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Gas flames, arc welding and heat input

Short answer

Oxyacetylene flames are neutral (steel welding), carburising (hard facing, aluminium), oxidising (brass/bronze). SMAW uses flux-coated consumable electrode; coating shields gas, slag, stabilises arc, alloys/deoxidises. Heat input = 0.8x24x200/5 = 768 J/mm; melted area = 0.5x768/10 = 38.4 mm2.

(a) Describe the three types of oxyacetylene flame and their uses. (b) Explain the principle of shielded metal arc welding (SMAW) and the functions of the electrode coating. (c) In arc welding at 24 V and 200 A the travel speed is 5 mm/s. The heat transfer efficiency is 0.8. Find the heat input per unit length of weld, and the cross-sectional area of weld metal that can be melted if the melting factor is 0.5 and the unit melting energy of the steel is 10 J/mm3^3.

(a) Oxyacetylene flames

Acetylene burns with oxygen in a torch. The flame has an inner cone (about 3100 °C at its tip) and an outer envelope. The ratio of oxygen to acetylene decides the type of flame:

FlameO2_2 : C2_2H2_2AppearanceUse
Neutralabout 1 : 1Clear, sharp inner coneWelding steel, cast iron; most work
Carburising (reducing)Excess acetyleneFeather around the inner coneHard facing, aluminium, Monel
OxidisingExcess oxygenShort, pointed inner coneBrass, bronze (prevents zinc loss), cutting

(b) SMAW

A consumable flux-coated electrode is held in a holder, and an arc (typically 20-30 V, 50-300 A) is struck between the electrode tip and the work. The arc heat melts the electrode core wire and the base metal, and the molten core wire is transferred to the weld pool as droplets. The coating burns and forms slag, which covers the weld. The current can be AC or DC (electrode negative or electrode positive).

Functions of the coating:

  1. Produces a shielding gas (CO2_2, H2_2O) that protects the molten metal from oxygen and nitrogen in the air.
  2. Forms slag that protects the cooling weld and slows its cooling.
  3. Stabilises the arc (potassium and sodium compounds).
  4. Adds alloying elements and iron powder (which increases deposition).
  5. Acts as a deoxidiser that refines the weld metal, and helps shape the bead.

(c) Heat input and melted area

Arc power:

P=VI=24×200=4800 WP = VI = 24\times200 = 4800\ \text{W}

Heat input per unit length of weld, with heat transfer efficiency f1=0.8f_1 = 0.8:

H=f1 VIv=0.8×48005=768 J/mmH = \frac{f_1\,VI}{v} = \frac{0.8\times4800}{5} = 768\ \text{J/mm}

The fraction of the transferred heat used for melting is f2=0.5f_2 = 0.5. Energy balance for melting:

f2 H=Um Awf_2\,H = U_m\,A_w Aw=0.5×76810=38.4 mm2A_w = \frac{0.5\times768}{10} = 38.4\ \text{mm}^2

Answer: heat input =768= 768 J/mm (0.768 kJ/mm); melted weld area =38.4= 38.4 mm2^2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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