Heat transfer in casting, welding and machining

Why it matters

Casting, welding and machining are all heat-transfer problems at heart. How fast a casting freezes decides its grain size, shrinkage defects and riser size; how much heat a welding arc puts in decides penetration, heat-affected zone (HAZ) width, distortion and cracking; and where the heat of cutting goes decides tool life, surface integrity and dimensional accuracy. Production engineers use simple heat-balance models of each to set process parameters.

Key ideas

Casting — solidification.

• Molten metal loses its superheat, then its latent heat of fusion, then sensible heat of the solid. Most of the time to freeze is spent removing latent heat.
• In a sand mould the sand has a much lower thermal conductivity than the metal, so the mould is the controlling resistance: heat flows by conduction from the solidifying metal into the sand. In a metal (permanent or die) mould the metal–mould interface resistance (air gap) often controls instead. Convection in the liquid and radiation from the mould surface are secondary.
• Chvorinov's rule: total solidification time is proportional to (V/A)ⁿ, where V/A is the casting modulus (volume over cooling surface area) and n ≈ 2. The mould constant B depends on the metal (latent heat, pouring temperature) and the mould material (conductivity, density, specific heat); it is found experimentally.
• Consequences: thin sections freeze first; a riser must have a larger modulus than the section it feeds so that it freezes last (a common design rule is t_riser ≈ 1.25 × t_casting or more). Chills (metal inserts) raise local heat extraction to promote directional solidification. Faster cooling gives finer grains and better strength; cooling that is too fast risks misruns, cold shuts and hot tears.

Welding — heat input and flow.

• Arc power is V·I. Only part reaches the workpiece: the heat transfer efficiency η_h is roughly 0.2–0.5 for oxy-fuel, 0.6–0.8 for GTAW and 0.8–0.95 for SMAW, GMAW and SAW (take values from your text or data book). Part of the heat that arrives melts metal (melting efficiency η_m); the rest conducts away into the base metal.
• Heat input per unit length H = η_h·V·I/v (J/mm) is the controlling parameter. High H gives deep penetration but a wide HAZ, coarse grains, more distortion and lower toughness. Low H with thick, conductive plate gives fast cooling, which in hardenable steels can form martensite and cause hydrogen cracking.
• Preheating reduces the temperature difference between weld and base metal, lowering the cooling rate; it is used for thick sections, high-carbon and alloy steels, and highly conductive metals like copper and aluminium.
• Thermal conductivity matters: aluminium and copper conduct heat away quickly, needing more heat input or preheat; stainless steel conducts poorly, concentrating heat and distortion.

Machining — where cutting heat goes.

• Almost all cutting power F_c·v becomes heat, generated in three zones: the primary shear zone (largest share), the secondary zone at the tool–chip interface (friction and secondary shear) and the tertiary zone at the tool–work flank contact.
• Most of the heat (often 70–90% at normal speeds) leaves with the chip; the tool and the workpiece take the rest. The fraction going to the chip rises with cutting speed, but the tool–chip interface temperature still rises with speed, so tool wear (diffusion, softening, crater wear) accelerates — the basis of Taylor's tool-life equation.
• Heat in the workpiece causes thermal expansion and dimensional error; in grinding, where the specific energy is very high and much of the heat enters the work, it causes burn, tempering and residual stress.
• Cutting fluids remove heat by convection (water-based soluble oils are best coolants) and reduce friction (neat oils are better lubricants). At very high speeds the fluid may not reach the interface, which is why carbide and ceramic tools are often run dry.

Formulas

Q = −k·A·(dT/dx)

• Fourier's law. k in W/m·K, A in m², dT/dx in K/m, Q in W. Applies to conduction into a mould or through a tool.

Q = h·A·(T_s − T_∞)

• Newton's law of cooling. h in W/m²·K.

t_s = B·(V/A)ⁿ, with n ≈ 2

• Chvorinov's rule. t_s in s or min, V in m³ (or mm³), A in m² (or mm²), B in s/m² (or min/mm²) — an empirical mould constant. For the same metal and mould: t₂/t₁ = [(V/A)₂/(V/A)₁]².

H = η_h·V·I/v

• Weld heat input per unit length. V in volts, I in amperes, v (travel speed) in mm/s, H in J/mm.

A_w = η_m·η_h·V·I/(u·v)

• Weld bead cross-section (mm²); u = unit energy to melt the metal (J/mm³), a material value from a data book.

