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EDM principle and ECM metal removal rate

Short answer

EDM uses pulsed spark between tool and workpiece in dielectric fluid, melting/vaporising material. ECM is anodic dissolution in electrolyte, governed by Faraday's law. For pure iron, MRR = 33.1 mm³/s (1.99 cm³/min) and feed rate = 4.97 mm/min for a 400 mm² cavity.

(a) Explain the working principle of electrical discharge machining (EDM) with a sketch. State its advantages and limitations. (b) Explain the principle of electrochemical machining (ECM). In ECM of pure iron (atomic weight 55.85, valency 2, density 7.86 g/cm3^3) a current of 1000 A is used with a current efficiency of 90 percent. Calculate the volumetric material removal rate and the tool feed rate if the cross-sectional area of the machined cavity is 400 mm2^2.

(a) EDM

In EDM, the work and the shaped tool (electrode) are both immersed in a dielectric fluid (kerosene, transformer oil or deionised water) and kept a small gap apart (0.01-0.5 mm). A pulsed DC supply produces spark discharges across the gap; each spark generates a temperature of 8000-12000 degree C in a tiny spot, melting and vaporising a small crater in the work. The dielectric flushes out the debris and re-insulates the gap. A servo system keeps the gap constant as the tool advances, and the tool shape is reproduced in the work.

   +---- pulse generator ----+
   |                         |
 tool (+/-)   spark gap    work (-/+)
   |__ dielectric fluid + flushing __|
          servo feed (down)

Advantages: machines any electrically conductive material regardless of hardness (hardened steel, carbide); produces complex cavities, dies and small deep holes; no cutting force and no burr; good accuracy.

Limitations: only conductive materials; slow metal removal; tool wear and a recast (heat-affected) layer; high power consumption; very sharp inside corners are not possible.

(b) ECM

In ECM, the work is the anode and the tool is the cathode. A low-voltage (10-20 V), high-current DC (current density about 20-200 A/cm2^2, total current up to several thousand amperes) flows through a fast-flowing electrolyte (NaCl or NaNO3_3 solution) in a gap of about 0.1-0.5 mm. The work metal dissolves according to Faraday's law, and the electrolyte carries the hydroxide sludge away. The tool does not wear.

By Faraday's laws, the mass removed per second is

m˙=η I Awz F\dot{m} = \frac{\eta\,I\,A_w}{z\,F}

Volumetric rate:

V˙=η I Awz F ρ\dot{V} = \frac{\eta\,I\,A_w}{z\,F\,\rho}

with Aw=55.85A_w = 55.85 g/mol, z=2z = 2, F=96 485F = 96\,485 C/mol, ρ=7.86\rho = 7.86 g/cm3^3, η=0.9\eta = 0.9, I=1000I = 1000 A.

V˙=0.9×1000×55.852×96 485×7.86=0.0331 cm3/s=33.1 mm3/s\dot{V} = \frac{0.9\times1000\times55.85}{2\times96\,485\times7.86} = 0.0331\ \text{cm}^3/\text{s} = 33.1\ \text{mm}^3/\text{s}

This equals 33.1×60=1988 mm3/min33.1\times60 = 1988\ \text{mm}^3/\text{min} (about 2.0 cm3^3/min).

Tool feed rate for the cavity cross-section of 400 mm2^2:

v=V˙A=33.1400=0.0828 mm/s=4.97 mm/minv = \frac{\dot{V}}{A} = \frac{33.1}{400} = 0.0828\ \text{mm/s} = 4.97\ \text{mm/min}

Answer: volumetric removal rate =33.1= 33.1 mm3^3/s (1.99 cm3^3/min); feed rate =4.97= 4.97 mm/min.

✓ Answer checked by Claude Opus 5.5 (high effort)

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