Shell Electrical Engineer Handbook

8/10/2019 Shell Electrical Engineer Handbook

1/16

10.06.24 - 010

1

Theory

Electrical engineering

Basic concepts in electrical engineering

We cannot imagine our present society without electricity. We only have to look

around to list a number of applications such as: radio, television, vacuum

cleaner, coffee-maker etc. The number of applications in which electricity is

used is still growing; for a case in point, we only have to think of computers.

The development of 'electricity' has taken place mainly over the past 100 years.

A big advantage of electricity is that it can be generated in a central location and

conveyed with relatively small losses. To keep these losses low, electrical

energy must be conveyed at a high voltage. For large distances, voltages of up to

380 kV (380,000 Volts) are used nowadays.

On the consumers' side, these high voltages are not suitable for things such as to

power machines. These voltages will have to be reduced. This can be done

easily by means of transformers, but for this purpose only alternating current can

be used. Direct voltages cannot be converted to a lower or higher voltage by

means of a transformer.

Apart from the conveyance of energy, electricity can also be used for the transfer

of information: radio, television and the telephone are examples of this. Here,

the objective is to minimise any distortion of the data in transfer.

In contrast to the conveyance of electricity as a source of energy, in which only

one frequency is used (50 or 60 Hertz depending on the country), with electricity

as a data carrier, a wide range of frequencies is used. These versatile

applications of electricity are based on a number of laws. A number of these

laws and concepts from electrical engineering are examined in the following

lessons.

Contents of the lesson

1 Electrical quantities: current, resistance and voltage

2 Ohm's law

3 Specific resistance, line resistance

4 Voltage loss across line resistance

5 Electrical power, unit of power

The copyright in this material is vested in Shell Global Solutions International B.V., The Hague, The Netherlands and Shell Netherlands Raffinaderij B.V. All rightsreserved. Neither the whole or any part of this document may be reproduced, stored in any retrieval system or transmitted in any form by any means (electronic,mechanical, reprographic, recording or otherwise) without the prior written consent of the copyright owner.


2/16

Theory / 10.06.24 - 010

2

Lesson

1. Electrical quantities: current, resistance andvoltage

1.1 Electric current, unit of current

An electric current in a conductor is a flow of electrons due to a potential

difference. This potential difference can be supplied by a battery, a generator or

the mains.

By the currentIwe mean the charge Qpassing through a cross section per unit

of time.

The unit of current is the ampere. This is the current in the case of a charge of

1 coulomb per second passing a cross section, so 1 A = 1 C/s (1 coulomb is the

electric charge of 6.3 * 1018

electrons).

Other indications of current are:

1 A (micro-ampere) = 0.000001 A or 1 * 10-6 A

1 mA (milli-ampere) = 0.001 A or 1 * 10-3

A

1 kA (kilo-ampere) = 1000 A or 1 * 103A

Current can be measured with an ammeter.The symbol of an ammeter in a circuit is a circle containing a capital A (see

figure 1).

5575-010-001

Figure 1

Symbol of ammeter

1.2 Resistance, unit of resistance

A good conductor conducts an electric current easily; in other words, a good

conductor offers little resistance.

The resistance of a certain conductor depends not only on the material of the

conductor but also on the dimensions of the conductor.

The unit of resistance is the ohm (). This is the resistance of a column of

mercury with a height of 106.3 cm and a cross section of 1 mm2at 0 C.

A column of mercury twice as long has a resistance of 2 . A column of

mercury with a height of 106.3 cm and a cross section of 3 mm2has a resistance

of 1/3 .

- current

- current

- ampere

- resistance

- ohm


3/16

Theory / 10.06.24 - 010

3

In engineering, resistors are produced varying from very small values to very

large ones, for instance 0.1 to 10 M(mega-ohm). (M = mega = 1,000,000

= 106).

The symbol of a resistorR (R= resistance) in a circuit is a rectangle containing acapital R, if necessary with an index, for instanceR1, R2, etc. (see figure 2).

5578-010-002

Figure 2

Symbol of resistance

1.3 Voltage, unit of voltage

An electric current is caused by a voltage. The higher the voltage, the bigger the

electric current through a conductor at a certain resistance.

This can be compared with the flow of a river. A river flows as a result of a

difference in altitude. This could be referred to as the 'potential difference' of the

river flow. The bigger the difference in altitude, the larger the flow of the river

through a certain river bed. A wide, deep river bed has a low flow resistance. A

narrow, shallow river bed has a high flow resistance.

