Heating Elements

Design of heating element Let P= Electrical input R= resistance of elements ,Ω ; ρ = Resistivity of element ,Ω/m and mm2 ;d= Diameter of wire, mm; l = length of element,m Heat Input Dissipating surface of the wire = π dl X 10-3m2 Heat input of wire surface qrad = qrad = Heat radiante per unit surface of the wire = 5.7 X 10-8 e ŋ (T12- T24) W/m2 Here the term ŋ has been include to take into consideration the effective value of emissivity . We have ŋ = Radiating efficiency = 1 for single element and the value may go down to 0.5 for multiple element units And e = 0.9 for heating elements

From () and () we have

But

Using equation () and () the length and diameter of wire can be calculated Ex 1. A 250 V , 1.0 kW single element electric furnace is to employ a nichrome resistance wire operating at 10000c. Estimate a suitable diameter and length of wire. Take radiating efficiency = 1, emissivity = 0.9 and the resistivity of wire = 0.424 Ω mt at 10000 C. The ambient temp in 200 C. Sol:- Heat radiated qrad = 5.7 X 10-8 e ŋ (T12- T24) W/m2 = 5.7 X 10-8 X 0.9 X 1 [(1000+273)- (273+20)] =135 X 103 W/m2

Total heat radiated 1000 Or Resistance R But

= qrad (π dl X 10-3) = 135 X 10-3 (π dl X 10-3) = 424 dl

dl = 2.31 --------------------------------------------------------(1) = V2/ P = (2502/1000) = 62.5 Ω R= = 62.5 Ω -------------------------------------(2)

So , solving (1) & (2) we have , diameter of the wire d = 0.272 mm and length of wire l = 8.5 m.