Made By

Mr. AVINASH JURIANI

M.tech-Manufacturing

14MT000354

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

2. Electrochemical machining is performed to remove material froman 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

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

4.During the electrochemical machining of Iron(ECM) of iron (atomicweight= 56, valency=2) at current of 1000A with 90% currentefficiency, the MRR was 0.26gm/s.If titanium (atomic weight= 48,valency=3) is machined by the ECM process at the current of2000A 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

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

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

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

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

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/ccSolution

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.

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.

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.

12.During electrochemical machining of iron with a copperelectrode workng in a 5(N) NaCl soln in water ,an equillibriumgap of 0.0125cm has been achieved at a current density of 150Amp/cm^2.If the operating voltage has been 10V,determine thefeed rate(f) of the tool.

Given,Sp. Resistance of NaCl solution ,p=3ohm-cmValency at which iron is dissoluted,V=2Density 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.minThus,

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 havef=0.020 cm/s

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.

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

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%

16.For ECM of steel which is used as the electrolyte

◦ kerosene

◦ NaCl

◦ Deionised water

◦ HNO3

Ans: Nacl so as to create anodic dissolution

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

18.ECM cannot be undertaken for

◦ Steel

◦ Nickel based superalloy

◦ Al2O3

◦ Titanium alloy

Ans: Al2O3 as it is ceramic

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