P_c = F_c·v; u_s = P_c/MRR; MRR = f·d·v (turning)

• Cutting power (W), specific cutting energy (J/mm³). F_c in N, v in m/s, f and d in mm.

ΔT_chip = x·P_c/(ρ·c·MRR)

• Mean chip temperature rise when fraction x of the heat goes to the chip. ρ·c in J/m³·K, MRR in m³/s.

Worked examples

Example 1 (standard): Chvorinov's rule. Given: a 100 mm cube of a steel casting solidifies in 10 min in a sand mould. Find the solidification time of a 200 mm × 100 mm × 25 mm plate of the same steel in the same mould (n = 2).

1. Cube: V/A = a/6 = 100/6 = 16.67 mm
2. Plate: V = 200 × 100 × 25 = 500 000 mm³; A = 2(200 × 100 + 200 × 25 + 100 × 25) = 55 000 mm²; V/A = 9.09 mm
3. t₂ = t₁·[(V/A)₂/(V/A)₁]² = 10 × (9.09/16.67)² = 10 × 0.2975 Answer: t ≈ 2.98 min — the thin plate freezes in under a third of the cube's time.

Example 2 (GATE level): sizing a riser. Given: the plate above is fed by a cylindrical side riser with height equal to diameter (H = D), cooling from all its surfaces. The riser must take 25% longer to solidify than the plate. Find D.

1. Riser modulus: V = πD²·D/4 = πD³/4; A = 2·πD²/4 + πD·D = 1.5πD²; V/A = D/6
2. Requirement: t_r = 1.25·t_p, so (V/A)_r = √1.25 × (V/A)_p = 1.118 × 9.09 = 10.16 mm
3. D = 6 × 10.16 = 61.0 mm Answer: D = H ≈ 61 mm

Example 3 (standard): weld bead size. Given: GMAW at 20 V, 200 A, travel speed 5 mm/s, η_h = 0.85, η_m = 0.5, unit melting energy u = 10 J/mm³ (data-book value for the steel).

1. H = η_h·V·I/v = 0.85 × 20 × 200/5 = 680 J/mm
2. Melting rate = η_m·η_h·V·I/u = 0.5 × 0.85 × 4000/10 = 170 mm³/s
3. A_w = 170/5 = 34 mm² Answer: H = 680 J/mm; weld cross-section ≈ 34 mm²

Example 4 (GATE level): chip temperature in turning. Given: F_c = 1000 N, v = 2 m/s, f = 0.25 mm, d = 2 mm; 75% of the heat goes to the chip; work material ρ·c = 3.6 × 10⁶ J/m³·K.

1. P_c = F_c·v = 1000 × 2 = 2000 W
2. MRR = f·d·v = 0.25 × 10⁻³ × 2 × 10⁻³ × 2 = 1.0 × 10⁻⁶ m³/s (so u_s = 2 J/mm³)
3. ΔT_chip = 0.75 × 2000/(3.6 × 10⁶ × 1.0 × 10⁻⁶) = 1500/3.6 Answer: ΔT_chip ≈ 417 K above ambient (a mean value; the tool–chip interface is hotter).

Common mistakes

• Using V/A with area in the wrong units, or including faces that do not cool (a face joined to a riser or another section).
• Treating Chvorinov's exponent as 1, or forgetting the square when taking ratios.
• Using arc power V·I as heat delivered, forgetting η_h, or mixing travel speed in mm/min and mm/s.
• Assuming most cutting heat goes into the tool; it goes mostly into the chip.
• Assuming higher cutting speed lowers tool temperature because the chip carries more heat — the interface temperature still rises.
• Using °C in a T⁴ radiation term when estimating losses from a mould or casting surface.

For GATE PI

Expect Chvorinov's-rule ratio numericals and riser sizing with modulus (cube, cylinder, sphere, plate), weld heat input and melting-rate numericals with heat transfer and melting efficiencies, cutting power, specific energy and heat-partition questions, and conceptual questions on preheating, HAZ, chills, directional solidification and the role of cutting fluids. Practise computing V/A quickly for standard shapes.

Quick check

1. What is V/A for a sphere of diameter D?
2. If the modulus of a casting doubles, by what factor does its solidification time change (n = 2)?
3. Arc welding at 30 V, 200 A, travel 10 mm/s, η_h = 0.8: what is the heat input per mm?
4. Where does most of the heat generated in metal cutting go?

Answers: 1. D/6 2. Four times longer 3. 480 J/mm 4. Into the chip