Question

(A) (SI units) In an ECM operation the frontal working area of the electrode is 1555 mm2. Applied current = 1200 amps and voltage = 12 volts. The material being cut is nickel (valence = 2). Consult Table 25.1 for the specific removal rate of nickel. (a) If the process is 90% efficient, determine the rate of metal removal in mm3/min. (b) If the resistivity of the electrolyte = 150 ohm-mm, determine the working gap.

Answer 1

you have not provided me table 25.1 but anyway i know what is there in the table

ok i am considering standard values of density and atomic mass of nickel

given

current =1200 ampere atomic mass= 58gm valency=2 faraday's constant=96500

density=8.9gm/cm3 area= 1555mm2     resistivity=150ohm-mm

let's apply the formula for metal removal rate

metal removal rate=(currentatomic mass) (valencyfarady constantdensity)

MRR= (120058)(2965008.9)= 0.04 cm3/s

convert it into mm3/min

MRR = 2400 mm3/min

Now for working gap apply the formula

current / area = voltage (restivityworking gap)

1200/1555 = 12/(working gap150)

WORKING GAP = 0.1 mm

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