Electric Refe Booj

Electrical, Electronics Engineering Department

Solve problems in complex polyphase power circuits Volume 1 of 1 UEUNEEG049B Version No Date Refer to:

1 12/2008 DMcR

2 06/2009 DK

3 05/10 DK

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Chisholm Institute of TAFE Dandenong Campus Stud Road DANDENONG 3175 Tel: +61 3 9212 5200 Fax: +61 3 9212 5232

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS

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Required Skills and Knowledge

E2.8.9.3 Polyphase power circuit analysis Evidence shall show an understanding of polyphase circuit analysis to an extent indicated by the following aspects: a) Complex power 1∅ and 3∅ b) Balanced, unbalanced 3∅ c) Impedance of 3∅ loads d) Measurement in 3∅ circuits e) Line voltage drops f) Neutral current T2.11.49 Polyphase circuit analysis Evidence shall show an understanding of polyphase systems and its application towards the calculation of circuits’ conditions to an extent indicated by the following aspects: • Structure of a three phase system encompassing: Components Sequence of phases Balanced and unbalanced load conditions • Calculations of phase and line voltages for a specified phase sequence • Calculation of voltage, current, power, power factor in a three phase system encompassing: Balanced three phase systems Unbalanced delta connected loads Unbalanced four-wire star connected loads Unbalanced three-wire star connected loads Unbalanced four-wire star connected systems • Calculation of the neutral displacement voltage in unbalanced three-wire star connected systems • Selection and connection of meters to confirm calculations

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Table of contents Required Skills and Knowledge .............................................................................................3 Unit 1 – Complex Power Waveforms .....................................................................................7 Introduction .................................................................................................................7 Revision of basic ac theory .........................................................................................8 Objectives .................................................................................................................11 Power measurement fundamentals........................................................................... 11 Complex power waveforms ....................................................................................... 12 Student Exercise 1.................................................................................................... 15 Inductive circuit ......................................................................................................... 17 Student Exercise 2.................................................................................................... 19 Capacitive circuit....................................................................................................... 19 Student Exercise 3.................................................................................................... 21 Resistive-inductive circuits and resistive-capacitive circuits ...................................... 22 Student Exercise 4.................................................................................................... 26 Student Exercise 5.................................................................................................... 28 Review ......................................................................................................................29 Power factor and power wave frequency .................................................................. 30 Complex power components..................................................................................... 31 Reactive power ......................................................................................................... 32 Student Exercise 6.................................................................................................... 35 Power triangle ........................................................................................................... 36 Summary ..................................................................................................................39 Calculation of complex power from voltage and current phasors .............................. 41 Student Exercise 7.................................................................................................... 45 Review ......................................................................................................................46 Power measurement fundamentals........................................................................... 47 The wattmeter ........................................................................................................... 48 Student Exercise 8.................................................................................................... 54 High voltage and high current power measurement connections .............................. 55 Student Exercise 9.................................................................................................... 58 Determining the power factor .................................................................................... 59 Student Exercise 10 .................................................................................................. 61 Review ......................................................................................................................62 Check your progress 1 .............................................................................................. 63 Unit 2 - Three Phase Systems ............................................................................................. 65 Introduction ...............................................................................................................65 Objectives ................................................................................................................. 66 Phasor notation......................................................................................................... 67 Student Exercise 1.................................................................................................... 74 Three-phase voltage generation ............................................................................... 75 Student Exercise 2.................................................................................................... 81 Comparison of three-phase and single-phase systems ............................................. 82 Review ......................................................................................................................84 Phase sequence ....................................................................................................... 85 Student Exercise 3.................................................................................................... 88 Dangers of changing phase sequence ...................................................................... 89 Review ......................................................................................................................91 Check your progress 2 .............................................................................................. 92 Glossary of terms...................................................................................................... 93 Formulae ..................................................................................................................94 5


Unit 3 - Three phase four wire star connected balanced loads ............................................ 97 Introduction ...............................................................................................................97 Objectives .................................................................................................................98 Three-phase four wire star-connected balanced loads ............................................... 99 Calculation of currents in star-connected balanced loads......................................... 100 One-line equivalent circuits for star-connected balanced loads .............................. 104 Three-phase three wire star-connected balanced loads.......................................... 105 Student Exercise 1.................................................................................................. 108 Effect of phase reversal on star load currents ......................................................... 109 Student Exercise 2.................................................................................................. 111 Calculation of power in star-connected balanced loads ............................................ 112 Reverse phase sequence ....................................................................................... 114 Student Exercise 3.................................................................................................. 115 Review .................................................................................................................... 116 Three-phase delta-connected balanced loads ........................................................ 117 Calculation of currents in delta-connected balanced loads ........................................ 118 Student Exercise 4.................................................................................................. 125 Effect of phase reversal on delta load currents ....................................................... 126 Student Exercise 5.................................................................................................. 130 Calculation of power in delta-connected balanced loads ........................................... 131 Student Exercise 6.................................................................................................. 133 Review .................................................................................................................... 134 Check your progress 3 ............................................................................................ 135 Glossary of terms.................................................................................................... 136 Formulae ................................................................................................................ 137

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Unit 1 – Complex Power Waveforms Introduction In this section we are going to look at power in single phase alternating current ac circuits and the ways it differs from power in dc circuits. We will develop and use formulae for calculating power in circuits with combinations of the basic components of resistance, capacitance and inductance. From these calculations we will be able to derive the three related concepts of power in ac circuits, that is, real power, reactive power and apparent power. We also examine the use of power triangles and the calculation of power factor, and the principles and calculations of power measurement. The section notes will cover these points in more detail, however, a textbook can be used to great advantage if it has worked examples and examples with answers, which you should look at if you have difficulties with any particular topic. The use of calculus is not necessary for this subject and your section notes will show you how to handle these topics without calculus, and show you the depth to which each topic should be covered. Before we start to understand the concepts of power in ac circuits, you must have a good working knowledge of basic ac theory which you have studied in previous modules. This basic ac theory is also covered in textbooks. Topics of particular importance are: • sinusoidal current and voltage • complex numbers • complex impedance • series and parallel circuits using complex impedances Before commencing this section, try the following ’Revision of basic ac theory’ exercises which will help you revise this work. Note In this module, the symbol ’(t)’ has been used to emphasise the fact that a particular quantity is time dependent. For example, the expression for a voltage may be given as v(t) = 100 sin 314t volts. In some texts, the ’(t)’ symbol is omitted and the expression would simply be shown as v = 100 sin 314t rather than v(t) = 100 sin 314t volts. You should treat the two forms of the expression as being the same. Also, where instantaneous voltages or current are expressed in the form v(t) = Vm sin ( t + ), the phase angle has on some occasions been expressed in degrees for convenience eg. v(t) = 100 sin (314t + 30°). It is important to remember that before calculating instantaneous values that all phase angles expressed in degrees should be converted to radians.

The magnitude of all phasor quantities are expressed in rms

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Revision of basic ac theory 1 The current in the circuit of Figure 1 is i(t) = 2 sin 500 t amperes. Calculate the total applied voltage v(t).

2 A series circuit of two components has the following applied voltage and resulting current: v(t) = 150 cos (200t - 30°) volts i(t) = 4.48 cos (200t - 56.6°) amperes Find the value of the two components of the circuit. 3 Convert the following complex numbers from polar to rectangular form,

4 Convert the following complex numbers from rectangular form to polar form.

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS 5 Solve, giving answers in rectangular form:

6 Three impedances, Z1 = 5 + j5 ohms Z2 = -j8 ohms Z3 = 4 ohms are connected in series to an unknown voltage source V as shown in Figure 2. Find I and V if the voltage drop across Z3 is:

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS 7 A four branch parallel circuit has Z1 = j5 ohms Z2 = 5 + j8.66 ohms Z3 = 15 ohms Z4 = —j10 ohms and a supply voltage of

volts, as shown in Figure 3.

Find the total current from the supply and the equivalent impedance of the circuit.

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Objectives

The following list of objectives is for your guidance when learning the work in this section. Do not be concerned if there are terms or concepts stated here which you do not understand. These will become dear as you progress through the section. At the completion of the section, review these objectives, you should understand them and be able to perform the tasks set out here. • Sketch voltage, current and power waveforms for resistive, capacitive, inductive, resistive-capacitive and resistive-inductive circuits. • Explain and calculate the frequency of the power wave. • Write the equation of the power wave in the time domain form, given the current and voltage expressions. • Resolve the power waveform into its average and alternating components. • Explain the significance of the area under the power waveform. • Explain the meaning of the terms real and average power, reactive power, apparent power and power factor. • Calculate the values of real, reactive and apparent power and power factor for single phase ac circuits. • Construct power triangles for single phase ac circuits.

Power measurement fundamentals • Describe the construction and principle of operation of the electrodynamic instrument. • Calculate the meter constant of a wattmeter when used with voltage and current instrument transformers. • Calculate the power factor of an ac circuit form the readings of the wattmeter, voltmeter and ammeter.

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Complex power waveforms Resistive circuit Consider the circuit in Figure 4

Figure 4 If the voltage v(t) is a time function, then the current in the circuit will be a time function also. As in the case of dc circuits, where the power was equal to the product of the voltage and current in the circuit, so it is with ac circuits except that the voltage and current vary with time. So with ac power, the product of voltage and current at any instant of time is called the instantaneous power and is expressed as p = vi watts where v = instantaneous value of voltage i = instantaneous value of current Depending on the signs of v and i, the value of p may be positive or negative. We consider that if the instantaneous value of p is positive then there is a transfer of energy from the supply to the load. If the instantaneous value of p is negative then there is a transfer of energy from the load to the source.

