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close this bookIntroduction to Electrical Engineering - Basic vocational knowledge (Institut für Berufliche Entwicklung, 213 p.)
close this folder4. Electrical Energy
View the document4.1. Energy and Power
View the document4.2. Efficiency
View the document4.3. Conversion of Electrical Energy into Heat
View the document4.4. Conversion of Electrical Energy into Mechanical Energy
Open this folder and view contents4.5. Conversion of Electrical Energy into Light
View the document4.6. Conversion of Electrical Energy into Chemical Energy and Chemical Energy into Electrical Energy

4.3. Conversion of Electrical Energy into Heat

The electrical energy in a resistor through which current passes (W = I2 · R · t) is converted, into heat energy. According to the law of conservation of energy, the consumed electrical energy is equal to the generated heat energy. This correlation is also clearly described by the energy units of the international system of units SI. For the heat energy, the unit J (joule) 1) is used.

1) Joule, English physicist 1818 - 1889

The following holds

1 Ws = 1 J
1 kWh = 3.6 MJ

The formerly used unit for heat energy, cal (calorie), is no longer permitted; 1 Ws = 0.239 cal.

Since any conductor (wire) has a resistance, in every conductor through which current passes electrical energy is converted in heat energy and, consequently, the conductor is heated. This phenomenon is a disturbing factor in supply lines to electrical equipment and in windings of motors and transformers. Therefore, lines included in an electrical installation should not be subjected to high currents or any value. When the current is too high, the rise in temperature will become excessive so that the insulation may be destroyed or the line catch fire. Depending on the method of the installation of the line, the maximum permissible current intensity for a certain cross-sectional area of the conductor can be drawn from special Tables. Table 4.3. gives some information in this respect.

Table 4.3. Power Bating of Moisture-proof Cables

Cross-sectional area of 2 conductor in mm2

1.5

2.5

4

6

maximum continuous current in A


copper

20

27

36

47


aluminium

-

21

29

37

rated safety current in A


copper

20

25

35

50


aluminium

-

20

25

35

voltage drop per 100 m of distance in V


copper

47.6

38.5

32.2

28


aluminium

-

47.3

41

34.8

For example, a multi-core moisture-proof cable of copper whose individual conductors have a cross-sectional area of 1.5 mm2 may be loaded with a current of up to 20 A. Considerable voltage drops occur which are dependent on the length of the conductor and which may be so detrimental that in many cases the choice of a greater cross-sectional area is necessary. For the above line, a length between feed point and consumer of 100 m would cause the extremely high voltage drop of 47.6 V when loaded with the maximum permanent current.

Example 4.5.

A load having a current input of 15 A is to be connected to a moisture-proof cable of copper of a length of 70 m. The voltage drop should not exceed 20 V. Determine the required cross-sectional area of the line.

Given:

S = 70 m
I = 15 A
U = 20 V
rCu = 0.0178 (mm2 · W)/m

To be found:

A in mm2

Solution:

From the Table, the minimum permissible cross-sectional area of the conductor can be drawn which is A = 1.5 mm2. Check that the voltage drop in the line will not become too great.

R = (r · l)/A
A = (r · l)/R with R = U/I
A = (r · l · I)/U

For the length, twice the distance s must be used taking the outgoing line and the return line into consideration.

A = (0.0178 W · mm2 · 140 m · 15 A)/(m · 20 V)
A = 1.87 mm2

A cross-sectional area of 2.5 mm2 of the conductor must be chosen.

For windings of motors and transformators, the heat dissipation from the winding wires to the environment is worse than in installation lines. Therefore, the individual wire cross-sections should be loaded with an adequately lower current than installation lines. For this purpose, there are Tables, too, from which the required conductor cross-sectional area can be drawn and in which frequently the maximum permissible current density S is stated.

S = I/A;
[S] = A/mm2

(4.6)

where:

S

current density

I

current intensity

A

cross-sectional area of the conductor

In screwed, plugged or clamped connections high transition resistances occur due to insufficient contact pressure or dirty surfaces, these connections will be subjected to particularly high rises in temperature which may lead to the destruction of the connection or the line may catch fire.

Therefore, great care must be taken when preparing such connections and one must see that the contact resistance is kept low. All connections have to be checked at regular intervals and when an impermissible rise in temperature is observed, the cause must be removed immediately.

