Heat flow, power density and weld metallurgy

Power density, heat-transfer and melting factors, unit melting energy, weld cooling rate and preheat, and the fusion zone, HAZ and cracking metallurgy of welds.

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Why it matters

Only part of the energy a welding source delivers actually melts metal; the rest heats the surroundings and sets the cooling rate. That cooling rate decides whether the heat-affected zone of a steel turns into soft ferrite–pearlite or hard, crack-prone martensite. Energy balances, power density and cooling-rate estimates let you choose a process, a travel speed and a preheat before the first weld is made.

Key ideas

Power density is the power transferred to the work per unit area of the heated spot. A low-density source (oxy-fuel flame, about 10 W/mm²) spreads heat widely, melts slowly and gives wide heat-affected zones and distortion. Arc processes are roughly 50 W/mm² and up; laser and electron beam reach about 10⁴ W/mm² and more, giving deep, narrow "keyhole" welds at high speed with small HAZ. Above roughly 10⁵ W/mm² metal vaporises — the regime used for cutting and drilling. (Figures are order-of-magnitude; exact values depend on spot size.)

Energy balance of a weld. Of the power P from the source, a fraction f₁ (the heat-transfer factor, or arc efficiency) enters the work; of that, a fraction f₂ (the melting factor) melts the metal in the fusion zone — the rest is conducted into the base metal. f₁ is about 0.1–0.3 for oxy-fuel, roughly 0.7 for GTAW and up to about 0.9–0.95 for consumable-electrode arc processes such as SMAW, GMAW and SAW; take values from your data book. f₂ rises with power density and travel speed (less time for heat to leak away) and is low for metals of high thermal conductivity such as copper and aluminium.

Unit melting energy U_m is the energy to raise a unit volume from room temperature to melting and melt it. Groover's approximation U_m = K·T_m² (T_m in kelvin) gives about 10 J/mm³ for steel.

Heat flow and cooling. The moving heat source produces a steep temperature gradient ahead of the pool and a long cooling "tail" behind it. For a thick plate (three-dimensional flow), Rosenthal's solution gives the centreline cooling rate at a temperature T_c as inversely proportional to heat input and proportional to (T_c − T₀)², where T₀ is the initial (preheat) temperature. So:

  • Higher heat input → slower cooling, coarser grains, softer HAZ, more distortion.
  • Preheat raises T₀ and sharply reduces cooling rate — the main defence against HAZ hardening and hydrogen cracking.
  • Thick sections cool faster than thin ones (three-dimensional versus two-dimensional heat flow). Cooling time between 800 °C and 500 °C (t₈/₅) is the usual measure for steels.

Zones of a fusion weld.

  • Fusion zone (weld metal): a cast structure. Grains grow epitaxially from the partially melted base-metal grains and then as columnar grains toward the pool centreline, following the direction of maximum temperature gradient. Composition is a mix of filler and diluted base metal.
  • Partially melted zone: a thin band where only low-melting constituents melt; it can show liquation cracking.
  • Heat-affected zone (HAZ): not melted but transformed. In a carbon steel, going outward: coarse-grained zone (just below melting, grain growth, highest hardenability), fine-grained (normalised) zone, intercritical zone (between A₁ and A₃, partial transformation) and subcritical zone (tempering or spheroidising). In cold-worked or precipitation-hardened alloys the HAZ is softened by recrystallisation or over-ageing.
  • Unaffected base metal.

Cracking and its control. Hydrogen-induced (cold) cracking needs three things together: hydrogen (moist electrodes, grease), a hard susceptible microstructure (martensite from high carbon equivalent and fast cooling) and tensile residual stress. Remedies: low-hydrogen electrodes baked dry, preheat and interpass temperature, controlled heat input, post-weld heat treatment. Hot (solidification) cracking occurs along the centreline in the last liquid to freeze, promoted by S and P and by deep, narrow beads.

Carbon equivalent combines the effect of alloying elements on hardenability into one number: below about 0.40 a steel is usually easy to weld; above about 0.45–0.50 preheat and low-hydrogen practice are normally specified (follow your welding code).

Formulas

PD = P / A PD = power density (W/mm²), P = power transferred to the surface (W), A = heated spot area (mm²).

U_m = K · T_m² U_m = unit melting energy (J/mm³), K = 3.33 × 10⁻⁶ J/(mm³·K²), T_m = melting temperature (K).