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Now let us consider the power expression p = v i applied to the circuit in Figure 4 with v(t) = Vm sin volts, where Vm is the maximum value of voltage. The resulting current will be:

where 1m = maximum value of current. Note: Remember expresses the cyclic nature of the sine and cosine wave as an angular velocity in radians per second. Since 2π radians correspond to 1 cycle frequency (f) is linked to angular velocity by the relationship

= 2π f Substituting v(t) and i(t) into p = vi, we have: p(t)

= Vm sin t x Im sin t = Vm Im sin2 t watts

Since sin2x = ½ (1 - cos 2 x),

This expression for p has two terms:

a constant

and a sinusoidal variable Let us look at this expression for instantaneous power in graphic form 13


Figure 5

From Figure 5 we can see that the wave of i(t) is in phase with the wave of v(t) as the load is a pure resistance. The resulting power wave, shown as a broken line, is produced by multiplying the instantaneous values of the current wave and the voltage wave together. When the current and voltage waves are at a maximum, so is the power wave and when the current and voltage waves are negative then their product produces a positive value for the power wave. Thus the power wave in the resistive case always remains positive. Note: You must memorise the explanation and shapes of the waves in Figure 5 to the point where you can redraw the waves and explain their relationship to one another. Now try Student Exercise 1.

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Student Exercise 1 Sketch the curves of v(t) = 10 sin 20t and i(t) = 5 sin 20t on the same axes. Calculate the values of instantaneous power for;

and sketch the corresponding power wave on the same axes.

Average power (resistive circuit) Let us now determine the average value of the instantaneous power in a resistive circuit.

From the expression

• the first term

has a constant value;

• the second term zero.

is a cosine wave having an average value of

The negative sign indicates that it is an inverted cosine wave.

watts.

Therefore the average power in the circuit is

Note: You must memorise this expression for average power for the resistive circuit.

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Example A circuit has a voltage supply of v(t) = 280 sin (314t) connected across a 20 ohm resistance. What is the current in the circuit and the average power? Solution

Average power

= 280x14 2 = 1960 watts

As most ammeters and voltmeters are calibrated to indicate rms (root mean square) values, it is useful to convert the peak values of V and I to rms values to calculate average power.

Since

and

From Ohm's law, Vrms = IrmsR. Therefore: P = I2rmsR watts Now calculate the average power of the above example using rms values for V and I. You should get the same answer.

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Inductive circuit Suppose we replace the resistor in Figure 4 with a pure inductance with the same voltage v(t) = Vm sin t across it. Now the current in the circuit will lag the voltage by 90° or

radians.

The resulting current will have the form:

This result can only be obtained in theory as all practical coils have some resistance. The effect of this resistance will be discussed later. We can now use the same expression for instantaneous power:

Where

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Figure 6 shows, graphically, the expression for instantaneous power for a purely inductive circuit.

Figure 6 Note that: • When the instantaneous value of either v(t) or i(t) is negative then the value of p(t) is negative. • When the instantaneous value of both v(t) and i(t) is negative then the value of p(t) is positive. • When either value of v(t) or i(t) is zero then the value of p(t) is zero. In this case the power wave swings equally in the positive and negative directions. For the half-cycle of the wave that is positive, power is being delivered to the inductance from the supply. For the negative half-cycle the energy stored in the magnetic field is released back to the supply. Thus the net power transfer to the load in the ideal inductive case is zero.

The average value of

is zero.

Note that the value of p(t) is negative. If you look at Figure 6 you will see that the power wave is an inverted sine wave, hence the negative sign. Note: You must memorise the explanation and shapes of the waves in Figure 6 to the point where you can redraw the waves and explain their relationship with one another. Now try Student Exercise 2.

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Student Exercise 2 Sketch the curves of

on

the same axes.

Calculate the values of instantaneous power at the positions:


Capacitive circuit Suppose we replace the resistor in Figure 4 with a capacitor with the same voltage v(t) = Vm sin t volts across it. Then the resulting current in the circuit will lead the voltage by 90° or The expression for the current will be:

lf we use the expression for instantaneous power p = vi, then

Using the proof as in the inductive case, the resulting power equation is:

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radians.

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Figure 7 shows graphically the expression for instantaneous power for the capacitive circuit.

Figure 7 For the capacitive circuit the power wave again swings equally in the positive and negative directions. In the positive half-cycle the capacitor absorbs energy from the supply and in the negative half-cycle the stored energy in the capacitor is released back to the supply. The net power transfer to the capacitor is zero. As before for the inductance, the average value of

Note: You must memorise the explanation and shapes of the waves in Figure 7 to the point where you can redraw the waves and explain their relationship to one another. Now try Student Exercise 3.

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Student Exercise 3 Sketch the curves of

on the same axes.

Calculate the values of instantaneous power at the positions:


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Resistive-inductive circuits and resistive-capacitive circuits The circuits I have discussed have been single element loads, either resistive, inductive or capacitive. Circuit components are usually a combination of the three basic elements. A coil has resistance and inductance and circuits are usually made up of a combination of components, so you must be able to calculate the power in these combined circuits. If a voltage v(t) is connected across a circuit containing resistance and inductance as shown in Figure 8, then the resulting current will lag the voltage by an angle which is more than 0° and less than 90° or

radians.

Figure 8 If the same voltage v(t) is connected across a circuit containing resistance and capacitance, in place of the inductance, then the resulting current will lead the voltage by more than 0° and less than 90°. In either case let us call this angle (theta) and consider it a negative angle for a resistive -inductive circuit and a positive angle for a resistive-capacitive circuit. Then for any circuit combination with a voltage of v(t) = Vm sin terminals, there is a current flow of i(t) = Im sin (

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t+

) amps.

t volts across the

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS For the power in this circuit,

p = vi watts p(t) = Vm sin

t x Im sin (

t+

) watts

This formula reduces down to:

The value of p(t) has two terms:

a constant of

a sinusoidal term of

Note that the second component has a peak to peak value of

which is the same as for the purely resistive, inductive and capacitive circuits. As for the previous circuits, the sinusoidal term has an average value of zero. Therefore the average power in a combination circuit is:

Note: You must memorise the formula

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Figure 9 shows the waveforms of voltage, current and power for a resistive—capacitive = 45°. circuit where

Figure 9 The power wave differs from the previous circuits as it is neither all on the positive side of the horizontal axis nor half above and half below the axis. The position of the wave between these two extremes is determined by the ratio of the reactance, either capacitive or inductive, to the resistance of the circuit. In Figure 9, the angle

is 45° lead and cos 45° = 0.707.

You will recall that the average power in the circuit is:

But

Then,

In this case, since cos

= 0.707,

Then This indicates that the average power is 70.7% of the product of rms voltage and rms current. If you examine Figure 9 you will see that where either the current wave or voltage wave is zero, so too is the power wave.

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS If the angle between I and V becomes greater, then the power wave becomes lower and is less. the value of cos Figure 10 shows the waveforms for wave.

= 60° - note the change in the position of the power

Figure 10 These principles apply to both resistive-inductive and resistive—capacitive circuits. The power wave for such circuits (which are 'passive’ circuits) can only be between the resistive case (above the horizontal axis) and the purely reactive case (half above and half below the horizontal axis). If the power wave had more negative area than positive area, then this would mean the load was supplying average power to the source, which is impossible for a passive load. Note: You must memorise the explanation and shapes of the waves in Figure 9 so that you can redraw the waves and explain their relationship to one another. Since readings taken by meters of ac voltages and currents are rms values, it is more convenient to have the power formula with rms values. Note: You must memorise the formula P = EI cos

using rms values.

This is used for both capacitive and inductive circuits. = cos As cos value of power.

, both positive values and negative values of

However it is important to know the sign of the angle Now try Student Exercise 4.

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will give a positive

for the particular circuit.

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Student Exercise 4 Sketch the curves of v(t) = 10 sin 20t and i(t) = 5 sin (20 -45°) on the same axes. Calculate the values of instantaneous power at the positions of:

and sketch the corresponding power wave on the same axes. Using P = VI cos

, calculate the average power.

Let us now work through the following example. Example Find the power delivered by the supply in the circuit shown in Figure 11.

Figure 11

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Solution Step 1: Convert circuit to ohms impedances.

Figure 12 Step 2: Find the total impedance of the circuit.

That is, the resultant Z has an inductive component. Step 3: From Ohm`s law,

Hence I lags E by 21.4°. Step 4:

Now try Student Exercise 5. 27


Student Exercise 5 1

A circuit has a supply voltage of v(t) = 150 sin i(t) = 10 sin ( t- 30°) amperes

t volts and a current of

What is the average power in the circuit?

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A circuit consists of a 10 ohm resistance in series with a capacitor of -j5 ohms. What is the average power in the circuit if the supply voltage is v(t) = 200 sin314t volts?

3

A circuit has a coil with a resistance of 35 ohms and an inductance of 3 henry. A voltage of v(t) = 100 sin300t volts is connected to the coil. What is the average power taken by the coil?

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Review So far in this section you have covered the topics which have explained the form of complex power. These concepts have been demonstrated through the explanation of power waveforms for the various types and combinations of circuit components. The circuits covered were the resistive, capacitive and inductive circuits and the combinations of resistive—capacitive and resistive—inductive circuits. In each case you have covered the construction of the complex power wave and the formula which represented each wave. You have learnt that in the resistive circuit the power expression is;

which indicates a constant or dc component and an alternating component at twice the supply frequency. By firstly examining the inductive and capacitive cases, which have only alternating components, you finally analysed the most common case of combined circuits of resistance-inductance and resistance- capacitance. These combined circuits produced the most important formula which you will use in ac power calculations:

In the remainder of this section we will be covering the topics of power factor, real (true) power, reactive power and apparent power.

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Power factor and power wave frequency Power factor

In the formula P = VI cos

the term cos

has great significance.

It is called the power factor, abbreviated pf. The angle is the angle between the current and the voltage, and the sign of the angle indicates whether the circuit is inductive with a lagging current or capacitive with a leading current. Note: Memorise the following conditions. • When the angle

is -ve, I lags V, circuit is inductive.

• When the angle

is +ve, I leads V, circuit is capacitive.

Power wave frequency When we derived the formula for the power wave in the resistive circuit, we arrived at the formula:

This formula has a second term of:

This term has a frequency of 20 radians/ second which is twice the supply voltage radians / second. frequency of The same condition exists in the inductive and capacitive circuits. We can say then that the frequency of all ac power waves is twice the frequency of the supply voltage. For example, If the supply frequency was 50 Hz then the power wave frequency would be 100 Hz.

