- The size and length of heating element for given temperature is calculated by knowing electrical input and input voltage.
- The heating element may be circular or rectangular in shape.
- The rectangle shape is generally used in toasters, ovens, vulcanizers etc.
- As the heat will be dissipated from the heating element at higher temperature, it is assume that all the heat is dissipated by radiation.

### Circular Shape

Let V be the supply voltage and R be the resistance of heating element. The Electrical input
voltage P = V

^{2}/ R………..( 1 )
Where R is the
resistance of heating element and V is the supply voltage

Resistance R =
ρL / a

= ρL / ( πd

^{2}/ 4 ) ( circular heating element )
= 4ρL / ( πd

^{2})……………….( 2 )
From equation ( 1
) and ( 2 )

Electrical input
power P = V

^{2}/ [ 4ρL / ( πd^{2}) ]
= ( πd

^{2 }V^{2}) / 4ρL ………( 3 )**OR**

L / d

^{2}= π^{ }V^{2}/ 4ρP …..…………( 4 )
Where

P = Electrical
input power / phase

V = Operating
voltage / phase

R = Resistance
of the heating element in ohm

ρ = Resistivity
of the heating element in ohm – meter

L = Length of
heating element in meter

d = Diameter of
circular heating element in meter

### Heat dissipated per unit area according to Stephen’s law

H = 5.72 ke[ ( T

_{1}/100 )^{4}– ( T_{2}/100 )^{4}] ………… ( 5 )
Where

T

_{1}= Absolute temperature of heating element in Kelvin
T

_{2}= Absolute temperature of charge in Kelvin
e = Emissivity
and

k = Radiant
efficiency

The surface area
of circular heating element S = πdL

Total heat
dissipated = πdL × H…………….( 6 )

**At steady state temperature**

Power input =
heat dissipated

P = πdL × H [ From equation ( 1
) and ( 6 ) ]

( πd

^{2 }V^{2}) / 4ρL = πdLH [ From equation ( 3 ) ]
d / L

^{2}= 4ρH / V^{2}………………..( 7 )
The length and
diameter of heating element is find out by solving equation ( 4 ) and ( 7 ).

**Ribbon ( Rectangular ) type element**

Let w be the
width and t be thickness of heating element.

Power input P =
V

^{2}/ R
= V

^{2}/ ( ρL / a )
= V

^{2}/ ( ρL / wt ) ( area = wt )
= V

^{2}wt / ρL**OR**

L / w = V

^{2 }t / ρP……….( 8 )
The surface area
of rectangular element S = 2Lw

Total heat
dissipated = 2Lw × H

At steady state
temperature

Power input =
heat dissipated

P = 2wLH

Lw = P /2H………….(
9 )

The length and
width of heating element can find out by solving equation ( 8 ) and ( 9 ).

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