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58.2 PRACTICE PROBLEMS

1.We try an EDM process where the copper tool has a mass of 200g before beginning and 180g after. The iron workpiece drops from 3.125kg to 3.096kg, but has rounded corners.

a)What is the tool wear factor?

b)If the tool was cylindrical to begin with, draw sketches of the electrode before and after.

2.What are the selection criteria for choosing between machining and EDM?

ans. EDM is particularly useful when dealing with internal cuts that are hard to get tools into. Machining tends to work best with external cuts. EDM is suitable for removal of smaller amounts of material at a much slower rate.

58.3 REFERENCES

Ghosh, A., Manufacturing Science, Ellis Horwood Ltd., Chichester, UK, 1986.

Lascoe, O.D., Handbook of Fabrication Processes, American Society of Materials, Metals Park, Ohio, 1988.

59. ELECTROCHEMICAL MACHINING (ECM)

•The physics - an electrode and workpiece (conductor) are placed in an electrolyte, and a potential voltage is applied. On the anode (+ve) side the metal molecules ionize (lose electrons) break free of the workpiece, and travel through the electrolyte to the electrode (a cathode; has a -ve charge; a surplus of electrons).

•NOTE: in EDM an arc was used to heat metal, here the metal dissolves chemically.

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cathode electrode

electrolyte

current density (closer sections have higher density)

anode work

electrode is moved down to cut

•Variation in the current density will result in work taking the electrodes shape.

•The electrode is fed with a constant velocity, and the electrolyte is fed through the tool. The tool is designed to eliminate deposition of the ionized metal on the electrode.

electrolyte is pumped

using high pressure

insulation

Vs

electrolyte splashes out

•Supply V = 8 to 20V, I = >1000A.

•Electrode gap is typically 0.1 to 0.2 mm.

•mrr is about 1600mm3/min. per 1000A, OR 3KWhr for 16000 mm3 (not very efficient, 30 times

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more than standard machining techniques).

•mrr is independent of material hardness.

•Good for low machinability, or complicated shapes.

•Very little tool wear,

•Forces are large with this method because of fluid pumping forces.

•Faraday’s laws state that,

m =		Itε
		------
		F
m =		weight (g) of a material
I	=	current (A)
t	=	time (sec)
ε	=	gram equivalent weight of the material

F= constant of proportionality - Faraday (96,500 coulombs)

•The basic principle is shown below

Cu	Fe
	could be a battery, or electroplating, or ECM bath
NaCl + H2O electrolyte

• The chemical reaction between an electrode and the electrolyte leads to electrons being added, or removed from the electrode metal. This addition/subtraction leads to a voltage potential.

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For Iron
Fe	Fe2 +	+ 2e–	V2	= –0.409V
For Copper
Cu	Cu2 +	+ 2e–	V1	= 0.304V

• To make a battery.

	I
Cu	+	-	Fe
		V
+	-	+	-
	V1
		V2

V = V1 - V2

V = 0.304V - (-0.409V) = 0.713V

			2e–
		2e–
2( H2O)			Fe +	+
H	2	+ 2( OH) –	Fe( OH) 2
		2( OH) –	= 2e–

• To do electrolysis.

I

Fe	+	-	Cu
		V
	electron
	flow

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In the iron (anode) - the iron dissolves, and 2 electrons are pulled out by the battery.

Fe → Fe++ + 2e–

In the copper (cathode) - extra electrons are pushed in

by the battery and they go to the electrolyte. The electrons liberate hydrogen gas, and create a base.

2e + 2H2O → H2 + 2( OH) –

hydroxides, and the radical iron react to create an insoluble precipitate that falls to the bottom.

Fe++ + 2( OH) – → Fe( OH) 2

The electrodes and electrolyte are chosen so that deposition does not occur at either electrode. (In electrolysis we would want deposition at one of the electrodes.)

• The mrr is,

AI

cm

3

0.1035 10

–2

1

cm

3

Q =

=

----------

----------

------------------------------ρ

---------------

--------------

ρ ZF sec

∑

x

Z

A sec

--------i

i

Ai

i

for alloys

• e.g.

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Using ECM remove 5cm3/min from an iron workpiece. What current is required?

