

2

As v1 � v2 you can neglect v1, which gives

voutlet ¼ffiffiffiffiffiffiffiffiffiffi2gh1

p¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 � 9:81 � 0:500

p¼ 3:13 m=s ð4Þ

Now you can easily calculate Aoutlet and doutlet!

Hint A24

HINT A306

(A152)

When the excess temperature is gone and the solidification

starts, the temperature profile will be different.

The dominant heat flow goes to the plate and radiation is

responsible for only a minor part of the heat transport. The

solid layer yL will therefore be thicker than the layer YL. If

the thermal conduction in the layer YL is neglected you

obtain the temperature profile illustrated in the figure.

Two coordinate systems with opposite directions are

introduced.

Calculate the solidification times for the two solid layers

as functions of the thickness yL and YL .

Hint A213

HINT A307

Exercise 5.12c

Consider the cooling process and set up a heat balance.

Hint A272

HINT A308

(A99)

From Table 4.4 on page 67 in Chapter 4 you find

erf(z) ¼ erf(l) ¼ erf(0.5) ¼ 0.5205. From Figure 4.15 on

page 72 you find the approximate valueffiffiffip

plel

2 � 1:2 for

l¼ 0.5. Then you obtainffiffiffip

plel

2ð0:410 þ erfðlÞÞ ¼ 1:2�ð0:410 þ 0:52Þ � 1:12, which is far too low.

Try a higher l value, e.g. l¼ 0.8 and some other values.

Hint A126

HINT A309

(A241)

10 ¼ 2:5ffiffit

p)

ffiffit

p¼ 4 ) t ¼ 16 min

Now you can calculate vfront from Equation (7) in

Hint 167:

vfront ¼2:5 � 10�2

2 �ffiffiffiffiffiffiffi60t

p ¼ 2:5 � 10�2

2 �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi60 � 16

p ¼ 4:0 � 10�4 m=s

What will next step be?

Hint A229

HINT A310

(A54)

(A132)

h ¼ 1:0 � 103 W=m2

K; h4 ¼ 700 W=m2

K; D ¼ 2:0 m;

a ¼ 60 � 10�3 m.

These values and material constants, given in the text,

are inserted into Equation (18):

vmax ¼p� 2:0� 1:0� 103ð1083� 100Þ

2� 0:060� 8:94� 103 � 206� 1031

2� 1

81þ 700

1000

� �� ¼ 0:097m=s

T T TL

T i metal

yL T w

0 yL Y L 0

T 0 YL

1 h1 = 0.500 m

2 h2 = 0

Tundish

•

•


Answer

(a) The maximum casting rate is 0.10 m/s.

(b) A more careful calculation gives also vmax � 0:10 m=s.

The approximation in (a) is obviously reasonable.

HINT A311

(A243)

The casting has solidified completely when yL is equal to

half the thickness d of the casting:

ttotal ¼rð��HÞ

hðTi metal � T0Þd

2ð4Þ

Find an expression of the casting rate.

Hint A139

HINT A312

(A66)

yL is the thickness of the solid shell at a given time.

All points on a curve are pairs of T and yL but you are

only interested in the pair which fulfils the condition

T ¼ TL. This pair corresponds to the ‘knee’ on each

curve. Hence you can read the yL value at the ‘knee’ of

each curve.

Read corresponding Ti metal and yL values and calculate

the h value for each curve with the aid of equation (6) in

Hint A66.

Hint A220

HINT A313

(A111)

ametal is the thermal diffusion constant of the metal (com-

pare Example 4.2 on page 67 in Chapter 4) and l is the

fifth constant which has to be determined from Equation

(4.36) on page 71 in Chapter 4:

cmetalp ðTL �T0Þ

��H¼

ffiffiffip

plel

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffikmetalrmetalc

metalp

kmouldrmouldcmouldp

sþ erfðlÞ

!ð3Þ

How can you solve l from Equation (3)?

Hint A204

HINT A314

(A177)

1

�3T3

� T

Ti

¼ � 2esB

rLcLp R

t ) 1

T3� 1

Ti3¼ 6esB

rLcLp R

t

Solve T and calculate the excess temperature.

Hint A203

HINT A315

(A135)

Fourier’s first law gives the heat flow through the cylinder

element:

dQ

dt¼ �k � 2prL

dT

drð1Þ

Integration over the whole cylinder with radius R gives

dQ

dt

ðr

R

dr

r¼ �k � 2pL

ðTmelt

Ti

dT

which gives

dQ

dtln

r

R

� �¼ �k � 2pLðTmelt � TiÞ ð2Þ

where Ti is the temperature at the interface between the

solid metal and the mould.