H_w = f₁ · f₂ · H and R_Hw = f₁ · f₂ · P = U_m · A_w · v H = total energy (J), H_w = energy that melts metal (J), R_Hw = rate of melting energy (W), A_w = weld cross-sectional area (mm²), v = travel speed (mm/s). Volume melted per second = A_w · v.

H_net = f₁ · V · I / v Net heat input per unit length (J/mm).

R = 2π · k · (T_c − T₀)² / H_net Thick-plate centreline cooling rate (K/s) at temperature T_c: k = thermal conductivity (W/(mm·K)), T₀ = initial plate (preheat) temperature, H_net in J/mm. Valid for thick plates where heat flows in three dimensions.

CE = C + Mn/6 + (Cr + Mo + V)/5 + (Ni + Cu)/15 IIW carbon equivalent, element contents in mass %.

Worked examples

Example 1 — melting factor and travel speed (standard). GTAW of steel at 20 V and 200 A. f₁ = 0.7, f₂ = 0.5, T_m = 1760 K, weld cross-section A_w = 20 mm². Find the travel speed.

  1. U_m = K·T_m² = 3.33 × 10⁻⁶ × 1760² = 3.33 × 10⁻⁶ × 3 097 600 = 10.3 J/mm³.
  2. Power P = V·I = 20 × 200 = 4000 W.
  3. Power used for melting R_Hw = f₁·f₂·P = 0.7 × 0.5 × 4000 = 1400 W.
  4. Volume melted per second = 1400 / 10.3 = 135.7 mm³/s.
  5. v = R_Hw / (U_m·A_w) = 135.7 / 20 = 6.8 mm/s (about 410 mm/min).

Example 2 — cooling rate and preheat (GATE level). A thick steel plate (k = 0.028 W/(mm·K)) is welded with net heat input 1.0 kJ/mm. Find the centreline cooling rate at 550 °C with the plate at 25 °C, and with a 200 °C preheat.

  1. R = 2π·k·(T_c − T₀)² / H_net, H_net = 1000 J/mm.
  2. No preheat: R = 2π × 0.028 × (550 − 25)² / 1000 = 0.17593 × 275 625 / 1000 = 48.5 K/s.
  3. With preheat: R = 2π × 0.028 × (550 − 200)² / 1000 = 0.17593 × 122 500 / 1000 = 21.6 K/s.
  4. Preheat cuts the cooling rate by more than half (ratio (350/525)² = 0.444), moving the HAZ away from martensite.

Example 3 — power density. An arc delivers 4 kW to a spot 5 mm in diameter; a laser delivers 2 kW to a spot 0.4 mm in diameter.

  1. Arc: PD = 4000 / [(π/4)(5²)] = 4000 / 19.63 = 204 W/mm².
  2. Laser: PD = 2000 / [(π/4)(0.4²)] = 2000 / 0.1257 = 15 900 W/mm² — about 78 times higher, which is why the laser gives keyhole penetration with a tiny HAZ despite half the power.

Example 4 — carbon equivalent. C 0.20, Mn 1.20, Cr 0.20, Mo 0.05, Ni 0.10, Cu 0.10 (mass %): CE = 0.20 + 1.20/6 + 0.25/5 + 0.20/15 = 0.20 + 0.20 + 0.05 + 0.013 = 0.46 — preheat and low-hydrogen electrodes advisable.

Common mistakes

  • Using T_m in °C in U_m = K·T_m²; it must be kelvin.
  • Applying f₁ but forgetting f₂ (or vice versa) when finding melted volume.
  • Thinking higher power density means more heat input; it means more concentrated heat — total heat input per length is often lower for laser and EB welds.
  • Expecting higher heat input to raise cooling rate; it lowers it.
  • Using the temperature difference to the first power in the thick-plate cooling-rate formula; it is squared.
  • Treating the HAZ as melted metal; it never melts, it transforms in the solid state.

For GATE PI

Expect numericals on unit melting energy, f₁ and f₂, melted volume and travel speed, heat input, power density and comparisons between processes, and the effect of preheat on cooling rate. Conceptual MCQs cover HAZ sub-zones, epitaxial and columnar growth, hydrogen cracking, carbon equivalent and why high-power-density processes give narrow HAZs.