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Complex power components Apparent power In previous subjects you have learnt that power in a dc circuit is the product of V and I. You have now learnt that average power in a resistive ac circuit is the product of the rms values V and I. But since this product is not the average power in all ac circuits, it is referred to as the apparent power. In reactive circuits, the average power is less than the apparent power. The symbol S is used for apparent power and its section is the volt ampere or VA. S = VI volt ampere or VA Since V = IZ then

Real or true power In the section on power in a resistive-inductive circuit we have shown that the average power in an ac circuit is: P = VI cos watts An alternative method of calculation would be: P = I2R watts This average power is also known as the real power or true power, as it is the actual power dissipated in the circuit and is equivalent to the dc circuit power. Real power has the symbol P and its unit is the watt.

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Reactive power The reactive power is that power which circulates between the supply and the reactive component during each positive part of the power wave and returns to the source during each negative part of the wave. The symbol for reactive power is Q and the unit is the volt ampere reactive (VAr). The component of power that is reactive is caused by the reactive component of current whose phasor is at right angles to the resistive component of current. Thus the expression for Q contains Itotal sin , the reactive componentof current. Refer to Figure 13, which illustrates the case of an RC circuit.

Figure 13 Reactive power Q = VI sin

volt amperes reactive

The reactive power is sometimes called a quadrature component as it results from the reactive component of current which leads or lags the voltage by 90°—hence the use of the letter Q as the symbol.

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Figure 14 below shows that the voltage VX across the reactive component in a series RL or RC circuit can be expressed as V sin . Then; Q = VI sin = I X VX = Ix D( (where X is either XL or XC) Q =12X VAr

In this case you must remember that I is the total current.

Figure 14 Example A load whose impedance is

ohms is supplied by a source whose

voltage is Find: (a) the apparent power supplied to load (b) the real power supplied to load (c) the reactive power supplied to load.

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Solution We must find the current in the circuit by Ohm’s law,

(a)

Apparent power S = VI = 300 X 3 = 900 VA

(b)

Real power P = VI cos

Where

= -36.9° P = 300 x 3 x cos -36.9° = 300 x 3 x 0.7997 = 719.7 W

Alternatively, we can find the real power P using the formula P = I2R.

Therefore R = 80 Ω and X = 60 Ω. P = 32 X 80 = 720 W (c)

Reactive power Q = VI sin = 300 x 3 x sin -36.9° = 300 x 3 x 0.600 = -540 VAr

[Neglect the minus sign for the moment, this will be discussed later.] Alternatively, we can find the reactive power Q using the formula Q = I2X. From part (b), X = 60 Ω Q = (3)x60 = 540 VAr [This approach gives only the magnitude of Q and neglects the sign.] 34

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Student Exercise 6 Using the formulae given to you in the last section, 'Complex power components', calculate the answers for the following and check with the results given at the end of the section.

1

Find the apparent power, true power and reactive power when a voltage of volts is supplying a load whose impedance is Z = 10 + j20 ohms.

2

Find the apparent power, true power and reactive power when a current of amperes is measured in a circuit whose impedance is Z = 35 + j25 ohms.

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Power triangle In the previous three sections you have learnt of the three types of power in an ac circuit: • apparent power • true power • reactive power As shown in Figure 14 the voltages across the components cannot be added arithmetically to obtain the total voltage V. Similarly the resulting complex power components cannot be added arithmetically. Firstly let us consider a circuit containing resistance and inductance, as shown in Figure 15. Note that the three power forms are scalar quantities ie. They have magnitude only and are not directional. However, for any particular circuit the relative values for true, reactive and apparent power can be represented on a power triangle.

Figure 15

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS The phasor relationship between the voltage V and the current I is shown in Figure 16 with I lagging V by the angle °.

Figure 16 Let us now take the phasor I and, using V as a 0° reference, draw the horizontal and vertical components as shown in Figure 17.

Figure 17 We will now take the component phasors of Figure 17 and multiply each by the voltage V with the resulting diagram in Figure 18.

Figure 18 37

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS From the previous work we can recognise that the horizontal component VI cos the real power. The vertical component VI sin the apparent power.

is P,

is Q, the reactive power and the inclined phasor VI is S,

This construction for an inductive circuit can be repeated for a capacitive circuit with a leading current and the results are shown in Figure 19.

Figure 19 The reactive power, or quadrature power, in both Figure 18 and 19 is at right angles to the real power.

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Summary You need to memorise the following three formulae. Real power = Voltage X in-phase component of current P = VI cos

watts

Apparent power = Voltage X current S = VI volt amperes

Reactive power = Voltage X quadrature component of current Q = VI sin

volt amperes reactive

Note: The explanation of the power triangle used here is different to that used in the result being that inductive VAr are shown in a downwards direction and capacitive VAr in an upwards direction. A number of texts use this approach. However, other texts use an alternative approach which results in the opposite convention for reactive VAr. (If you take the impedance triangle and multiply all sides by I2 you will see that this indicates that inductive VAr are upwards and capacitive VAr are downwards.) Both conventions are acceptable. The result will be the same whichever convention is used, as long as the same convention is used throughout the problem. For the remainder of this section, and in subsequent sections, I will use the convention that inductive VAr have positive values and are represented in an upward direction in the power triangle. I suggest that you follow the same convention as you work through the sections.

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Now let’s examine the relationships of S, Q and P. We can consider S as the complex power and as it is represented as a phasor, it can be also represented in both rectangular and polar forms.

For inductive load:

For capacitive load:

From Figure 19 we can derive the following relationships:

We can also derive the following two relationships from Figure 19:

Note: You need to memorise the relationships between S, Q and P.

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Calculation of complex power from voltage and current phasors To obtain the relationship between the complex power S and the voltage and current phasors we must consider the voltage phasor and the resulting current phasor caused by the load Z.

For an inductive load:

is positive for a lagging power factor.

where I Z I represents the modulus of Z and

We can represent the current in the load by the phasor I, where ' The voltage phasor across the load is

This is shown in Figure 20.

Figure 20

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS The magnitude of the apparent power, S, is the product of the magnitudes of V and I. The phase angle associated with the value of S is the angle, , between V and I, that is, the difference between the angles GV and Bi in Figure 20. If the current and voltage phasors were multiplied in the normal way, the resultant angle would be the sum of the individual angles, not the difference. Therefore, to obtain the correct result, the sign of the current phase angle is reversed (ie the conjugate, I* , of the current is used). The resultant phase angle for S now becomes the difference between the angles of V and I, that is °. Therefore, using the conjugate of the current in the calculation of S leads to the correct phase angle for S and the correct values for the true power and reactive power components of S. Referring now to Figure 20, the conjugate of the current Phasor,

S = VI* VA

Note: You must memorise this expression. In the expression for the complex power S, a positive imaginary part (+jQ) corresponds to a positive impedance angle °, and represents the reactive power in an inductive load with a lagging power factor. If the imaginary part of the complex power S is negative (-jQ), it corresponds to a negative impedance angle, °, and represents the reactive power in a capacitive load with a leading power factor.

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SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Example A network has an equivalent impedance Z = 3 + j4 and an applied voltage v(t) = 42.5 sin (1000t + 30°). Find: (a) the apparent power, S (b) the real power, P (c) the reactive power, Q (d) the power factor

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Since is a positive angle the current must lag the voltage and the power factor is "lagging". 1

The complex power S used the conjugate of I, thus the current angle used was +23.1°. If we hadn`t used the conjugate of I, then the resulting angle would have been 30° — 23.1° = 6.9°. Since the correct value for impedance angle, answers for P and Q if we used 6.9°.

is 53.1°, we would get the wrong

3 4

The real and reactive powers were obtained directly from the rectangular form of the complex power S.

5

The power factor must indicate whether it is leading or lagging. This is determined by the sign of the angle in the polar form of the complex power S. ln this case the angle is +23.1° so the power factor is lagging. If it had been -23° then the power factor would be leading.

Now try answering Student Exericse 7.

44


Student Exercise 7

45


Review In the second part of this section you have covered the topics of power factor, apparent power, real power and reactive power. These topics will be used extensively in your work in the further study of ac theory. The understanding of power factor and hence therelationships of current with voltage in ac systems will be essential in both power and electronic subjects. Your understanding of the formulae for; apparent power S = VI volt amps real power P = VI cos

watts

reactive power Q = VI sin

VAr

will be important in further sections and other subjects covering ac machines, transformers and power electronics to name a few. You could say that these concepts with respect to ac circuits are as important as Ohm’s law is to dc circuits. At this stage it would be to your advantage to review the objectives at the beginning of this section to ensure you can perform the tasks listed.

46


Power measurement fundamentals Introduction In the previous part you learnt about power in variously configured ac circuits and how to resolve the power into its components of apparent power, true power and reactive power. The concept of power factor was also explained and you learnt how to calculate its value in various circuits. This unit develops ac complex power from the practical side by firstly discussing measurement of complex power using wattmeters, ammeters and voltmeters. You will then learn the application of these meters i.n circuits which have high levels of voltage, current and power as occurs in high-voltage industrial and commercial supply systems. In these larger applications you will learn of the significance of the power factor of installations and power supply systems and how to calculate its value from meter readings. The need to improve efficiency of the power system by correcting the power factor will be explained and you will finally leam how to combine various ac circuits to determine the overall complex power used from the supply. Many texts, which you may have or may be able to borrow from a library, have information on wattmeters and power measurement, but the notes given in this unit should be sufficient for the requirements of this subject. By all means, however, consult: Electric Circuit Analysis by S A Boctor Fundamentals of Electric Circuits by David A Bell.

47


The wattmeter The wattmeter is the instrument used to measure the average power dissipated in a given load. It contains two coils, with a pair of terminals for each coil. One terminal of each of the coils, designated respectively as the voltage and current coils, is marked with a t polarity mark. Figure 21 illustrates the general arrangement of such a wattmeter.

Figure 21:

Electrodynamic wattmeter

These instruments are called electrodynamic or dynamometer instruments. They are similar to a permanent magnet moving-coil instrument, except that the permanent magnet is replaced by a coil which forms an electromagnet. The action of this type of instrument depends upon the magnetic force exerted between the fixed and moving coils carrying current. ln Figure 22 the upper diagram shows a sectional elevation through fixed coils FF and the lower diagram represents a sectional plan on the line XX. The moving coil M is carried by a spindle S and the controlling torque is exerted by spiral hairsprings H, which also serve to lead the current into and out of the moving coil.