Besides the given examples of an undesired development of heat, in many cases the heating effect of the electrical current is ingeniously used. In order to protect lines and equipment from currents of impermissible high values, various types of safety devices are used. As to the safety fuse (Fig. 4.2.), a particularly thin wire is used inside the fuse which is heated up to the melting point when the rated current of the fuse is reached.


Fig. 4.2. Sectional view of a fuse

1 - Fusible wire
2 - Ceramic body
3 - Springy clamping plate
4 - Quartz sand filling
5 - Foot contact
6 - Head contact

As a consequence, the circuit is interrupted. The springy small identification plate is detached and thus shows that the fuse is blown. As to the bimetal safety device, a strip of two different metals on which a resistance winding is arranged, through which the current to be controlled flows, is used. In the case of heating, the bimetallic strip shown in Fig. 4.3. is bent and, consequently, the circuit is interrupted by means of a switching mechanis not shown in this illustration. The heated strip is bent because the metals used expand to different degrees when subjected to heat. When, for example, the lower metal in Fig. 4.3. will bend to a higher degree than the upper one, the strip will bend upwards.


Fig. 4.3. Bimetal safety device

1 - Bimetallic strip
2 - Contact
3 - Carrying body

In heat-generating electrical devices usually a spiral of resistance wire is embedded in a ceramic carrier body. In this way, heating cartridges of different shapes are made which are used for electrical soldering irons, boiling plates, immersion heaters and heating inserts for other heat-generating electrical appliances.

Another important use of electric heat is in the process of spot welding (Fig. 4.4.). Two electrodes of copper properly press the parts of sheet metal together. Then current is switc&ed on which heats the joint area to such an extend that the sheet metal parts are welded together. The great heat involved in an electric arc is utilised for technical purposes. Thus, in properly closed vessels a high-grade melt of a metal can be prepared (Fig. 4.5.) or arc welding can be carried out, a process which is widely used today.


Fig. 4.4. Spot welding

1 - Copper electrodes
2 - Sheet-metal parts to be welded


Fig. 4.5. Arc furnace

1 - Electrodes
2 - Melt
3 - Electric arc
4 - Melting crucible
5 - Vessel

Example 4.6.

In a thermal storage water heater, 5 1 of water (this equal to a mass m of about 5 kg) with a temperature of 20 °C is to be heated, up to the boiling point. The heating coil has a power of 1000 W. For energy conversion, an efficiency of 0.9 is assumed. The specific heat of the water c is drawn from a table. After which time will the water boil after switching on?

Given:

V = 5 l
m = 5 kg
J1 = 20 °C
J2 = 100 °C
P = 1000 W
h = 0.9
c = 4.19 kJ/(K · kg)

To be found:

t in min

Solution:

Starting point of the solution is the fact that the required heat energy must be equal to the electrical energy converted into heat which corresponds to the supplied electrical energy multiplied by the efficiency.

WQ = We1 · h

With WQ 3 m c (J2 - J1) and Wel = P · t we have
m · c (J2 - J1) = P · t · h

Now the equation is inverted with respect to the quantity to be found, namely,

t = [m · c (J2 - J1)]/(P · h)
t = [5 kg · 4.19 kJ · (100 °C - 20 °C)]/(K · kg · 1 · kW · 0.9)
t = (5 · 4.19 Ws · 80 K)/(K · 1 W · 0.9)
t = 1862 s
t = 31 min

After 31 minutes the water will boil in the thermal storage water heater.

In each resistor through which current passes, electrical energy is converted, into heat energy. The rise in temperature involved, is not desired, in all electrical installations and windings of motors and transformers. Advantage is taken of this heat in all heat-generating devices (soldering iron, boiling plate, hardening furnace, melting furnace) and in the process of electric welding. The action of safety devices frequently is based on the heating when current passes through them.

Questions and problems:

1. Describe further examples of the utilisation of the heat effect of electrical current!

2. Why are fuses or other safety devices required in circuits?

3. Describe and substantiate the consequence of excessively high contact resistances!

4. Explain the mode of action of the fuse and the bimetallic safety device!

5. Calculate the voltage drop in a 50 m long lead of copper cores having a cross-sectional area of 1.5 mm2 to a load with a current input of 15 A!

6. By means of an immersion heater of 1000 W, 51 of water are heated from an initial temperature of 20 °C. What is the temperature of the water after 15 min when an efficiency of 0.9 is assumed for the energy conversion?