The unit of voltage is the volt (V). This is the voltage that must be applied across

a resistance of 1 to obtain a current of 1 A in that wire.

Voltages can be measured with a voltmeter. The symbol of a voltmeter in acircuit is a circle containing a capital V (see figure 3).

5578-010-003

Figure 3

Symbol of voltmeter

The symbol of a (direct) voltage supply in a circuit is a long line (+ pole) and a

short line (- pole) at right angles to the conductor (see figure 4).

5578-010-004

Figure 4

Symbol of direct voltage supply

- voltage

- volt


4/16

Theory / 10.06.24 - 010

4

2. Ohm's law

From the above, it is clear that the currentIdepends on:

- the voltage V;- the resistanceR.

The current flows from a point with a high voltage (+) to a point with a lower

voltage (-). This can again be compared with the flow of a river, which

invariably flows from a point at a higher altitude to a point at a lower altitude

(see figure 5).

5578-010-005

Figure 5

Electrical circuit with one voltage supply and one resistor

Measurements show that the currentIis directly proportional to the voltage V

and inversely proportional to the resistanceR, i.e.:

- if Vbecomes twice as big,I also becomes twice as big;

- ifRbecomes twice as big,Ibecomes twice as small.

The equation is:

R

VI=

Then:

V = I * R or I

VR=

In which:

I = current in ampere (A)

V = voltage in volt (V)

R = resistance in ohm ()

This is Ohm's law. Once two of the three quantities in the equation are know, the

third quantity can be calculated.

- Ohm's law


5/16

Theory / 10.06.24 - 010

5

Example 1

The filament in a light bulb has a resistance of 6 . The light bulb is connected

to a 12 V battery.

Calculate the current through the light bulb.

Answer

5578-010-006

Figure 6

Circuit diagram for example 1

The current through the light bulb isR

VI= =

!

12V= 2 A

Example 2

An electrical circuit contains a 3.8 kresistor. Through it flows a current of

2.5 mA.

Calculate the voltage across this resistor.

Answer

Figure 7

Circuit diagram for example 2

First we convert the units of the given quantities:

3.8 kbecomes 3800 or 3.8 * 103

2.5 mA becomes 0.0025 A or 2.5 * 10-3A


6/16

Theory / 10.06.24 - 010

6

The equation is:

V = I * R

consequently:

V = 3800 * 0.0025

V = 9.5 V

or:

V = 3.8 * 103* 2.5 * 10-3

V = 9.5 V

Question 1

The mains voltage is 220 V. Calculate the resistance of a light bulb if the current

flowing through it is 0.27 A.

3. Specific resistance, line resistance

If a voltage supply is connected between the ends of a conductor, a current will

flow through this wire. The magnitude of the current depends on the voltage

applied and the resistance (according to V = I * R).

When we measure the current and the voltage (see figure 8), the resistance of the

wire can be calculated.

5578-010-008

Figure 8

Circuit for the determination of a resistance

R = resistance of the conductor

V = voltmeter

A = ammeter

The resistance of a conductor is determined by:

- the material of the conductor;- the length of the conductor;- the area or cross section of the conductor.


7/16

Theory / 10.06.24 - 010

7

The specific resistance () of a material is the resistance of a 1 m wire of this

material with a cross section of 1 mm2. The of copper is 0.0175 mm

2/m, i.e.

a copper wire 1 m in length and with a cross section of 1 mm2has a resistance of

0.0175 .

Table 1 gives the specific resistance of some materials.

Table 1

Specific resistances of different materials

material specific resistance (mm2/m)

aluminium 0.03

gold 0.022

copper 0.0175

mercury 0.941

lead 0.21brass 0.065

silver 0.016

tungsten 0.045

iron 0.12

carbon 100 - 1000

glass 1013

- 1016

rubber 1010

- 1015

If the length of a wire is doubled, the resistance is also doubled, as this comes

down to connecting two identical resistors in series.

ConclusionThe resistance of a wire is directly proportional to the length of the wire.

If the area of a cross section is doubled, the resistance is halved.

Conclusion

The resistance of a wire is inversely proportional to the area of the cross section

of the wire.

In addition, the resistance of a wire depends on the specific resistance. This is a

material constant.