48


The deflecting torque is due to the interaction of the magnetic fields produced by the currents in the fixed and unmoving coils. Figure 23 shows the magnetic field due to current flowing through the fixed coils which are arranged with half the turns in each side, to give an even flux in the vicinity of the moving coil. Figure 24 shows the magnetic field due to the current flowing in the moving coil only. By combining these magnetic fields you can see that when currents flow simultaneously through the fixed and moving coils, the resultant magnetic field is distorted as shown in Figure 25 and the effect is to exert a clockwise torque on the moving coil.

49


Since the moving coil is carrying current at right angles to the magnetic field produced by the fixed coils, the deflecting force on each side of the moving coil M is proportional to: (current in M) x (density of magnetic field due to current in F) That is, the force is proportional to: current in M x current in F Thus the deflecting torque is proportional to the mean value of the product of the current in the fixed coils (which represents the current being measured) and the current in the moving coil (which represents the voltage being measured). The fixed coils are connected in series with the load and the moving coil is connected in series with a non-inductive resistor R which is similar to the multiplying resistor in a dc voltmeter. This series circuit is connected across the supply so that the current through the moving coil is proportional to and practically in phase with the supply voltage V. These connections are shown in Figure 26.

50


If we again consider the forces acting on the moving coil, the instantaneous force on each side of the moving coil M is proportional to: • (instantaneous current through F) x (instantaneous current through M); therefore • (instantaneous current through load) x (instantaneous p.d. across load); therefore • instantaneous power taken by load. This means that the average deflecting force on M is proportional to the average value of the power. Note: When the instrument is used in an ac circuit, the moving coil, due to its inertia, takes up a position where the average deflecting torque over one cycle is balanced by the restoring torque of the spiral springs. Hence the instrument can be calibrated to read the average value of the power in an ac circuit. At the beginning of this section on wattmeters, I mentioned that the connections to the meter have been polarity marked. If the incorrect polarity is used between the current coil and the voltage coil, then the magnetic fields will produce a torque in the reverse direction which will force the meter needle in the wrong direction. If this happens when reading power in a single-phase circuit, it is necessary only to reverse the connection to one coil to obtain a reading. Reversing both coil connections will result in the reverse direction of both the fields. In three-phase connections, the polarity of the coils is far more important, and this will be discussed in detail in later units.

51

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Note: For correct connection of the wattmeter, the current must enter the current coil polarity marked t at the same time that the voltage coil polarity marked i is connected to the higher potential point of the load, usually the active lead. General notes on wattmeters • They usually have a good level of accuracy (typically 1%). • Readings are not affected by different wave shapes, as they read average values. • Readings are not affected by frequency of the supply up to 400 Hz. • They are called transfer instruments since they will work satisfactorily on ac and dc supplies. • They can be affected by stray magnetic fields. • They have low sensitivity at low scale readings. • They have to be specially made for low power factor readings below 0.3 pf as these low power factors give inaccurate readings on ordinary wattmeters. The low power factor wattmeter has a weaker control torque (usually provided by a hairspring or a taut band). Electrodynamic type wattmeters are now being replaced by electronic digital meters which have the same input requirements but usually have better flexibility and accuracy than the mechanical types. Precautions When using wattmeters, because the reading is a function of both voltage and current, it is very easy to burn out the coils by overloading them accidentally. Always make sure that the meter voltage range is greater than the supply voltage. If you are reading 300 volt supply, for example, do not use the 250 volt range, use the 500 volt range. Always know what the current is in the circuit you are measuring. Use an ammeter in series with the wattmeter current coil. Do not use a current range less than the current in the circuit. For example, if the ammeter is reading 3 amperes then you must use the 5 amperes range. You can make a mistake if the power factor is low so that, although you may be using the correct current and voltage ranges, the wattmeter reading is low. To make the reading more readable up the scale, you will be tempted to decrease the current or voltage ranges, which will increase the percentage reading but will overload the coils in the instrument. Do not be tempted.

52

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Wattmeter meter constants Most laboratory wattmeters, depending on their manufacturer, have the following current and voltage ranges: • current ranges: 5 A, 2.5 A, 1.25 A • voltage ranges: 500 V, 250 V, 100 V, 50 V, 25 V The scale on the instrument is divided into 100 equal divisions. Note: To interpret the instrument reading, the ranges for voltage and current must be multiplied together to obtain a scale constant, and this must be multiplied by the percentage of scale. On some wattmeters a table of scale constants is presented on the instrument itself, or on the lid of the instrument case. Example A wattmeter has its voltage range set at 250 V and current range set at 2.5 A. The reading on the scale is 67 on the 0-100 scale. Scale constant

= voltage range x current range = 250 x 2.5 = 625

Actual reading

= scale constant x % of full scale reading 67 = 625 x = 418.8 watts

Now try student exercise 8.

53



Meter ranges are 500 volts and 1.25 amperes and the meter reads 35% full scale. What is the actual reading in watts?

2

Meter ranges are 100 volts and 5 amperes and the meter reads 98% full scale. What is the actual reading in watts?

3

Meter ranges are 250 volts and 2.5 amperes and the actual power in the load is 400 watts. What is the percentage scale reading on the meter?

54


High voltage and high current power measurement connections In the previous section the ranges on a common wattmeter were stated. The maximum ranges were 5 amperes and 500 volts. For most laboratory or domestic installations these ranges are adequate, but for commercial and industrial loads or for electricity supply systems, they are totally inadequate. We must extend the ranges of the meters to accommodate the higher levels of voltage and currents used. To make larger coils for higher current and higher voltages is not practical and would be very expensive. It is not economical to extend the range of ac meters using current shunts and voltmeter multipliers. The methods which can be used to extend the ranges are as follows. High voltage measurement 11,000 volts is a popular voltage for distributing power in cities and to large industrial consumers. We use a small transformer with its primary connected to the 11 000 volts and its secondary, which has a voltage of 110 volts, connected to the meter. Provided the ratio of the transformer is taken into consideration when calculating the power, the problem of the high voltage is overcome. The transformer used is called a voltage transformer (VT) or a potential transformer (PT). The term voltage transformer is the name common in modern textbooks, and will be used in these units. The connection diagram showing polarity markings, which must be correct as mentioned in the previous section, is shown in Figure 27.

55

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Note: Most voltage transformers use 110 volts on their secondary in order to standardise meter ratings. This means that the same meter can be used in many different primary voltage applications. High current measurement Most applications require current in excess of 5 amperes, and the device used to extend the range of the 5 A meter is called a current transformer (CT). This transformer differs from the voltage transformer as it has a very few turns (sometimes even only one) on the primary and many turns on the secondary. The ratio of the current transformer is always given as the current ratio. Secondary current outputs depend on the type of current transformer, but are usually 5 ampere or 1 ampere. Primary currents can be from tens of amperes to thousands of amperes, depending on the application. An example of a current transformer would be one having a 500:5 ratio. Again, the ratio must be considered when the power in the circuit is calculated. The connection diagram showing polarity markings, which must be correct as mentioned in the previous section, is shown in Figure 28.

56

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Precaution lf a current transformer is inadvertently disconnected on the secondary side (meter removed) while current is flowing in the primary side (in the main circuit), then a very high voltage (up to thousands of volts) can be generated across the secondary terminals. This creates a serious safety hazard and can damage equipment. The secondary must always be short-circuited to allow current to circulate. The lower the resistance in the secondary circuit, the better is the performance of the current transformer.

Example The wattmeter is connected as shown in Figure 29. The meter has a current range of 5 amperes and a voltage range of 110 volts. The scale is 0-100 and has a reading of 45%.

Now try the exercise in Student Exercise 9.

57


Student Exercise 9

In Figure 30, the wattmeter has a voltage range of 250 volts and a current range of 1 ampere. The scale reading is 27% of full scale. What is the power dissipated in the load?

58


Determining the power factor In the earlier part of this section you learnt how to calculate the power factor of a load either by knowing the complex impedance of the load, that is the resistance and the reactance, or by knowing the phase angle by which the current either lags or leads the voltage. In practice the complex impedance of the load may not be known. Meters which measure phase angles are not common, so we must use another method. In most instances you will have available a voltmeter, an ammeter and a wattmeter, and you can calculate the power factor. from the formula P = VI cos Since P = VI cos

,

Power factor, cos Where:

P = true power = wattmeter reading V = rms voltage = voltmeter reading I = rms current = ammeter reading

Example In Figure 31, the voltmeter reads 240 volts, the ammeter reads 5.4 amperes and the wattmeter reads 960 watts. Determine the power factor of the load.

59

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Note: This result does not indicate whether the power factor is leading or lagging. This information can only be gained by having some knowledge of the type of load. If the load is all induction motors then it can be deduced that the power factor is lagging. lf the load is a mixed load, both capacitive and inductive, then the lead or lag of the power factor cannot be determined directly. In the section ’Power measurement in three phase circuits’, (Section 6), a method for determining the lead or lag in this unknown case is detailed. Now try Student Exercise 10.

60


Student Exercise 10

Figure 32 In Figure 32 the voltmeter reading is 270 volts, the ammeter reading is 8.7 amperes and the wattmeter reading is 1470 watts. What is the power factor of the load?

In Figure 33 the ammeter on the 0-5 ampere scale shows a reading of 80% full scale deflection. The voltmeter on the 110 volt range shows 100% full scale deflection. The wattmeter is set on the 150 volt range and the 5 ampere range and indicates 42% of the full scale deflection. Determine: (a) the power absorbed by the load (b) the load power factor (c) the component values of R and L of the load.

61


Review In the latter part of this section we have covered the construction and use of the wattmeter. It is important to understand the reasons for taking precautions when using the wattmeter so as not to overload either the voltage coil or the current coil. The application of the wattmeter to high levels of current and voltage through the use of current transformers and voltage transformers has been explained. You must be familiar with the circuit connections of the various components for the measurement of power in high current and voltage systems. You must also understand and be able to calculate meter constants using the various ratios so that you can calculate the power on these systems. An important aspect of the use of wattmeters was shown in the calculation of the power factor of a circuit. as it provides a useful You must understand the application of the formula method of calculating power factor from the readings of a voltmeter, ammeter and wattmeter. ln a later section of this module we will be covering the concept of power factor correction, and the calculations involved in the combining of different complex loads into a single total load, to work out the total complex power.

