CHAPTER 4 GENERATOR INTERNAL IMPEDANCE AND REGULATION

4.1       GENERAL

This chapter examines what goes on inside a generator.  First we will examine the normal ‘ohmic’ voltage drop due to load current in the main windings.  Also we shall consider how the emf which is generated in the armature by the rotating field system undergoes certain changes.  They cause it to be less than the emf due to the pure field system and so reduce still further the voltage that appears at the generator terminals.

Take first the simpler case of a d.c. generator.

4.2       D.C. GENERATOR

In a d.c. generator the current which flows from the machine to the external load and back again also flows through the internal resistance of the generator, as shown in Figure 4.1.

FIGURE 4.1

D.C. GENERATOR - INTERNAL RESISTANCE

If the excitation causes an emf E in the generator armature, the open-circuit voltage V₀ will be equal to E.  But if a load current I flows, it flows also through the generator’s internal resistance r causing an internal voltage drop equal to I.r.  When the load is applied therefore the terminal voltage drops from V₀ to a lower value V, where

                                                                   V = V₀ - I.r                                                          ….(i)

The reduction of terminal voltage (V₀ - V) is the ‘regulation’ of the generator under the stated load conditions.  Its value is given by the drop of voltmeter reading when the load is applied (assuming no change in the exciting field).

Since V₀ = E, equation (i) can also be written

V = E - I.r

                                              or                 E = V + I.r                                                         ….(ii)

4.3       A.C. GENERATOR

A similar situation occurs with a.c. generators, but the effect is more complicated, as shown

in Figure 4.2.

FIGURE 4.2

A.C. GENERATOR - INTERNAL IMPEDANCE

In this case the ‘load’ may be an impedance which draws both active and reactive current, and the internal resistance of the d.c. generator of Figure 4.1 is replaced by a generator internal impedance z, which consists of both resistance r and reactance x.

The same principles apply as for the d.c. generator - namely that the terminal voltage V₀ is reduced to a lower value V by the internal ‘voltage drop’ caused by the load current flowing through the internal impedance z.

but these are now all vector quantities.

The numerical difference between no-load terminal voltage and the terminal voltage on load, namely V₀ - V, is still the regulation of the generator under the stated load condition.  Its value is given by the drop of voltmeter reading when the load is applied (assuming no change in the excitation).

When considering what goes on inside a generator it would be logical to take first the generated emf E; then to examine the voltage drop due to the load current flowing through the generator’s internal impedance and finally to arrive at the terminal voltage V.  However, because the emf of a machine cannot actually be measured (except on open circuit), it would be more convenient to start with the terminal voltage V - for that alone is what the external load sees - and then to work backwards through the internal voltage drop, adding it to V, to arrive at the emf E needed to produce the required terminal voltage V.  Expressed as an equation (similar to equation (ii)):

 

                                                               E = V + I.z (vectorially)                                       ….(iii)

This gives a measure of the excitation needed to produce a given voltage at the terminals for any given load current, and, in particular, to find the maximum excitation required from the AVR and exciter to maintain full rated terminal voltage at maximum rated load.

It is known that, if an a.c. voltage V is applied across a resistance R, a current equal to V/R (by Ohm’s Law) will flow through it and will be in phase with V.  Similarly, if a voltage V is applied across a pure reactance X, a current equal to V/X (by Ohm’s Law for a.c.) will flow

through it and will lag 90^o on V.

FIGURE 4.3

COMMON CURRENT IN RESISTOR AND REACTOR

If, instead of starting with a common voltage V, a common current I had been flowing through both a resistance R and a pure reactance X, as shown in Figure 4.3, then the voltage V_R equal to I.R and developed across R, will be in phase with I.  The voltage Vx, equal to I.X and developed across X, will lead the current I, since the current must lag on the voltage.  This is shown in Figure 4.3, where the voltage vectors are in blue, and the currents in red.  The voltages V_R and V_X are respectively the voltage drops in R and X due to the common current I flowing through them.

