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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+// A Texbook on POWER SYSTEM ENGINEERING
+// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
+// DHANPAT RAI & Co.
+// SECOND EDITION
+
+// PART IV : UTILIZATION AND TRACTION
+// CHAPTER 2: HEATING AND WELDING
+
+// EXAMPLE : 2.1 :
+// Page number 724-725
+clear ; clc ; close ; // Clear the work space and console
+
+// Given data
+P = 15.0*10**3 // Power supplied(W)
+V = 220.0 // Voltage(V)
+T_w = 1000.0 // Temperature of wire(°C)
+T_c = 600.0 // Temperature of charges(°C)
+k = 0.6 // Radiatting efficiency
+e = 0.9 // Emissivity
+
+// Calculations
+rho = 1.016/10**6 // Specific resistance(ohm-m)
+d_square = 4*rho*P/(%pi*V**2) // d^2 in terms of l
+T_1 = T_w+273 // Absolute temperature(°C)
+T_2 = T_c+273 // Absolute temperature(°C)
+H = 5.72*10**4*k*e*((T_1/1000)**4-(T_2/1000)**4) // Heat produced(watts/sq.m)
+dl = P/(%pi*H)
+l = (dl**2/d_square)**(1.0/3) // Length of wire(m)
+d = dl/l // Diameter of wire(m)
+T_2_cold = 20.0+273 // Absolute temperature at the 20°C normal temperature(°C)
+T_1_cold = (H/(5.72*10**4*k*e)+(T_2_cold/1000)**4)**(1.0/4)*1000 // Absolute temperature when charge is cold(°C)
+T_1_c = T_1_cold-273 // Temperature when charge is cold(°C)
+
+// Results
+disp("PART IV - EXAMPLE : 2.1 : SOLUTION :-")
+printf("\nDiameter of the wire, d = %.3f cm", d*100)
+printf("\nLength of the wire, l = %.2f m", l)
+printf("\nTemperature of the wire when charge is cold, T_1 = %.f°C absolute = %.f°C \n", T_1_cold,T_1_c)
+printf("\nNOTE: Slight changes in the obtained answer from that of textbook is due to more precision here")