Introduction to Manufacturing Technology –Lecture 7

Instructors: (1)Shantanu Bhattacharya, ME, IITK, email: bhattacs@iitk.ac.in(2)Prof. Arvind Kumar, ME, IITK email: arvindkr@iitk.ac.in

Review of last lecture• Electrochemical machining basic

principle.• Why is the ECM process a die sinking

process?• Electrolyte flow in an ECM system.• Electrochemistry of ECM process.• Ion-Ion, Ion Solvent interaction.

(Debye Huckle Equations)

The Electrode Double layer

+

+

-

-

•For the case of an electrode dipping into a solution of an electrolyte, the total charge on the electrode surface must balance the total solution charge (opposite charge).•There is an electrostatic interaction between the electrode charge and solution charge.•Ions from the solution may approach the electrode surface only so far as their inner solvation shell.•The surface array of the ions is thus cushioned from the electrode surface by a monolayer of solvent. Outer

Helmholtz Plane

Electrode surface

•The line drawn through the center of such ions at this distance of closest approach marks a boundary known as the outer Helmholtz plane.•The size of the ions forming the outer Helmholtz plane are larger and the total no. of ions needed to do a complete charge balance cannot be all fitted on this plane. •The remaining charges are all held with increasing disorder in the outward direction into the solution. This layer is known as the diffused part of double layer.

The Electrode Double Layer•The variation of the potential Ψ, with distance from the electrode surface is shown in the figure below.•Thus all the charge which neutralizes that on the electrodes is held in a region between the outer Helmholtz plane and the bulk of the electrolyte solution.•This situation is very similar to that in Debye Huckle theory for 3-D in a ion solution.•In the case of the planar electrode only 1-D needs to be considered (normal to electrode surface)

As x ∞ , Ψ 0

+

+

-

-

Outer Helmholtz

Plane

Electrode surface

X

Ψ

δ

Ψ0

+

+

-

-

Outer Helmholtz

Plane

Electrode surface

X

Ψ

δ

Ψ0

Capacitance Modeling of the diffuse layer

•There are effectively two components which make up the total potential drop across the interface; Ψ0

Across the diffuse part and Ψ-Ψ0 across the fixed part.•The total capacitance of the double layer, C, is made up of that due to the inner layer, which we may designate CH and that due to the diffuse layer CD

Electrochemical Machining• The gram equivalent weight of the metal is given by ε = A/Z, where A is the atomic weight and Z is the valency of the ions produced.

•The rate of mass removal is given by:

m = AI/ZF

If density of the anode material is ρ, the volumetric removal rate is given by

Q = AI/ ρZF cm3/ sec,

WhereA = gram atomic weight of the metallic ions,

I = current (amperes)

ρ = Density of the anode (gm/ cm3 ),

Z = valency of the cation,

F = 96,500 coulomb

Numerical ProblemsIn an electrochemical machining process with a pure iron work-piece, a removal rate of 5 cm3/ min. is desired. Determine the current required.

Numerical Problem•In the actual ECM process, there are many other factors which affect the removal rate.•The process is seldom as ideal as we have described.•The actual removal rate may vary slightly from that obtained theoretically from equation for Q

•The theoretical and the actual removal rates with nickel as work material are shown in the figure and they vary slightly.•The reason is that the theoretical removal is based on Divalent dissociation.

Multivalent dissociation•So, as the current is more, i.e., the dissolution takes place at a higher potential difference, trivalent dissolution also occurs.•Therefore, at larger currents the theoretical values tend to be more than the actual ones. •Sometimes the dissolution valence also depends on the electrolytes. (For eg.: Copper dissolves in mono-valent form in chlorine solutions, whereas in nitrate solutions, the dissolution takes place in divalent state.

•The table on left shows some imp. Data on some elements

Anode made of Alloy•When the anode is made up of an alloy instead of pure metal, the removal rate can be found out by considering the charge required to remove an unit volume of each element.

• If the atomic weights and the valencies of corresponding ions entering the electrolyte are A1, A2, A3……. and Z1, Z2, Z3…….. Respectively, and the composition by weight of the alloy is x1% of element 1, X2% of element 2,……., then a volume v cm3 of the alloy would contain vρxi/ 100 gram of the i th element, where ρ is the overall density of the alloy in gm/ cm3 . The charge required to remove the ith element in volume v is given by:

(v ρ xi/ 100). (Zi F/ Ai)

The the volume of the alloy removed per unit charge is

Q = (100/ ρF) ( 1/ Σ(xiZi/ Ai)) cm3/ amp-sec or

Q = [(0.1035 X 10-2)/ ρ] ( 1/ Σ(xiZi/ Ai)) cm3/ amp-sec

Numerical ProblemsThe Composition (%) weight of the Nimonic 75 alloy is given here:

Calculate the removal rate (in cm3/min.) when a current of 1000 amp is passed. Use the lowest valency of dissolution of each element.

Numerical Problems

Numerical Problems