. A workpiece in the form of a bar 100 mm in diameter is to be turned down to 70 mm
diameter for 50 mm of its length. A roughing cut using maximum power and a depth
of cut of 12 mm is to be followed by a finishing cut using a feed of 0.1 mm and a
cutting speed of 1.5 m/s.
It takes 20 s to load and unload the workpiece and 30 s to set the cutting
conditions, set the tool at the beginning of the cut and engage the feed.
The specific cutting energy for the material is 2.3 GJ/m3
, and the lathe has a
3-kW motor and a 70% efficiency. Estimate
1. The machining time for the rough cut
2. The machining time for the finish cut
3. The total production time for each workpiece
Answers
Answer:
I think the answer is machining time for the rough cut
make me brain list
Concept:
The power needed by the machine = (Cutting Energy x MRR) /Efficiency
Speed of machine can be defined as
Speed, N = (Kv)/(πD)
where K is constant, v is cutting speed and D is the initial diameter after the rough cut.
Total production time can be defined as the sum of Loading time, upload time, Rough cut Time, and finishing time.
Given:
Diameter of workpiece = 100mm
Reduced diameter = 70 mm
Length = 50 mm = 0.05 m
Cutting speed = 1.5 m/s
Loading time = 20 sec
Unload time = 30 sec
Specific cutting energy = 2.3 GJ/m³ = 2.3 × 10⁹ J/m³
Motor power = 3kW = 3 × 10³ W
Efficiency = 70% = 0.7
Final cutting depth. = 100 - 12 = 88 mm
Find:
The machining time for the rough cut, the machining time for the finish cuT, and the total production time for each workpiece.
Solution:
Power needed by machine, P = (Cutting Energy × MRR) /Efficiency
3 × 10³ = (2.3 × 10⁹) × MRR/ 0.7
MRR = (3×10³ × 0.7) / (2.3 × 10⁹)
MRR = 0.913×10⁻⁶ m³/s
MRR = volume removed / cutting time
So, cutting time = volume removed / MRR
cutting time = (π(D₁²-D₂²)L)/ (0.913×10⁻⁶ m³/s)
D₂ is cutting depth.
cutting time = (π(0.1²-0.088²)0.05)/ (0.913×10⁻⁶ m³/s)
Cutting time = 388 sec
Again cutting down,
cutting time = (π(D₁²-D₂²)L)/ (0.913×10⁻⁶ m³/s)
D₂ is cutting depth.
cutting time = (π(0.088²-0.076²)0.05)/ (0.913×10⁻⁶ m³/s)
Cutting time = 338.59 sec
No further depth is cut because a further cut requires a more amount of material is required.
So, Required Time for cut = 388.8 sec + 338.59 sec = 727.39 sec
(b) Speed, N = (Kv)/(πD)
N = (60000 × 1.5)/(76π) = 377
CT = (L + A)/(f×N)
CT = (50 + 0)/(0.15×377) = 0.884 min
CT = 53 sec
So, the time required for finishing the cut is 53 sec.
(c) Total production time (Loading time + upload time + Rough cut Time + finishing time)
Total = 20 + 30 + [388.8 + 338.59] + 53 = 830.4 sec
Hence, the machining time for the rough cut is 727.39 sec, the machining time for the finish cut is 53 sec and the total production time for each workpiece is 830.4 sec.
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