The equation is:

A

lR *= or R * A =* l

In which:

R = resistance in

l = length in m

A = area (cross section) in mm2

= specific resistance in mm2/m

- specific resistance


8/16

Theory / 10.06.24 - 010

8

If 3 of the 4 quantities in this equation are known, the 4th quantity can be

calculated. Remember to use the right units.

ExampleCalculate the resistance of a 2-core copper cable with a length of 1.14 km. Each

core has a cross section of 2.5 mm2.

Answer

The total length of the cores is 2 * 1140 = 2280 m (one supply core and one

return core).

R * A =* l R =A

l*

R =5.2

2280*0175.0

R = 16

Question 2

If the resistance of a cable with a cross section of 1.2 mm2is 2.8 , calculate the

resistance of a cable of identical length with a cross section of 4 mm2.

Question 3

The resistance of a piece of wire with a diameter of 1.5 mm is 4.2 .

Calculate the resistance of a wire of identical length in the same material with a

diameter of 2.5 mm.

4. Voltage loss across line resistance

Let a device be connected to a voltage supply V with a resistance Rb(load

resistance) via a long line with resistanceR1(line resistance), divided into two

parts by the supply and return line (see figure 9).

5578-010-009

Figure 9

Line resistance in series with load resistance

The current now flows through both resistancesR1and Rb. A voltage lossis

consequently generated acrossR1, leaving a lower voltage forRb.

- load resistance

- line resistance


9/16

Theory / 10.06.24 - 010

9

Example 1

Given V = 12 V,R1= 2andRb= 18, calculate the voltage loss across the

wire resistance and the residual voltage acrossRb.

Answer

V = I * RtotV = I * (R1+Rb)

20

12

182

12

b1=

+=

+=

RR

VI = 0.6A

Voltage loss acrossR1is:

VR1= I * R1VR1= 0.6 * 2 = 1.2 V

Voltage acrossRbis:VRb= 12 - 1.2 = 10.8 V

Example 2

A 2-core supply cable 100 m in length with a core cross section of 4 mm2causes

a voltage loss of 7 V, with an absorbed current of 8 A.

Calculate the specific resistance.

Answer

The wire resistance:

V = I * R1

R1= =87

IV

The total length of the core is 2 * 100 = 200 m

The specific resistance is then calculated as follows:

R * A =* l

200

4*8

7

=

Which can also be written as:

200

4*

8

7= = 0.0175 mm

2/m (copper)

Question 4

A motor is located at a distance of 55 m of the supply point, where the voltage is

220 V. The copper connecting cable has a cross section of 2 * 2 mm2. The

motor absorbs a current of 9.4 A. Calculate:

a. the resistance of each core;

b. the voltage loss in both cores;

c. the voltage across the motor terminals.


10/16


11/16

Theory / 10.06.24 - 010

11

In figure 10d, finally, as a result of increasing the voltage, the current has also

doubled. The power will now be 2 * 2 = 4 times as high (compare figures 10a

and 10d).

Question 5

What will the power be if in figure 10d, the battery voltage is increased to 4 V?

We can conclude that the power loss in for instance a cable increases

proportionally to the square of the increase in current. The heat will then not be

able to be removed fast enough, leading to a further increase in temperature and

the possibility of the insulation melting or burning.

Example

A two-core cable has a length of 25 meters. The cross section of the copper

cores is 10 mm2. The current flowing through the cores is 50 A. What is the

voltage loss and the power loss in the cable?

Answer

To be able to calculate this voltage loss, we first have to calculate the resistance

of the cable:

R * A =* l

=== 0875.010

50*0175.0*

A

lR

The voltage loss now is:

V = I * R = 50 * 0.0875 = 4.375 V

The power lost in the cable is:

P = V * I = 4.375 * 50 = 218.75 W

From these equations and examples, a number of conclusions can be drawn,

which are of importance in for instance the use of welding cables:

- If a welding cable is made longer, the resistance increases proportionally.

For an identical welding current, the voltage loss in the cable will increase.The voltage at the end of the cable (on the welding tongs) will therefore be

reduced.

- If the welding cable used for a particular welding current is too thin, itsresistance will be too high. The welding current will cause an excessive

voltage loss in the cable, which in turn will reduce the welding voltage. In

addition, due to the excessive current, the cable can overheat and burn.

Question 6

Through a welding cable a 100 A current flows. This current is increased to 200

A.

a. What can we say about the power loss in this cable?

b. What will happen to this power?

