5/7/2025

Ministry of Higher Education and Scientific Research

Higher Institute of Engineering, El-Shorouk City

Electrical Power & Machines Eng. Department

Utilization of Electrical Energy

[EPM 481]
4th year Power
Chapter3: Electric Heating
By
Dr. Ahmed Abdelbaset

E-mail: a.abdelbasset@sha.edu.eg
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INTRODUCTION

Heat plays a major role in everyday life. All heating requirements in

domestic purposes such as cooking, room heater, immersion water
heaters, and electric toasters and also in industrial purposes such as
LECTURE 1: welding, melting of metals, tempering, hardening, and drying can be
met easily by electric heating, over the other forms of conventional
heating.
Chapter 3: Electric Heating Heat and electricity are interchangeable. Heat also can be produced
by passing the current through material to be heated. This is called
electric heating; there are various methods of heating a material but
electric heating is considered far superior compared to the heat
produced by coal, oil, and natural gas.

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ADVANTAGES OF ELECTRIC HEATING (vi) High efficiency

In non-electric heating, only 40–60% of heat is utilized but in electric heating 75–
(i) Economical 100% of heat can be successfully utilized. So, overall efficiency of electric heating
They do not require much skilled persons; therefore, maintenance cost is less. is very high.

(ii) Cleanliness (vii) Automatic protection

Protection against over current and over heating can be provided by using fast
Since dust and ash are completely eliminated in the electric heating, it keeps control devices.
surroundings cleanly.
(viii) Heating of non-conducting materials
(iii) Pollution free The heat developed in the non-conducting materials such as wood and porcelain
As there are no flue gases in the electric heating, atmosphere around is is possible only through the electric heating.
pollution free; no need of providing space for their exit. (ix) Better working conditions
(iv) Ease of control No irritating noise is produced with electric heating and also radiating losses are
In this heating, temperature can be controlled and regulated accurately either low.
manually or automatically. (x) Less floor area
Due to the compactness of electric furnace, floor area required is less.
(v) Uniform heating
With electric heating, the substance can be heated uniformly, throughout (xi) High temperature
whether it may be conducting or non-conducting material. High temperature can be obtained by the electric heating except the ability of the
material to withstand the heat.
(xii) Safety The electric heating is quite safe

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MODES OF TRANSFER OF HEAT

The transmission of the heat energy from one body to another because of the
temperature gradient takes place by any of the following methods:
1. conduction,
2. convection, or 2. Convection
3. radiation. The heat transfer takes place from one part to another part of substance or
1. Conduction fluid due to the actual motion of the molecules. The rate of conduction of
The heat transfers from one part of substance to another part without the heat depends mainly on the difference in the fluid density at different
movement in the molecules of substance. The rate of the conduction of heat temperatures.
along the substance depends upon the temperature gradient. Ex: Immersion water heater.
The mount of heat absorbed by the water from heater through convection
depends mainly upon the temperature of heating element and also depends
partly on the position of the heater. Heat dissipation is given by the following
expression.

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3.Radiation
The heat transfers from source to the substance to be heated without ESSENTIAL REQUIREMENTS OF GOOD HEATING ELEMENT
heating the medium in between. It is dependent on surface.
The materials used for heating element should have the following properties:
Ex: Solar heaters.
The rate of heat dissipation through radiation is given by Stefan's Law. •High-specific resistance
Material should have high-specific resistance so that small length of
wire may be required to provide given amount of heat.
• High-melting point
It should have high-melting point so that it can withstand for high
temperature, a small increase in temperature will not destroy the
element.
• Low temperature coefficient of resistance
From above Equation, the radiant heat is proportional to fourth powers
of the temperatures, it is very efficient heating at high temperature.
For accurate temperature control, the variation of resistance with the
operating temperature should be very low. This can be obtained only
if the material has low temperature coefficient of resistance
• Free from oxidation
The element material should not be oxidized when it is subjected to
high temperatures; otherwise the formation of oxidized layers will
From Equation, the radiant heat is proportional to the difference of fourth shorten its life.
power of the temperature, so it is very efficient heating at high temperature.
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DESIGN OF HEATING ELEMENTS

