Numericals on Electrochemical Machining

1. Made By
Mr. AVINASH JURIANI
M.tech-Manufacturing
14MT000354

2. 1.Calculate the MRR when copper is electrochemically
machined under following conditions:
V= 18volt
I= 500A
Atomic weight= 56
Valency= 2
ρ= 7.8gm/cm3
Solution
Use MRR= eI/Fρ
Here e= atomic weight/valency= 56/2= 28
F= 96500= Faraday constant
Hence MRR= (28*500)/(96500*7.8)
= 0.0186cm3/s

3. 2. Electrochemical machining is performed to remove material from
an iron surface of 20mm*20mm under the following conditions:
Inter electrode gap, l= 0.2mm
Supply voltage, V= 12volt
Specific resistance of electrolyte, ρs= 2Ω cm
Atomic weight of Iron, A= 55.85
Valency of Iron, Z= 2
Faraday’s constant, F= 96540 coulombs
Find MRR
Solution
Use R= ρsl/A
V=IR
Here MRR= eI/Fρ
Now, R= (2*10-1*0.2)/(20*20)= 10-4Ω
We get I= 12*104A
Again e=55.85/2=27.925
Hence MRR= (27.925*12*104)/96540
= 34.725gm/s

4. 3.Calculate MRR of Nimonic alloy(Co-Ni-Cr)
% Atomic Weight Valency
Co 18 58.93 2
Ni 62 58.71 2
Cr 20 51.99 6
Solution
I= 500 A d=8.28gm/cm3
Now, 1/e= Σ%*valency/atomic weight
=(0.18*2/58.93)+(0.62*2/58.71)+(0.2*6/51.99)
=0.050311
e= 19.876
Hence , MRR= eI/Fρ
= (19.876*500)/(96500*8.28)
= 0.0124cm3/sec

5. 4.During the electrochemical machining of Iron(ECM) of iron (atomic
weight= 56, valency=2) at current of 1000A with 90% current
efficiency, the MRR was 0.26gm/s.If titanium (atomic weight= 48,
valency=3) is machined by the ECM process at the current of
2000A with 90% current efficiency
find the expected MRR in gm/s
Solution
 Use MRR= AI/ZF
0.26= 56*900/F*2
F= 96923
Now for Titanium
A= 48
Z= 3
I= 0.9*2000= 1800A
MRR= 48*1800/3*96923= 0.3

6. 5.While Removing Material from Iron (atomic weight= 56,
valency= 2 & density= 7.8g/cc by ECM, MRR was 2cc/min is
desired.
Find the current in A for achieving this MRR
Solution
 Use MRR=AI/ρZF
Plug in all values in above formula
2/60= 56*I/7.8*2*96500
hence I= 448A

7. 6.In an ECM operation, a square hole of Dimensions
5mm*5mm is drilled in a block of copper. The current
used is 5000A. Atomic weight of copper is 63 & valence
of dissolution is 1. Faraday’s constant is 96500coulomb.
Find MRR
Solution
Plug all values in
MRR= AI/ZF= 5000*63/1*96500= 3.364gm/sec

8. 7.Composition of a Nickel superalloy is as follows:
Ni = 70.0%, Cr = 20.0%, Fe = 5.0% and rest Titanium
Calculate rate of dissolution if the area of the tool is 1500 mm2
and a current of 2000 A is being passed through the cell. Assume
dissolution to take place at lowest
valency of the elements.
ANi = 58.71 ρNi = 8.9 νNi = 2
ACr = 51.99 ρCr = 7.19 νCr = 2
AFe = 55.85 ρFe = 7.86 νFe = 2
ATi = 47.9 ρTi = 4.51 νTi = 3

10. 8.In ECM operation of pure iron an equilibrium gap of 2 mm
is to be kept. Determine supply voltage, if the total
overvoltage is 2.5 V. The resistivity of the electrolyte is 50 Ω-
mm and the set feed rate is 0.25 mm/min.
Solution
Where h*= steady state gap
c= constant
f= feed

11. 9.In a certain electro-chemical dissolution process of iron ,a metal
removal rate of 2 cc/min was desired.Determine the amount of
current required for the process.Assume :
Atomic wt of iron,M=5.6gm
Valence at which dissolution occurs,V=2 F=1609
amp.min
Density of iron,P=7gm/cc
Solution
At. Wt of iron,M=56gm
Valency of iron dissolution,V=2
So, Gram equivqlent weight of iron,E=M/V=56/2=28
Amount of iron dissolved,W=EIT/FP
Thus, MRR in cc/min=EI/FP
2=28I/1609*7.09
I=896 Amp.

