Manufacturing Processes II ECM & EDM Equations and Numerical Problems

Derivations and Solving Numerical Problems

As per Faradays Law Volume of metal removed is proportional to current V I and also to the time V t or V I t or V = C I t Where C is a constant called Specific removal rate constant . (For Aluminum (3) C = 3.44E-2 mm3/Amp-sec or 1.26E-4 in3/Amp-min)

As per Ohms Law I = E/R is well known. Under ECM conditions, here R = gr/A, where g is gap between electrodes, r is the resistivity of the electrolyte and A is the surface area between work and tool. Thus I = EA/gr. (Also we can write I/A = E/gr) Substituting I in Faradays EqnVolume removed V = C t ( EA/gr)

Volume removed V = C t ( EA/gr) Linear Travel V/A = CtE/gr, (dividing by A ) Feed (mm/sec) fr = CE/gr , (dividing by t ) Also fr = CI/A , (replacing E/gr for I/A ) Metal Removal Rate MRR = fr A Time to complete an Op t = Thickness/fr

Q1. An ECM operation is to be used to cut a hole into a plate of aluminum that is 12mm thick. The hole has a rectangular crosssection, 10 x 30 mm sq. The ECM operation is to be accomplished at a current of 1200 Amps. Efficiency is 90 %. Determine feed rate and time required to cut through. C fro aluminum is 0.0344 mm cu/Amps-sec. Sol: Feed fr = CI/A = .0344 (1200)/(10x30) = 0.1367 mm/sec At 90% Eff : fr = 0.9 (0.1367) = 0.1307 mm/sec Time to machine = Thickness/fr = 12/ 0.1307 = 91.8 Sec

Q2. The frontal working area of the electrode is 2000 mm sq in a certain ECM operation in which the applied current = 1800 amps and the voltage = 12 volts. The material being cut is nickel (valence =2), whose specific removal rate C is 3.42E-2. (a) If the process is 90% efficient, determine the rate of metal removal in mm^3/min. (b) If the resistivity of the electrolyte is 140 ohm-mm, determine the working gap?

Q3. A 3.5 in diameter through-hole is to be cut in a block of pure iron(2) (C= 0.000135 in cu /Amp-min) by electrochemical machining. The block is 2.0 in thick. To speed the cutting process, the electrode tool will have a center hole which will produce a center core of 3 in diameter that can be removed after the tool breaks through. The outside/inside diameters of the electrode are undersized / oversized to allow for overcuts. If the efficiency of the ECM operation is 90%, what current will be required to complete the cutting operation in 20 minutes?

A 3.5 in diameter through-hole is to be cut in a block of pure iron(2) (C= 0.000135 in cu /Amp-min) by electrochemical machining. The block is 2.0 in thick. To speed the cutting process, the electrode tool will have a center hole which will produce a center core of 3 in diameter that can be removed after the tool breaks through. The outside/inside diameters of the electrode are undersized / oversized to allow for overcuts. If the efficiency of the ECM operation is 90%, what current will be required to complete the cutting operation in 20 minutes?

Derivations and Solving Numerical Problems

The hardness and strength of the work material are not factors in EDM, since the process is not a contest between tool and work. The melting point of the work material is an important property. Weller has determined that MRR = KI/Tm1.23 where MRR is metal removal rate K is constant (664 in SI & 5.08 in US) I is discharge current Tm is melting temperature in o C of o F

Q1. An electric discharge machining operation is being performed on tungsten. (a) Determine the amount of metal removed in the operation after one hour at a discharge amperage = 20 amps. (b) If the work material were tin, determine the amount of material removed in the same time. Use metric units and express the answer in mm^3. Tm for tungsten is 3410 o C and . Tm for tin is 232 o C.

Q2. Suppose the hole in Students Assignment Problem solved for ECM machining, were to be cut using EDM rather than ECM. Using a discharge current = 20 amps (which would be typical for EDM), how long would it take to cut the hole? Tm for iron is 2802 deg F.