- rolled up

condition

- welding cables


12/16

Theory / 10.06.24 - 010

12

Summary

Indicated by the letter Expressed in the following unitVoltage V volt (V)

Current I ampere (A)

Resistance R ohm ()

Power P watt (W)

- Ohm's law:

V = I * R orR

VI= or

I

VR=

- The power in an electrical circuit depends both on the voltage and on the

current:

P = V * I orP = I2* R or

R

VP

2

=

- The specific resistance is the resistance of a wire with a length of 1 meter

and a cross section of 1 mm2. The letter used to indicate the specific

resistance is the Greek letter rho (). The unit of is mm2/m.

The resistance of a wire can be calculated using the following equation:

R * A =* l orA

lR *=


13/16

Theory / 10.06.24 - 010

13

Test

ExercisesDo not send in your answers for correction

1. The current through an electric shaver is 0.02 A if the appliance is connectedto the mains voltage of 220 V.

Calculate the resistance of the appliance.

2. The heating element of a soldering iron has a resistance of 550 .

Calculate the current through the element when the soldering iron is

connected to the mains voltage of 220 V.

3. A bicycle light marked as 8 V / 0.4 A is connected to a dynamo which at acertain speed of rotation generates a voltage of only 4 V.

Calculate the current at this voltage.

4. The resistance of a 80 cm length of wire with a diameter of 0.8 mm is 0.6 .Calculate the wire resistance of a 2-core cable in the same material with a

length of 6 km and a diameter of 2.5 mm.

5. Verify the definition of the ohm by means of a resistance calculation.

6. The total resistance of a 2-core copper cable is 15 ; the cross section of

each core is 4 mm2. Calculate the length of the cable.

7. A 12 V car light has a power of 48 Watt. What will be the current throughthe light? Calculate the resistance of the light.

Answers to the questions in the lesson

1.27.0

220==

I

VR = 814.8

2. The resistance is inversely proportional to the cross section, so 8.2*42.1=R

= 0.84 .

3. The cross section of a wire is 2

4DA

= (D= core diameter). The resistance

is inversely proportional to the cross section, so

2.4*

5.2*4

5.1*4

2

2

=R = 1.512


14/16

Theory / 10.06.24 - 010

14

4.

5578-010-011

a. The resistance of one core: (R1) is:

5.2

55*0175.0* ==

A

lR =0.385

b. The voltage loss in one core is:

V = I * R = 9.4 * 0.385 = 3.619 V

In the two cores the voltage loss therefore is 2 * 3.619 V = 7.238 V

c. The residual voltage across the terminals is therefore 220 - 7.238 =212.762 V

5. P = V * IIf V = 4 V, according to Ohm's law, I = 4 A. The power P will then be:

P = V * I = 4 * 4 = 16 W

6. a. P = I 2* R

If the current in the cable is doubled, the power loss is quadrupled.

b. The power loss is converted to heat.

Answers to the exercises

1.02.0

220==

I

UR = 11,000

2.550

220==

R

UI = 0.4 A

3. From the given 8 V / 0.4 A, the resistance of the light bulb can be

calculated:

4.0

8==

I

VR = 20

At a voltage of 4 V, the current is:

20

4

== R

V

I = 0.2 A


15/16

Theory / 10.06.24 - 010

15

4. From the given resistance, the specific resistance of the wire material can be

calculated:

R * A =* l =l

DR

lAR

2

4*

*

= (lin meters!)

=8.0

8.04

*6.0 2

= 0.377 mm2/m

Now the wire resistance can be calculated, in which l= 12,000 m (2 wires!):

25.2*4

000,12*377.0*

==A

lR = 921.6

5 A

l

R *=

1 = column of mercury with a height of 106.3 cm and a cross section of

1 mm2.

1

063.1*941.0=R = 1 (QED)

6. Resistance of 1 core is therefore2

15= 7.5

R * A =* l

ARl

*=

=0175.0

4*5.7 = 1714 m

7. P = V * I 12

48==

V

PI = 4 A

4

12==

I

VR = 3

Problems and assignments

Answer and send in for correction

1. An immersion pump is earthed by means of a length of iron wire. The lengthof the iron wire is 6 meters and the cross section is 4 mm

2. What is the

resistance of the wire?

2. An electrical appliance has a power of 2200 Watt.

a. What is the current absorbed from the mains if the mains voltage is 220

V?

b. What is the resistance of this appliance?


16/16

Documents

Shell Electrical Engineer Handbook