Welding Formulas:
Q1:
A pressure of 20 volts is applied across ends of a wire, and a current of 5
Amperes flow through it. Find the resistance of the wire in ohms:
Ans: R= V÷I = 20÷5= 4 ohms
Q2:
A welding resistor has a resistance of 0.1 ohms. Find the voltage drop across
it when a current of 150 Amperes is flowing through it.
Ans: V= I X R, i.e, V= 150 X 0.1= 15 volts drop
Q3:
A welding generator has an output of 80 volts, 250 Amperes. Find output in
Kilowatts and Joules per second:
Ans:
V= 80
A= 250
W= 80 X 250 = 20000
KW= 20000/1000= 20 KW
20000 J/s
Q4:
How do you find Chromium equivalent if Si= 0.8% and Cr = 19.0%?
Ans: Formula:
% Cr + % Mo(1.5 X %Si) + (0.5 X % Nb)
% Cr + % Mo(1.5 X % Si) + ( 0.5 X %Nb)
% 19.0 + 1.2 + 0
% 20.2
Q5: How
do you find Nickel equivalent if C = 0.03%, Mn = 0.7%, Ni = 10.0% ?
Ans= Formula=
% Ni + (30 X % C)
+ ( 0.5 X % Mn)
% 10.0 + 30 X 0.03 + 0.5 X 0.7
% 10.0 + 0.9 + 0.35
% 11.25
Q:
What is definition of dilution?
Ans: When two metals are fusion welded together by metal
arc, TIG, MIG or submerged Arc processes, the final composition consists of an
admixture of parent plate and welding wire.
Formula:
% percentage dilution =
weight of parent metal in weld
X 100
Total
weight of weld
If there are 15 parts by weight of parent plate in75
parts by weight of weld metal then the dilution is 15/75 X 100 = 20%
Average values of dilution for various processes are:
Metal Arc: 25-40%
SAW: 25-40%
MIG (S.T) 25-40%
MIG (D.T) 15-30%
TIG 25-50%
Q6:
What will be the approximate composition of the final weld if there is 40%
dilution? With 40% dilution the plate will contribute 40% and the welding wire
60%. If 9% Nickel plate, 80% composition welding wire Nickel, 20% chromium:
Nickel Plate = 40 / 100 X 9 = 3.6
80% Ni welding wire= 60/100 X 80 = 48
20% Cr welding wire = 60/100 X 20 = 12
Iron = 40/100 X 91 = 36.4
So, %
Nickel = 3.6 + 48 = 51.6 %
Chromium = 12 %
Iron = 36.4%
Q7: A plate of an alloy of composition 70% Ni, 30 % Cu is to be welded to a
plate of alloy steel of composition Cr-18%,
Ni-12%, Fe-70% using a wire of composition 75%, Ni-15%, Cr-8% Fe.
Assuming 30% dilution, what will be the approximate
composition of the final weld?
With 30% dilution each plate
will contribute 15% and the welding wire 70%.
Nickel: 15/100 X 70= 10.5
A plate of alloy steel:
15/100 X 12 = 1.8
Welding wire: 70/100 X 75 = 52.5
So, Nickel is: 10.5 + 1.8 +
52.5 = 64.8%
Chromium:
15/100 X 18 = 2.7
70/100 X 8 = 10.5
So, Chromium is = 2.7 + 10.5
= 13.2
Copper:
15/100 X 30 = 4.5
Iron:
15/100 X 70 = 10.5
70/100 X 8 = 5.6
10.5 + 5.6 = 16.1
Q8:
How do you find Arc energy (KJ/mm)?
Ans: Arc energy (KJ/mm) = arc voltage X welding
current
Welding speed (mm/s) X 1000
Q9:
How do you find Heat Input by Japanese standard?
Ans:
Heat = Ampere X Volts X 60
Travelling speed
Heat= 170 X 25 X 60 = 17000 Joules/cm
15
If you require in KJ then,
Heat = 170 X 25
X 0.06 = 17 KJ/cm
15
Q10:
What is Arc Energy ?
Ans: ARC Energy = (KJ/mm) = Volts X Ampere
Travel
speed (mm/sec) X 1000
MAG
Process:
Volts = 24
Ampere = 240
Travel Speed = 300 mm/ per minute
Arc Energy (KJ/mm) = Volts X Amperes
Travel
speed (mm/sec) X 1000
= 24 X
240 X 60
300 X 1000
= 345600
300000
Arc Energy = 1.152 or 1.2 KJ/mm
Important:
Thermal Efficiency factors:
SAW (Wire Electrode) :
1.0
MMA (Covered Electrode): 0.8
MIG /MAG: 0.8
FCAW (with or without gas shield): 0.8
Tungsten Inert Gas (TIG): 0.6
Plasma: 0.6
Heat Input (KJ/mm) = volts X Amperes X 60 X K
Travel
speed (mm/sec) X 1000
Volts = 24
Amp= 240
v = 300 mm/per minute
Heat Input (KJ/mm) = Volts X Amperes X 60 X K
Travel
speed (mm/sec) X 1000
Heat Input (KJ/mm) = 24 X 240 X 60 X 0.6
300 X 1000
Heat Input (KJ/mm) =
207360 / 300000 = 0.6912 KJ/mm
Q10:
How do you find Heat Input by American Standard?
