6.2 Joule Heating, Electrode Stickout Preheating & Circuit Voltage Distribution

4. Joule's First Law and Electrode Stickout Resistive Preheating

Joule's First Law defines the rate at which electrical energy is irreversibly converted into thermal energy within a resistive conductor:

$$P = I^2 R = \frac{V^2}{R}$$ $$Q = \int_0^t I^2 R \, dt = I^2 R t \quad (\text{for constant } I, R)$$

Parasitic Joule Losses in Welding Cables

Welding cables carry massive currents ($200\text{ to }600+\text{ A}$). If cables are undersized or excessively long, parasitic Joule losses ($I^2 R_{\text{cable}}$) cause:

• Severe terminal voltage drop, starving the arc and destabilizing metal transfer.
• Thermal degradation of cable insulation (standard synthetic rubber is rated for $60^\circ\text{C}, 75^\circ\text{C},\text{ or }90^\circ\text{C}$).
• Runaway resistance growth: copper has a positive temperature coefficient of resistance ($\alpha_{\text{Cu}} \approx +0.00393 / ^\circ\text{C}$), such that heating from $20^\circ\text{C}$ to $80^\circ\text{C}$ increases cable resistance by $23.6\%$, further compounding voltage drop.

Contact Tip Interface Resistance

The electrical contact between the copper alloy contact tip and the advancing electrode wire occurs across microscopic asperities. Sliding contact resistance ($R_{\text{ct}}$) typically ranges from $2\text{ to }8\text{ m}\Omega$ in new tips, but escalates above $30\text{ m}\Omega$ as the bore erodes oval from mechanical wear and micro-arcing. This produces local Joule overheating, micro-fusing of the wire inside the bore (burnback), and erratic feeding.

Functional Joule Heating: Wire Extension (Stickout) Preheating

In continuous wire processes (GMAW, FCAW, SAW), the electrode extension ($L_{\text{ext}}$, or stickout) is the length of unmelted filler wire projecting beyond the contact tip to the arc.

     Contact Tip
     +---------+
     |  Copper |   Wire Feed Speed (v_w)
     |  Alloy  | =====>
     +----+----+-----------------------------------
          |    | Electrode Wire (ER70S-6, Dia d)   |
          |    |                                   |  <-- Resistive Preheating Zone
          +----+-----------------------------------+      (Joule Heating: Q = I^2 * R_ext * t)
               |<----------- L_ext --------------->|
                                                   +--/--- Weld Pool
                                                   | ARC |

The total melting rate of a continuous consumable electrode is expressed by the fundamental Lesnewich equation:

$$\text{MR} = \alpha I + \beta L_{\text{ext}} I^2$$

where:

• $\alpha I$ represents arc heating at the cathode or anode fall region ($[\text{kg}/(\text{h}\cdot\text{A})]$).
• $\beta L_{\text{ext}} I^2$ represents Joule resistance preheating along the length of the extension ($[\text{kg}/(\text{h}\cdot\text{mm}\cdot\text{A}^2)]$).

The temperature rise $\Delta T$ of the wire solid mass before it reaches the arc is:

$$\Delta T = \frac{\rho_e L_{\text{ext}} I^2}{A^2 \cdot v_w \cdot \rho_m \cdot c_p}$$

where $\rho_e$ is electrical resistivity, $A$ is wire cross-sectional area, $v_w$ is wire feed speed, $\rho_m$ is metal density, and $c_p$ is specific heat capacity.

Increasing stickout increases $R_{\text{ext}}$, preheating the wire to over $800^\circ\text{C}-1000^\circ\text{C}$ before it reaches the arc. Consequently, less arc energy is required to complete fusion, allowing dramatic increases in deposition rate at fixed current, or requiring lower current to melt wire at a fixed wire feed speed.

---

5. Comprehensive Worked Numerical Example: Welding Circuit Voltage Distribution & Stickout Preheating

Problem Statement

A robotic GMAW cell operates with an AWS A5.18 ER70S-6 solid steel wire of diameter $d = 1.20\text{ mm}$ ($0.00120\text{ m}$) at a welding current $I = 320\text{ A}$ DC. The power supply terminal voltage is maintained at $V_{\text{source}} = 32.0\text{ V}$.