62


Check your progress 1 1 A circuit has an impedance of Z = 10+ j6 Ω and an applied voltage of v(t) = 50 sin 314t volts. Draw the power triangle by determining the values of P, Q, S and the power factor. 2 A circuit consists of R = 10 ohms in parallel with total supply current of 5 amperes rms.

ohms and has a

Find the complete power triangle by determining the values of P, Q, S and the power factor. 3 An impedance carries a current of 18 amperes rms which results in 3500 VA and a power factor of 0.76 lagging. Find the impedance. 4 (a) Determine the total watts, VAr and VA of a system which has the following loads connected: • Load 1: 250 VA at 0.5 pf lagging • Load 2: 180 W at 0.8 pf leading • Load 3: 300 VA and 100 VAr lagging (b) Determine the circuit's overall pf. 5 For the circuit in Figure 34, calculate the total power delivered to the load and the load power factor. • Voltmeter reads 100 V on 110 V range • Ammeter reads 3.8 A on 5 A range • Wattmeter reads 25% on 250 V and 5 A ranges

63


64


Unit 2 - Three Phase Systems Introduction In this and subsequent sections you will be introduced to topics related to three-phase supply systems. In this particular section you will cover the basic concepts of phasor notation which will explain the convention of phasors which will be used in subsequent sections in threephase calculations. The generation of three-phase voltages will be described so that you will understand the advantages of a three-phase supply system. By using phasor diagrams, you will be shown how to construct three-phase voltage phasors for use in later sections. Phase rotation will be detailed and the dangers of reversing or using incorrect rotation with three-phase electrical equipment, particularly three-phase motors, will be outlined. In conclusion, we will examine the types of phase sequence indicators which may be used to check phase rotation to avoid danger to yourself and to equipment. In this section and further sections the letters A, B and C will be used to describe the threephase voltages in calculations. Some texts use the colour code red, white and blue, or R, W and B. It does not matter what notation is used provided the theory is correctly applied when doing calculations.

65


Objectives The following list of objectives is for your guidance when learning the work in this section. Do not be concerned if there are terms or concepts stated here you do not understand. These will become clear as you progress through the section. At the completion of this section, review these objectives—you should be able to understand them and: • discuss the advantages of polyphase over single-phase supply systems; • describe the advantages of three-phase systems in commercial applications (eg motor starting, sizes of three-phase equipment, noise, vibration, cost); • discuss the application of phasor notation and its double subscript notation; • state the convention for positive and negative phase sequence; • describe the standard ABC phase rotation sequence and the effect of changing phase sequence on rotating machinery; • describe the methods of indicating phase sequence, and the use of phase rotation indicator; • explain the dangers of connecting incorrect phase sequence to SCR control equipment and some machine drives such as compressors, fans conveyors, lathes, drills.

66

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Phasor notation Until now you have probably referred to voltages with respect to the component with which they are associated. But voltages always reflect potential differences between any two points or nodes in a circuit. If you have denoted these according to the name of the element as shown in Figure 1, then the voltage would be V1 or VR across R1.

The voltage can also be denoted according to the symbols given to the two points. In the case of Figure 1, VAB is the voltage at point A with respect to B. This indicates that, if VAB is positive, then the potential at A is higher than the potential at B. This is called the double subscript notation. In an ac circuit the notation is applied by assuming that the voltage is in the positive half cycle and the current is flowing from source to load. In some circuits it may be convenient to choose one point in the circuit as a reference point, and refer all voltages to this common reference point. If this point is assigned the reference voltage value of 0 volts, some points in the circuit will have voltages above this point and some below this point. This point is called the earth point. In some textbooks it is called the ground point. In these cases in double subscript notation the second subscript is zero and in some cases is omitted, leaving a single subscript which assumes the reference to be earth or zero potential. In the cases we will be discussing in later sections on three-phase systems, a common reference point is called the neutral which may or may not be connected to earth or zero potential. This will be detailed in later sections. Double subscript notation also gives us a logical system of seeing how voltages sum together. Let us examine Figure 2.

67


If you consider the current is entering point A so that A is considered at higher potential than point D, then using double subscript notation:

This is Kirchhoff s voltage law. To be correct the first letter of the sequence on the left side of the equation must be the first letter on the right side. Also the last letter on the last voltage on the left side must be the last letter on the right side. Note: Because we have assumed that the voltage VAD is positive, as point A is higher than point D, then: V A D = -V D A This applies to any other voltage in this sequence or any voltage indicated by the double subscript notation. Also

where VA and VD are both earth referred. Figure 3 shows the phasor relationship of this concept.

Graphical representation If we wish to solve the first equation by graphical methods or even simply to display the relationships between voltages, there are two methods available: • funicular diagram • polar diagram. Without being given the formal name, you have already used both diagrams before, but the construction and uses of each diagram will now be explained.

68


Funicular diagram This is the diagram you used when adding together the power triangles in Section 1. The basic construction is to take each phasor and place the tail of the next phasor to the head of the previous phasor as in the sequence of Figure 4.

Polar diagram The polar diagram will give the same answer as the funicular diagram except that, because of the different construction, other benefits are available. Taking the same phasors as Figure 4, use the method of the polar diagram in Figure 5. The first stage is to sum the first two phasors using the parallelogram of addition. Using the resultant of that, add it to the third phasor to obtain the resultant. The last diagram shows the complete polar diagram without the construction lines.

69


Advantages of funicular and polar diagrams Let us now examine the differences between the two diagrams to determine the advantages and so be in a position to decide which one to use in a problem. If a resultant value is all that is required as was the case with power triangles, the funicular diagram is simple, clear and has adequate information for the problem. So, if you want a resultant of several phasors, use the funicular diagram. When examining the last diagrams of Figure 4 and Figure 5, you will be able to see a significant advantage of the polar diagram. This advantage is that the relationship of each phasor with any other phasor or the resultant, in both magnitude and angle, is far more obvious than it is with the funicular diagram. At this stage this may not appear very useful but in the next section, when we become involved with three-phase diagrams, the difference will become very obvious. 70

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Note: The convention used to indicate angular displacement is the same in any diagram. We always use positive angles from the horizontal right-hand side of the origin in a counterclockwise direction (see Figure 6).

I Example 1

Solution Draw VAB at Then draw VCD connecting D to B since both C and A are considered as the higher potentials. Then resultant is the diagonal of the parallelogram, shown as VEF in Figure 7.

71


Example 2 If V AN = 240 <0 volts and VBN = 240 <-120 is common to both original voltages.

volts Using a polar diagram find VAB if N

Solution

-VBN may be represented by a phasor which is 180° phase displaced from V NB . The polar summation of VAN – VBN is illustrated in Figure 8.

72

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Example 3 A simple similar example is to find the length between two points, one at 10 metres from a zero point and the other at 5 metres in the opposite direction, or -5 metres. The distance is 10 - (-5) = 15 metres. Now have a look at Figure 9 to see how it is done graphically.

Now try the exercise in Student Exercise 1.

73

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Student Exercise 1 Construct a phasor diagram of the two vectors VCN = 200
74

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Three-phase voltage generation One method of generating three voltages for use in a three-phase system is to use three separate generators coupled to a single prime mover. The prime mover could be a diesel engine, steam turbine or the like, as shown in Figure 10. The generators would be connected to the shaft and mechanically displaced from one another by 120°, so that the voltages are not in phase but spaced evenly around the 360° of rotation.

To provide a common reference point, one side of each generator winding would be earthed. This system could be combined so that the return line is only one conductor as shown in Figure 11.

This method would be expensive and physically large—the normal method to generate three-phase voltage is to have one generator with three separate windings spaced at 120° so that they are balanced around the 360° of rotation. These angles are usually called 'electrical degrees' as this allows for a generator to be a multipole type, meaning that each winding would occupy a smaller physical angle in the generator. One revolution (360°) of a six-pole generator would produce one cycle of voltage for each phase, each one being displaced from its adjacent voltages by 120°.

75


This single generator with three-phase windings is represented diagrammatically in Figure 12.

Star connection If the windings are located at 120° apart around the generator the voltages will be spaced at 120° in their phase relationship as shown in Figure 13.

76


The centre point is called the neutral point and if earthed, is considered at zero voltage. These voltage phasors when displayed as a time function of e = Em sin ( t + ) are shown in Figure 14.

In Figure 14 the wave shapes for each phase are identical but displaced by 120°. If you take a value of at any point along the horizontal axis, the addition of the instantaneous value of the three voltages will equal zero. Consider the phasors in Figure 13: Refer to Section 1 for revision of polar to rectangular conversion:

The connection shown in Figure 12 is called a star connection. Some textbooks use the term 'wye (Y) connection'. In this module the term 'star connection' will be used.

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Phase-to-phase voltages These voltages are sometimes referred to as 'line-to-line voltages' or simply 'line voltages', as they are usually measured on the lines or conductors supplying a load. If we again consider Example 2 detailed in the previous section, with Figure 8 as the method, and reproduce this figure for each phase voltage in a three-phase diagram, we will obtain a result shown in Figure 15.

78

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS The complete diagram of phase-to-neutral and phase-to-phase (line) voltages is shown in Figure 16.

Delta connection The connection we have considered has been the star connection. If we call each phase winding, A to A1 B to B1 and C to C1 in the star connection, we joined together A1 and B1 and C1 to form the star point. If we now connect each phase in turn to the next so that A1 is connected to B, and Bl connected to C, and C1 connected to A, we have a complete loop called a delta as shown in Figure 17.

79

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS If we connect a wire to each point of the delta we have a three-phase supply with the same wave shapes as Figure 15. This system has only phase-to-phase (line) voltages EAB EBC ECA as the phase voltages are equal to the line voltages.

The phase relationship of the line voltages remains the same as the star connection and are shown in Figure 18.

Note: Where a three-phase system is referred to as a particular voltage, for example a 415 V system, the value of voltage referred to is always the phase-to-phase (line) voltage. This applies to star or delta load connections or generator connection. To calculate the phase-to-neutral value you must divide by Vs.

Now try the exercises in Student Exercise 2.

80



Sketch a three-phase supply voltage phasor diagram and then sketch the resulting voltage wave shape with respect to time, indicating the phase differences and letter the individual phases.

