7.4 Maecker Plasma Jet, Degree of Ionization & Anode Heat Partitioning

4. Thermal Plasma Profiles & The Maecker Axial Plasma Jet

Arc Temperature Distribution

Spectroscopic temperature mapping reveals that electric welding arcs are intensely non-isothermal:

• Cathode Region ($z < 0.5\text{ mm}$): Peak axial temperature reaches $18,000\text{ to }25,000\text{ K}$ in pure argon directly beneath the sharpened tungsten tip, where current density is concentrated ($J > 10^4\text{ A/cm}^2$).
• Mid-Column: Temperature relaxes to $12,000\text{ to }15,000\text{ K}$ along the centerline, tapering radially outward to $6,000\text{ K}$ at the luminous plasma boundary.
• Anode Boundary: Near the liquid weld pool, temperature drops to $7,000\text{ to }10,000\text{ K}$.

                    (-) Cathode Tip (Tungsten)
                         |   |   |
                         |   |   |
                         +---+---+   <-- Max Current Density (J_max), Peak Temp ~22,000 K
                             *
                           /   \     <-- Radial Magnetic Pinch: F_em = J x B
                          /     \        Generates Stagnation Pressure: P_mag = mu_0*I^2 / (4*pi*R_c^2)
                         /       \       Axial Pressure Gradient (dP/dz) Accelerates Fluid Downward
                        / Plasma  \
                       /   Jet     \ <-- Maecker Plasma Jet Velocity: 100 - 500 m/s
                      /             \
      ===============+===============+===============  Weld Pool Surface (Anode)
                     |               |                 P_mag is low (diffuse attachment)
                     |  Penetration  |                 Jet Momentum Depresses Liquid Puddle
                     |    Profile    |
                     +---------------+

The Maecker Effect (Magnetically Driven Axial Plasma Jet)

In 1955, H. Maecker published the fundamental theoretical derivation explaining why arc plasma does not drift passively like a candle flame, but acts as a high-velocity, rigid fluid jet.

1. Current Streamline Convergence: Because current must enter the narrow cathode spot ($R_c$), current streamlines converge tightly toward the tip. The local current density ($J = I / \pi R^2$) surges at the cathode and disperses at the workpiece ($R_a > R_c$).
2. Lorentz Magnetic Pinch Force: The axial electric current produces a self-induced azimuthal magnetic field ($B_\theta$). The interaction between current density and magnetic field generates an inward radial Lorentz pinch force: $\mathbf{F}_{\text{pinch}} = \mathbf{J} \times \mathbf{B}$
3. Magnetic Stagnation Pressure: Integrating this radial body force from the arc perimeter to the centerline yields the magnetic pressure on the arc axis: $P_{\text{mag}}(0) = P_{\text{ambient}} + \frac{\mu_0 I^2}{4 \pi R^2}$ where $\mu_0 = 4\pi \times 10^{-7}\text{ H/m}$ is the magnetic permeability of free space.
4. Axial Pressure Gradient & Jet Velocity: Because the radius at the cathode tip ($R_c$) is far smaller than at the anode attachment ($R_a$), the magnetic stagnation pressure at the cathode is immensely higher than at the anode ($P_{\text{mag}, c} \gg P_{\text{mag}, a}$): $\frac{\partial P}{\partial z} < 0$ This steep axial pressure gradient continuously sucks fresh shielding gas into the cathode constriction zone, ionizes it, and violently accelerates it axially toward the workpiece pool. Applying Bernoulli's equation along the central streamline: $\Delta P_{\text{mag}} = P_{\text{mag}, c} - P_{\text{mag}, a} = \frac{\mu_0 I^2}{4 \pi} \left( \frac{1}{R_c^2} - \frac{1}{R_a^2} \right) = \frac{1}{2} \rho_{\text{plasma}} v_{\text{jet}}^2$ $v_{\text{jet}} = \sqrt{\frac{\mu_0 I^2}{2 \pi \rho_{\text{plasma}}} \left( \frac{1}{R_c^2} - \frac{1}{R_a^2} \right)}$

Plasma jet velocities routinely exceed $100\text{ to }500\text{ m/s}$ (subsonic to near-sonic in hot plasma). The aerodynamic stagnation pressure of this jet ($q = \frac{1}{2}\rho v^2$) impacts the molten pool, digging a deep depression that drives the classic finger-like penetration characteristic of DCEP GMAW and DCEN GTAW.