Consider now the voltage drops due to a load current I flowing through the internal resistance r and reactance x of a generator armature.

FIGURE 4.4

INTERNAL VOLTAGE DROPS

In Figure 4.4 a generator is supplying an external load Z (resistance R and reactance X) with a current I.  The generator is being excited to an emf E, and it has an internal resistance r and reactance x.  The terminal voltage while load current is flowing is V.

0

The external impedance Z causes the load current / to lag by an angle φ on the terminal voltage, cos φ being the power factor of the load. The load current I is then flowing through both r and x (x is assumed typically to be larger than r).

FIGURE 4.5

TERMINAL VOLTAGE AND EMF IN A GENERATOR

As already explained, the voltage drop v_r due to current I flowing through r is I.r and is in phase with I; it is represented by the vector OP.  The voltage drop v_x due to current I flowing through x is I.x and leads I by 90^o.  It is represented by the vector OQ.

 

The total voltage drop I.z in the generator is the vector sum of v_x and v_r namely (I.x + I.r)

=OQ + OP.

 

Since E = V+ I.z (equation (iii)),

                                                            E = V + OQ + OP

 

                                                               = V + CS + ST (Since CS = OQ and ST = OP)

 

and is so shown in Figure 4.5, the resultant vector E being in red.

This shows that, to achieve a terminal voltage V on load, the emf must not only be greater than V numerically but must also in general lead on V in phase.  This angle of lead is called the ‘power angle’ or ‘load angle’ for that particular load and is given the symbol λ.  It represents the advance in the rotor’s pole position as compared with its no-load position.  The power angle λ should not be confused with the load’s phase angle φ.

It should be noted that r and x are the resistance and reactance (in ohms) per phase in a 3-phase system.  This assumes a star connection, and V in the formulae is the phase voltage (line voltage divided by √3).

Figure 4.5 gives the relationship between the emf E and the terminal voltage V for a given load current.  It was evolved by starting with V and evaluating the corresponding E, but equally it shows that, given an emf E, at the stated load current the terminal voltage will be V - that is, it can be used either way.

For example, if the terminal voltage has to be maintained at a steady value V at all loads, then the emf (and so the excitation) can be determined for any stated load current, and in particular for the maximum rated load.  From this the size of the excitation system can be deduced.

FIGURE 4.6

TERMINAL VOLTAGE AND EMF IN A GENERATOR

AT CONSTANT NO-LOAD EXCITATION

A special case of this is at no load, where there is no load current I and therefore no internal voltage drops I.r and I.x.  The vector E then lies on top of V and is equal to it; that is to say, at no load the terminal voltage is identical with the emf, as shown in Figure 4.6(a).  If V₀ is the terminal voltage at no load and E₀ the corresponding emf at no load,

then V₀ = E₀.  This is the only situation where an emf can be actually measured.

As before, V and E are the phase voltage and phase emf.

4.4       REGULATION

If the no-load terminal voltage V₀ is the system voltage (as it normally will be), then E₀ is also equal to the system voltage.  If, as load is applied, the emf were maintained at its no-load value E₀ (that is, if there were no change in the excitation), then the situation would be as shown in Figure 4.6(b), which is a repetition of Figure 4.5 but everywhere reduced in scale.  Since the magnitude E is unaltered from its no-load value E₀, the on-load terminal voltage vector V, the current vector and both the volt-drop vectors will be reduced in proportion but will retain their relative sizes and positions.

In Figure 4.6(b), therefore, V₀ is the no-load terminal voltage and V the terminal voltage when a stated load current is applied.  The numerical difference between V₀ and V is the drop of terminal voltage, as measured by voltmeter, when the load is applied with no change in the excitation; it is the regulation of the generator at that load.

The regulation of a generator is usually expressed as a percentage of the system voltage when the generator is supplying full rated load current at rated power factor.  Then the percentage regulation is

at rated output.