The circuit parameters are:

• Total welding cable run: $25.0\text{ m}$ electrode lead + $25.0\text{ m}$ work return lead ($L_{\text{cable}} = 50.0\text{ m}$ total) of 2/0 AWG stranded copper cable. Resistance is $0.260\text{ m}\Omega/\text{m}$ at operating temperature ($R_{\text{cable}} = 0.0130\ \Omega$).
• Contact tip sliding interface resistance: $R_{\text{ct}} = 5.0\text{ m}\Omega = 0.0050\ \Omega$.
• Ground clamp-to-workpiece contact resistance: $R_{\text{clamp}} = 4.0\text{ m}\Omega = 0.0040\ \Omega$.
• Base metal plate resistance: negligible ($R_{\text{work}} \approx 0$).
• Electrode extension (stickout): $L_{\text{ext}} = 20.0\text{ mm} = 0.020\text{ m}$.
• Average effective electrical resistivity of the steel wire over its preheat range: $\rho_e = 7.50 \times 10^{-7}\ \Omega\cdot\text{m}$.
• Wire Feed Speed: $WFS = 10.0\text{ m/min} = 0.1667\text{ m/s}$.
• Steel physical properties: density $\rho_m = 7850\text{ kg/m}^3$, specific heat $c_p = 620\text{ J/(kg}\cdot^\circ\text{C)}$.

Calculate:

1. The electrical resistance of the wire extension ($R_{\text{ext}}$).
2. The total series resistance of all circuit components excluding the arc ($R_{\text{loop}}$).
3. The total parasitic/resistive voltage drop across all components ($V_{\text{drops}}$).
4. The actual voltage available across the welding arc ($V_{\text{arc}}$).
5. The power dissipated as waste heat in cables and connections versus power utilized in the arc.
6. The temperature rise ($\Delta T$) of the electrode wire from Joule preheating upon entering the arc.

Step-by-Step Solution

Step 1: Compute wire cross-sectional area and extension resistance ($R_{\text{ext}}$)

$$A = \frac{\pi d^2}{4} = \frac{\pi (0.00120\text{ m})^2}{4} = 1.131 \times 10^{-6}\text{ m}^2$$ $$R_{\text{ext}} = \frac{\rho_e L_{\text{ext}}}{A} = \frac{(7.50 \times 10^{-7}\ \Omega\cdot\text{m})(0.020\text{ m})}{1.131 \times 10^{-6}\text{ m}^2} = 0.01326\ \Omega = 13.26\text{ m}\Omega$$

Step 2: Calculate total non-arc series loop resistance ($R_{\text{loop}}$)

$$R_{\text{loop}} = R_{\text{cable}} + R_{\text{ct}} + R_{\text{ext}} + R_{\text{clamp}}$$ $$R_{\text{loop}} = 0.0130\ \Omega + 0.0050\ \Omega + 0.01326\ \Omega + 0.0040\ \Omega = 0.03526\ \Omega = 35.26\text{ m}\Omega$$

Step 3: Calculate total parasitic voltage drop ($V_{\text{drops}}$)

$$V_{\text{drops}} = I \cdot R_{\text{loop}} = 320\text{ A} \times 0.03526\ \Omega = 11.28\text{ V}$$

Individual voltage drops:

• Cables: $V_{\text{cable}} = 320 \times 0.0130 = 4.16\text{ V}$
• Contact tip: $V_{\text{ct}} = 320 \times 0.0050 = 1.60\text{ V}$
• Stickout: $V_{\text{ext}} = 320 \times 0.01326 = 4.24\text{ V}$
• Ground clamp: $V_{\text{clamp}} = 320 \times 0.0040 = 1.28\text{ V}$

Step 4: Determine actual arc voltage ($V_{\text{arc}}$) Applying Kirchhoff's Voltage Law:

$$V_{\text{arc}} = V_{\text{source}} - V_{\text{drops}} = 32.00\text{ V} - 11.28\text{ V} = 20.72\text{ V}$$

