5.4 Arc Radiation Losses, Preheat Maintenance & PWHT Boundary Conditions

Arc Column Radiation Losses and Process Efficiency

The welding electric arc is a thermal plasma column operating at peak temperatures between $10,000\text{ K}$ and $25,000\text{ K}$. The plasma radiates intense electromagnetic energy across the ultraviolet (UV-A, UV-B, UV-C), visible, and infrared spectra.

                    ARC COLUMN RADIATION INTERCEPTION

                                  Gas Nozzle / Cup
                                   ┌─────────────┐
                                   │  Tungsten   │
                                   │  Electrode  │
                                   └──────┬──────┘
                                          │
                      Radiant Loss        ▼       Radiant Loss
                      to Ambient      /───────\   to Ambient
                     ◄───────────────│ Arc     │──────────────► (10% - 25% of P_elec)
                                      \ Plasma/ 
                     ─────────────────┴───────┴─────────────────
                      Direct Radiation to Molten Pool & Workpiece

Radiation Partitioning in the Arc

The total electrical power input delivered to the welding arc is:

$$P_{\text{elec}} = V \cdot I$$

This power is partitioned into three primary mechanisms:

1. Anode and Cathode Heat Input ($q_{\text{electrode}}, q_{\text{workpiece}}$): Direct charge carrier transfer (electron condensation/emission) and convective heat transfer from plasma jets.
2. Radiation Dissipated to the Workpiece ($q_{\text{rad,plate}}$): Photons emitted by the arc column that intercept the weld pool and plate, governed by the geometric view factor $F_{\text{arc-plate}}$.
3. Radiation Lost to the Environment ($q_{\text{rad,lost}}$): High-energy photons radiating laterally into the atmosphere, which heat the surroundings, degrade torch nozzles, and pose ocular (arc eye) and skin burn hazards to personnel.

Impact on Arc Thermal Efficiency ($\eta$)

The magnitude of radiant loss directly dictates the arc thermal efficiency ($\eta = q_{\text{net}} / P_{\text{elec}}$) across different welding processes:

• Gas Tungsten Arc Welding (GTAW, $\eta \approx 0.60 - 0.70$): GTAW features a completely open, unshielded arc column with high plasma temperatures and a large view factor to ambient surroundings. Up to $20% - 25%$ of total arc power is lost directly as radiation to the environment.
• Gas Metal Arc Welding (GMAW, $\eta \approx 0.80 - 0.85$): The metal transfer droplets absorb internal radiation, and the consumable wire continuously carries heat into the puddle, reducing radiant losses to $10% - 15%$.
• Shielded Metal Arc Welding (SMAW, $\eta \approx 0.75 - 0.85$): The decomposing flux coating forms a protective gaseous envelope and molten slag shelf that partially shrouds the arc column.
• Submerged Arc Welding (SAW, $\eta \approx 0.95 - 1.00$): The arc is completely buried beneath a thick blanket of granular mineral flux. Almost zero radiant energy escapes into the ambient environment; virtually $100%$ of radiative and convective energy is trapped and recycled into the weld pool and flux slag.

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Combined Boundary Conditions in Preheat Maintenance and PWHT

In practical welding fabrication—particularly on thick-wall pressure vessels (ASME Section VIII, Div 1) and heavy structural joints (AWS D1.1)—engineers must calculate the total heat dissipation to size preheat heating systems and design Post-Weld Heat Treatment (PWHT) soak protocols.

The Linearized Effective Heat Transfer Coefficient ($h_{\text{eff}}$)

At the boundary surface, convection and radiation act in parallel. The total boundary heat flux is the direct sum of both modes:

$$q''_{\text{total}} = q''_{\text{conv}} + q''_{\text{rad}} = h_{\text{conv}} (T_s - T_\infty) + \varepsilon \sigma \left( T_s^4 - T_{\text{sur}}^4 \right)$$

Assuming the surrounding radiating environment is at ambient temperature ($T_{\text{sur}} = T_\infty$), the radiation equation can be algebraically linearized by factoring the difference of fourth powers:

$$T_s^4 - T_\infty^4 = \left( T_s^2 - T_\infty^2 \right)\left( T_s^2 + T_\infty^2 \right) = (T_s - T_\infty)(T_s + T_\infty)\left( T_s^2 + T_\infty^2 \right)$$