In practice, of course, the situation of Figure 4.6(b) would be immediately detected by the AVR, which senses only the terminal voltage.  Finding V reduced below the system voltage, it would raise the excitation and so the emf E.  This would increase the scale of Figure 4.6(b) again until it approached that of Figure 4.5 and the terminal voltage V once more reached the system voltage.  As will be seen from Figure 4.5, the emf will then be much greater than the system voltage.

4.5       PRACTICAL APPLICATION

The foregoing is the full treatment for determining the emf and the regulation of a generator, taking into account the power factor of the load current and both the internal resistance and reactance of the machine.

In practice the internal resistance of an a.c. generator - particularly of a large one - is negligibly small compared with its reactance.  In other words, we can neglect r as compared with x, which means that the internal voltage drop I.r can be neglected as compared with I.x.  The full Figure 4.5 then reduces to the simpler Figure 4.7 where the vector ST (=I.r) has disappeared.

FIGURE 4.7

DIAGRAM NEGLECTING INTERNAL RESISTANCE

 

The vector CS, equal to I.x, is then the principal element in determining the difference in magnitude between E and V, and so in determining the regulation.

In Figure 4.7 the angle between the vector CS and the horizontal is also φ, so that the numerical difference between E and V is approximately CY, or CS sin φ.  So the difference between the emf and the terminal voltage is approximately I.x sin φ.  But I.sin φ is the quadrature component (I_q) of the load current, so the regulation can be taken as I_q.x.

YS, the other component of CS, namely CS cos φ (= I.x cos φ), has virtually no effect on the voltage difference between E and V, but it does affect the power angle λ.  But I cos φ is the in-phase component (I_p) of the load current, so the power angle depends mainly on the in-phase, or active power, component of the load current.  The power angle is given approximately by

tan λ	=	SY
		OY
	=	CS.cos φ
		OC + CY
	=	I.x cos φ
		V + I.x sin φ
	=

As before, x is the reactance (in ohms) per phase and V is the phase voltage.

To sum up, with the approximations made (that r may be neglected), the regulation or voltage drop within a loaded generator depends principally on the quadrature component of the load current - that is, on the kvar loading only - and on the generator reactance.

The in-phase component of the load current - that is, the kW loading - affects principally the power angle and has little effect on the voltage drop.

A notable case of this occurs when starting large motors.  On switching on, not only is the starting current heavy but it is also drawn at a low power factor.  In other words a very large reactive load is thrown onto the system - a situation causing the maximum voltage drop in the generators.  This will cause a noticeable dip as evidenced by the dimming of lights until the AVR has had time to take charge and restore the voltage.

4.6       SYNCHRONOUS REACTANCE

There is one further matter which has a very marked effect on the value of excitation needed to produce the required emf.  It was shown in Figure 2.4 of Chapter 2 that, when a load current - particularly a reactive load current - passes through the armature of a generator, the magnetic field of that current opposes that of field poles and demagnetises them; this is the phenomenon of ‘armature reaction’.  Consequently, with this effect present, the net field flux is reduced and, for a given exciting current, less emf is generated.

FIGURE 4.8

ARMATURE REACTION AND SYNCHRONOUS REACTANCE

All that has been said regarding the relationship between the emf E and the terminal voltage V still stands.  Up until now the emf E has been considered to reflect exactly the field exciting current, but, because of demagnetisation, this is no longer true and the net flux is now less than that produced by the excitation.  Consequently the emf E actually generated by the reduced flux is less than would occur with the given exciting current alone.  This effect is shown in Figure 4.8(a).  E_ex represents the emf that would be generated by the exciter field alone, but E is the reduced emf due to the demagnetisation of the field.  This results in a smaller V and a smaller I.  The heavy reduction in V is as if the exciter emf E_ex had remained unaltered but the generator had had a much higher reactance X_d than its real one x.  This is indicated by the blue vectors in Figure 4.8(a).