12. 10. While machining tungsten,a MRR of 0.98cc/min has been
observed at 1000amp.Evaluate the valency at which the metal has
been electrochemically dissoluted.
Assume:
Atomic wt of tungsten,M=186gm F=1609amp.min
Density of tungsten,P=19.4gm/cc
Solution
Let its valency be V.
Thus
Gram equivalent weight of tungsten, E=M/V=186/V
Amount of tungsten dissolved, W=EIT/FP
MRR in cc/min=EI/FP
0.98=(E*100)/(1609*19.4)
E=30.59
Again
V=186/E=186/30.59=5.88=6.

13. 11.Calculate the MRR(m^3/min)in an anodic dissolution process of
chromium if the current density available has been
250Amp/cm^2.
Assume
At. Wt of chromium,M=52gm
Valency of chromium,V=2
Density of chromium,P=7.2gm/cm^3
Solution
Gram equivalent wt of chromium,E=M/V=26
Amount of chromium dissolved,W=EIT/FP
Thus,MRR in cm^3/min=EI/FP
=(26*25)/(1609*7.2)=0.561 cm^3/min.

14. 12.During electrochemical machining of iron with a copper
electrode workng in a 5(N) NaCl soln in water ,an equillibrium
gap of 0.0125cm has been achieved at a current density of 150
Amp/cm^2.If the operating voltage has been 10V,determine the
feed rate(f) of the tool.
Given,
Sp. Resistance of NaCl solution ,p=3ohm-cm
Valency at which iron is dissoluted,V=2
Density of iron, P=7.8gm/cm^3
Solution
We know that,
E=M/V=5/2=2.5
AQ, v=10V, h=0.0125cm,F=1609 amp.min
Thus,
Feed rate, f
=(v*MRR)/(P*h*I) (Again, MRR/I=E/FP )
=(v*E)/(P*h*F*P)
Substituting the values in the equation we have
f=0.020 cm/s

15. 13.In a Certain electro chemical dissolution process of iron,a metal
removal rate of 3cm^3/min was desired.Determine the amount of
current req for the process,assuming:
Atomic wt of iron,N=56gm
Valency,V=2 (E=N/V=56/2=28)
Density of iron,P=7.8gm/cm^3
Solution
MRR=3cm^3/min
Now, MRR=(EI/FP)
Substituting the corresponding values in the above equation we
have
3=(28*I)/(1609*7.8)
Solving,
we get, I=1344.6 Amp.

16. 14. In electrochemical machining of pure iron a material removal rate
of 600 mm3/min is required. Estimate current requirement.
Given-
atomic weight of iron(A)=56
valency (v)=2
F =96500 coulomb
density of iron(ρ)=7.8 gm/cc
Solution
MRR=600 mm3
/min
=10 mm3
/sec
=10 x10-3 cc/sec
MRR=AI/Fρv
10 X10-3 =(56xI)/96500X7.8X2
I =268.82 amp

17. 15. A researcher conducts ECM on binary alloy (density 6000kg/m3)
of iron (atomic weight(A1) 56, valency(v1) 2) and metal P
(atomic weight(A2) 24, valency(v2) 4) faradays constant 96500
coloumbs/mole. Volumetric material removal rate of the alloy is
50mm3 /s at a current of 2000A. What is the percentage of the
metal P in the alloys.
Solution
F=96500 , ρ=6000 kg/m3
A1 =56 , v1=2 , MRR=50 mm3/sec
A2 =24 , v2=4 , i=2000A
X1+X2=1 ---- (1)
MRR=[i/Fρ{(X1v1/A1)+(X2v2/A2)}]
(X1*2/56)+(X2*4/24)=2000/6*96500*.05 ---- (2)
after solving we get
X1=75% X2=25%

18. 16.For ECM of steel which is used as the electrolyte
◦ kerosene
◦ NaCl
◦ Deionised water
◦ HNO3
Ans: Nacl so as to create anodic dissolution

19. 17.MRR in ECM depends on
◦ Hardness of work material
◦ atomic weight of work material
◦ thermal conductivity of work material
◦ ductility of work material
Ans: atomic weight of work material

20. 18.ECM cannot be undertaken for
◦ Steel
◦ Nickel based superalloy
◦ Al2O3
◦ Titanium alloy
Ans: Al2O3 as it is ceramic

21. 19.Commercial ECM is carried out at a combination of
◦ low voltage high current
◦ low current low voltage
◦ high current high voltage
◦ low current low voltage
Ans: low voltage high current