# Electric Machine Dynamics

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[below] Permanent-Split-Capacitor, Normal-Slip, Non-Synchronous, single-phase motors for industrial instrumentation and precision applications (Bodine Electric)  ### Intro

The discussion of electric machines presented in the preceding sections was limited to the steady-state operating conditions. That is, the machine has either been running at a given condition for a long time or adjusting from one operating condition to another very slowly so that the energy imbalance between the electrical and mechanical elements would be insignificant during the transition period.

However, when the change from one operating condition to another is sudden, changes in the stored magnetic energy and the stored mechanical energy in terms of the inertia of the rotating members don’t occur instantaneously. As a result, a finite time, known as the transient (dynamic) period, is needed to restore the energy balance from the initial to the final conditions. For example, if the applied voltage to a motor changes suddenly, the machine will undergo a transient period prior to attaining its new steady-state condition. In fact, such events occur quite often when machines are driven by electronic drives, as they generate periodic discontinuous voltage and current waveforms.

Another condition that may cause a change in the dynamic equilibrium of a machine is a sudden change in its load.

In Section 2 we pointed out that the magnetization characteristic of a magnetic material deviates from a straight line relationship at high values of magnetic flux density. It simply means that the permeability of a magnetic material is dependent upon the flux density in it. Only at low flux densities, where the magnetization characteristic follows a fairly straight-line relationship, the permeability of the magnetic material can be considered as constant. In this straight-line (linear) region, the flux density in the magnetic material is directly proportional to the applied magnetomotive force (mmf). For instance, a 10% increase in the applied mmf gives rise to an almost 10% increase in the flux density or the flux in the magnetic material. A machine that operates in the linear region of its magnetization characteristic can be treated as a linear device. For such a machine, it’s possible to obtain a closed-form solution to characterize its behavior under all operating conditions (steady-state or transient). As the applied mmf is increased, a point is reached at which the flux density in the magnetic material begins to deviate from the linear relationship. This is the point, called the knee of the magnetization curve, at which the magnetic material begins to saturate. Subsequent increase in the applied mmf results in further saturation of the magnetic material. In this region, the permeability of the magnetic material depends greatly upon the flux density in it. As the flux density increases the permeability decreases. When the magnetic material becomes completely saturated, the permeability of the magnetic material approaches that of the air and the magnetic material behaves almost like a nonmagnetic material. It’s quite difficult if not impossible to obtain a closed-form solution for a machine that operates in the saturated (nonlinear) region. We can, however, determine the behavior of a machine that operates in the saturated region by using various numerical techniques. One such technique based upon the forth-order Runge-Kutta algorithm is described later in this section.

We can also approximate the magnetization characteristic of a machine by many piecewise linear segments. The operation of the machine in each segment can then be deemed linear. Under such linear approximations, it’s possible to obtain an approximate closed-form solution of a machine using differential equations or Laplace transformation techniques. In this section, we have used the Laplace transformation technique to shed light on the dynamic behavior of direct -current (dc) machines.

### DC Machine Dynamics

When a voltage is suddenly applied across the field winding of a dc generator, the current in the field winding builds up slowly from zero to its final steady -state value owing to the high inductance of the field winding. An exactly similar situation takes place when the generator is coupled to a prime mover. It takes time for the speed of the armature of a generator to change from zero to its rated value. How does a dc generator behave when the load is removed suddenly or the generator develops a short-circuit? To answer these questions, we have to understand the dynamic behavior of a dc generator.

Similar questions can also be raised for a dc motor. For example: What happens when the voltage is suddenly applied across the armature winding of a dc motor? How long does it take for the current to reach its steady-state value? How fast does it readjust its speed when the torque is changed suddenly? To answer some of these questions, let us first reexamine the steady-state behavior of a dc motor.

