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DESIGN AND CONSTRUCTION OF AN INDUCTION FURNACE (COOLING SYSTEM)

DESIGN AND CONSTRUCTION OF AN INDUCTION FURNACE (COOLING SYSTEM). Presented by MG THANT ZIN WIN Roll No: Ph.D-M-7. Supervisors : Dr Mi Sanda Mon Daw Khin War Oo.

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DESIGN AND CONSTRUCTION OF AN INDUCTION FURNACE (COOLING SYSTEM)

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  1. DESIGN AND CONSTRUCTION OF AN INDUCTION FURNACE(COOLING SYSTEM) Presented by MG THANT ZIN WIN Roll No: Ph.D-M-7 Supervisors : Dr Mi Sanda Mon Daw Khin War Oo 19th Seminar 27.10.2004

  2. Calculating Heat Transfer Rate of the Side of Induction Coil By substituting the desire values in the above equation, where, kA = 32 W/mC at 230 kB = 0.166 W/mC at 51C, kC = 12.604 W/mC at 27C, Ts,1 = 1600C Ts,4 = 74C from practical measuring data, L1 = 0.46228 m r1 = 0.12 m r2 = 0.193336 m r3 = 0.194836 m r4 = 0.1973336 m

  3. Calculating Heat Generation Rate Inside the Induction Coil due to the Electrical Resistance Heating In R value at 20˚C, 0.003183 Ω In R value at 60˚C, 0.00365455 Ω where, α = temperature coefficient for copper = 0.004 Ω/Ω/˚C Heat generation rate is

  4. Total Heat Transfer Rate Total heat transfer rate = Heat transfer rate passing through the refractory lining + Heat generation rate due to the electrical resistance heating = (53.6997 + 13.1929) KW = 66.8926 KW

  5. Flow velocity, Finally, Flow velocity of induction coil = 2.7321 m/s #

  6. Defining the Volume Flow Rate of Each Branches Q4 Q2 Q5 Q7 Q3 Capacitor Bank QT Q1 Q9 Q6 Q11 Control Panel Q8 Q10 Cooling Tank Fig – Pipe line connections of Unido Inducton Furnace

  7. From equation (1) & (2) From continuity equation, By substituting the relationship of V1 and V2 in the continuity equation, V1 and V2 is known So, Q1 = 0.099078416 Q2 = 0.590921584 First approximation or calculate Q1 & Q2 again by replacing the pervious formulas. Q1 = 0.0986022294 # So, Q2 = 0.591397706 #

  8. Solve Each Branches using Microsoft Excel • QT = 0.69 m3/min • Q1 = 0.099078416 m3/min • Q2 = 0.590921584 m3/min • Q3 = 0.349195157 m3/min • Q4 = 0.241726427 m3/min • Q5 = 0.022667796 m3/min • Q6 = 0.219058631 m3/min • Q7 = 0.034942438 m3/min • Q8 = 0.184116193 m3/min • Q9 = 0.018227606 m3/min • Q10 = 0.165888587 m3/min • Q11 = 0.165888587 m3/min Induction Coil

  9. Volume Flow Rate of Induction Coil • Q5 = 0.022667796 m3/min for furnace No1. • Q9 = 0.018227606 m3/min for furnace No2. Flow velocity from calculating result is Flow velocity from heat transfer result is So, its velocity V5 and V4 values are sufficient.

  10. THANK YOU

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