# A Primer on Ferrous Foundry Practice & Metallurgy by Rajendra Prasad.G.M. - HTML preview

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4.2

2

Steel chill mould

4.3

3

Graphite mould

5.1

4

Copper shot bonded

6.3

5

Steel shot bonded

9.0

6

Silicon carbide bonded

10.4

7

Magnesite bonded

11.2

8

Alumina bonded

12.1

9

Chrome ore bonded

13.4

10 Zircon bonded

13.8

11 Olivine bonded

15.8

12 Silica sand bonded

17.0

13 Thermit-silica bonded

31.0

14 Exothermic compound( 45 mm thick) 48.0

15 ---------do-------------(80 mm thick)

90

Now that you know the volume of the sphere, surface area of the sphere and

Solidification time in minutes, you can calculate the CONSTANT.

25

2

Solidification time in minutes = k x ( modulus in cm)

Now going back to our cube- 200 x 200 x 200 (all dimensions in mm)

We had got a sand feeder of 240 mm dia x 240 mm height.

Since cube is only 200 mm and feeder dia being 240 mm dia. there will be a

20 mm Projection of feeder on each side. This projection will be in touch

with the mould material, which will result in faster cooling. This faster cooling

will result in premature solidification of feeder, which result in defective

casting. To avoid this problem THE TOP PORTION of the cube is made bigger

by adding a metallic pad at the top of the cube( becomes a part of the casting)

so as to accommodate 240 mm dia feeder. This would give a SOUND casting.

But it would increase the fettling work on the casting viz. cutting and grinding

that extra portion which adds to cost of production.

This problem can be solved by INSERTING a BREAKER CORE (BRC) in

between casting and feeder.

As BRC is made of sand , it’s dimensions are always governed by sand

feeder.

Circular opening of BRC is 40 % of sand feeder dia and it’s thickness is 10 %

of sand feeder dia. As our sand feeder dia is 240 mm dia. The opening of the

Breaker core is 240 x 0.4 = 96 mm dia. Thickness of BRC is 240 x 0.1 =

24mm. Along the thickness there is a TRIANGLE WHOSE BASE IS 24 mm.

Angle of the top of the triangle is 90 degrees. DIA of BRC from inside edge to

the opposite end is 96 mm.

Now look at the benefits that will accrue.

Earlier you had to cut 240 mm dia & pads then grind them.

At the minimum 0.785 x 2.4 x 2.4 =4.52 dm2 + was to be cut & ground.

Now 0.785 x 0.96 x 0.96 = 0.723 dm2 is to be cut &ground.

0.723 / 4.52 x100 = 15.99 % or 16 %. Your savings in cutting &grinding will

be more than 84%.you save on gas, grinding wheel, time and labour charges.

Note : no matter what feeder you use, sand, insulating or exothermic your

BRC dimension will remain that of sand feeder as BRC is made of SAND.

Now we change gears.

A Sand feeder will have an available metal of 10%, in other words, if a sand

feeder has 10 Kgs of liquid metal in it, only 1.0 Kg of liquid metal comes down

to feed the shrinkage. This 10% metal is called AVAILABLE METAL.

This figure in case of an INSULATING SLEEVE is 20%. The same in case

of an EXOTHERMIC SLEEVE is 30 %.

In order to IMPROVE the YIELD i.e. CASTING WEIGHT / TOTAL METAL

used for getting a SOUND CASTING, we need to reduce the metal used in

FEEDERS. This is achieved by using insulating & exothermic sleeves.

This reduces the total metal used per casting & thus improving the YIELD.

An insulating feeder will have a MODULUS EXTENSION FACTOR(MEF)

of 1.3 to 1.5 i.e. sand feeder dia / 1.3= diameter of insulating sleeve.

In our case 240 / 1.3 = 185 mm dia x 185 ht

26

A sand feeder of 240 mm dia x 240 mm ht will have 82.47 Kg of metal.

An insulating feeder with a MEF of 1.3 will give 185 mm dia x185 ht. This

takes a metal of 37.77 Kg .82.47-37.77 = 44.7 Kg of liquid metal savings.

It works out to 44.7 / 82.47 x100 = 54.2 % savings of liquid metal.

An insulating sleeve will have an available metal of 20 %

37.77 x 0.2 = 7.554 Kg of liquid feed metal is available against our requirement

of 3.8 Kg.

If you use an exothermic sleeve, it will have a MEF of 1.5

240 / 1.5 =160 mm dia x160 ht

0.785 x ( 1.6)3 x 7.6 = 24.4 Kg.

An exo sleeve gives anywhere between 25 to 30 % available metal.

24.4 x 0.25 = 6.1Kg.

Now let us look at what we need and what we get.

Required metal = 3.8 Kg.

Sand feeder 240 dia x240 = 82.47 Kg—available metal = 8.24 Kg

Insulating feeder-185 dia x185 ht =37.77 Kg available metal = 7.55Kg.