Can you get another expression of the heat flow dQ=dt?

Hint A256

HINT A316

(A163)

h � 2pRLðTi � T0Þ ¼�k � 2pLðTmelt � TiÞ

lnr

R

or

Ti ¼T0hR ln

r

R� kTmelt

hR lnr

R� k

ð6Þ

Now Ti is expressed in known quantities. How do you

get the solidification time t as a function of r?

Hint A265

HINT A317

(A283)

The solidification time can be calculated by means of

Equation (4.48) on page 74 in chapter 4:


yL

h1 þ h

2kyL

� �ð2Þ

The temperature of the cooling water T0 ¼ 100 �C.


Inserting the material values given in the text into Equa-

tion (2) gives the total solidification time:

t ¼ 7:88 � 103 � 272 � 103

1470 � 100� 0:145

313

� 1 þ 313

2 � 30� 0:145

� �¼ 1273 s ¼ 21:2 min

The answer is given in Hint A228

Hint A228

HINT A318

(A242)

�Tmelt ¼ Tcrystal � Tfront ¼ ðTL � TfrontÞ � ðTL � TcrystalÞ

Application of the growth law gives

�Tmelt ¼ ðTL �TfrontÞ� ðTL �TcrystalÞ ¼vfront

m� vcrystal

mð5Þ

The value of m is given in the text. vcrsytal is equal to

dr=dt and should be left as it is. To find the value of

�Tmelt you must calculate vfront in one way or other. How?

Hint A218

HINT A319

(A235)

Answer

Region 2:

1. The thickness of the shell grows and heat has to be

transported a longer distance than before. The tempera-

ture of the shell in contact with the mould decreases

with increasing thickness of the solidified shell. This

leads to a decreasing heat flow.

2. The cooling of the solidified shell leads to shrinkage of

the solid shell and an air gap between the mould and

shell. The width of the air gap increases gradually

with decreasing temperature of the solid shell. Hence

h decreases with increasing distance from the top of

the mould because air is a very poor thermal conductor.

This effect is the main reason for the decrease in the

heat flux.

HINT A320

(A287)

You eliminate dr=dt ¼ vgrowth between Equation (1) and the

relation given in the text, which gives

h ¼ rð��HÞ � 10�11

r0ðTL � T0Þr

lden2

ð2Þ

where lden is given by the curve in the text.

Now you know h as a function of known quantities and r

and lden. To find the h values you have to use a numerical

method and use the diagram in the text to find the l values.

Make a table with the headings r0 � r, r, lden, h and z. Read

the dendrite arm distances from the diagram and calculate

the heat transfer coefficients for some values of y ¼ r0 � r,

e.g. 10, 20, 30, 40, 50 and 60 mm.

Hint A170

HINT A321

(A253)

There is one complication, which must be handled before

Equation (1) can be used. The equation is easy to use if

the solidification is symmetrical, which is not the case

here with two different heat transfer coefficients. As an

approximation you keep the symmetry and use a weighted

average value of h for all four sides:

hav ¼3�hrollerþhbelt

4¼3�1000þ700

4¼925W=m

2K ð2Þ

Consider a volume element of the square casting of

length dl along the cooled steel belt. The heat balance

can be written as

ða � 2yÞdlrð��HÞ dy

dtSolidification heat flux

¼ adlhavðTL � T0ÞHeat flux from the metal

to the water from the

surface of the casting

ð3Þ

y

a

y

a


The variables y and t are separated. Integrate Equa-

tion (3).

Hint A263

HINT A322

(A222)

(A148)

Required material data for aluminium are: TL ¼ 660 �C,

rL ¼ 2:36 � 103 kg=m3; cL

p ¼ 900 J=kg K and

Lf ¼ 0:001056 � 0:158 � 2:36 � 103 � 900

300

� ln660 þ 30 � 20

660 � 20

� �¼ 1:18 � 0:0459 ¼ 0:0542 m

Answer

(a) Lf ¼ constantv � rL � cL

p

hln

TL þ �Tð Þi � T0

TL � T0

where

constant ¼ asina

4 � 1 þ sina2

� � ¼ 0:0011

(b) The estimated value of the maximum fluidity length

for aluminium is 5.4 cm at an excess temperature of

30 �C.