Defining the linearized radiation heat transfer coefficient ($h_{\text{rad}}$):

$$h_{\text{rad}} = \varepsilon \sigma (T_s + T_\infty)\left( T_s^2 + T_\infty^2 \right)$$

Then the total heat loss can be expressed in terms of a single combined effective heat transfer coefficient ($h_{\text{eff}}$):

$$q''_{\text{total}} = \left( h_{\text{conv}} + h_{\text{rad}} \right)(T_s - T_\infty) = h_{\text{eff}} (T_s - T_\infty)$$ $$h_{\text{eff}} = h_{\text{conv}} + \varepsilon \sigma (T_s + T_\infty)\left( T_s^2 + T_\infty^2 \right)$$

                 THERMAL RESISTANCE NETWORK (PREHEAT & PWHT)

             Bare Plate Surface                     Insulated Surface
     Hot Plate (Ts) ───[ 1 / h_eff ]───► T_∞   Hot Plate (Ts) ───[ t_ins / k_ins ]───(T_so)───[ 1 / h_eff,o ]───► T_∞
          Total Resistance: R = 1 / h_eff           Total Resistance: R_total = (t_ins / k_ins) + (1 / h_eff,o)

Thermal Insulation Design for Preheat and PWHT Soaking

When preheating thick sections to $200^\circ\text{C} - 250^\circ\text{C}$ or conducting PWHT soaking at $600^\circ\text{C} - 650^\circ\text{C}$ per ASME Section VIII UCS-56, bare steel surfaces dissipate massive thermal power, causing severe through-thickness temperature gradients and tripping thermal gradient limits. Applying flexible ceramic fiber blankets (refractory alumina-silica wool) inserts a high thermal resistance in series with the outer convective-radiative boundary:

$$R''_{\text{ins}} = \frac{t_{\text{ins}}}{k_{\text{ins}}}$$ $$R''_{\text{total}} = R''_{\text{ins}} + R''_{\text{boundary,outer}} = \frac{t_{\text{ins}}}{k_{\text{ins}}} + \frac{1}{h_{\text{eff,outer}}}$$ $$q''_{\text{insulated}} = \frac{T_s - T_\infty}{R''_{\text{total}}} = \frac{T_s - T_\infty}{\frac{t_{\text{ins}}}{k_{\text{ins}}} + \frac{1}{h_{\text{eff,outer}}}}$$

where:

• $t_{\text{ins}}$ = Insulation blanket thickness ($\text{m}$)
• $k_{\text{ins}}$ = Thermal conductivity of the ceramic fiber blanket (typically $0.06 - 0.12\text{ W/(m}\cdot\text{K)}$)
• $h_{\text{eff,outer}}$ = Effective boundary coefficient at the cooler outer blanket surface

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Comprehensive Worked Numerical Example: Preheat Maintenance Power & Insulation Sizing

Problem Statement

A heavy-wall cylindrical pressure vessel shell ($2.25\text{Cr}-1\text{Mo}$ alloy steel, SA-387 Grade 22) has an outer surface area of $A_s = 4.0\text{ m}^2$ in the weld joint preheat band. The welding procedure specification (WPS) mandates a continuous preheat and interpass temperature of $T_s = 220.0^\circ\text{C}$ ($493.15\text{ K}$). Ambient shop air and surrounding walls are at $T_\infty = T_{\text{sur}} = 20.0^\circ\text{C}$ ($293.15\text{ K}$). Under quiescent shop conditions, the natural convection heat transfer coefficient is $h_{\text{conv}} = 11.5\text{ W/(m}^2\cdot\text{K)}$.

Evaluate two operational configurations:

1. Configuration 1 (Uninsulated Bare Plate): The vessel surface has a heavy, dark mill-scale oxide film with emissivity $\varepsilon = 0.82$.
   • Calculate convective heat flux ($q''{\text{conv}}$), radiative heat flux ($q''{\text{rad}}$), total combined heat flux ($q''{\text{total}}$), effective heat transfer coefficient ($h{\text{eff}}$), and the total electrical heating power ($Q_{\text{bare}}$ in $\text{kW}$) required just to maintain preheat against ambient heat loss.
2. Configuration 2 (Insulated Plate): The preheat band is wrapped with a $t_{\text{ins}} = 50.0\text{ mm}$ ($0.050\text{ m}$) thick ceramic fiber blanket ($k_{\text{ins}} = 0.065\text{ W/(m}\cdot\text{K)}$). The outer blanket surface has an effective heat transfer coefficient $h_{\text{eff,outer}} = 10.0\text{ W/(m}^2\cdot\text{K)}$.
   • Calculate the total thermal resistance ($R''{\text{total}}$), the reduced heat flux ($q''{\text{ins}}$), outer blanket surface temperature ($T_{\text{so}}$), the required electrical heating power ($Q_{\text{ins}}$ in $\text{kW}$), and the percentage of energy saved.