We already know the governing equations for the torque developed by and the angular velocity of a separately excited or a shunt motor under steady-state condition as … and … where K, is the machine constant, I, is the armature current, QP is the flux per pole, R, is the effective resistance of the armature circuit, and V, is the source voltage. [above: DC armature]

From these equations, it’s obvious that the motor speed depends upon the applied voltage V,, the effective armature resistance R,, and the flux per pole QP The flux per pole is controlled by increasing or decreasing the shunt-field winding current +. The build-up of current \$ depends upon the resistance Rf and the inductance Lf of the field winding. As you may have guessed, it seems as if there are three ways to control the speed of a dc motor while it supplies the required torque. However, controlling the speed by changing either the applied voltage V, and/or the effective resistance of the armature circuit R, is regarded as one method of speed control. This method is generally called the armature-control method.

When the speed of a dc motor is controlled by changing the flux in the motor, it’s referred to as the field-control method. What follows is a discussion of the dynamic behavior of a dc motor when these methods are used for speed control.

====1 A separately excited dc motor with a constant field current.

Armature-Controlled DC Motors:

The switch S connects a separately excited dc motor to a dc voltage supply. Owing to the sudden application of the armature voltage, the motor experiences a transient state whose duration is essentially governed by the parameters of the motor and the load. When the field current is kept constant or it’s a permanent magnet motor, we can mathematically represent the variation of the armature current by the following first-order differential equation:

...where R, and L, are the armature resistance and inductance, respectively. In this model ia(t) is the armature current while V, is the applied voltage across the motor terminals Ko,(t) is the back electromotive force (emf) of the motor (e,), where K = &aDp is a constant because of uniform field current +, and o,(t) is the angular velocity of the rotor.

Similarly, the torque developed by the motor can be expressed as ...

Here, Ki,(t) is the developed torque (Td), and TL is the load torque. In this equation D and J are the viscous friction coefficient and the moment of inertia of the rotating members can be rearranged in the matrix form as:

In the above equations the dynamics of a dc motor with a constant excitation is expressed in the state-space form, where w,(t) and i,(t) constitute the state variables and V, and TL are the input variables.

The solutions of the above equations provide us with the variations of the motor speed, and the armature current, i, as a function of time. Since ---- are a set of linear differential equations, they can be solved either by Laplace transformation technique or by a numerical method. In this section we use both techniques to determine w,(f) and #). Laplace Transformation Method Laplace transformation allows us to transform linear differential equations into algebraic equations.

Taking the Laplace transform, we obtain:

...where I,(s), a&), V,(s) and TL(s) are the Laplace transforms of i,(f), w,(f), V,, and T,, respectively. On the other hand, i,(O) and o,(O) are the initial values of the armature current and the angular velocity of the motor at t = 0. When we solve Eqs. (4) and (5) simultaneously, we get ...

If the motor is at standstill when it’s energized at t = 0, i,(O) and w,(O) are zero.

The inverse Laplace transforms yield the motor speed and the armature current in time domain, as illustrated by the following examples.

EXAMPLE 1: A 240-V, permanent magnet, dc motor takes 2 A whenever it operates at no load.

Its armature winding resistance and inductance are 1.43 R and 10.4 mH, respectively. The flux per pole is 5 mWb and the motor constant K, is 360. The moment of inertia is 0.068 kg.m^2. If the motor is suddenly connected to a 240-V dc source while operating at no load, determine its speed and armature current as a function of time.

SOLUTION: The no-load data of the motor helps us determine the friction coefficient, D, as outlined below Since the motor requires 2 A at no load, its no-load speed is

At no load, the developed torque ...is primarily for the rotational losses. The friction coefficient D can then be calculated as …

Prior to the application of armature voltage the motor speed and armature current are zero. That is, at t = 0, w,(O) = 0 and i,,(O) = 0. In addition, the load torque is zero because the motor operates at no load.

In order to determine the inverse Laplace transform of n, (s), we expand n,(s) into partial fractions as ....

....where A, B, and C can now be determined by the root-substitution method. Thus, ...

Finally, we can take the inverse Laplace transform of ...and get the angular velocity in rad/s as ...