HINT A323

(A183)

RAl ¼sffiffiffi2

p

rAlgh¼ 1:5 �

ffiffiffi2

p

2:7 � 103 � 9:81 h¼ 8:00 � 10�5

h


Hint A175

HINT A324

(A285)

According to Equation (3.9) on page 33 in Chapter 3, you

obtain

Alower

Aupper

¼ Asprue

Aupper

¼ffiffiffiffiffiffiffiffiffiffiffiffiffihcasting

htotal

rð3Þ

or

Aupper ¼ Alower

ffiffiffiffiffiffiffiffiffiffiffiffiffihtotal

hcasting

s

¼ 1:10 � 10�4 �ffiffiffiffiffiffiffiffiffi0:28

0:23

r¼ 1:21 � 10�4 m2

Calculate the diameters of the circular areas and give the

answer.

Hint A90

HINT A325

(A195) ðTL

TLþð�TÞi

� dT

T � T0

¼ðtf

0

4h 1 þ sina2

h iasinarLcL

p

dt

or

4h 1 þ sina2

h iasinarLcL

p

tf ¼ � lnTL � T0

TL þ ð�TÞi � T0

where ð�TÞi is the initial excess temperature and tf the

solidification time.

How do you get Lf when tf is known?

Hint A222

HINT A326

(A196)

Separate the variables and integrate Equation (1):ðd=2

0

dyL ¼ hðTL � T0Þrð��HÞ

ðt

0

dt ð2Þ

d ¼ 2hðTL � T0Þrð��HÞ tsol


1

tsol

¼ 2hðTL � T0Þrdð��HÞ ð3Þ

where d is the thickness of the slabs. Find the casting rate.

Hint A143

HINT A327

Exercise 4.3b

Calculate the solidification time for a steel plate of the same

size and shape as that in Exercise 4.3a.

Hint A286


HINT A328

(A140)

A is the area of the five ingot walls at a distance of 10 cm

from the surface:

A ¼ 2 � 0:20 � 1:40 þ 2 � 0:40 � 1:40 þ 0:20 � 0:40 m2

or A ¼ 1:76 m2.

Next you have to find an expression of ð�TÞmelt. Start

with the definition of ð�TÞmelt.

Hint A225

HINT A329

(A117)

The heat balances for sides 1 and 3 are identical and can be

written as



or, with the aid of Equation (8) and the designations in Hint

123, you obtain

½a � ðy þ y4Þ�dlrð��HÞ dy

dt¼ adlhðTL � T0Þ ð14Þ

Introduce Equation (12) into Equation (14) and integrate

the latter equation.

Hint A206

HINT A330

(A279)

According to the energy law, the two expressions must be

equal. This gives the relation

4pr2 dr

dtNrð��HÞ ¼ A


p

pt


Equation (4) offers the possibility of finding a relation

between N, r and t. Derive this relation.

Hint A236

HINT A331

Exercise 4.11a

The figure shows that the Al alloy has a solidification inter-

val. At temperatures T � Tliquidus the alloy is molten. The

solidification is complete at temperatures T � Tsolidus.

Within the intermediate temperature interval there are two

phases, solid and liquid. The melt has an initial excess

temperature.

Draw a sketch of the temperature distribution of the

ingot and set up equations for the heat flow.

Hint A264

HINT A332

(A108)

The contact between the melt and the sand mould is poor

and Ti � TL ¼ 660 �C. The area is equal to the total area

of the cube ¼ 6 � 0:252 m2.

t1 ¼ 2:7 � 103 � 0:253 � 1:18 � 103 � 50

2 � 6 � 0:252 � ð660 � 20Þ

� 2

� p0:63 � 1:61 � 103 � 1:05 � 103

¼ 79 s ¼ 1:3 min

Set up a heat balance for the solidification process.

Hint A113

HINT A333

(A157)

The general equation with poor contact between the mould

and solid metal is Equation (1). Hence it is valid for Figure

4.17 on page 73 in Chapter 4 (left figure here).

Figure 4.27 on page 86 in Chapter 4 (right figure here)

is a special case which is valid when hyL=k � 1 and can

150 – 10 = 140 cm

60 – 20 = 40 cm

40 – 20 = 20 cm


be neglected in Equation (1). In this case Equation (1)

gives


1 þ h

kyL

þ T0 � TL � T0 þ T0 ¼ TL

In this special case, the temperature in the metal is inde-

pendent of yL and equals the liquidus temperature, in agree-

ment with Figure 4.27.


Hint A146

HINT A334

(A298)

400pðTL � T0Þ ¼ rð��HÞ 10�12

lden2

ð4Þ

Now you are very close to the final answer!

