Extract information from graph below and solve for power spe

Extract information from graph below and solve for power, specific heat capacity, latent heat of fusion, and conservation of energy. Consider the data given in Figure 1 which shows graphically the optimum rates at which lasers of different powers cut through sheets of mild-steel. This figure is rich in qualitative as well as quantitative information. This information can easily be understood if you read the graph in a systematic manner SHEET THICKNESS(m 26810 12 14 400 10 160 120 280 240 O 160 120 1000 80 500 W 0 02 0.3 0.5 0.6 SHEET THICKNESS (in) NGZZLE 0.045 in. 0.051. 0.06\". NCREASING A. RYGEN PRESS 2- DECKEASING PRESSURE Source: J. Powell, CO2 Laser Cutting, to be published in Eng- lish by Carl Hanser Verlag, Munich, F.R.G, in 1990 Figure 1. Cutting speeds for sheets of mild steel using CO2 laser.

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%.

 Extract information from graph below and solve for power, specific heat capacity, latent heat of fusion, and conservation of energy. Consider the data given in

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