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Degree of Ionization, the Saha Equation & Shielding-Gas Influence

AWS B5.16 Clause 8.3.2 calls out arc temperature and degree of ionization (shielding gas influence) as a distinct Part 3 sub-topic. Arc plasma is only partially ionized: at GTAW column temperatures the argon plasma still contains far more neutral atoms than ions, and the ionized fraction is what sets electrical conductivity.

The degree of ionization $\alpha$ is the fraction of atoms that have lost an electron:

$\alpha = \frac{n_i}{n_i + n_0}$

where $n_i$ is ion number density and $n_0$ is neutral number density. Thermal (equilibrium) ionization is governed by the Saha equation, whose dominant terms are the plasma temperature $T$ and the first ionization potential $V_i$ of the species:

$\frac{n_e n_i}{n_0} \propto T^{3/2} \exp\!\left(-\frac{eV_i}{kT}\right)$

Two consequences drive every exam question on this topic:

1. Ionization is exponential in $-V_i/T$. A species with a low ionization potential ionizes far more readily at the same temperature. Argon ($V_i = 15.76\text{ eV}$) ionizes far more easily than helium ($V_i = 24.59\text{ eV}$), which is exactly why a helium-shielded arc needs a much higher arc voltage to sustain the same current, and why helium delivers higher heat input per ampere.
2. Low-ionization-potential metal vapour dominates the column. Iron ($7.90\text{ eV}$), aluminum ($5.99\text{ eV}$), sodium ($5.14\text{ eV}$) and potassium ($4.34\text{ eV}$) ionize at a fraction of the argon energy. Even a few percent of metal vapour or an alkali flux coating raises the ionized fraction, raises conductivity, and lowers the arc voltage needed to carry the current — the physical reason SMAW electrode coatings contain potassium and sodium silicates for AC arc stabilization.

Because ionization is a thermal equilibrium, degree of ionization is also strongly non-uniform across the column: on the arc axis, where argon plasma runs near $10{,}000$–$20{,}000\text{ K}$, $\alpha$ may approach unity, while a few millimetres radially outward the temperature collapses below $\sim 7{,}000\text{ K}$ and $\alpha$ falls by orders of magnitude. That steep radial gradient is what confines current to a narrow conductive core and produces the measured current density profile.

Exam Trap: "Higher ionization potential means a hotter arc." Helium has the highest ionization potential of the common shielding gases and does produce the hottest, highest-voltage arc — but not because it ionizes easily. It is the opposite: helium resists ionization, so the column must run at a higher electric field (and therefore higher voltage and power) to carry the same current. Answer choices that claim helium "ionizes more easily than argon" are always wrong.

5. Comprehensive Worked Numerical Example: Maecker Jet Velocity & Anode Heat Partitioning

Problem Statement

An automated DCEN Gas Tungsten Arc Welding (GTAW) torch operates in pure argon at a welding current $I = 280\text{ A}$ with an arc length $L_{\text{arc}} = 4.50\text{ mm}$.

The system physical and geometric parameters are:

• Cathode spot radius on the ground $2%\text{ thoriated}$ tungsten tip: $R_c = 0.70\text{ mm} = 7.00 \times 10^{-4}\text{ m}$.
• Anode arc attachment radius on the carbon steel workpiece: $R_a = 2.80\text{ mm} = 2.80 \times 10^{-3}\text{ m}$.
• Average plasma gas density along the core centerline: $\rho_{\text{plasma}} = 0.040\text{ kg/m}^3$ (argon at $\sim 16,000\text{ K}$).
• Cathode fall voltage: $V_c = 8.00\text{ V}$.
• Anode fall voltage: $V_a = 3.20\text{ V}$.
• Plasma column electric field: $E_p = 1.30\text{ V/mm}$.
• Carbon steel workpiece electronic work function: $\Phi_a = 4.50\text{ eV}$ ($4.50\text{ V}$ energy equivalent).
• Electron gas thermal kinetic temperature exiting plasma: $T_e = 12,000\text{ K}$.
• Permeability of free space: $\mu_0 = 4\pi \times 10^{-7}\text{ H/m} = 1.2566 \times 10^{-6}\text{ H/m}$.
• Boltzmann's constant: $k_B = 1.3806 \times 10^{-23}\text{ J/K}$; Elementary charge: $e = 1.6022 \times 10^{-19}\text{ C}$.

