Extract information from graph below and solve for power spe
Solution
Solution:
1. Optimum cutting speed : From figure 1, for 0.25\" sheet thickness and 1500 W CO2 laser, optimum cutting speed is approximately 70\"/min.
2. r = 2.5 ft. So, perimeter = 2*3.14157*r= 31.416 ft = 377\".
Light energy delivered = (377/50)*1500*60 =678600 kJ
3. Mass of the metal that melted out = 377\"x2.54x0.2x0.25x2.54x7.86 = 955.87 gm.
4. Energy spent to raise the temperature to the melting point = mCpdT = (955.87/1000)x620x(1515-20) =886 kJ
5. Energy absorbed by the metal surface = 10% of the laser power = 678600x10% = 6786 kJ.
6. Fraction of the laser light spent in raising the temperature of the metal to its melting point & melting it
= 100x886/6786 = 13.06%
7. Energy spent for melting = Latent heat of fusion x mass of the metal = 247 x (955.87/1000) = 236.1 kJ
So, energy dispersed by conduction, convection and radiation = 6786 - (886 + 236.1) = 5664 kJ.
Fraction that dispersed = (5664/6786)x100 = 83.46%
So, the efficiency of the CO2 laser = 100 - 83.46 = 16.54%.