Step 5: Power distribution breakdown

• Total electrical power from source: $P_{\text{total}} = V_{\text{source}} \cdot I = 32.0\text{ V} \times 320\text{ A} = 10,240\text{ W} = 10.24\text{ kW}$
• Arc thermal power: $P_{\text{arc}} = V_{\text{arc}} \cdot I = 20.72\text{ V} \times 320\text{ A} = 6,630.4\text{ W}$ ($64.75\%$ of total power)
• Stickout preheating power: $P_{\text{ext}} = I^2 R_{\text{ext}} = (320)^2 \times 0.01326 = 1,357.8\text{ W}$ ($13.26\%$)
• Waste heat in cables: $P_{\text{cable}} = I^2 R_{\text{cable}} = (320)^2 \times 0.0130 = 1,331.2\text{ W}$ ($13.00\%$)
• Waste heat in contact tip: $P_{\text{ct}} = (320)^2 \times 0.0050 = 512.0\text{ W}$ ($5.00\%$)
• Waste heat in ground clamp: $P_{\text{clamp}} = (320)^2 \times 0.0040 = 409.6\text{ W}$ ($4.00\%$)

Step 6: Compute temperature rise of wire from Joule preheating Compute mass flow rate of wire ($\dot{m}$):

$$\dot{m} = A \cdot v_w \cdot \rho_m = (1.131 \times 10^{-6}\text{ m}^2) \times (0.1667\text{ m/s}) \times (7850\text{ kg/m}^3) = 1.480 \times 10^{-3}\text{ kg/s}$$

Using the thermal energy balance $P_{\text{ext}} = \dot{m} c_p \Delta T$:

$$\Delta T = \frac{P_{\text{ext}}}{\dot{m} \cdot c_p} = \frac{1357.8\text{ W}}{(1.480 \times 10^{-3}\text{ kg/s})(620\text{ J/kg}\cdot^\circ\text{C})} = \frac{1357.8}{0.9176} = 1479.7^\circ\text{C}$$

Engineering Insight: The calculated $\Delta T \approx 1480^\circ\text{C}$ indicates that Joule heating alone brings the steel wire almost directly to its solidus/liquidus melting temperature ($1450^\circ\text{C}-1520^\circ\text{C}$) before it even enters the arc! This illustrates why electrode extension is such a potent variable in continuous wire processes: modest changes in stickout alter the balance between resistance preheating and arc column melting.

---

6. Real-World Engineering Scenarios & Exam Pitfalls

Practical Industrial Scenario

On a shipyard fabrication berth, welders using flux-cored arc welding (FCAW) experienced intermittent porosity, cold lap, and poor penetration. Welding procedure specifications (WPS) required $28\text{ V}$ and $280\text{ A}$. The welding power source display in the substation indicated $28.5\text{ V}$. However, the leads were $60\text{ m}$ long ($120\text{ m}$ total loop) of 1/0 AWG cable ($R = 0.39\text{ m}\Omega/\text{m}$, total $R = 0.0468\ \Omega$), coiled in tight loops on steel staging.

Diagnostic Investigation: The resistance of the coiled, overheated leads plus worn magnetic ground clamps totaled $R_{\text{loop}} = 0.065\ \Omega$. At $280\text{ A}$, the voltage drop across the cables was:

$$V_{\text{drop}} = 280\text{ A} \times 0.065\ \Omega = 18.2\text{ V}$$

The actual voltage at the welding arc was only $28.5\text{ V} - 18.2\text{ V} = 10.3\text{ V}$. The welders were attempting to operate in short-circuit mode under parameters qualified for globular/spray transfer. Installing 4/0 AWG uncoiled leads and replacing worn spring clamps restored arc voltage to $26.8\text{ V}$, eliminating all lack-of-fusion defects.

Common CWEng Exam Traps

Exam Trap 1: Polarity & Penetration Reversal Between Processes A major exam trap asks which polarity produces deepest penetration. For GTAW, the correct answer is DCEN (straight polarity, workpiece is anode). For GMAW, the correct answer is DCEP (reverse polarity). Candidates who memorize "DCEN always penetrates deepest" will get consumable electrode questions wrong.

Exam Trap 2: Faceplate Voltage vs Arc Voltage Examination problems frequently supply a voltage reading measured at the power supply terminals and ask for heat input. If cable length and resistance are given, you must subtract the $I R$ cable and contact drops before calculating true heat input. Using power source terminal voltage will dramatically overestimate heat input.

Exam Trap 3: Treating the Arc Column as an Ohmic Resistor Never apply Ohm's Law ($V = IR$) to solve for arc column behavior under changing current. The arc is an ionized gas whose resistance drops as current rises due to increased ionization density. Doubling current does NOT double arc voltage; arc voltage remains relatively constant or changes only slightly.