Introduction to Electrical Engineering - Basic vocational knowledge (Institut für Berufliche Entwicklung, 213 p.) |
(introduction...) |
Preface |
1. Importance of Electrical Engineering |
2. Fundamental Quantities of Electrical Engineering |
2.1. Current |
2.2. Voltage |
2.3. Resistance and Conductance |
3. Electric Circuits |
3.1. Basic Circuit |
3.2. Ohm’s Law |
3.3. Branched and Unbranched Circuits |
3.3.1. Branched Circuits |
3.3.2. Unbranched Circuits |
3.3.3. Meshed Circuits |
4. Electrical Energy |
4.1. Energy and Power |
4.2. Efficiency |
4.3. Conversion of Electrical Energy into Heat |
4.4. Conversion of Electrical Energy into Mechanical Energy |
4.5. Conversion of Electrical Energy into Light |
4.5.1. Fundamentals of Illumination Engineering |
4.5.2. Light Sources |
4.5.3. Illuminating Engineering |
4.6. Conversion of Electrical Energy into Chemical Energy and Chemical Energy into Electrical Energy |
5. Magnetic Field |
5.1. Magnetic Phenomena |
5.2. Force Actions in a Magnetic Field |
5.3. Electromagnetic Induction |
5.3.1. The General Law of Induction |
5.3.2. Utilisation of the Phenomena of Induction |
5.3.3. Inductance |
6. Electrical Field |
6.1. Electrical Phenomena in Non-conductors |
6.2. Capacity |
6.2.1. Capacity and Capacitor |
6.2.2. Behaviour of a Capacitor in a Direct Current Circuit |
6.2.3. Types of Capacitors |
7. Alternating Current |
7.1. Importance and Advantages of Alternating Current |
7.2. Characteristics of Alternating Current |
7.3. Resistances in an Alternating Current Circuit |
7.4. Power of Alternating Current |
8. Three-phase Current |
8.1. Generation of Three-phase Current |
8.2. The Rotating Field |
8.3. Interlinking of the Three-phase Current |
8.4. Power of Three-phase Current |
9. Protective Measures in Electrical Installations |
9.1. Danger to Man by Electric Shock |
9.2. Measures for the Protection of Man from Electric Shock |
9.2.1. Protective Insulation |
9.2.2. Extra-low Protective Voltage |
9.2.3. Protective Isolation |
9.2.4. Protective Wire System |
9.2.5. Protective Earthing |
9.2.6. Connection to the Neutral |
9.2.7. Fault-current Protection |
9.3. Checking the Protective Measures |
The electrical energy is used advantageously for the drive of machines, for lighting and other purposes. Here, a study is made to find out how the quantities current, voltage and resistance discussed above can be used to determine the converted electrical energy or the available electrical power. For this purpose we should remember the comparison of the electrical circuit and the circulation of water. When a pump is used to pump water from a vessel arranged at a lower level into a vessel at a higher level, energy must be expended. When the water is allowed to flow from the upper to the lower vessel, the energy expended originally, actually fed into the water, will be released. The magnitude of the energy of the water is dependent on the weight of the water and the difference between the two levels. The following holds
W = G · h
where:
G |
weight |
h |
difference in height |
When comparing the weight G with the quantity of electricity Q and the height with the voltage U, we have for the electrical energy
W = Q · U
Since Q = I · t, we have for the electrical energy
W = U · I · t
The energy stored in the upper water basin can flow down in different times. The energy conversion related to time is called power.
P = W/t = (U · I · t)/t
P = U · I
This is illustrated by an example. When I fill the upper water basin by means of a hand pump, so I can do this during a long period of time without particular effort or in a very short time exerting myself. The shorter the time, the greater the energy I have to expend or, in other words, the higher the power attained.
The unit of power is expressed as
[P] = [U] · [I]
[P] = V · A and from 1V · A = 1W follows
[P] = W
The product V · A is called watt in honour of the English physicist James Watt (1736 - 1819).
Thus, for the unit of the electrical energy we have
[W] = [U] · [I] · [t]
[W] = V · A · s
[W] = W · s
The energy unit Nm used in mechanics and the energy unit J used in heat engineering are of the same magnitude as the Ws
1 Ws = 1 Nm = 1 J
where:
W |
watt |
s |
second |
N |
Newton |
m |
metre |
J |
joule |
Since a watt-second is a very small energy unit and in most cases the operating times of electrical equipment amounts to many hours, the kWh (kilowatt-hour) is also used as energy unit. Thus, we have
1 kWh = 3,600,000 Ws = 3.6 MWs
With the help of Ohm’s law, the equations (4.2.) and (4.3.) can be written in the following form
W = U · I · t = I^{2 · }R^{ · }t_{2} = (U_{2}/R) · t
P = U · I = I^{2 · }R = U^{2}/R
In accordance with the great variety of electrical devices used in practice, the magnitude of the power input ranges from very small values to very great values. Table 4.1. shows some examples.
Table 4.1. Power Input to Selected Electrical Devices
Device |
mean power input |
electronic pocket computer |
µW |
headphone |
mW |
loudspeaker |
5 W |
incandescent lamp |
60 W |
ventilator |
100 W |
motors at machines |
1 kW |
motor of electric locomotive |
1 MW |
power station generator |
100 MW |
Example 4.1.
In the supply line for a soldering iron connected to 220 V, a current of 0.2 A is measured. What is the power input to the soldering iron? What is the amount of energy converted within 8 hours?
Given:
U = 220 V
I = 0.2 A
t = 8 h
To be found:
P in W
W in kWh
Solution:
P = U · I
P = 220 V · 0.2 A
P = 44 W
W = U · I · t
W = 220 V · 0.2 A · 8 h
W = 352 Wh
W = 0.352 kWh
The energy input to the soldering iron is 44 W. Within 8 hours, an energy of 0.552 kWh is converted.
Example 4.2.
An electrical hardening furnace having a resistance of 20 W requires a current input of 5 A. Calculate the electrical energy consumed within a period of 24 h.
Given:
R = 20
I = 5 A
t = 24 h
To be found:
W in kWh
Solution:
W = I^{2} · R · t
W = 5A · 5A · 20W · 24h
W = 12 kWh
Within 24 hours, the hardening furnace consumes 12 kWh of electrical energy.
Example 4.3.
The circuit of a washing machine with the rating-plate markings P = 2 kW and U = 220 V is to be provided with fuses. Find out whether a fuse with the rated current of 6 A will be sufficient.
Given:
P = 2 kW
U = 220 V
To be found:
I in A
Solution:
P = U · I
I = P/U
I = 2000 W/220V
I = 9.09 A
The current input to the washing machine is 9.09 A; therefore, a fuse with the rating of 6 A will not suffice. A fuse having a rated, current of 10 A must be used.
The electrical energy is calculated, according to W = U · I · t and the electrical power according to P = U · I. As unit for the energy, the Ws has been laid down, while the greater unit kWh may be used when required. The unit of power is W. The relation with other energy units is 1 Ws = 1 Nm = 1 J.
Questions and problems:
1. Derive from the relations P = U · I and W = U · I t further formulas, taking the resistance into account.2. Gather information about the price of 1 kWh.
3. How great is the energy in Nm corresponding to one kWh?
4. Using the markings on the rating-plate of various technical devices (P and U), determine the resistance of these devices.
5. What is the current input to a washing machine connected to 220 V having a power input of 2.2 kW?
6. What is the time for which the washing machine mentioned in problem 5 has been operated when the electric meter indicates a consumption of 5 kWh?