Exothermic sleeve 160 dia x 160 ht =24.43 Kg available metal = 6.1 Kg.

All the three figures are more than required---3.8 Kg.

In any foundry, an Induction furnace is the costly equipment with a limited

capacity. So our endeavour should always be to get more Kg of Casting per

ton of liquid metal. This is called YIELD IMPROVEMENT.

This is achieved by changing from SAND FEEDERS to INSULATING

FEEDERS to EXOTHERMIC FEEDERS.

Let us say a Sand feeder of dia D has a solidification time of ( a) ,an Insulating

feeder of dia D will have a solidification time of ( b ), an Exothermic feeder

of dia D will have a solidification time of ( c).

If you look at the solidification times of all the three-- c > b > a. As our casting

being same we need a certain solidification time so as to get a sound casting.

For argument’s sake let us say a sand feeder with a solidification time ( a) is

adequate for our casting, then an insulating feeder with the same solidification

time of (a) is adequate for our purpose. But solidification time of an insulating

feeder of same dia as that of sand feeder has a longer solidification time ( b).

In order to get a solidification time of ( a) in an insulating feeder, we need to

use a smaller insulating feeder to get a solidification time of ( a).Thus saving

in amount of liquid metal that goes into an insulating feeder. The argument

holds good for an exothermic feeder, which will be even smaller than an

insulating feeder for the same solidification time of ( a ).Thus we save on

weight of liquid metal that goes into feeder per Kg of casting.

There by our total liquid metal requirement per Kg of casting comes down.

Hence we can produce more weight of castings per ton of liquid metal.

27

Let us take a relook at the difference between H / D =1.0 & H / D =1.5

Mf in H/D=1.0

Weight

H/D=1.5

Weight

Weight

Cm

Dia in mm

in Kg

Dia in mm

in Kg

Difference

1

60

1.15

54

1.22

0.07

2

120

9.5

107

10

0.5

3

180

30.5

160

34

3.5

4

240

75

214

82

7

5

300

143

266

150

7

This is extracted from Wlodawer’s DIRECTIONAL SLOIDIFICATION OF

STEEL CASTINGS.

Conclusion: For same Mf –H / D = 1.5 consumes more metal.

Hence reduces YIELD + increases conversion cost of returns to melt.

It gives you an advantage of less cutting and grinding cost as the dia (h/ d = 1.5)

is less .Eventual positive VALUE ADDED has to guide you as to which feeder

to be used.

Left to me I would rather use H / D = 1.0,but I am unable to put my finger on

the reason.

( I am 58 yrs old and have 30 years of experience , if I am having a dilemma,

what about youngsters and those who have been working in foundries but

have no proper exposure to foundry technology. That is foundry for you

I had not thought of this problem in this direction in last 30 years, why

all of a sudden when I am writing a book. Could it be a DIVINE

INTERVENTION ?)

Let us digress a bit.

1) One Prof JOHN BARDEEN had come to I.I.Sc, long time back, you are

probably wondering as to who this BARDEEN is. He is only person on face

of the earth who has got TWO NOBEL PRIZES IN THE SAME SUBJECT-

PHYSICS ( SEMI-CONDUCTORS AND SUPERCONDUCTORS).When

some body asked him a question, with out even batting his eye lid he said

that he does not know the answer.

2)

Prof

EINSTEIN

was

with

INSTITUTE

OF

STUDIES,PRINCETON in USA. One day a little girl from the

neighbourhood went to him and asked him if he would teach her Physics.

Einstein in turn asked her as to how much she would pay him. The girl said

my weekly pocket money is 10\$ and she would pay him 5\$ per week. Einstein

is supposed to have said that no university in the world had paid him half of

it’s income, so he will teach her free. Einstein was already a famous and

well respected man. One day the girl’s mother went to Einstein and

apologized to him about her daughter disturbing him and told him that she

28

would stop her daughter from disturbing him. Einstein is supposed to have

told her not to do any such thing, as he was learning more from the girl

rather than her learning from him.

3) In BRITAIN, there was very well known Chemist by name HENRY

CAVENDISH. One day Cavendish saw a small, poor boy tinkering with

some thing on road side. CAVENDISH asked that boy if he would go with

him to his place where he could do what he liked and he would be provided

with food and shelter. The boy readily agreed and went with him.

One evening Cavendish was stirring some solution. Then his wife came and

reminded him of a party for which they had been invited ( Cavendish & his

family use to live up-stairs).Cavendish told the boy to keep stirring and he

would come back soon. When they returned from the party it was late in the

night. Cavendish straight away went to bed. Next morning when Cavendish

went to the lab what he saw surprised him. The boy was still stirring. That

boy later on became a famous scientist himself. He is none other than

Back to our work.

Solidification times of 100 mm dia x 100 mm ht feeders in minutes

(bottom is a cooling surface. No treatment means sand feeder with no

top cover )

No

Top Insulation Side Insulation Top

&

Side

Treatment

only

only

Insulation

Steel

5

13.4

7.5

43

Copper

8.2

14

15.1

45

Aluminium 12.3

14.3

31.1

45.6

2

Solidification time in minutes = 2.1 x ( Mf) where Mf is in cm.