Step-by-Step Solution

Configuration 1: Bare Uninsulated Workpiece

Step 1: Compute Convective Heat Flux ($q''_{\text{conv}}$)

$$\Delta T = T_s - T_\infty = 220.0^\circ\text{C} - 20.0^\circ\text{C} = 200.0^\circ\text{C} = 200.0\text{ K}$$ $$q''_{\text{conv}} = h_{\text{conv}} \cdot \Delta T = 11.5\text{ W/(m}^2\cdot\text{K)} \times 200.0\text{ K} = 2300.0\text{ W/m}^2$$

Step 2: Compute Radiative Heat Flux ($q''_{\text{rad}}$) Convert temperatures to Kelvin: $T_s = 220.0 + 273.15 = 493.15\text{ K}$, $T_{\text{sur}} = 20.0 + 273.15 = 293.15\text{ K}$.

$$T_s^4 = (493.15)^4 = 5.9135 \times 10^{10}\text{ K}^4$$ $$T_{\text{sur}}^4 = (293.15)^4 = 7.3636 \times 10^9\text{ K}^4$$ $$T_s^4 - T_{\text{sur}}^4 = 5.9135 \times 10^{10} - 0.7364 \times 10^{10} = 5.1771 \times 10^{10}\text{ K}^4$$ $$q''_{\text{rad}} = \varepsilon \cdot \sigma \cdot \left( T_s^4 - T_{\text{sur}}^4 \right) = 0.82 \times \left( 5.670 \times 10^{-8}\text{ W/(m}^2\cdot\text{K}^4) \right) \times 5.1771 \times 10^{10}\text{ K}^4$$ $$q''_{\text{rad}} = 0.82 \times 2935.42\text{ W/m}^2 = 2407.04\text{ W/m}^2$$

Step 3: Compute Total Heat Flux ($q''{\text{total}}$) and Effective Coefficient ($h{\text{eff}}$)

$$q''_{\text{total}} = q''_{\text{conv}} + q''_{\text{rad}} = 2300.0 + 2407.04 = 4707.04\text{ W/m}^2$$ $$h_{\text{rad}} = \frac{q''_{\text{rad}}}{\Delta T} = \frac{2407.04\text{ W/m}^2}{200.0\text{ K}} = 12.04\text{ W/(m}^2\cdot\text{K)}$$ $$h_{\text{eff}} = h_{\text{conv}} + h_{\text{rad}} = 11.5 + 12.04 = 23.54\text{ W/(m}^2\cdot\text{K)}$$

Notice: Radiation accounts for $51.1%$ of total heat dissipation, exceeding convection even at the modest preheat temperature of $220^\circ\text{C}$!

Step 4: Compute Total Electrical Power Required for Bare Plate ($Q_{\text{bare}}$)

$$Q_{\text{bare}} = q''_{\text{total}} \times A_s = 4707.04\text{ W/m}^2 \times 4.0\text{ m}^2 = 18,828.16\text{ W} = 18.83\text{ kW}$$

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Configuration 2: Insulated with $50\text{ mm}$ Ceramic Fiber Blanket

Step 5: Calculate Thermal Resistances ($R''{\text{ins}}, R''{\text{total}}$)

$$R''_{\text{ins}} = \frac{t_{\text{ins}}}{k_{\text{ins}}} = \frac{0.050\text{ m}}{0.065\text{ W/(m}\cdot\text{K)}} = 0.76923\text{ m}^2\cdot\text{K/W}$$ $$R''_{\text{boundary,outer}} = \frac{1}{h_{\text{eff,outer}}} = \frac{1}{10.0\text{ W/(m}^2\cdot\text{K)}} = 0.10000\text{ m}^2\cdot\text{K/W}$$ $$R''_{\text{total}} = R''_{\text{ins}} + R''_{\text{boundary,outer}} = 0.76923 + 0.10000 = 0.86923\text{ m}^2\cdot\text{K/W}$$

