Heating Elements

Design of heating element
Let P= Electrical input
R= resistance of elements ,Ω ; ρ = Resistivity of element ,Ω/m and mm
2
;d= Diameter of wire, mm;
l = length of element,m
Heat Input













Dissipating surface of the wire = π dl X 10
-3
m
2

Heat input of wire surface q
rad
=




















q
rad
= Heat radiante per unit surface of the wire
= 5.7 X 10
-8
e ŋ (T
1
2
- T
2
4
) W/m
2

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 1000
0
c. 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 1000
0
C.
The ambient temp in 20
0
C.
Sol:- Heat radiated q
rad
= 5.7 X 10
-8
e ŋ (T
1
2
- T
2
4
) W/m
2

= 5.7 X 10
-8
X 0.9 X 1 [(1000+273)- (273+20)]
=135 X 10
3
W/m
2

Total heat radiated = q
rad
(π dl X 10
-3
) = 135 X 10
-3
(π dl X 10
-3
)
1000 = 424 dl
Or dl = 2.31 --------------------------------------------------------(1)
Resistance R = V
2
/ P = (250
2
/1000) = 62.5 Ω
But 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.