Let us find out the geometrical modulus of the cylinder.

Let us take the case of STEEL with no treatment

It has 3 cooling surfaces. Bottom, top and side.

As the dimension of the feeder are 100 mm dia x 100 ht—for our calculation

purpose 1 dm dia x 1dm ht.

2

3

Volume = 0.785 x ( 1.0) x 1.0 = 0.785 dm

2

2

Surface area 1&2 ( top + bottom) = 2 x 0.785 x ( 1.0 ) = 1.57dm

2

3 ( side) = 3.14 x1.0 x 1.0 = 3.14 dm

2

Mf = V / SA = 0.785 /4.71 =0.1666 dm or 1.67 cm.

29

2

Solidification time in minutes = 2.1x ( 1.67) = 5.85 minutes.

But what we are getting in our case is only 5.0 minutes

Why is this difference?

As the temp of a thing raises( after it looks bright red) radiation losses become

4

predominant. Radiation losses are proportional to T .Here T is a temperature

which is not centigrade but Kelvin or Absolute.

( Information: zero degree Kelvin is supposed to be an Ideal temperature which

is still not achieved. 0 degree centigrade = 273 degree K)

Let us assume that we pour our steel at 1640 degree centigrade which is equal to

1640 +273 =1913 degree Kelvin .( T )

4

Heat loss due to radiation is proportional to (1913) ---it is indeed a huge loss.

Though we have assumed that heat loss from bottom and top are same, it is not

true. Heat loss from top is far, far more than what heat is being lost from

bottom.

2

So the top Surface area is not 0.785dm but more. This change of area leads to

2

an area which is more than 0.785dm .This area is called APPARENT

SURFACE AREA. This becomes evident when you cover the top.

Solidification time has gone up to 13.4 minutes.

THE FEEDER( BE IT SAND,INSULATIG OR EXOTHERMIC ) COVER

IS QUITE EFFECTIVE. THIS INCREASES SOLIDIFICATION TIME OF

FEEDER. SO WE CAN USE A SMALLER FEEDER THEREBY

INCREASING THE YIELD.

Calorific Values of certain materials which can be used as ANTI-PIPING

Compounds, which can reduce RADIATION LOSS from liquid metal from

the top are given here.

Material Calorific value in KCals / Kg

Dry cow dung cake powder 2100

Dry fire wood powder 4500

Coconut coir dust 4200

Coconut shell powder 7800

For a given casting, we can’t alter the volume, but we can alter the APPARENT

SURFACE AREA by using various materials whose list is given earlier .

If you observe here a 150 mm diameter STEEL sphere has a solidification time

of 17.00 minutes in normal SAND MOULD. But this solidification time for the

Same casting can be VARIED from 4.2 minutes to 90.0 minutes.

As a rule, we attempt to hasten the solidification time of casting and delay the

solidification time of the feeder. This is done by altering the APPARENT

SURFACE AREA. If APPARENT SURFACE AREA(ASA) OF A SAND

30

CASTING WITH A SAND FEEDER is 1.0.The ASA can be altered to less than

1.0 or more than 1.0.

In a normal sand casting with sand feeder, with Mf = 1.2 x Mc, as the liquid

metal begins to become solid, a vacuum is created in the casting, as a result of

vacuum in the casting and the atmospheric pressure acting on top of the liquid

metal in feeder, liquid metal from feeder is sucked into casting. This process

goes on during the entire period of solidification.

When casting is solidifying in the mould, liquid metal is also getting solidified

in the feeder, as a result we get a CONICAL shrinkage in feeder. At Mf = 1.2

Mc THE TIP OF THE CONICAL SHRINKAGE STAYS INSIDE THE

FEEDER, DOES NOT ENTER THE CASTING. This results in a SOUND

CASTING.

Supposing I use STEEL CHILLS to extract heat from casting faster,

solidifying metal in feeder would begin to suck liquid metal from feeder sooner

than later.

As a result, height of the shrinkage cone in feeder decreases and it’s base

at the top ( opening of the shrinkage WIDENS).We can go on altering the rates

of HEAT EXTRACTION FROM THE MOULD AS WELL AS FEEDER TO

SUCH AN EXTENT THAT WE CAN TOTALLY ELIMINATE THE

SHRINKAGE CONE AND MAKE IT FLAT. This condition will give you

BEST condition of FEEDING AND HIGHEST YIELD.

What is this YIELD ?.YIELD is the total weight of good casting obtained per,

every 100 Kg of liquid metal poured into mould. Higher is the YIELD more

Now let us take an example and do a feeder calculation.

Let us take a plate (ignoring the length of the casting, if width ( W ) is equal 5 x

Thickness ( T ) or more, it is called, a PLATE, if T = W ,it is called a BAR.