2

Sketch a three-phase star-connected generator connection diagram and name all voltages.

3

Sketch a three-phase delta-connected generator connection diagram and name all voltages.

81

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Comparison of three-phase and single-phase systems In the comparison of three-phase and single-phase systems three basic areas will be covered: • generation • distribution • utilisation. Generation Three-phase generators for the same output power use less copper in their conductors and less iron in their magnetic circuits and are hence cheaper than single-phase generators. When a single-phase generator is producing power from its prime mover it can only generate power in pulses. You will recall the power wave for a resistive load in Section 1, particularly in Figure 5. This wave showed a power pulse at twice the supply voltage frequency. This means that the prime mover must supply power to the generator in the same manner. his means that there is considerable stress put on the machinery, shafts and bearings, causing vibration and noise. In the three-phase generator, there is power being generated from all three phases, evenly distributed around the shaft. The result is a continuous flow of power from the prime mover to the generator since total power of the three phases at any instant is constant, so there is much less vibration and noise. Less vibration means less wear on the generator, prime mover and associated equipment. Distribution There is less copper or aluminium used in power lines in a three-phase system to transfer the same amount of power than in a single-phase system. Utilisation Industrial, commercial and domestic customers purchase the power from electricity supply authorities and use it for their own purposes. The first main advantage in the three-phase system is the availability of two voltage levels without the need of transformers. In the normal supply system we have 415 volts between any two phases, or 240 volts between any phase and neutral. This means that in small loads, such as domestic supply to houses, 240 volts is adequate to service the house. 240 volts is considered a reasonably safe voltage if used sensibly, compared with a higher voltage such as 415 volts. At the same time if a factory or shopping centre needs supply, then there is three-phase 415 volts available to operate large machinery and the like. The second main advantage of the three-phase system is the fact that there is a rotating magnetic field available.

82

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS For motors this means that the motor is self-starting. Single-phase motors require auxiliary starting windings and switches which increase their cost and size. The rotating magnetic field, rather than a pulsating magnetic field in the single-phase case, gives a constant torque to the motor. This constant torque means that machinery being driven by the motors does not suffer vibration so it is much smoother in operation and is quieter. For the same shaft output power, a three-phase motor is smaller, more efficient, runs at a better power factor, and is cheaper and quieter than a single-phase motor. Also a three-phase motor usually has a higher starting torque.

83

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Review So far in this section you have covered the topic of phasor notation and its double subscript. This concept is particularly important in calculations related to three-phase systems so that the precise relationships of voltages and currents are known/ and to reduce ambiguity. The use of phasors was developed further with the explanation of funicular and polar diagrams. Three-phase voltage generation followed, with a description of the connection of generators in two basic methods. The first is the star connection where the common ends of the three windings are connected to form a star with 120° spacing between the voltages, resulting in a phase-to-phase voltage of Vs times the phase-to-neutral voltage. The second connection is the delta, where a loop is formed of the windings which leads to a set of three voltages 120° apart, having only phase-to-phase values. The need for three-phase systems was described by emphasising the differences between three-phase systems and single-phase systems. These differences were covered under the headings of generation, distribution and utilisation. In the remainder of this section you will cover the concept of phase sequence and the precautions to be taken when the phase sequence is reversed.

84


Phase sequence You will recall the phasor diagram of Figure 14 where all phase-to-neutral voltages and phase-to-phase voltages were shown. This diagram is reproduced with less construction detail in Figure 19.

In the section 'Phaser notation' where this diagram was explained, no mention was made of the sequence, in which the phasors reached their maximum value. If we consider that the rotation of these phasors is anticlockwise as shown by the arrow, then if you were positioned at the top right-hand corner as indicated, the phase-to-neutral phasors would reach their maximum value in the order of EAN, EBN, ECN or ABC. You will also notice that the phase-to-phase voltages pass in a sequence of EAB, EBC and ECA or AB, BC, CA which has the same repeating relationships as ABC. This sequence by convention, is called the positive phase sequence. Now let us consider the same phasors in Figure 19 but if the prime mover of the generator is rotated in the opposite direction then the arrow would be in the clockwise direction. The phasors would then appear to you as passing in the sequence of E^, ECN and EBN or ACB. This is called the reverse or negative phase sequence. Because the generators on the power system cannot be run in reverse, but the actual phase conductors can be interchanged either by accident or by design, the reverse or negative phase sequence can be achieved by exchanging (transposing) any two phases. You can see this result with the phasors in Figure 20 where EBN and ECN have changed positions from where they were in Figure 19. 85


In Figure 20 the direction of rotation is still anti-clockwise, but because B and C have been interchanged, the phase sequence is EAN, ECN and EBN or ACB. Some textbooks refer to the negative phase sequence as CBA, which is the same sequence as ACB since the series can begin with any letter provided the overall sequence is the same. For example,

Let us now construct the phase-to-phase voltages from the phase-to-neutral voltages shown in Figure 20 with negative phase sequence:

Then EAB = EAN - EBN as for positive phase sequence. But taking the above values for EAN and EBN then

The magnitude of EAB is the same as the positive sequence but the angle is now negative.

86


For the negative phase sequence phase-to-phase voltages, all the angles have changed relative to the positive sequence phase-to-phase voltages. All the negative phase sequence voltages are shown in Figure 21.

Figure 21 These results apply irrespective of which two phases are transposed. You must also note that the phase-to-phase voltages also now have a reverse or negative sequence. When drawing the phasor diagram for phase-to-neutral and phase-to-phase voltages in ABC sequence and CBA sequence, a simple way to remember where the phase-tophase voltage EAB is with respect to E^ is as follows: • In positive phase sequence ABC EAB has a positive angle to EAN (see Figure 19). • In negative phase sequence CBA, EAB has a negative angle to EAN (see Figure 21). Now try the exercise in Student Exercise 3.

87


Student Exercise 3 Draw the phasor diagram of the three-phase voltages if EAN = 300 <0 phase sequence is negative.

and the

Determine the magnitudes and angles of all three phase-to-phase voltages.

88


Dangers of changing phase sequence Motors Most machinery must run in the designed direction of rotation. If you reverse the direction of a drill, the drill bit will not cut the material. A power saw, if run in the reverse direction, will not cut. With most machinery, if the direction is reversed, a significant level of danger exists to the machine itself and the operators, or just people in the area where the machine is in operation. If a pump is reversed, the flow may be reversed and pressures might build up to dangerous levels. If you replace a three-phase motor or even just disconnect a motor, and then reconnect it to the supply, it is essential that before you couple the motor to the load, the direction of rotation is checked. This can be done by momentarily switching the motor on, if it is safe to do so, and checking the shaft rotation. A more satisfactory way is to check the phase rotation of the supply at the motor before disconnection with a phase rotation indicator. When reconnecting the motor the phase rotation is again checked to see if it is the same as before. Note: If it is found that the three-phase motor is running in the incorrect direction, transposing any two-phase connections will reverse the phase sequence of the supply to the motor and so will reverse the direction of the motor. Silicon controlled rectifiers More use is now being made of silicon controlled rectifiers in controlling the speed of large motors. In the control circuits of these devices, voltages are used which are derived from other phase voltages than the one the device is actually controlling. The chosen phase relationship of these control voltages with the main phase voltages is essential to the control system operation. If the phase sequence is changed/ then the wrong voltages are used to control the SCRs so that the devices cease to operate correctly, with possible damage to the control equipment or the machinery involved.

89

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Phase rotation indicator The use of a phase rotation indicator has been mentioned. These instruments vary between manufacturers but they are basically a meter which has three leads colour coded, for example, red, white, blue, so that the leads can be recognised. On the instrument some form of indication of the rotation of the phases is shown. The earlier types had a small meter in the form of a very small motor with a disc on the shaft which had an arrow in it. When the leads were connected the disc would rotate to indicate the direction of the phase rotation. Later types have two lamps on the front, one with ABC or RWB beside it, the other with CBA or RBW beside it. When the leads are connected, one of the lamps would light up indicating phase direction. These two meters are shown in Figure 22.

90


Review In the second part of this section you have covered the topic of phase sequence. Standard conventions regarding phase sequence and the definition of positive phase sequence (ABC) were discussed. The reverse sequence of CBA was demonstrated and the effect of changing the phase sequence on rotating machinery and SCR controlled equipment was explained. Because of the obvious need to know the phase sequence, the operation and use of phase sequence indicators was explained. The correct use of these indicators should ensure that the dangers associated with inadvertent phase reversal can be avoided. At this stage it would be to your advantage to review the objectives at the beginning of this section to ensure that you can perform the tasks listed.

91


Check your progress 2

1 Using a funicular diagram, determine the sum of the following three phasors:

Check the result using a mathematical method. 2 A three-phase generator has a phase-to-neutral voltage on A phase of EAN = 150 <0 . (a) It is connected in star. Draw the complete voltage phasor diagram of the generator showing magnitudes and angles of all voltages. The phase sequence is positive. (b) The same generator is connected in delta. Draw a complete phasor diagram of all voltages showing magnitudes and angles. The phase sequence is positive. 3 Repeat Question 2 for a negative phase sequence. 4 List the possible dangers if the following loads have t heir motor connected with a reverse phase sequence: (a) (b) (c) (d)

a pump with non-return valves a conveyor belt a bench grinder automatic controlled lathe.

92


Glossary of terms angular displacement

the angular difference between two phasors expressed in degrees or radians

delta connection

three-phase connection involving a closed loop of all three phases

double subscript notation

a notation used to identify the voltage between a particular point in a circuit and the point of reference for that voltage

earth point

usually the frame of an appliance or a stake in the ground. It is considered as having zero volts.

funicular diagram

a method of adding phasors in a polygon to find the resultant.

ground point

see earth point.

neutral

the terminal or conductor in an electrical system with respect to which the voltages of all active conductors are of equal magnitude in normal operation. In MEN systems the neutral is earthed (at zero voltage).

phase

one voltage of a multi-voltage system.

phase rotation

set by the direction of rotation of the generators and the sequence of the voltages of the three phases.

phase rotation indicator

an instrument to measure the phase rotation on a supply system.

phase-to-phase voltage

the voltage between the line terminals of a polyphase generator or load.

line voltage

same as phase-to-phase voltage.