• High-mechanical strength
The material should have high-mechanical strength and should withstand
for mechanical vibrations. By knowing the voltage and electrical energy input, the design of the heating
• Non-corrosive element for an electric furnace is required to determine the size and length
The element should not corrode when exposed to atmosphere or any of the heating element. The wire employed may be circular or rectangular
other chemical fumes. like a ribbon. The ribbon-type heating element permits the use of higher
• Economical wattage per unit area compared to the circular-typeelement.
The cost of material should not be so high.
1. Circular-type heating element
CAUSES OF FAILURE OF HEATING ELEMENTS Initially when the heating element is connected to the supply, the temperature
goes on increasing and finally reaches high temperature.
Heating element may fail due to any one of the following reasons.
1. Formation of hot spots. Let V be the supply voltage of the system and
R be the resistance of the element, then
2. Oxidation of the element and intermittency of operation. electric power input

3. Embrittlement caused by gain growth. If ρ is the resistivity of the element, l is the

length, ‘a’ is the area, and d is the diameter of
4. Contamination and corrosion. the element, then:

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Therefore, power input 1 ∴ Total heat dissipated = surface area × H = Hπdl.

Under thermal equilibrium,

By rearranging the above equation, we get: 2
Power input = heat dissipated
where P is the electrical power input per phase (watt), V is the operating
voltage per phase (volts), R is the resistance of the element (Ω), l is the length
P = H × πdl.
of the element (m), a is the area of cross-section (m*m), d is the diameter of Substituting P from Equation (1) in above equation:
the element (m), and ρ is the specific resistance (Ω-m)

According to Stefan's law, heat dissipated per unit area is

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4
where T1 is the absolute temperature of the element (K), T2 is the absolute
temperature of the charge (K), e is the emissivity, and k is the radiant
efficiency. By solving Equations (2) and (4), the length and diameter of the
The surface area of the circular heating element: wire can be determined.

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2. Ribbon-type element Example 1:

A 4.5-kW, 200-V, and 1-φ resistance oven is to have nichrome wire heating
Let ‘w’ be the width and ‘t’ be the thickness of the ribbon-type heating elements. If the wire temperature is to be 1,000°C and that of the charge
element. for ribbon or rectangular element, a = w × t 500°C. Estimate the diameter and length of the wire. The resistivity of the
nichrome alloy is 42.5 μΩ-m. Assume the radiating efficiency and the
emissivity of the element as 1.0 and 0.9, respectively.

Temperature of the source (T1) = 1,000 + 273 = 1,273 K.

Temperature of the charge T2 = 500 + 273 = 773 K.

5 According to the Stefan's law,

The surface area of the rectangular element (S) = 2 l × w.

∴ Total heat dissipated = H × S = H × 2 lw.
∴ Under the thermal equilibrium,
Electrical power input = heat dissipated
P = H × 2 lw
By solving Equations (5) and (6), the
6 length and width of the heating
element can be determined.

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Example 2: A20-kW, 230-V, and single-phase resistance oven employs

nickel—chrome strip 25-mm thick is used, for its heating elements. If the
wire temperature is not to exceed 1,200°C and the temperature of the
charge is to be 700°C. Calculate the width and length of the wire. Assume
The heat dissipation is given by: the radiating efficiency as 0.6 and emissivity as 0.9. Determine also the
temperature of the wire when the charge is cold.

Let ‘w’ be the width in meters, t be the thickness in meters, ‘l’ be the
length also in meters and A= w*t. Then:

By solving Equations (1) and (2):

Substitute the value of ‘d’ in Equation (2):

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The total amount of the heat dissipation × the surface area of strip = power supplied Example 3: Determine the diameter and length of the wire, if a 17-kW, 220-
P=H×S V, and 1-φ resistance oven employs nickel-chrome wire for its heating
elements. The temperature is not exceeding to 1,100°C and the temperature
= H × 2 lw (S = surface area of strip = 2lw) of the charge is to be 500°C. Assume the radiating efficiency as 0.5 and the
emissivity as 0.9, respectively.

For a circular element:

From Equations (1) and (2):

and
When the charge is cold, it would be at normal temperature, say 25°C.

At steady temperature, crucial power input = heat output:

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Solving Equations (1) and (2), we get:

With my best wishes With my best wishes

Substitute the value of ‘d' in Equation (2) gives:

Dr.Ahmed Abdelbaset

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