Let us take a casting 250 x 250 x 50 ( all dimensions in mm )

3

Let us find out the volume. V = 2.5 x 2.5 x 0.5 = 3.125 dm .

2

Surface areas—there are 6 surfaces.1) 2.5 x 2.5 x 2 = 12.5 dm

2

2) 2.5 x 0.5 x 4 = 5.0 dm

2

TOTAL = 17.5 dm

Mc = V / SA = 3.125 / 17.5 = 0.1785 dm

Mf = 1.2 x Mc = 1.2 x 0.1785 = 0.2143 dm

We all ready know that if H / D = 1.0, Mf = 0.2d.

0.2d = 0.2143, d = 0.2143 / 0.2 = 1.0715 dm dia x 1.0715 mm ht.

1.0715 dm ( 107.15 mm) For sake of convenience assume it to be 1.1 dm

We have taken care of Modulus of the casting.

Now we need to take care of FEED METAL requirement.

Our V = 3.125 dm3 = 3.125 x 7.8 ( density) = 24.375 Kg = 24.4 Kg

31

A liquid steel poured, at 1640 degrees, has liquid-liquid shrinkage ( because of

superheat )and a solidification shrinkage ( because of Latent heat ) has a total

shrinkage of 6.0 % ( this varies from alloy to alloy ) in Plain C steel.

24.4 Kg say 25 Kg x 0.06 = 1.5 kg i.e. 1.5 Kg of liquid metal has to come

down from feeder ( this is called Available Metal )

From Modulus calculation, our dia of sand feeder is 110 mm dia x 110 ht.

Let us find out the total metal in this feeder.

3

0.785 x ( 1.1) x 7.6 = 7.94 Kg. Assuming that a sand feeder has 10 % available

metal, this sand feeder of 110 mm dia x 110 mm ht has 7.94 Kg.

It can give 7.94 x 0.1 = 0.794 kg. But our requirement is 1.5 Kg.

If 10 %-------- 1.5 Kg

100 %--------- 1.5 / 10 x 100 = 15.00 Kg. Our feeder should have 15.00 Kg

of metal to deliver 1.5 Kg of FEED METAL or AVAILABLE METAL.

3

15 = 0.785 x ( d ) x 7.6

D3 = 15/ 0.785 x 7.6 = 2.514

3

D3 = 2.514 dm ( 1.35 x 1.35 x 1.35 = 2.46 not adequate)

( 1.40 x1.40 x 1.40 = 2.744 is adequate )

D = 140 mm dia x 140 mm ht. This 140 mm dia x 140 mm ht sand feeder has

adequate feed metal to give a sound casting.

INFORMATION: In case of our cube, feeder size obtained by METHODS

calculation itself had more than adequate feed metal, but in the case of the

plate it is not so. Hence you need to CALCULATE FEEDER DIMENSION

from both angles. Whichever FEEDER is BIGGER should be USED.

2

Solidification Time in minutes = 2.1 ( MODULUS in cm ) --- for sand castings

of Steel casting.

Now by using different materials, we can alter the solidification times. The

Materials details are already given. For given casting (Which can’t be changed )

We can only change materials of FEEDERS so that these materials DELAY the

Solidification .Hence increase the solidification time. But we want the same

Solidification time as our casting as remained the SAME.

Let us view it this way.

T( solidification time in minutes ) = 2.1 ( M sand in cm )2

2

T (solidification time in minutes ) = 2.1 ( a x M sand in cm )

2

T (solidification time in minutes ) = 2.1 ( b x M sand in cm )

HERE a and b are CONSTANTS for insulating materials(a) and exothermic

materials( b ). Insulating materials are like your WOOLLEN SWEATER ,they

do not allow your body heat to be lost, so you stay warm irrespective of out side

temp. In other words insulating materials DELAY heat loss. As we are

interested in the time the sand feeder took to solidify, we can REDUCE the

SIZE of INSULATING FEEDER to such an EXTENT that it solidifies in the

32

same TIME as our ORIGINAL SAND FEEDER. This factor (a) is called

MODULUS EXTENSION FACTOR( MEF )FOR INSULATIG SLEEVES IT

THAT IS IF OUR ORIGINAL SAND DIA WAS 100 mm, THE

INSULATING FEEDER SIZE WILL BECOME 100 / 1.4 = 71.4 mm dia x

71.4 mm ht. As we can’t get each and every size we want, we have to accept the

nearest available size which is 75 mm dia x 75 mm ht. Unlike sand feeders

( which used to give a feed metal of 10 % of it’s total weight) insulating

feeders are supposed to give a feed metal of about 20% or more of it’s own

weight.

Let us imagine , on New Years eve you are in Kashmir, it is so cold there ,your

sweater alone is inadequate to keep you warm. Then you light a CAMP FIRE to

get additional heat so that you stay warm. Exothermic materials are like camp

fire. These materials not only generate heat, they are also insulating as a result

of this their solidification times are much longer than a SAND OR

INSULATING FEEDER OF SAME SIZE. Hence exothermic sleeves can be

smaller than an insulating sleeve for same solidification time.