Calculate:

1. The magnetic stagnation pressure at the cathode tip ($P_{\text{mag}, c}$) and at the anode surface ($P_{\text{mag}, a}$).
2. The theoretical maximum axial velocity ($v_{\text{jet}}$ in $\text{m/s}$) of the Maecker plasma jet.
3. The total electrical voltage drop across the arc ($V_{\text{arc}}$) and the total electrical input power ($P_{\text{total}}$).
4. The total thermal power delivered directly into the workpiece anode pool ($Q_{\text{anode}}$) by electron absorption, and the percentage of total arc power it represents.

Step-by-Step Solution

Step 1: Compute magnetic stagnation pressures Calculate cathode magnetic pressure: $P_{\text{mag}, c} = \frac{\mu_0 I^2}{4 \pi R_c^2} = \frac{(4\pi \times 10^{-7}\text{ H/m})(280\text{ A})^2}{4 \pi (7.00 \times 10^{-4}\text{ m})^2}$ $P_{\text{mag}, c} = \frac{10^{-7} \times 78,400}{4.90 \times 10^{-7}} = \frac{78,400}{4.90} = 16,000\text{ Pa} = 16.00\text{ kPa}$ Calculate anode magnetic pressure: $P_{\text{mag}, a} = \frac{\mu_0 I^2}{4 \pi R_a^2} = \frac{10^{-7} \times 78,400}{(2.80 \times 10^{-3}\text{ m})^2} = \frac{78,400 \times 10^{-7}}{7.84 \times 10^{-6}} = \frac{7,840}{7.84} = 1,000\text{ Pa} = 1.00\text{ kPa}$ Compute the net axial magnetic pressure differential: $\Delta P_{\text{mag}} = P_{\text{mag}, c} - P_{\text{mag}, a} = 16,000\text{ Pa} - 1,000\text{ Pa} = 15,000\text{ Pa} = 15.00\text{ kPa}$

Step 2: Calculate Maecker plasma jet velocity ($v_{\text{jet}}$) Using Bernoulli's relationship along the central streamline: $v_{\text{jet}} = \sqrt{\frac{2 \Delta P_{\text{mag}}}{\rho_{\text{plasma}}}} = \sqrt{\frac{2 \times 15,000\text{ Pa}}{0.040\text{ kg/m}^3}} = \sqrt{\frac{30,000}{0.040}} = \sqrt{750,000} = 866.0\text{ m/s}$

Engineering Note: In real arcs, viscous boundary dissipation and momentum exchange reduce this idealized frictionless velocity by approximately $50%\text{ to }60%$, yielding real peak jet velocities of $350\text{ to }450\text{ m/s}$. This jet still moves at supersonic speeds relative to ambient air ($343\text{ m/s}$), creating significant dynamic pressure ($q = 15\text{ kPa}$) that depresses the liquid weld pool.

Step 3: Calculate total arc voltage and arc input power Compute column voltage drop: $V_p = E_p \cdot L_{\text{arc}} = (1.30\text{ V/mm})(4.50\text{ mm}) = 5.85\text{ V}$ Compute total arc voltage: $V_{\text{arc}} = V_c + V_p + V_a = 8.00\text{ V} + 5.85\text{ V} + 3.20\text{ V} = 17.05\text{ V}$ Compute total electrical power: $P_{\text{total}} = V_{\text{arc}} \cdot I = 17.05\text{ V} \times 280\text{ A} = 4,774.0\text{ W} = 4.774\text{ kW}$