Quick check

  1. Unit melting energy of aluminium (T_m = 933 K)?
  2. Why does copper have a low melting factor?
  3. Name the three conditions for hydrogen cracking.
  4. If preheat halves (T_c − T₀), by what factor does the thick-plate cooling rate change?
  5. Which HAZ sub-zone of a steel weld is usually the hardest?

Answers: 1. 3.33 × 10⁻⁶ × 933² = 2.90 J/mm³. 2. Its high thermal conductivity drains heat away from the pool. 3. Diffusible hydrogen, a hard susceptible microstructure and tensile stress. 4. It falls to one-quarter. 5. The coarse-grained zone next to the fusion line.

Try answering each one aloud before you open it.

  1. 1.What is heat flow in the context of welding?Concept

    Heat flow in welding refers to the transfer of thermal energy from the heat source, such as a welding torch, to the workpiece. This process affects the temperature distribution within the material, influencing the weld quality, microstructure, and mechanical properties. Understanding heat flow is crucial for controlling the welding process and ensuring a strong, defect-free joint.

  2. 2.Explain the concept of power density in welding.Concept

    Power density in welding is the amount of power (energy per unit time) delivered per unit area of the workpiece. It is a critical factor in determining the heat input and the resulting thermal cycle of the weld. High power density can lead to rapid heating and cooling, affecting the microstructure and mechanical properties of the weld. It is typically measured in watts per square meter (W/m²).

  3. 3.What is weld metallurgy and why is it important?Concept

    Weld metallurgy involves the study of the microstructural changes and phase transformations that occur in the material during welding. It is important because these changes can significantly affect the mechanical properties, such as strength, toughness, and corrosion resistance, of the welded joint. Understanding weld metallurgy helps in selecting appropriate welding parameters and materials to achieve desired weld quality.

  4. 4.Why is preheating used in welding certain materials?Application

    Preheating is used in welding to reduce the cooling rate of the weld and the surrounding base metal. This helps in minimizing the risk of cracking, especially in materials that are prone to hardening or have high carbon content. Preheating also reduces thermal stresses and improves the ductility of the weld, leading to a more reliable joint.

  5. 5.What happens if the power density is too low during welding?Application

    If the power density is too low during welding, the heat input may be insufficient to properly melt the base material and filler metal. This can result in poor fusion, incomplete penetration, and weak welds. Additionally, low power density can lead to excessive heat spread, causing distortion and affecting the dimensional accuracy of the welded components.

  6. 6.How does heat flow affect the microstructure of a weld?Application

    Heat flow affects the microstructure of a weld by influencing the cooling rate and thermal gradients within the material. Rapid cooling can lead to the formation of martensite or other hard phases, while slower cooling may result in a more ductile microstructure. The heat-affected zone (HAZ) is particularly sensitive to these changes, and its properties can significantly impact the overall performance of the weld.

  7. 7.An arc delivers 5 kW to a heated spot 6 mm in diameter. What is the power density, and how does it compare with a laser?Numerical

    Spot area = (π/4) × 6² = 28.3 mm², so power density = 5000/28.3 ≈ 177 W/mm² (1.77 × 10⁸ W/m²). A laser focused to a few tenths of a millimetre reaches around 10⁴ W/mm² or more, roughly a hundred times higher, which is why it melts a deep narrow keyhole quickly and leaves a much smaller heat-affected zone.

  8. 8.What is the effect of high cooling rates on weld metallurgy?Application

    High cooling rates in weld metallurgy can lead to the formation of hard and brittle microstructures, such as martensite, especially in steels. This can increase the risk of cracking and reduce the toughness of the weld. Controlling the cooling rate is essential to achieve a balance between strength and ductility in the welded joint.

  9. 9.Explain why certain alloys require post-weld heat treatment.Application

    Certain alloys require post-weld heat treatment to relieve residual stresses, reduce hardness, and improve ductility. This process helps in homogenizing the microstructure and enhancing the mechanical properties of the weld. It is particularly important for alloys that are susceptible to cracking or have undergone significant phase transformations during welding.

  10. 10.A process needs at least 100 W/mm² at the work surface to melt steel efficiently, and the source delivers 3 kW to the work. What is the largest spot diameter you can use?Numerical

    Maximum area = P/PD = 3000/100 = 30 mm². For a circular spot, d = √(4A/π) = √(4 × 30/π) = 6.2 mm. A larger spot would spread the same power too thinly, so more heat would conduct away before melting occurred, lowering the melting factor and widening the HAZ.

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