MEF of EXO sleeve can be as high as 1.5 to 1.7.

For a 100 mm dia sand feeder can be replaced by 100 / 1.6 = 62.5 mm dia x

62.5 mm ht exo sleeve. like earlier we have to take the nearest available size.

As the sizes of insulating sleeves and exo sleeves are much less than sand

feeder, they consume lesser amount of liquid metal than what sand feeders

would have taken. Hence Total requirement of liquid metal per casting is

reduced. It results in increasing of yield. In other words you pour more number

of castings per ton of molten metal.

Never trust any body’s material ( such as sleeves, mould paints, metal

generating exothermic compounds etc, ) unless a free trial is done and a good

casting is obtained. In the event casting gets rejected, will he make good the loss

you have incurred.(If you deeply look at it, it will have many hidden things such

as loss of capacity, failure to meet commitments etc)

I am personally of the opinion, FOSECO’S materials are best. Let me hasten to

add I have nothing to do with FOSECO in any form.

FOSECO is an acronym for FOUNDRY SERVICES COMPANY OF U.K.

They have plants all over the world. For a stringent condition if you want a

material, they can import it from ENGLAND,GERMANY,FRANCE OR USA.

But they are bit expensive, to over come this problem have an annual contract.

Feeling bit heavy:

There are two laws. For your own good learn to respect them.

1) MURPHY’S LAW:ANY THING THAT CAN GO WRONG WILL GO

WRONG.

2) PARKINSON’S LAW : EVERY BODY REACHES HIS LEVEL OF

INCOMPETENCY SOONER OR LATER.

33

As far as I am concerned CONSTANT THINKING AND STUDY are the only

solutions. Our Ex-President, Dr ABDUL KALAM in his book “ WINGS OF

FIRE” states that TO AVOID FAILURES ANTICIPATE THEM.

Now take an example of a plate ,whose dimensions are 1000 mm x 200 mm

x 25 mm.

Volume of this plate = 10 dm x 2.0 dm x 0.25 dm

3

= 5.0 dm

3

3

Weight of the casting = 5.0 dm x 7.8 Kg / dm

= 39.00 Kg.

2

Surface area of the casting = 10 x 2 x 2 = 40 dm

2

10 x 0.25 x 2 = 5 dm

2

2.0 x 0.25 x 2 = 1.0 dm

2

Sum of all these surface areas are = 40 + 5 + 1 = 46dm

Modulus of the casting = V / SA = 5.0 / 46.0 = 0.109 dm

Mf = 1.2 x Mc = 1.2 x 0.109 = 0.139 dm

Mf = 0.2 x d = 0.139

d = 0.139 / 0.2 = 0.695 dm = 0.70 dm = 70 mm dia x 70 mm ht

3

3

Here V = 5.0 dm x 7.8 Kg / dm = 39.00 Kg.

39 x 0.06 = 2.34 Kg feed metal required.

3

Weight of feeder = 0.785 x (0.7) x 7.6 = 2.046 Kg.

Sand feeder = 2.046 x 0.1 = 0.204 Kg available Metal

Insulating feeder = 2.046 x 0.2 = 0.409 Kg available Metal

Exothermic feeder = 2.046 x 0.3 = 0.612 kg available Metal

None of the above three have enough available Metal to give a Sound casting.

Now let us find out a feeder which has enough feed metal.

Feed metal required is 2.34 Kg.

If 10 % is 2.34 Kg

100 % ---- 2.34 /10 x 100 =23.40 Kg is the total metal in the sand feeder.

23.4 = 0.785 x d3 x 7.6 = 5.966 x d3

d3 =23.4 / 5.966 = 3.922 dm3

d = 1.6dm (closest value) =160 mm dia x 160 mm ht sand feeder. Though this

sand feeder has ENOUGH FEED METAL, this will not give you a SOUND

CASTING.

Having made this statement, the onus is on me to tell you as to how to get a

sound casting.

Here I need to INTRODUCE A NEW CONCEPT CALLED “ FEEDING

DISTANCE.”

1) Now a days, virtually, every one uses a CELL PHONE. When you make a

call on CELL you get an answer, at times, THE PERSON YOU ARE TRYING

TO REACH IS BEYOND REACH or OUT OF REACH.IT MEANS THAT

THAT PERSON IS BEYOND THE REACH OF SIGNALS FROM THE NET

34

WORK TOWER.DOES THIS MEAN THAT THE TOWER HAS A REACH,

BEYOND WHICH IT DOES NOT WORK? SHALL WE CALL THIS REACH

UPTO WHICH THE PERSON IS REACHABLE A “ FEEDING DISTANCE “

of the SIGNALS.