Q =		5	cm3			= 5	cm3		1min	=	5 cm3
			---------min				---------min		-----------------		60----- ----------sec
									60 sec
I	= ?
A = 56g (periodic table)
ρ				g
	= 7.8			---------		(from handbook)
				cm	3
Z = 2 (from chemical reaction)
F = 96, 500 coulombs
	5	=				56I
	-----			7.8--------------------------------------		( 2) ( 96, 500)
	60

I= 2240A

•Actual rates may vary from theory as other factors come into effect.

Theory

for nickel

mrr (g/min)

Current (A)

This can be caused by other chemical reactions also taking place. e.g. production of Fe3+ instead of Fe2+ ions can result when a

larger current or different electrolyte is used.

• The table below shows various materials and relevant properties,

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metal	gram atomic	valency of	density
	weight	dissolution	(g/cm3)
aluminum	26.97	3	2.67
chromium	51.99	2/3/6	7.19
cobalt	58.93	2/3	8.85
copper	63.57	1/2	8.96
iron	55.85	2/3	7.86
nickel	58.71	2/3	8.90
tin	118.69	2/4	7.30
titanium	47.9	3/4	4.51
tungsten	183.85	6/8	19.3
zinc	65.37	2	7.13
silicon	28.09	4	2.33
manganese	54.94	2/4/6/7	7.43
carbon		2/4	1.8-2.1

• e.g.

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The composition of the Nimonic 75 alloy is given below. Find the mmr for a current of 1000A.

Q =	0.1035× 10–2			1
	------------------------------		---------------------
	ρ		∑	x Z
				--------
				i	i
				Ai
			i

%component

72.5Ni

19.5Cr

5.0Fe

0.4Ti

1.0Si

1.0Mn

0.6Cu

ρ =	100													100										g
		xi	=	----------------------------------------------------------------------------------------------------------------																	=	8.18		---------
				72.5		+	19.5	+			5.0		+	0.4	+	1	+	1	+	0.6				cm	3
∑		----		---------			---------			---------				---------		---------		---------		---------
		ρ i		8.9			7.19			7.86				4.51		2.33		7.43		8.96
	i
	0.1035× 10				–2													1
Q =			8.18------------------------------			----------------------------------------------------------------------------------------------------------------------------------------------------
						( 72.5) 2						( 19.5) 2			( 5) 2			( 0.4) 3		( 1) 4			( 1) 2			( 0.6) 1
						------------------	58.71			+		------------------51.99 + 55.85------------					+	---------------47.9		+ 28.09------------		+	54.94------------ +			--------------63.57
Q =	0.35× 10			–4 cm3
				--------------
				A sec
therefore, for 1 min, and 1000A
				= 0.35× 10					–4		( 1000) ( 60)						cm3
																	---------
																= 2.1 min

•While the current required is related to the metal removed, the voltage required depends upon,

-electrode potential.

-the current flow in and about the electrodes will disturb the normal distribution of voltage. Extra potential is required to overcome the effects.

-Ion collect near electrodes and impede ion transfer from the electrode to the electrolyte, also adding a potential.

-Some solid film forms on the surface of the electrode, also increasing resistance.

-electrolyte resistance,

	I	=	V – ∆ V
			----------------
	1		R
		=	10 → 1( Ω cm)
	--
	R

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required V
	anode potential
	activation polarization overvoltage
	ohmic overvoltage
	concentration polarization overvoltage
	electrolyte ohmic voltage	cathode
anode	concentration polarization overvoltage
		ohmic overvoltage
	activation polarization overvoltage
		cathode potential
		distance

• The feed of the electrodes has the following effects

+	-
	tool
	v
	y
	work

dy	=		KA( V – ∆ V)		1	– v
-----			-----------------------------ρ ZF		--
dt					y

K = conductivity of electrolyte( Ω –1)

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y(t)

																												for electrode velocity v = 0
								y0																																															t
4					gap
										y0 = 4												For electrode starting above/below equilibrium
																						electrode gap - eventually the electrode approaches
3					3																	the equilibrium gap.
2
					2
																									y
			1.5
1
			0.5
0			0
			2																	4																		6															t

• The ECM process will erode material in a radial direction, so care must be made in tooling design.

Produces

•As current flows through the electrolyte, it is heated, and conductivity decreases.

•Surface finish is affected by,

-selective dissolution

-sporadic breakdown of the anodic film