Materials Science and Manufacturing Engineering Miscellaneous
- During the electrochemical machining (ECM) of iron (atomic weight = 56, valency = 2) at current of 1000 A with 90% current efficiency, the material removal rate was observed to be 0.26 cc/s. If Titanium (atomic weight = 48, valency = 3) is machined by the ECM process at the current of 2000 A with 90% current efficiency, the expected material removal rate in gm/s will be
-
View Hint View Answer Discuss in Forum
MRR = AI ρZF 0.26 = 56 × 1000 × 0.9 ρ × 2 × 96500
ρ = 0.9956.Now, Expected MRR = 48 × 2000 × 0.9 0.9956 × 3 × 96500
= 0.30 gm/sCorrect Option: C
MRR = AI ρZF 0.26 = 56 × 1000 × 0.9 ρ × 2 × 96500
ρ = 0.9956.Now, Expected MRR = 48 × 2000 × 0.9 0.9956 × 3 × 96500
= 0.30 gm/s
- Match the following non-traditional machining processes with the corresponding material removal mechanisms:
Machining process
P. Chemical machining
Q. Electro-chemical machining
R. Electro-discharge machining
S. Ultrasonic machining
Mechanism of material removal
1. Erosion
2. Corrosive reaction
3. lon displacement
4. Fusion and vaporization
-
View Hint View Answer Discuss in Forum
P – 2, Q – 3, R – 4, S – 1
Correct Option: A
P – 2, Q – 3, R – 4, S – 1
- In abrasive jet machining, as the distance between the nozzle tip and the work surface increases, the material removal rate
-
View Hint View Answer Discuss in Forum
In abrasive jet machining the variation is as shown
MRR – Metal removal rate
NTD – Nozzle tip distance.Correct Option: D
In abrasive jet machining the variation is as shown
MRR – Metal removal rate
NTD – Nozzle tip distance.
- Electrochemical machining is performed to remove material from an iron surface of 20 mm × 20 mm under the following conditions:
Inter electrode gap = 0.2 mm
Supply voltage (DC) = 12 V
Specific resistance of electrolyte = 2 ohm- cm
Atomic weight of iron = 55.85
Valency of Iron = 2
Faraday's constant = 96540 Coulombs
The material removal rate (in g/s) is
-
View Hint View Answer Discuss in Forum
We know, R = ρl A
where, ρ = resistivity
= special resistance of electrolyte
= 2Ω
R =resistance of electrolyte between electrodes
l = interelectrode gap = 0.2mm
A = electrodes cross-sectional area = 20 × 20 mm2∴ R = (2 × 10) × 0.2 = 0.01Ω 20 × 20 ∴ Current, I = V = 12 = 1200 A R 0.01 Faraday of electricity, E = M v
where, E = equivalent weight
M = molecular weight
v = valencyM.R.R (gm/sec) = I × M F v = 1200 × 55.85 = 0.3471 96540 2 Correct Option: A
We know, R = ρl A
where, ρ = resistivity
= special resistance of electrolyte
= 2Ω
R =resistance of electrolyte between electrodes
l = interelectrode gap = 0.2mm
A = electrodes cross-sectional area = 20 × 20 mm2∴ R = (2 × 10) × 0.2 = 0.01Ω 20 × 20 ∴ Current, I = V = 12 = 1200 A R 0.01 Faraday of electricity, E = M v
where, E = equivalent weight
M = molecular weight
v = valencyM.R.R (gm/sec) = I × M F v = 1200 × 55.85 = 0.3471 96540 2
- A researcher conducts electrochemical machining (ECM) on a binary alloy (density 6000 kg/m3) of iron (atomic weight 56, valency 2) and metal P (atomic weight 24, valency 4). Faraday's constant = 96500 coulomb/mole. Volumetric material removal rate of the alloy is 50 mm3/s at a current of 2000 A. The percentage of the metal P in the alloy is closest to
-
View Hint View Answer Discuss in Forum
Given data:
ρ = 6000 kg /m3 = 6 gm/cc.
MRR (Q) = 50 mm3 /s
50 × 10–3 CC/SMRR = eI ρZF 50 × 10–3 = e × 2000 96500 × 6
e = 14.475
and for Iron Atomic weight
AFE = 56, Valency: rFe = 2
For metal P, at weight
(At)p = 24, Valency rp = 4
Now, let% of P in alloy = x.1 = x × VP + 100 − x × VFt e 100 AP 100 AFet 1 = x × 4 + 2 
100 − x 
14.75 100 27 56 100
x = 24.48 ≈ 25Correct Option: B
Given data:
ρ = 6000 kg /m3 = 6 gm/cc.
MRR (Q) = 50 mm3 /s
50 × 10–3 CC/SMRR = eI ρZF 50 × 10–3 = e × 2000 96500 × 6
e = 14.475
and for Iron Atomic weight
AFE = 56, Valency: rFe = 2
For metal P, at weight
(At)p = 24, Valency rp = 4
Now, let% of P in alloy = x.1 = x × VP + 100 − x × VFt e 100 AP 100 AFet 1 = x × 4 + 2 100 − x 14.75 100 27 56 100
x = 24.48 ≈ 25
