SME
Manufacturing Processes
Welding and joining
Use this for “A GMAW pass runs at 24 V and 220 A, 6 mm/s, with 85% heat transfer and 55% melting efficiency (Uₘ = 10 J/mm³). Find the heat input and the weld bead’s area.”
Chart
Hₘ = Earc − Qarc − Qc
880 − 132 − 336.6 = 411.4
The flows in add and the flows out take away: Hₘ = 411.4.
This is an example. Type your own values to start a new problem.
H (J/mm)
=
%
×
V (V)
×
I (A)
v (mm/s)
Aw (mm²)
=
%
×
H (J/mm)
Uₘ (J/mm³)
On the bars, the arc gives Earc = VI ÷ v per mm; f₁ of it enters the work as H, and f₂ of H melts the bead: Hₘ = Uₘ × Aw.
Earc
Arc energy
worked out (not typed) · J/mm
Qarc
Arc loss
worked out (not typed) · J/mm
Qc
Heat conducted away
worked out (not typed) · J/mm
Hₘ
Melting energy
worked out (not typed) · J/mm
Formulas
H
=
(
f₁
100
)
×
V
×
I
v
748
=
(
85
100
)
×
24
×
220
6
Aw
=
(
f₂
100
)
×
H
Uₘ
41.14
=
(
55
100
)
×
748
10
Earc
=
V
×
I
v
880
=
24
×
220
6
Qarc
=
(1
−
f₁
100
)
×
Earc
132
=
(1
−
85
100
)
×
880
Qc
=
(1
−
f₂
100
)
×
H
336.6
=
(1
−
55
100
)
×
748
Hₘ
=
(
f₂
100
)
×
H
411.4
=
(
55
100
)
×
748
Assumptions
•
The arc and the travel are steady, so each mm of weld gets the same heat.
•
f₁ is about 85–90% for GMAW and SAW, and 70–80% for stick welding (SMAW).
•
Uₘ comes from the metal’s melting point; for steel it is about 10 J/mm³.
Step-by-step
WE KNOW
V = 24 V, I = 220 A, v = 6 mm/s, f₁ = 85%, f₂ = 55%, Uₘ = 10 J/mm³
FIND
heat input (H), weld area (Aw), arc energy (Earc), arc loss (Qarc), heat conducted away (Qc), melting energy (Hₘ)
Step 1 · Find heat input (H)
Use H = (f₁ ÷ 100) × V × I ÷ v
The
arc’s
power
VI
spread
over
the
v
mm
it
travels
each
second;
f₁
of
it
reaches
the
work.
H
=
(
85
100
)
×
24
×
220
6
H
=
0.85
×
24
×
220
6
H
=
20.4
×
220
6
H
=
4,488
6
H
=
748
J/mm
Step 2 · Find weld area (Aw)
Use Aw = (f₂ ÷ 100) × H ÷ Uₘ
The
share
f₂
of
the
heat
input
melts
metal;
each
mm³
takes
Uₘ,
so
each
mm
of
bead
has
that
area.
Aw
=
(
55
100
)
×
748
10
Aw
=
0.55
×
748
10
Aw
=
411.4
10
Aw = 41.14 mm²
Step 3 · Find arc energy (Earc)
Use Earc = V × I ÷ v
The arc’s power over the travel speed: the energy it gives each mm of joint.
Earc
=
24
×
220
6
Earc
=
5,280
6
Earc = 880 J/mm
Step 4 · Find arc loss (Qarc)
Use Qarc = (1 − f₁ ÷ 100) × Earc
The part of the arc’s energy that never reaches the work: spatter, light and hot gas.
Qarc
=
(1
−
85
100
)
×
880
Qarc = (1 − 0.85) × 880
Qarc = 0.15 × 880
Qarc = 132 J/mm
Step 5 · Find heat conducted away (Qc)
Use Qc = (1 − f₂ ÷ 100) × H
The part of the heat input that flows into the cold metal around the weld.
Qc
=
(1
−
55
100
)
×
748
Qc = (1 − 0.55) × 748
Qc = 0.45 × 748
Qc = 336.6 J/mm
Step 6 · Find melting energy (Hₘ)
Use Hₘ = (f₂ ÷ 100) × H
The part of the heat input that melts the metal in the bead.
Hₘ
=
(
55
100
)
×
748
Hₘ = 0.55 × 748
Hₘ = 411.4 J/mm
Check
Put the numbers back in. Both sides should match.
748
=
(
85
100
)
×
24
×
220
6
✓
41.14
=
(
55
100
)
×
748
10
✓
880
=
24
×
220
6
✓
132
=
(1
−
85
100
)
×
880
✓
336.6
=
(1
−
55
100
)
×
748
✓
411.4
=
(
55
100
)
×
748
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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Use this for “A spot weld runs 10 kA through 0.1 mΩ for 0.2 s; 30% of the heat goes into the nugget (Uₘ = 10 J/mm³). Find the heat and the nugget’s volume.”