Phasor

a line drawn to represent a sinusoidally varying quantity in magnitude and relative phase.

polar diagram

a method of adding or displaying phasors where all phasors emanate from a single point.

prime mover

the device which supplies motive power to a generator (eg a diesel engine, a steam turbine etc).

star connection

a connection in which corresponding ends of the three phases are connected together at one point forming a three-point star.

vector

a value which has magnitude and direction 93


Formulae

Three-phase generated voltage

Calculation of phase-to-phase (line) voltages -

in a star system

-

in a balanced three-phase system star-connected:

-

in a balanced three-phase system delta-connected:

94


95


96


Unit 3 - Three phase four wire star connected balanced loads Introduction In Section 2, the previous section, you learnt of the generation of and the advantages of three-phase voltages. In this section you will learn of the connection of three-phase loads to the voltage systems previously covered. There were two basic connection of voltages in the three-phase system you have covered. They were the star and delta connections. In this section you will cover the same two connections, star and delta, as they apply to loads. The main consideration in this section is that the loads used are all balanced. This means they have the same impedance in each phase. This is important to realise as, when you progress to Sections 4 and 5, the loads will be unbalanced and will require more work to calculate the results. This section has two main parts: the first will cover star-connected systems, and the second will cover delta-connected systems. Each part will cover the calculation of currents in the loads and then investigate the effects of reversing the phase sequence. Each part is completed with the description and calculation of the power consumed by the respective loads. As was mentioned in Section 2, for three-phase calculations it is important to always refer to a reference voltage when describing voltages and currents in the load. The most common reference given is the phase-to-neutral voltage on A phase given as VAN = 0°. You should remember that the use of this reference is not compulsory as any reference can be used and in some textbooks various references are used. It will not alter the results but the answers will not be exactly the same if a difference reference is used, as the angles will refer to a different position. When supply voltages are referred to in a three-phase system, they are always assumed to be phase-to-phase voltages. If a system is said to be a three-phase star connected 400 volt system, the 400 volts would be a phase-to-phase voltage. Do not be concerned if, when in the course of calculations, in this and subsequent sections, your answers are not exactly the same as those given. When calculations are performed, answers are usually rounded up to two decimal places and this can lead to small differences in final results.

97


Objectives The following list of objectives is for your guidance when learning the work in this section. Do not be concerned if there are terms or concepts stated here you do not understand. These will become clear as you progress through the section. At the completion of this section, review these objectives—you should understand them and be able to: • draw and describe the connections to the three-phase three and four wire starconnected balanced loads; • calculate the phase and line currents for star-connected balanced loads; • draw the phasor diagrams showing graphical representation of voltages and currents for three-phase star-connected balanced loads; • show the effect of phase rotation reversal ABC to CBA on star-connected balanced loads; • calculate the power in a star-connected balanced load using 3(EphIph cos and *

El cos

)

;

• draw and describe the connections of the three-phase three wire delta-connected balanced loads; • calculate the load and line currents for delta-connected balanced loads; • draw the phasor diagrams showing graphical representation of voltages and currents of three-phase delta-connected balanced loads; • show the effect of phase rotation reversal ABC to CBA on delta-connected balanced load; • calculate the power in a delta-connected balanced load using 3(EphIph cos and

El cos

.

98

)


Three-phase four wire star-connected balanced loads In Section 2 you learnt the connections for a star-connected generator. A star-connected load has the same connections and the voltage drops across the impedances form a set of phasors in the same way as the generated voltages. In this section we are only going to consider balanced loads. These are loads which have the same impedances in each phase. They may be resistive or a combination of resistance and inductance or capacitance. The only essential feature is that the same impedance exists in each phase. For our purposes a three-phase motor is a balanced load as it is assumed that each phase has the same number of coils, so the resistance and inductance is the same in each phase. In Figure 1 we are going to consider a three-phase balanced load connected to a generator with three-phase conductors and a neutral conductor joining the star points of the generator and the load.

Figure 1 In Section 2 you were shown how three electrically independent generators could be combined to supply the load in the same way as the load is connected in Figure 1. We can make the assumption with a balanced load that what is happening in each phase does not affect what is happening in any other phase. We can also say that whatever happens in A phase will occur in B phase 120° later and again in C phase 240° later, assuming a positive phase sequence. The last condition that is taken is that there is no resistance in the connecting conductors (including the neutral conductor). This means that the two neutral points or star points are at the same potential.

99


Calculation of currents in star-connected balanced loads

We will now consider what the current values are in the circuit in Figure 2.

Figure 2 If we consider each phase in isolation for the calculation of each current, the result is as follows:

100


Now if we add , the resultant should flow in the neutral conductor since all three phases are flowing to the neutral point of the load.

This is demonstrated in the finicular diagram in Figure 3.

Figure 3 IA,IB and IC form a balanced set of phasors of equal magnitude and at 120° spacing since the angle of 6 is the same for each. The resultant must be zero. This condition of balanced currents can only exist for equal impedances in the load. You will notice that the current in each phase is the same current in the line supplying it so that the line currents are equal to the phase currents. Example 1 A four wire three-phase supply is connected to a balanced three-phase load, the connections being the three phases plus a neutral conductor. If VAN = 240 <0° and ZA = 10 IN

and the phase sequence is ABC, find IA IB and IC and

(the current in the neutral).

101


Solution

Figure 4

102

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Phasor Diagram

Figure 5 Power factor The power factor in a three-phase system must have the same meaning as the power factor in a single-phase system. It must be the cosine of the angle which the current lags or leads the voltage which causes that current. Because of the number of voltages both phase-to-neutral and phase-to-phase in a three-phase circuit, we must be particularly careful that we use the correct voltage with the current for which we are trying to find the power factor. In all cases we use the phase voltage and the respective phase current to determine the angle. In the previous example if we use IA and VAN the angle is 30°. So the power factor is cos 30° or 0.866 lag. If we had used the line voltage VAB as shown in Figure 5 the angle would be 60°. This would give the wrong answer. Note: The power factor of any load can be determined by the impedance angle of the load. In this example, the impedance was 10 <30°. The impedance angle was 30° so the power factor is cos 30° or 0.866. A positive impedance angle indicates an inductive load so the power factor must be lagging.

103


One-line equivalent circuits for star-connected balanced loads We showed in the first part of this section that all the currents can be obtained by considering each phase separately. Since each current was equal in magnitude and only varied in angle, then, if we are given one current and the phase sequence, we can calculate the other two currents. Example 2 Consider the previous Example 1 as a one-line equivalent circuit. Solution

Figure 6

104


Three-phase three wire star-connected balanced loads In the previous calculations we found that for a balanced load

This means that for all balanced loads there will be no current in the neutral. If the neutral conductor is removed, then the circuit will not act any differently than it did when the neutral was connected. This is not the case, as you will see in later sections, with a load which is unbalanced (ie where there are different impedances in different phases). Example 3 A 300 V three-phase three wire star-connected load has an impedance of 18 <-20° Q in each phase. Find the line currents and draw the phasor diagram. The phase sequence is ABC. Solution

Figure 7

105


We must use phase-to-neutral voltages. The 300 volts given is the phase-to-phase value.

106


Figure 8 In this example there was no difference to the previous example where there was a neutral conductor connected. The three wire system on a balanced load behaves the same as a balanced four wire system. Now try the exercises in Student Exercise 1.

107



Three identical impedances of 17 <-30° Q are connected to a 415 volt three- phase four wire supply. The load is connected in star and the phase sequence is ABC. Use VAN as reference. Find the line currents and draw the phasor diagram.

2

Three identical impedances are connected in star to a 500 volt three-phase, three wire supply. The line current for phase A is 15 <-30° A. Find the value of the impedance and the value of the two components which are connected in series to make up the impedance in each phase. The system is ABC phase sequence and the system frequency is 50 Hz.

108


Effect of phase reversal on star load currents

All the examples given have been using the positive phase sequence of ABC. We can now consider the effect on a three-phase four wire star-connected load with a negative phase sequence of CBA. Example 4 Consider Example 1 with a negative phase sequence. The supply voltage of VAN = 240 <0 volts and ZA = 10 <30 Ω. Solution

Figure 9

109

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS IA in both positive and negative sequence is the same phasor.

IB and Ic are in the reverse positions which is consistent with the voltage EB and Ec being in the reverse positions. The conclusion is that reversing the phase sequence on a balanced star-connected load does not alter the magnitudes of the currents, only that the currents appear in the same order as the voltage sequence (see Figure 10).

Figure 10 Now try the exercises in Student Exercise 2.

110


Student Exercise 2 Three identical impedances of 17 <-30° Ω are connected to a 415 volt three-phase four wire supply. The load is connected in star and the phase sequence is CBA. Use VAN as reference. Find the line currents and draw the phasor diagram.

Three identical impedances are connected in star to a 500 volt three-phase, three wire supply. The line current is 15 <-30° A in each line. Find the value of the impedance and the value of the two components which are connected in series to make up the impedance in each phase. The system is CBA phase sequence and the system frequency is 50 Hz.

111


Calculation of power in star-connected balanced loads

If you look at Figure 11 you will notice that the wattmeter is connected in one leg of the star load to measure the power in that leg.

Figure 11 Since we have assumed that the load is balanced, then the power dissipated in ZA (called Pph) will be Then the wattmeter uses IA the line current, which is equal to the phase current and VAN the phase-to-neutral voltage.

112


Remember, Z is the impedance angle of load in A phase. This is the angle between the current in A phase, IA and the phase-to-neutral voltage VAN. This applies if it is a three wire without a neutral connection or a four wire with a neutral connection on a balanced load.

From Section 1 we can now say for a three-phase load that the apparent power and the reactive power is

113


Reverse phase sequence We found in the previous section that when we reverse the phase sequence the current remains the same for the individual phase. Then the power formula is unaffected by phase reversal in the case of the star-connected balanced load. Example 5 In Example 1 we found that the phase currents were

The phase voltages were

Find the power in each phase and then prove that total power

Solution

Checking

Note: In each case the angle used was the angle between the phase current and the respective phase-to-neutral voltage. Now try Student Exercise 3. 114

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Student Exercise 3 1

Calculate the power in each phase and the total power using P = ELIL cos Z using the results you obtained in Student Exercise 1, Question 1.