Any communication is complete only when the communication is received by

the person to whom it was meant to be and understood by the receiver.

otherwise it is called “UNSOUND COMMUNICATION”

2)There was a train called “ MAHALAKSHMI EXPRESS” between

BANGALORE AND BOMBAY. This train used to travel between

BANGALORE and MEERAJ on METRE GAUGE. From MEERAJ to

BOMBAY on BROAD GAUGE. It was a connecting train for BOMBAY

bound Passengers. Obviously TRAIN from MEERAJ should leave after Train

from BANGALORE has arrived at MEERAJ. So train from BANGALORE was

FEEDING passengers to “ MAHALAKSHMI EXPRESS” at MEERAJ. For

some unknown reason if the train from MEERAJ left MEERAJ before the

arrival of the train from BANGALORE. BOMBAY bound passengers are

stranded at MEERAJ. So in effect train from BANGALORE has not been able

to feed Passengers to BOMBAY bound train at MEERAJ. The is a failure in

the System. This kind of thing happens in castings. The stranded

passengers have nowhere to go. These stranded passengers are compared

to CENTERLINE SHRINKAGE.

3) Here your is another Example. Please pay a greater attention to it and

imagine a bit. You will understand it. Hold your RIGHT PALM(partly

closed) in such a way that your THUMB is facing towards your FACE. This is

NEVER straight. If it were to be straight it would not be called a gradient.

inside to a very small degree. Now it is like a > with a large angle inside. As this

> moves it closes.

As regards to why it closes, I will tell you little later. When the slope is steep,

the angle inside the > is more. Assuming that the rate of closing is same, no

matter what is the angle, if the angle inside > is more ( which is less than 180

degrees.) > traverses a longer distance before the angle inside the becomes

ZERO. The initial inside angle of > is decided by the THERMAL

CONDUCTIVITY of the CASTING alloy. For a pure metal, which has got the

HIGHEST thermal conductivity ,any addition of different elements is an

impurity ,We add different alloying elements for our purpose, the Thermal

conductivity is REDUCED. It is this Thermal conductivity which decides the

angle of gradient. Lower is the Thermal conductivity ( in high alloy steels )

larger is the angle inside > . So it has to traverse a longer distance before the

angle becomes ZERO. In case of Plain Carbon steels & low alloy steels ,

35

Thermal conductivity is higher & hence inside angle of > is smaller.

So it traverses a shorter distance before the angle inside > becomes ZERO.

In a CASTING, thermal gradient ( SLOPE) starts at the end &it is steepest at

the end because of 1) flat portion & 2) 4 corners.

A thermal gradient, so started closes while traversing ,towards the feeder,

because the path available closes because of concurrent solidification taking

place from 4 sides. The point up to which this > traverses, from the END, before

the angle of > becomes ZERO is called END EFFECT. By the time gradient

becomes ZERO, an other gradient starts under the INFLUENCE of feeder &

this ends in the feeder itself. This distance is called FEEDER EFFECT.

Sum of these END EFFECT & FEEDER EFFECT IS CALLED TOTAL

FEEDING DISTANCE.

In case of Plain carbon steels, as thermal conductivity is high, hence the angle

of gradient is less. Hence it closes faster and thus a shorter feeding distance.

2.5T (END EFFECT) + 2.0T ( FEEDER FEEECT) = 4.5T

That is why in the above example 1000 mm x 200 mm x 25 mm, I said you

will never get a sound casting, even if your feeder has enough FEED METAL.

In our example, at both ENDS we will have 2.5 x 25 mm = 62.5 mm SOUND

PORTION.( per end) Then adjacent to FEEDER on both sides you will have

2.0 x 25 mm =50 mm sound portion(per side).So in effect we have a TOTAL

SOUND PORTION of 62.5 mm +62.5 mm + 50 mm + 50 mm = 225 mm

SOUND PORTION.OUT of 1000 mm length ,only 225 mm is sound.

The rest 1000 – 225 = 775 mm will have what are called CENTRE LINE

SHRINKAGE on either side.

When you look at the Thermal conductivity of stainless steel ( 18 Cr / 8 Ni ),

it is 1/ 3 rd of carbon steel. Does that mean that angle of gradient , in case of

Stainless steel, is 3 TIMES more. Hence it would traverse 3 TIMES longer

distance than Plain C steel before gradient’s angle becomes ZERO.

i.e. 4.5 T x 3 =13.5 T.

To CHECK this I poured a stainless steel ( 18 Cr / 8 Ni ) plate of 15 T.

on radiography, it was found that THE END EFFECT WAS 7.5 T &

THE FEEDER EFFECT WAS 6.0 T. Sum of these two are 13.5T.

……………….

T

[---------------6T----------------][----- 1.5T-----][---------------7.5T------------------]

6T is the FEEDER EFFECT,7.5 T is the END EFFECT,1.5T is the defective

zone, these defects are called CENTER LINE SHRINKAGE which is not

acceptable in radiography. Let us ask ourselves a QUESTION as to why did

36

immediately after 7.5T.So a parallel solidification front was running for 1.5T

& it resulted in centreline shrinkage. Another > started after this 1.5T.