Step 4: Calculate thermal power delivered to the workpiece anode First, convert electron thermal energy $\frac{5}{2} \frac{k_B T_e}{e}$ to equivalent volts: $V_{\text{thermal}} = \frac{5}{2} \frac{(1.3806 \times 10^{-23}\text{ J/K})(12,000\text{ K})}{1.6022 \times 10^{-19}\text{ C}} = \frac{4.1418 \times 10^{-19}}{1.6022 \times 10^{-19}} = 2.585\text{ V}$ Apply the complete anode power equation: $Q_{\text{anode}} = I \left( V_a + \Phi_a + V_{\text{thermal}} \right)$ $Q_{\text{anode}} = 280\text{ A} \times \left( 3.20\text{ V} + 4.50\text{ V} + 2.585\text{ V} \right)$ $Q_{\text{anode}} = 280\text{ A} \times 10.285\text{ V} = 2,879.8\text{ W} = 2.880\text{ kW}$ Calculate the fraction of arc power absorbed directly at the anode: $\eta_{\text{anode}} = \frac{Q_{\text{anode}}}{P_{\text{total}}} = \frac{2,879.8\text{ W}}{4,774.0\text{ W}} = 0.6032 \quad (60.32\%)$ Notice that electronic condensation alone ($I \cdot \Phi_a = 280 \times 4.50 = 1,260\text{ W}$) accounts for $43.8%$ of all heat transferred into the workpiece! (The remaining arc power is dissipated via cathode heating, column radiation, and radial convection).

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6. Real-World Engineering Scenarios & Exam Pitfalls

Industrial Case Study: Automated Orbital GTAW Arc Wander on Alloy 625

An aerospace supplier was performing automated tube-to-tubesheet GTAW on Inconel 625 heat exchangers using high-purity argon shielding. During a production run, operators observed severe arc wandering, erratic high-frequency starting, and tungsten tip erosion within two hours of operation. Welds exhibited lack-of-fusion defects at the joint root.

Investigation: The shop had mistakenly loaded EW-P (pure tungsten) electrodes instead of the specified EWLa-2 ($2%\text{ lanthanated}$). Under the required $180\text{ A}$ current, the pure tungsten electrode's high work function ($\Phi = 4.55\text{ eV}$) forced the tip temperature above $3700\text{ K}$ to satisfy the Richardson-Dushman relation. At this temperature, the tungsten melted, balling up into a large spherical hemisphere ($R_c$ increased from $0.5\text{ mm}$ to $1.8\text{ mm}$).

Because magnetic stagnation pressure scales inversely with radius squared ($P \propto 1/R_c^2$), the Maecker plasma jet collapsed by over $90%$. The arc lost its axial stiffness, allowed cathode spots to wander erratically across the melted ball, and failed to project heat into the root. Replacing the electrodes with EWLa-2 ($\Phi = 2.70\text{ eV}$) lowered the operating temperature to $3000\text{ K}$, preserved a sharp $30^\circ$ ground point, restored the Maecker jet velocity, and eliminated all lack-of-fusion defects.

Common CWEng Exam Traps

Exam Trap 1: Location of Peak Arc Temperature A perennial CWEng exam trap asks where the highest temperature in a welding arc occurs. Many candidates select "the center of the weld pool" or "the mid-point of the plasma column." The correct answer is immediately adjacent to the cathode tip, where current density ($J$) is maximum and thermal constriction generates temperatures exceeding $20,000\text{ to }25,000\text{ K}$.

Exam Trap 2: Shielding Gas vs Metal Vapor Ionization Dominance Questions frequently ask which element dominates electrical conductivity in a GMAW arc shielded with $100%\text{ Argon}$. While argon comprises $99+%$ of the gas volume, the correct answer is vaporized iron (or manganese). Because iron's ionization potential ($7.90\text{ eV}$) is roughly half that of argon ($15.76\text{ eV}$), metal vapor ionizes exponentially faster, providing the vast majority of free charge carriers in the conductive core.

Exam Trap 3: Work Function Role in Anode vs Cathode Candidates often confuse how work function affects each electrode. At the cathode, a low work function is desirable because it makes electron emission easier (Richardson-Dushman). At the anode, a higher work function actually increases heat input into the weld puddle because electrons release their work function upon condensing into the metal lattice ($Q = I \cdot \Phi_a$).