Indeed the effect of armature reaction is taken into account by assuming just this, as shown in Figure 4.8(b).  The generator is considered to have a larger, false reactance whose effect is to restore the relationship between a new and much increased excitation emf E_ex (not the actual E) and the terminal voltage to that of Figure 4.7 by using a higher value of x.  This higher, false reactance is called the ‘synchronous reactance’ of the generator.  Its symbol is X_d.  If used instead of the actual reactance x, it gives the same relationship between the excitation emf and terminal voltage as in Figure 4.7.

In fact, taking this artificial synchronous reactance into account, which is normally several times greater than the actual reactance x, many large generators have excitation systems some two to three times that required for the no-load value in order to maintain full terminal voltage at full rated output. Compare the relative sizes of E_ex and V(= E₀) in Figure 4.8(b).  It also causes a larger power angle λ.

The whole concept of ‘synchronous reactance is merely a device for taking into account the armature reaction of a generator when loaded.  The voltage drop within a generator is caused partly by the loss of emf due to this armature reaction, and partly by the normal impedance drop due to load current in the windings.  By using this wholly fictitious synchronous reactance, the total drop is considered to be due to an increased impedance drop alone.  Its use enables currents and voltages to be calculated by ordinary methods as if no demagnetisation were taking place.

It is always used in network ‘steady-state’ problems such as power flow and load sharing. It is also used in generator design to establish what excitation system will be needed - the sizes of the exciters and main field system, rotor cooling, etc.

4.7       PERCENTAGE REACTANCE

Throughout this chapter reactances per phase have been used directly (x or X_d) and are measured in ohms. In practice an alternative method is generally used, namely to express them as ‘percentage reactances’.

Instead of using Ohm’s Law I = V/X to determine currents, the generator is considered to be short-circuited and the excitation reduced so that the short-circuit current just equals the rated full-load current.  The reduced excitation is expressed as a fraction (or percentage) of the excitation at full rated load; percentage is then the ‘percentage reactance’ of the generator.

For example, if the rated full-load current of a generator were 1 000A, and on short-circuit this current were achieved with only 25% excitation, the generator would have a ‘percentage reactance’ of 25%.  If the full-load current is divided by the percentage reactance (expressed as a fraction), it gives the current on short-circuit with full excitation.  Thus the short-circuit current in this case   If the generator is rated 5 000kVA at full load, then on short-circuit it will deliver or 20MVA.

This method of expressing reactances is very useful, for it avoids having to deal with actual voltages and currents for any particular generator.  Most drawings give reactances only in percentage form.

In Chapter 9 percentage synchronous, transient and subtransient reactances are given for a typical large generator, together with formulae for converting from percentage to ohmic reactance and from ohmic to percentage.  These conversions, however, require knowledge of the rated voltage and current.

The very large synchronous reactances which arise as explained in para. 4.6, result in percentage synchronous reactances being well above 100% - typically nearer 200%.  This causes the current of a short-circuited generator to settle (in the 200% case) at only one-half the normal full-load current - that is to say, the generator current is actually reduced by a short circuit (assuming that the excitation does not change) - a result which perhaps may be surprising to some.

A consequence of this is that normal overcurrent protection of the generator may actually fail to work on short circuit.  This is investigated further in the manual ‘Electrical Protection’.

4.8       DIRECT AXIS AND QUADRATURE AXIS REACTANCES

In the foregoing discussion it was assumed that the machine concerned had a uniform air gap and was thus of the cylindrical rotor type.  All reactances therefore were ‘direct axis

It was explained in Chapter 2 that salient-pole generators, which form the majority in offshore and onshore installations, do not have a uniform air gap.  In consequence their reactances are of two types - direct axis and quadrature axis.

In practice, omitting the quadrature-axis quantities from the calculation has little effect on the magnitude of the excitation E needed to sustain a given load as described in the foregoing paragraphs.  The quadrature-axis reactance does, however, have a considerable influence on the power angle λ for that load.  Omitting it leads to a calculated angle smaller than actually occurs and may present stability problems.