Do you still remember “stranded passengers in Meeraj”

For getting a sound casting SOLIDIFICATION FRONT should be

CONICAL.

If a casting has 6 numbers of feeders, from the above conclusion we need to

have 6 numbers of conical fronts. The tip of the conicl front starts from the

END and base of the cone is in the FEEDER.

The management of formation of these conical fronts for each casting ensures a

SOUND CASTING having moved all the SHRINKAGE from casting to feeder

is METHODING. To facilitate this aspect we have the following tools at our

disposal, they are CHILLS( metallic & non metallic), pads ( metallic & non

metallic-insulating & exothermic), feeders- sand, insulating &exothermic.

TO PUT IT SUCCINCTLY, IT IS LIKE MANAGEMENT OF TRAFFIC

IN A BIG CITY. THE WORD TRAFFIC MANAGEMENT IS

INTENTIONALLY USED TO CONVEY TO YOU THAT YOU ARE

DIRECTING THE FLOW OF LIQUID METAL FROM FEEDER IN

ONE DIRECTION & MOVEMENT OF SHRINKAGE INTO FEEDER IN

AN OTHER DIRECTION BY EMPLOYING THE PRINCIPLES OF

HEAT TRANSFER.

Even to this day I do not know as to how 4.5 T was arrived at for C steels. Is it

only an experimental outcome or it has a theoretical basis.

A bar has a cross section of 1:1( T:W) & plate has a cross section of 1:5 or

more.

Variation in Feeding Distances between bar , bars, plate

Width

Thickness T

Total Feeding Distance

5T

100 mm

450 mm - 4.5 T

4T

100 mm

425 mm - 4.25T

3T

100 mm

400 mm - 4.0T

2T

100 mm

375 mm - 3.75T

1T

100 mm

275 mm - 2.75T

IN PLAIN CARBON STEEL CASTINGS.

For getting a sound casting. there should always a gradient, outside the

gradient is solid metal & inside > is always liquid metal. No parallel

solidification front is acceptable as it leads to centreline shrinkage.

For our 1000 mm x 200 mm x 25 mm plate, we need to treat like this.

1000 mm / 25 =40—4.5T +9.0 T + 9.0 T + 9.0 T + 9.0 T + 4.5 T = 45 T

37

So we need to keep 4 numbers of feeders which can take care of MODULUS &

FEED METAL. In between Feeders are there are NO EDGES, hence it

will give you only 2.0 T +2.0 T = 4.0 T. So we need to create an END by

keeping CHILLS in between FEEDERS. These chills, immediately above

them hasten the solidification rate and thus NARROWING the flat path

above the chill, which in effect acts as ENDS. So between the FEEDERS

we get 4.5 T +4.5 T = 9.0 T instead of 4.0 T

Earlier we have specified the solidification rates of Steels with various

moulding Materials .

When we put a metallic CHILL, it extracts heat faster from liquid steel, so the

solidification time is shorter.(as it’s heat absorption capacity is more than that of

sand)

Despite the fact, size, shape, volume & material has remained the same, the

Solidification rate has altered to a shorter time.

For a 150 mm dia steel sphere,

Solidification time with silica sand is-------------- 17 minutes

Solidification time with a STEEL CHILL is ------ 4.3 minutes.

2

Since solidification time in minutes = A constant x ( modulus in cm )

As the VOLUME has remained the SAME ,the ONLY thing that is changing is

the SURFACE AREA. This new surface area is called APPARENT

SURFACE AREA.

PRECAUTION : While using METALLIC CHILLS ensure that they are clean.

free from DUST,OIL,RUST or any other unwanted item which could affect

the QUALITY OF THE CASTING. If required grind, sand blast the

CHILL so that the CHILL is clean in the real sense of the term.

THERE ARE NO SHORT CUTS HERE. CHILL THICKNESS SHOULD

BE EQUAL TO CASTING THICKNESS.

The Chills used should be of same thickness as the Casting. Ensure that there

is a gap between two Chills to the extent of Chill thickness to prevent TEARS.

3

2

Let us take a CUBE of 100 mm. V = 1.0 dm ,SA = 5.0 dm ( on top feeder will

be placed, hence it is a non-cooling area )

Mc = V / SA = 1 / 5 = 0.2 dm

Supposing we keep a chill at the entire bottom, it is EQUIVALENT to

increasing the surface area BY THREE TIMES AS THERE IS NO LOSS

OF CONTACT AS THE CHILL IS AT THE BOTTOM.

2

Now our new surface area is 3x1 +4x1 = 7.0 dm .

Our new Mc = 1 / 7 = 0.14 dm instead of 0.2dm.

If we put chills in the sides, they lose the contact as the casting shrinks, so the

APPARENT SURFACE AREA is ONLY TWICE THE ACTUAL SAND

SURFACE AREA.

Now let us have 5 chills on all 5 sides. Bottom = 3 ,sides = 4x2 =8,the total

APPERENT surface area is 3 + 8 = 11

38

Our new Mc = 1 / 11 = 0.09 dm instead of 0.2 dm.