Ans: Root = 140 X 14.5 X 60 = 121800/ 2.300= 52.9
KJ/mm
1000
X 2.3
Filling = 160 X 16 X 60 = 153600/ 1500 = 102.4 KJ/
mm
1000 X 1.5
Capping = 150 X 15 X 60 = 135000 / 1500 = 90 KJ /mm
1000
x 1.5
Travelling
Speed :
Root: 140 X 14.5 X 60 = 121800/ 52950 = 2.3 X 25.4
= 58.4mm/min
1000 X
52.95
Filling: 160 X 16 X 60 = 153600 / 102400 = 1.5 X
25.4 = 38.1mm/min
1000 X
102.4
Capping: 150 X 15 X 60 = 135000 / 90000 = 1.5 X
25.4 = 38.1 mm/min
1000 X 90
Q11:
How do you calculate Welder Travel speed?
Ans:
Welder Travel= 4 inches
Time : 50
seconds
Formula: 4 / 50 =
0.08 inch per/sec
0.08 X 60 sec/per min= 4.8 inches /per min
Conversion formula:
4.8 inches X 2.54 = 12.1cm/min
12.1cm X 0.393 = 4.8 inches/min
Important:
Shield
Metal Arc Welding:
Electrode
Holder: Maximum Current Range Resistance: 150A – 500A Normal Range. If metal
thickness less than 5mm then Gap will be between 1mm -2mm and if thickness is
5mm – 8mm and the gap will be between 2mm-4mm without any special preparation.
Lap
weld:
(i)
Parent
metal should be proper cleaned
(ii)
Pipes
to have different diameters
(iii)
Not
recommended above 10mm thickness
(iv)
Metal
is wasted in providing a lap
Butt
Weld:
(i)
When
the material thickness exceeds 8mm, it is often difficult to achieve full
penetration with square edge preparation.
(ii)
Then
prepared by machining to V configuration.
(iii)
When
plate thickness above 20mm, it is better to have double V edge preparation.
This helps in achieving good quality weld without distortion.
Current
Adjustment:
If
plate thickness is between 5mm-60mm, the electrode size is recommended between
3mm-6mm.
SMAW
operation:
(i)
In a
welding transformer usually two settings for OCV are available say 80 V and
100V.
(ii)
The
OCV setting for a D.C. power source is usually 10-20% lower than in a welding
transformer of the same current ratings.
(iii)
It
is often observed that the weld bead becomes wide and peaky at the point of restarting;
this is usually due to over welding of the crater. This should be avoided , as
far as possible, because it is not only un-slightly but also can be the source
of weld defects like slag entrapment, porosity and cracks.
Electrode
Motions:
The width of the weld bead formed under normal welding
conditions in SMAW is between 1.5-2.5 times the diameter of the electrode; with
well penetrated and smooth passage of the deposited metal to the work piece
surface.
Cellulose
coated electrodes: E6010
These are usually light coated, all position electrodes
with a forceful penetrating arc. The weld metal is highly ductile.
Rutile
coated Electrodes: E6012
It is an all position electrode with good penetration and
quick freezing slag. Easy to operate.
Rutile
coated Electrodes: E6013
All positional electrode, gives low spatter and easy to
remove slag.
Rutile
plus Iron Powder: E7014
A medium heavy coated all position electrode containing
iron powder that enables the use of heavy current which consequently leads to
higher welding output with a deposition efficiency of up to 110%.
Rutile
plus Iron Powder: E7024
It is a heavy coated electrode with high deposition rate
for down hand butt and fillet welds as well as horizontal fillet welds. Very
easy to manipulate, low spatter rate, high welding current and deposition
efficiency is nearly 140%. It can be used as a touch electrode.
Rutile
plus Iron Powder: E7024
A super heavy coated iron powder electrode with a metal
recovery rate of about 210%, suitable for high speed welding of down hand butt,
fillet and horizontal fillet weld.
Acid
Coating: E 6020
A medium heavy coated electrode producing a fluid slag
for down hand horizontal, and vertical welding. For low carbon steel where high
strength and high quality weld.
1 comment:
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