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UPRVUNL JE EE 2015 Official Paper

Option 1 : Reduce the length of heating element

ST 1: General Knowledge

5609

20 Questions
20 Marks
20 Mins

**Design of Heating Element:**

Wires of circular cross-section or rectangular conducting ribbons are used as heating elements.

Under steady-state conditions, a heating element dissipates as much heat from its surface as it receives the power from the electric supply.

If P is the power input and H is the heat dissipated by radiation,

then P = H under steady-state conditions.

As per Stefan’s law of radiation, heat radiated by a hot body is given by

\(H = 5.72eK\left[ {{{\left( {\frac{{{T_1}}}{{100}}} \right)}^4} - {{\left( {\frac{{{T_2}}}{{100}}} \right)}^4}} \right]W/{m^2}\;\) …. (1)

where T_{1} is the temperature of the hot body in K and T_{2} that of the cold body (or cold surroundings) in K.

We know that,

\(P = \frac{{{V^2}}}{R}\) …. (2)

And, \(R = \rho \frac{l}{A}\) …. (3)

Where l is the length of material, A is the area of cross-section, and ρ is the resistivity of the material.

\(A = \frac{{\pi {d^2}}}{4}\) …. (4)

From equation (2), (3), and (4),

\(P = \frac{{{V^2}}}{{\left( {\frac{{4\rho l}}{{\pi {d^2}}}} \right)}} = \frac{{\pi {V^2}}}{{4l\rho }}\) …. (5)

The total surface area of the wire of the element = (πd) × l

If H is the heat dissipated by radiation per second per unit surface area of the wire then,

Heat radiated per second (P) = (πd) × l × H …. (6)

From equation (5) and (6),

\(\frac{{\pi {V^2}}}{{4l\rho }} = \pi d \times l \times H\)

\(\frac{d}{{{l^2}}} = \frac{{4\rho H}}{{{V^2}}}\) …. (7)

From equation (7) it is clear that that the resistivity (ρ) of the material is inversely proportional to the square of the length of material.

Hence, highly resistive material is chosen for the heating element in order to reduce the length of the heating element.