Electrochemical Machining

Numericals on Electrochemical Machining

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

2. 1.CalculatetheMRRwhencopperiselectrochemically
machinedunderfollowingconditions:
V=18volt
I=500A
Atomicweight=56
Valency=2
ρ=7.8gm/cm
3
Solution
UseMRR=eI/Fρ
Heree=atomicweight/valency=56/2=28
F=96500=Faradayconstant
HenceMRR=(28*500)/(96500*7.8)
=0.0186cm
3
/s

3. 2.Electrochemicalmachiningisperformedtoremovematerialfrom
anironsurfaceof20mm*20mmunderthefollowingconditions:
Interelectrodegap,l=0.2mm
Supplyvoltage,V=12volt
Specificresistanceofelectrolyte,ρ
s=2Ωcm
AtomicweightofIron,A=55.85
ValencyofIron,Z=2
Faraday’sconstant,F=96540coulombs
FindMRR
Solution
Use R= ρ
sl/A
V=IR
Here MRR= eI/Fρ
Now, R= (2*10
-1
*0.2)/(20*20)= 10
-4
Ω
We get I= 12*10
4
A
Again e=55.85/2=27.925
Hence MRR= (27.925*12*10
4
)/96540
= 34.725gm/s

4. 3.CalculateMRRofNimonicalloy(Co-Ni-Cr)
%AtomicWeightValency
Co18 58.93 2
Ni62 58.71 2
Cr2051.99 6
Solution
I=500A d=8.28gm/cm
3
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.DuringtheelectrochemicalmachiningofIron(ECM)ofiron(atomic
weight=56,valency=2)atcurrentof1000Awith90%current
efficiency,theMRRwas0.26gm/s.Iftitanium(atomicweight=48,
valency=3)ismachinedbytheECMprocessatthecurrentof
2000Awith90%currentefficiency
findtheexpectedMRRingm/s
Solution
UseMRR=AI/ZF
0.26=56*900/F*2
F=96923
NowforTitanium
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.CompositionofaNickelsuperalloyisasfollows:
Ni=70.0%,Cr=20.0%,Fe=5.0%andrestTitanium
Calculaterateofdissolutioniftheareaofthetoolis1500mm
2
andacurrentof2000Aisbeingpassedthroughthecell.Assume
dissolutiontotakeplaceatlowest
valencyoftheelements.
A
Ni=58.71ρ
Ni=8.9ν
Ni=2
A
Cr=51.99ρ
Cr=7.19ν
Cr=2
A
Fe=55.85ρ
Fe=7.86ν
Fe=2
A
Ti=47.9ρ
Ti=4.51ν
Ti=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.Determinethe 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
Valencyof iron dissolution,V=2
So, Gram equivqlentweight 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,aMRR of 0.98cc/min has been
observed at 1000amp.Evaluate the valencyat 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 valencybe 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.CalculatetheMRR(m^3/min)inananodicdissolutionprocessof
chromiumifthecurrentdensityavailablehasbeen
250Amp/cm^2.
Assume
At.Wtofchromium,M=52gm
Valencyofchromium,V=2
Densityofchromium,P=7.2gm/cm^3
Solution
Gramequivalentwtofchromium,E=M/V=26
Amountofchromiumdissolved,W=EIT/FP
Thus,MRRincm^3/min=EI/FP
=(26*25)/(1609*7.2)=0.561cm^3/min.

14. 12.Duringelectrochemicalmachiningofironwithacopper
electrodeworkngina5(N)NaClsolninwater,anequillibrium
gapof0.0125cmhasbeenachievedatacurrentdensityof150
Amp/cm^2.Iftheoperatingvoltagehasbeen10V,determinethe
feedrate(f)ofthetool.
Given,
Sp.ResistanceofNaClsolution,p=3ohm-cm
Valencyatwhichironisdissoluted,V=2
Density of iron, P=7.8gm/cm^3
Solution
Weknowthat,
E=M/V=5/2=2.5
AQ,v=10V,h=0.0125cm,F=1609amp.min
Thus,
Feedrate,f
=(v*MRR)/(P*h*I) (Again,MRR/I=E/FP)
=(v*E)/(P*h*F*P)
Substitutingthevaluesintheequationwehave
f=0.020cm/s

15. 13.In a Certain electro chemical dissolution process of iron,ametal
removal rate of 3cm^3/min was desired.Determinethe amount of
current reqfor the process,assuming:
Atomic wtof 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 mm
3
/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 mm
3
/
min
=10 mm
3
/
sec
=10 x10
-3
cc/sec
MRR=AI/Fρv
10 X10
-3
=(56xI)/96500X7.8X2
I =
268.82
amp

17. 15.AresearcherconductsECMonbinaryalloy(density6000kg/m
3
)
ofiron(atomicweight(A1)56,valency(v1)2)andmetalP
(atomicweight(A2)24,valency(v2)4)faradaysconstant96500
coloumbs/mole.Volumetricmaterialremovalrateofthealloyis
50mm
3
/satacurrentof2000A.Whatisthepercentageofthe
metalPinthealloys.
Solution
F=96500 , ρ=6000 kg/m
3
A
1=56 , v
1=2 , MRR=50 mm
3
/sec
A
2=24 , v
2=4 , i=2000A
X
1+X
2=1 ----(1)
MRR=[i/Fρ{(X
1v
1/A
1)+(X
2v
2/A
2)}]
(X
1*2/56)+(X
2*4/24)=2000/6*96500*.05 ----(2)
after solving we get
X
1=75% X
2=25%

18. 16.For ECM of steel which is used as the electrolyte
◦kerosene
◦NaCl
◦Deionisedwater
◦HNO
3
Ans: Naclso 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
◦Al
2O
3
◦Titanium alloy
Ans: Al
2O
3 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