Step 6: Calculate Reduced Heat Flux ($q''{\text{ins}}$) and Total Insulated Power ($Q{\text{ins}}$)

$$q''_{\text{ins}} = \frac{\Delta T}{R''_{\text{total}}} = \frac{200.0\text{ K}}{0.86923\text{ m}^2\cdot\text{K/W}} = 230.09\text{ W/m}^2$$ $$Q_{\text{ins}} = q''_{\text{ins}} \times A_s = 230.09\text{ W/m}^2 \times 4.0\text{ m}^2 = 920.36\text{ W} = 0.92\text{ kW}$$

Step 7: Compute Outer Blanket Surface Temperature ($T_{\text{so}}$)

$$T_{\text{so}} = T_\infty + q''_{\text{ins}} \cdot R''_{\text{boundary,outer}} = 20.0^\circ\text{C} + (230.09\text{ W/m}^2 \times 0.10000\text{ m}^2\cdot\text{K/W}) = 20.0 + 23.01 = 43.0^\circ\text{C}$$

Step 8: Calculate Energy Savings

$$\text{Power Reduction} = \frac{Q_{\text{bare}} - Q_{\text{ins}}}{Q_{\text{bare}}} \times 100\% = \frac{18.83\text{ kW} - 0.92\text{ kW}}{18.83\text{ kW}} \times 100\% = \frac{17.91}{18.83} \times 100\% = 95.12\%$$

Engineering Conclusion: Wrapping the preheat zone with $50\text{ mm}$ of ceramic insulation slashes power consumption from $18.83\text{ kW}$ to $0.92\text{ kW}$—a $95.1%$ energy reduction. Furthermore, the outer blanket temperature drops to $43.0^\circ\text{C}$ (safe for personnel touch), and severe through-thickness thermal gradients are completely eliminated.

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Real-World Engineering Scenarios & Exam Pitfalls

Industrial Case: Preheating Cracking on Cross-Country Gas Pipelines

During winter field construction of an API 5L X70 gas transmission pipeline in northern Alberta (ambient air $-25^\circ\text{C}$, wind speed $25\text{ km/h}$), induction preheat coils brought the pipe bevel to the WPS-specified $120^\circ\text{C}$. However, within 60 seconds after removing the induction bands to position the internal line-up clamp, root pass cracking occurred. Field investigation revealed that high forced convection ($h_{\text{conv}} \approx 85\text{ W/(m}^2\cdot\text{K)}$) and intense radiation rapidly chilled the root land from $120^\circ\text{C}$ to below $15^\circ\text{C}$ before the root bead could be deposited. The welding engineer resolved the failure by deploying heated shelter huts to keep wind speed under $5\text{ km/h}$ and implementing insulating thermal blankets on adjacent pipe rings, successfully maintaining the mandatory $120^\circ\text{C}$ interpass temperature.

Common Exam Traps

Exam Trap 1: Forgetting to Convert Celsius to Kelvin in Stefan-Boltzmann Calculations The single most frequent calculation error in thermal radiation is inserting Celsius temperatures directly into $(T_s^4 - T_{\text{sur}}^4)$. If you compute $(220^4 - 20^4) = (2.34 \times 10^9 - 1.6 \times 10^5) = 2.34 \times 10^9$, your result is smaller by a factor of 22 than the correct Kelvin value $(493.15^4 - 293.15^4 = 5.18 \times 10^{10})$! Always convert to Kelvin ($+273.15$).

Exam Trap 2: Assuming Thermal Radiation Is Negligible at Preheat Temperatures Many candidates erroneously assume radiation only matters at melting temperatures. As demonstrated in our worked example, on heavily oxidized steel ($\varepsilon = 0.82$) at $220^\circ\text{C}$, radiation accounts for over $51%$ of total heat dissipation! Never neglect radiation when calculating preheat power or cool-down rates.

Exam Trap 3: Ignoring Emissivity Drift During Optical Pyrometry When monitoring a stainless steel or aluminum weld pass, the surface transitions from shiny bare metal (low $\varepsilon$) to a heavily colored oxide film (high $\varepsilon$). If the inspector does not update the emissivity setting on an optical pyrometer, the instrument's reading will drift by more than $150^\circ\text{C}$, leading to false acceptances or rejections of interpass temperature compliance.