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Finding Metal Removal Rate

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In an electrochemical machining operation, the frontal working area of the electrode is 2.5in squared. The applied current=1500amps, and the voltage=12volts. The material being cut is pure aluminum, whose specific removal rate is 1.26 x 10^-4 in^3/amp-min.

If the electrochemical machining process is 90 percent efficent, determine the rate of metal removal in in^3/hr.

https://brainmass.com/engineering/mechanical-engineering/finding-metal-removal-rate-149355

SOLUTION This solution is FREE courtesy of BrainMass!

To express the rate of metal removal in terms of (in^3/hr) we calculate

(the rate of metal removal) = (current)*(specific removal rate) * (minutes per hour) *(efficiency)

= (1500amps)*(1.26 x 10^-4 in^3/amp-min) * (60 min / hr) * (90%/100%) =

= 10.2 in^3/hr

This is all the question asks now, so the voltage and working area are not needed, and this is the complete answer to your posting as it is written now.

However, it may be that you forgot to add the question about the working gap, given the resistivity 6.2 ohm-in of the electrolyte, so, just in case, I give the answer to this question as well:

the gap = (voltage)*(area) / [ (resistivity)*(current) ] = (12 V) * (2.5 in^2) / [ (6.2 ohm-in) * (1500 amps)] =

= 0.0032 in

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