# Design of Heating Element

Power SystemsUtilization of Electrical EnergyTransmission of Electric PowerDistribution of Electric Power

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The device which receives electrical energy as input and converts it into the heat energy is known as a heating element.

The wire employed for the heating element may be circular or rectangular. By determining the electrical input and its voltage, the size (or diameter) and length of the wire required as the heating element to produce the given temperature can be calculated.

## Design of Heating Element

Consider the heating element is in the circular shape. Therefore, the length and diameter of the heating element can be calculated as follows − According to the Stefan's law, the heat dissipated is given by,

$$\mathrm{\mathit{H}\:=\:5.72\:\mathit{Ke}\mathrm{\left [ \mathrm{\left ( \frac{\mathrm{T_{1}}}{100} \right )^{\mathrm{4}}}\:-\:\mathrm{\left ( \frac{\mathrm{T_{2}}}{100} \right )^{\mathrm{4}}} \right ]}\:\mathrm{Watt/m^{\mathrm{2}}}\:\:\:\cdot \cdot \cdot \mathrm{\left ( 1 \right )}}$$

$$\mathrm{\mathrm{Input\: electric\: power,}\mathit{P}\:=\:\frac{\mathit{V^{\mathrm{2}}}}{\mathit{R}}}$$

Where,

• V is the supply voltage, and

• R is the resistance of heating element

$$\mathrm{\because \mathit{R}\:=\:\frac{\rho \mathit{l}}{\mathit{a}}\:\mathrm{and}\:\mathit{a}\:=\:\frac{\pi \mathit{d^{\mathrm{2}}}}{\mathrm{4}}}$$

Where, 'ρ' is the resistivity of the wire material, 'l' is the length of the wire and 'd' is the diameter of the circular wire.

$$\mathrm{\therefore \mathit{P}\:=\:\frac{\mathit{V^{\mathrm{2}}\mathit{a}}}{\rho \mathit{l}}\:=\:\frac{\pi \mathit{d^{\mathrm{2}}\mathit{V^{\mathrm{2}}}}}{\mathrm{4\rho \mathit{l}}}}$$

If

$$\mathrm{\mathrm{Surface\: area\: of\: element}\:=\:\pi \:\times \mathit{d}\:\times \mathit{l}\:\mathit{m^{\mathrm{2}}}}$$

Thus, the heat input per m2 of the surface area is

$$\mathrm{\mathrm{Heat\: input\: per \:m^{\mathrm{2}}}\:=\:\frac{\pi \mathit{d^{\mathrm{2}}\mathit{V^{\mathrm{2}}}}}{\mathrm{4\rho \mathit{l}}}\:\times \:\frac{\mathrm{1}}{\pi \:\times \mathit{d}\:\times \mathit{l}}\:=\:\frac{\mathit{V^{\mathrm{2}}\mathit{d}}}{\mathrm{4\rho \mathit{l^{\mathrm{2}}}}}\:\:\:\cdot \cdot \cdot \mathrm{\left ( 2 \right )}}$$

The temperature of the heating element at the initial stage will raise gradually but after certain time it becomes constant. At this instant, the heat dissipated from the surface of the heating element is equal to input of the electrical energy. Therefore, equating eqns. (1) and (2), we get,

$$\mathrm{\mathit{H}\:=\:\frac{\mathit{V^{\mathrm{2}}\mathit{d}}}{\mathrm{4\rho \mathit{l^{\mathrm{2}}}}}}$$

$$\mathrm{\Rightarrow \frac{\mathit{l^{\mathrm{2}}}}{\mathit{d}}\:=\:\frac{\mathit{V^{\mathrm{2}}}}{\mathrm{4}\rho \mathit{H}}\:\:\:\cdot \cdot \cdot \mathrm{\left ( 3 \right )}}$$

$$\mathrm{\because \mathit{P}\:=\:\frac{\mathit{V^{\mathrm{2}}}}{\mathit{R}}\:\mathrm{and}\:\mathit{R}\:=\:\frac{\rho \mathit{l}}{\mathit{a}}}$$

$$\mathrm{\therefore \frac{\mathit{V^{\mathrm{2}}}}{\mathit{P}}\:=\:\mathit{R}\:=\:\frac{\rho \mathit{l}}{\mathrm{\left ( \pi \mathit{d^{\mathrm{2}}/\mathrm{4}} \right )}}}$$

$$\mathrm{\Rightarrow \frac{\mathit{V^{\mathrm{2}}}}{\mathit{P}}\:=\:\frac{\mathrm{4\rho \mathit{l}}}{\pi \mathit{d^{\mathrm{2}}}}}$$

$$\mathrm{\therefore \frac{\mathrm{1}}{\mathit{d^{\mathrm{2}}}}\:=\:\frac{\pi \mathit{V^{\mathrm{2}}}}{4\rho\mathit{P}}\:\:\:\cdot \cdot \cdot \mathrm{\left ( 4 \right )}}$$

Hence, for the given voltage and power, the length and diameter of the heating element can be determined by using the eqns. (3) and (4).