2

Calculate the power in each phase and the total power, using P = ELIL cos Z using the information in Student Exercise 1, Question 2.

115


Review So far in this section you have covered the analysis of three-phase three wire and four wire star-connected balanced loads. You were shown how to draw and describe the connections of these loads. We followed this part with the calculation of the currents in the lines and phases of the two systems and because of the balanced loads, found that the currents were of equal value and spaced at 120° apart. One other important feature of a balanced load was found to be that the three currents summed to zero, so there was no current in the neutral wire. This resulted in the same currents for both the three and four wire systems. The simplified, and shorter, approach using a single-phase diagram was demonstrated and the construction of the phasor diagram showed a valuable method of displaying the results of the calculations. The effect of phase reversal on both systems only resulted in the current angles changing, so that they remained in the same angular relationship to the phase voltages which were causing them. When calculating the power in the star-connected balanced load, it was shown that the power per phase, which is one third of the total power, may be calculated from the formula

The angle 6 is always the impedance angle of the load. In the remainder of this section we will be covering the explanation of the delta-connected balanced load. I will show you how to calculate the line and phase currents, how to calculate power and power factor in the delta connection. The effects of phase reversal on the line and phase currents will also be analysed.

116


Three-phase delta-connected balanced loads In the first two parts of this section you were shown the star-connected load. In this part we can assume that the supply voltages are the same as for the star three wire case. Let us consider Figure 12.

Figure 12 The most significant difference with voltages for the balanced delta load is that there is no neutral point, so there are only phase-to-phase voltages. Each voltage is across one leg of the delta but it is also across two other legs in parallel with the first leg. For example, VAB is across ZAB which is in parallel with ZCA and ZBC. This point will be further considered when we calculate the currents flowing in the delta phases and the lines connected to the delta. When we consider the supply voltage phasor diagram as in Figure 13, we do not have phase-to-phase voltages. But it is convenient at times to use the same reference in the diagrams as we did with the star-connected loads. We can use VAN the phase-toneutral voltage of A phase. This voltage does not strictly exist on a delta, but to use the reference point on the vector diagram can be convenient to keep the diagrams consistent with star and delta loads, as both may be connected to a single supply. These multiple load calculations will not be required in this course but we will maintain the convention.

Figure 13 117


Calculation of currents in delta-connected balanced loads

When we consider the delta-connected load, there is a significant difference to the starconnected load. In the star-connected load the line currents and the phase current were equal as they were the same currents. We now compare the star case with the delta by looking at Figure 14.

Figure 14 Let us look at this diagram in individual sections: firstly, the supply voltages which cause currents to flow through the impedances.

These phase currents also form a balanced set of phasors as they did in the star connection. They have equal magnitudes and are out of phase with each other by 120°.

118

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS The line currents are obtained by using Kirchoff's current laws at each node of the delta.

The relationships of IA, IB and IC must be memorised. In memorising these relationships, you will note that there is a very good symmetry about each. When finding IA the other two vectors start and finish with letter A and have the sequence B and C in the middle. The other groups have the same relationship IB has B at the beginning and end, and C and A in the middle. We have said that as this is also a balanced load as in the connection considered,

To find the relationship between IA and IAB, that is the relationship between the line current and the phase current in a delta load, consider the following,

119

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS These are phase currents in the load. Now consider Figure 15.

Figure 15

Then in general, the expression is

As we will see in the next part, the phase sequence will determine the angle between the line and phase currents. The general expression IL =

Iph must be memorised.

Let us now consider an example of a delta-connected load.

Example 6 A three-phase three wire 415 volt ABC system supplies a delta-connected load whose phase impedance is 60 <45° Ω. Find the phase currents and line currents in this system and draw the phasor diagrams,

120

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Firstly, let us draw the connection diagram.

Figure 16 Now let us consider the following diagram (Figure 17) of voltages supplying the load using an "imaginary" VAN as reference.

121


Figure 17

To calculate the phase currents IAB, IBC and ICA.

122

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS To find the line currents

The phasor diagram for all currents is shown in Figure 18.

Figure 18 123


Consequently once finding IA you could follow the same procedure, giving:

This method is far quicker and achieves the same results. Note: We will discuss the effect of changing the phase sequence in the next part. Power factor For the same reasons as when we calculated power factor in the star-connected load, we must consider the voltage which causes the current in the load. With a delta load there is only one voltage, the phase-to-phase voltage, but there are two currents. If we consider the impedance in A phase then we have VAB as the voltage which causes IAB to flow and you will recall the formula:

So the angle of the power factor is the angle between V AB and I AB As in the star case and for a single-phase circuit the impedance angle z will give the current its angular relationship with the applied voltage. Thus the power factor for a delta circuit can be determined by considering the impedance angle and applying the power factor formula

Now try the exercises in Student Exercise 4. 124

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Student Exercise 4 1

A three-phase three wire 300 volt ABC phase sequence system supplies a balanced delta load. Each phase impedance is made up of a resistance of 35 ohms and an inductance of 0.25 henrys in series. The system frequency is 50 Hz. Use VAN as reference. Draw the diagram of the load connections. Determine the phase currents, line currents and draw the complete phasor diagram.

2

A three-phase star-connected generator having phase sequence ABC and a phase-to-neutral voltage VAN of 200 volts supplies a balanced delta load. The phase impedance of the delta load is 35 <-20 . Determine the phase currents, line currents and power factor and draw the complete phasor diagram. Use VAN as reference.

125


Effect of phase reversal on delta load currents

In the previous two parts of this section you have only considered the phase sequence ABC or the positive phase sequence. Now let us consider the results of reversing the phase sequence on the currents flowing in the delta load. Firstly consider the voltages in Figure 13 but apply the phase sequence CBA or negative sequence. These phasors are shown in Figure 19.

Figure 19 You were shown in Section 2 how to construct the phasors of Figure 19. If you are unsure of the construction refer to the part, 'Phase Sequence', in Section 2 to refresh your memory. These voltage constructions are important as you will get the wrong currents if you use the wrong voltages. To determine the effect of phase reversal on the currents we must consider the effect of the angular change of the voltage.

The phase-to-phase voltages are now

To find the relationship between IA and IAB with the reverse phase sequence in the delta load, consider the following discussion which is the same as for the positive sequence case. 126


If we let

then in CBA sequence

and These are the phase currents in the load. If we now consider Figure 20

Figure 20

We will now solve Example 7 using only the short method. Example 7 is the same as Example 6 except that the phase sequence is reversed. Example 7 A three-phase three wire 415 volt CBA system supplies a delta-connected load whose phase impedance is 60 <45 Find the phase currents and line currents in this system and draw the phasor diagram.

127


Solution The voltage phasor diagram is as shown in Figure 21.

Figure 21

128

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS The phasor diagram of voltages and currents is shown Figure 22.

Figure 22 Note that the line current IA is now leading the phase current IAB by 30°. All other magnitudes are the same as the ABC case. Now try the exercises in Student Exercise 5.

129


Student Exercise 5 A three-phase three wire 300 volt CBA phase sequence system supplies a balanced delta load whose phase impedances are made up of a resistance of 35 ohms and an inductance of 0.25 henrys in series. The system frequency is 50 Hz. Use VAN as reference. Draw the diagram of the load connections. Determine the phase currents, line currents and draw the complete phasor diagram.

130


Calculation of power in delta-connected balanced loads With delta-connected loads, as it was with star-connected loads, the average power dissipated in each phase is the same and is called Pph. The total power dissipated by the three phase balanced load is P

= 3Pph

The wattmeter is connected as shown in Figure 23 to measure the single-phase power.

Figure 23

131

SOLVE PROBLEMS IN COMPLEX POLYPHASE POWER CIRCUITS Also we know that

Z

is the impedance angle of the load Z.

This formula is identical to the formula used to calculate the three-phase balanced star load total power. So, providing we know the line current and phase-to-phase (line) voltage and the impedance angle, we can calculate the power of the load without having to know how it is connected.

Example 8 Calculate the power dissipated in the load of Example 7. Solution Power dissipated in each phase

Now try the exercises in Student Exercise 6.

132



Calculate the phase power and total power dissipated in the load of Question 1 of Student Exercise 4.

2

Calculate the phase power and total power in the load of Question 2 of Student Exercise 4

133


Review

In the second part of this section we have covered the basic topic of delta-connected balanced loads. In that coverage I described the connection of a three-phase three wire delta-connected balanced load. Using this connection, the formula for the phase and line currents were developed which were found to be in the basic form of

In constructing the phasor diagrams the above formulae were illustrated graphically. It was shown that phase sequence reversal from ABC to CBA affected the phase relationship of the line currents with the phase-to-phase voltages. Again, the magnitudes of the currents were unchanged by the phase reversal. The power in the delta was shown to have the same formula

as the star relationship, taking care to select the line currents, phase-to-phase voltage and the impedance angle of the load. At this stage it would be to your advantage to review the objectives at the beginning of this section to ensure that you can perform the tasks listed.

134


Check your progress 3 A three-phase star-connected balanced load with a phase impedance of 22 <40° ohms is connected to a three-phase four wire supply of 440 volts, 50 Hz, CBA phase sequence. Calculate: (a) (b) (c) (d)

the phase and line currents power in each phase the total power using two methods the power factor.

Draw the connection diagram and complete phasor diagram. Use VAN as reference. A supply of 220 volts, 50 Hz, ABC phase sequence supplies a delta load which has a phase impedance consisting of a resistance of 15 ohms in series with an inductance of 0.1 henrys. Calculate: (a) the phase and line currents (b) power consumed by the load by two different methods. Draw the connection diagram and complete phasor diagram. Use VAN as reference. Calculate the phase and line currents of Question 2 if the phase sequence is reversed. Use VAN as reference.

135


Glossary of terms Balanced loads

a three-phase load which has the same impedance in each phase.

equivalent circuit

a circuit which has the same effect on a supply but usually has different components and component arrangements to the original circuit.

Kirchhoff's current law

the algebraic sum of currents at a node is zero.

phase power

the power consumed in one phase of a three-phase load.

total power

the sum of the powers in the three phases of a three-phase circuit.

136


Formulae

Star connection

Delta connection

The power formulae are the same as for star connection.

137

Electric Refe Booj

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