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
ADVANCED
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
MICHAEL FARADAY.
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
Add 1+2+3 = 4.71dm
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.
IN ANY FERROUS FOUNDRY,IN ORDER TO GET THE BEST OUT OF
THE FEEDER( BE IT SAND,INSULATIG OR EXOTHERMIC ) COVER
TOP AT LEAST WITH “ PADDY HUSK.” INCIDENTALLY PADDY HUSK
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
Paddy Husk 3800
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
efficient is your Methoding practice.
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
IS ABOUT 1.3 TO 1.5.
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