By reducing the Mc you have been able to REDUCE THE DIA OF THE

FEEDER required, but you must ensure that it has adequate FEED METAL.

( There won’t be any change in feed metal requirement. It still remains

7.8 Kg x 0.06= 0.468 Kg.)

What is talked about is EXTERNAL CHILLS.

We have INTERNAL CHILLS also. According to Wlodawer 0.3 Kg of

internal chill makes 1.0 dm3 of material defect free. Precautions to be taken

for INTERNAL CHILLS remain same as external chill.

Unfortunately many people are afraid of internal chills needlessly. Metals have

no mind of their of their own , hence they do not become cunning like human

beings and misbehave. It is our business to understand as to how metals behave

and treat them accordingly.

Here I would like to share an experience of mine. We were making Ni-Hard

Casings ( WHITE IRON – BRITTLE ).This casting had 12 Mild steel inserts

for drilling holes.( Ni-Hard is too hard you can’t drill holes).We were finding

cracks around the M.S. Inserts. Our rejection rate was almost 70 %.

At that time I was in charge of MELTING & MOULDING. To make up for

the rejected castings, we were making moulds again and pour metal into the

moulds ,again the same story repeated. There was no solution in sight.

I was driven to the wall and felt helpless. Since no help was coming from

any side, I started doing some IDLE thinking. A thought came which

was some thing like this.

As M.S. ( Mild Steel) has a high melting point, it did not become soft

enough to accommodate the contraction of Ni-Hard liquid metal while

solidifying, so I thought if I put CARBON on M.S, this C would diffuse into

M.S. thus reducing it’s melting point and softens a bit thereby accommodating

the contraction of Ni-hard which in turn would reduce cracks. But how to put C

into M.S.? I dipped the M.S. Inserts in furnace oil, torched with Oxygen

-acetylene flame thoroughly so that absolutely no wetness was left ,only a dry

carbon soot was left on the inserts. Then we made the castings with those

inserts.

BELIEVE me the problem got sorted out .

So I suggest when ever you have a problem for which you have no solution,

leave it to your SUB-CONSCIOUS MIND. Your are likely to get a solution.

A small story—one Organic Chemist wanted to know the BENZENE ring.

No matter how much he tried ,he could not get it. One night he had a

DREAM.

The content of the dream was, he saw a SNAKE WITH IT”S TAIL IN ITS

MOUTH. Next day he wrote the BENZENE ring. Famous author Cohello

in his book “ALCHEMIST” writes that if you WILL something intensely

the WHOLE OF COSMIC ENERGY CONVERGES TO FULLFIL YOUR

39

WILL.

Let me say some thing about ourselves i.e. INDIANS.

1) we don’t play as a group (I am not talking of games)

2) I know every thing and I don’t need to ask any body.( attitude)

3)After getting into a job we hardly do any intellectual work. In actuality

that is the time to do that kind of work as you are financially secure.

“ AN EXPERT IS ONE WHO KNOWS MORE &MORE ABOUT LESS &

LESS”

Thomas Alva Edison is supposed to have succeeded in perfecting

th

incandescent BULB in his 250 attempt. Some body asked him if he was a

failure. Edison’s answer was “I am the only person in the world who knows

249 ways how not to make a bulb” THAT INDEED IS EXPERTIZE

WLODAWER’S “DIRECTIONAL SOLIDIFICATION OF STEEL

CASTINGS”

As he was practicing foundryman, he is most reliable.

Before we go any further let us do a RECAP

1)Check your modulus calculations after you have done it for the first time( for

each drawings. We are all liable to make mistakes. Few extra minutes spent

on rechecking is better than loosing, repairing the casting.

2)Use preferably feeders of H / D = 1.0

3)Use exothermic sleeves up to 150 dia only.

4)you can use insulating sleeves up to any size.

5)After pouring is over cover the feeder with

b)Anti-piping compound, bought out or made in-house.

c)Use clean CHILLS, if required sand blast them before using.

d)Use special refractory sands like Zircon, Chromite sands where you cannot

use chills.

e) Establish the effectiveness of sleeves for yourselves. Do not take any body’s

word for granted.

f) Be careful with internal corners.( because heat concentration & hence

shrinkage, such area should be provided with adequate radius & put Zircon sand

to extract heat faster)

g)If there is a cylindrical casting, if it’s ID is less than 27 % OD .Make it

coreless, otherwise you will have lot of sand fusion problems.

h)Use vents carefully. vents purpose is to let the air in the mould to get out as

the metal level is raising .vent should solidify as soon as the metal enters vent

hole. It should be so small in dia that it should not give any shrinkage problem

to casting.

i)Breaker core (BRC) opening should be 40 % SAND FEEDER DIA & it’s

thickness should be 10% sand feeder dia.

40

j) Breaker cores should be of highly collapsible sand.(Binders used are Organic

so that they burn out & sand becomes loose resulting in easier removal of sand)