Hint A231

HINT A335

(A186)

t ¼ rð��HÞTL �T0

0:0875z

h¼ 2650� 371� 103 � 0:0875

ð853� 293Þ� 2:0� 103z ¼ 77z

tmax ¼ 77� 0:10 s

Answer

The solidification time at distance z is t ¼ 77z. The total

solidification time is about 8 s.

HINT A336

(A184)

Combine Equations (2) and (3) and solve Ti metal.

ks

TL � Ti metal

y¼ h1ðTi metal � T0Þ


Ti melt ¼TL � T0

1 þ h1

ks

y

þ T0 ð6Þ

How do you proceed?

Hint A237

HINT A337

(A290)

If you insert given numerical values into Equation (5), you

obtain

t ¼ 7:0 � 103 � 280 � 103

2 � 65 � 10�6 � ð1450 � 20Þrnþ1

2 � rn2

hav

The first value will be:

t ¼ 1:05 � 1010 � 1000 � 1012

26:5 � 105¼ 4:0 � 10�6s

The other values are calculated in the same way.

Now you know the solidification times of the different

intervals. How do you calculate z?

Hint A262

HINT A338

(A192)

The energy principle tells that the heat flows in Equations

(1) and (2) are equal:

�cprAddT

dt¼ hAðTstrip � T0Þ ð3Þ

How do you get the desired relation?

Hint A274

HINT A339

(A240)

Consider a wire element with height L, radius r and thick-

ness dr. The solidification heat is equal to the radiation heat

flow during the time dt through the outer cylinder with

height L and radius R. Compare Hint A288.

2pRLsBeðTL4 � T0

4Þdt ¼ r� 2prð�drÞLð��HÞ ð1Þ

RsBeðTL4 � T0

4Þðt

0

dt ¼ �rð��HÞð0

R

rdr

rnþ1 � rn hav rnþ12 � rn

2 t

(mm) (W/m2 K) (m2) (s)

55–45 26.5 � 105 1000 � 10�12 4.0 � 10�6

45–35 19.5 � 105 800 � 10�12 4.3 � 10�6

35–25 11.0 � 105 600 � 10�12 5.7 � 10�6

25–15 4.65 � 105 400 � 10�12 9.0 � 10�6

15–5 1.6 � 105 200 � 10�12 13.0 � 10�6


which gives

t ¼ rð��HÞr2

2sBeRðTL4 � T0

4Þ

� R

0

¼ rRð��HÞ2sBeðTL

4 � T04Þ

ð2Þ

Insert given data and material constants in Equation (2).

Hint A65

HINT A340

(A125)

In this case kCu ¼ 398 W=m K and h ¼ 400 W=m2

K.

s ¼ yL ¼ dmax

2¼ 0:10 m

Nu ¼ hs

k¼ 400 � 0:10

398¼ 0:10 � 1

Hence the temperature distribution can be assumed to be

that illustrated in the figure.

Set up a heat flux balance.

Hint A196

HINT A341

(A239)

The number of crystals is increased on inoculation. When N

is increased, the undercooling ðTE � TÞ decreases and the

risk of white solidification becomes reduced.

The main reason for inoculating cast iron is the possibi-

lity of avoiding white solidification. The cast iron may soli-

dify as grey iron through the influence of the additives. Cast

iron is inoculated with FeSi, for example, in order to

achieve grey solidification.

Mention another advantage.

Hint A80

HINT A342

(A143)

The temperature of the cooling water is assumed to be

100 �C. Inserting all the known values into Equation (4)

gives

vmax ¼ 2LhðTL � T0Þrð��HÞ

1

d¼ 2 � 2:5 � 400ð1083 � 100Þ

8:94 � 103 � 206 � 103d

¼ 1:07 � 10�3 m2=s

d¼ 64 � 10�3 m2=min

d

Plot the function in a diagram.

Hint A219

HINT A343

(A218)

y ¼ 2:5ffiffit

pð6Þ

where the thickness y is measured in centimetres and the

time t in minutes.

How do you obtain the growth rate of the solidification

front with the aid of Equation (6)?

Hint A167

HINT A344

(A91)

Use Equations (2) and (4) and eliminate t.

Hint A41

HINT A345

(A299)

Answer

Region III.

The central equiaxed crystal zone replaces the columnar

crystal zone approximately at the minimum of the curve

(page in Chapter 6). The number of crystals and their

total surface area increase in addition to the evolved heat

of fusion. The total heat of solidification exceeds the heat

losses and the temperature increases. The derivative of

the curve becomes positive but the slope decreases with

time.

Characterize region IV.

Hint A209


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