The graph of o,(t) is given.

The Laplace transform of the armature current is ...In terms of its partial fraction expansion, I,\$) can be written as ...Finally, we obtain the armature current as ...which is shown graphically.

====2 The motor speed as a function of time.

====3 The armature current as a function of time.

EXAMPLE 2: The motor given in Example 1 is coupled to a load having a torque of 18.58 N-m. Determine the variation of the motor speed as a function of time after the motor is suddenly energized at its rated voltage at t = 0.

====4 The variations in (a) the motor speed and (b) armature current as a function of time.

SOLUTION: Since the voltage is applied to the armature circuit at t = 0, the initial speed and the armature current are both zero.

The inverse Laplace transform of n,(s) yields ...

The Laplace transform of the armature current is calculated as ....or in terms of its partial-fraction expansion as ...

The inverse Laplace transform of Z,(s) yields ...

The variations in motor speed and armature current as a function of time are given.

For all practical purposes, the motor attains its steady-state operation after five time constants. Thus, this motor takes approximately 86.06 ms (based upon the largest time constant) to achieve its steady state.

EXAMPLE 3: The motor studied in Example 1 is suddenly energized with its rated voltage at t = 0 when it was at rest and coupled to a linear load of TL = 0.1 om.

Determine the variation of the motor speed as a function of time for t_2 0. Calculate the time needed to achieve the steady state.

SOLUTION: (a) Since the motor was at rest before it was energized at t = 0, the initial speed and the initial armature current are zero. Thus, from Eq. (8) …

…with T,(s) = 0.1 Q,(s), we get ...and after grouping the terms, we obtain ...

The roots of the polynomial in the denominator are ...Thus, in terms of partial-fraction expansion, n,(s) can be written as ....

The inverse Laplace transform yields, for t > 0 ....

The variation of the speed with time is also given in graphic form, we can calculate the Laplace transform of the armature current as ....or in partial-fraction expansion form as ....

Finally, we can obtain i,(t> from the inverse Laplace transform as ....

==== the variation of the armature current as a function of time.

The largest time constant of the above exponential terms is ....

To achieve a steady state, the time taken should, at least, be 57. Hence, ....

In Examples 1, 2, and 3, we deliberately used the same motor to give you the opportunity to observe the responses of the same motor under different operating conditions.

==== The variations in (a) the motor speed and (b) armature current as a function of time.

Exercises:

The armature winding of the motor given in Example 1 can withstand twice its rated current for 6 ms. Determine the amount of resistance to be connected in series with the armature winding for a safe start-up at no load.

A 240-V, 2-hp, 850-rpm permanent-magnet (PM) dc motor operates at its rated conditions for a long time. The motor constant is 625 and the flux per pole is 4 mWb. Determine the variation of speed and the armature current as a function of time if the load is suddenly removed from its shaft. The armature resistance and inductance are 2.6 R and 19 mH, respectively. The moment of inertia of the motor is 0.08 kg.m^2 while the friction coefficient D is 0.011 N-mss.

Field-Controlled DC Motors:

In an armature-controlled dc motor we kept the field current at a constant level and changed the supply voltage to adjust the motor speed to below its rated speed.

In a field-controlled dc motor, however, we will change the field current in order to obtain a motor speed higher than its rated speed.

Let us consider the separately excited dc motor, operating in the linear region of its magnetization characteristic so that the flux produced by the motor is proportional to the field current (i.e, Qp = k&). Consequently, the back emf and the developed torque in the motor can be expressed as ...where, K, = K, Kf is a constant of proportionality.

As the field voltage, Vf, or total resistance, Rf, in the field circuit is adjusted to change the field current, \$, the following equation governs the variation of if:

....where Lf is the inductance of the field circuit.

the equations below: As 1 changes, so do the back emf and the developed torque as described by the equations below:

====6 A separately-excited dc motor with a change in its field voltage at t = 0.

In ---, we note that om(t), i,W, and \$(t) are the state variables since they dictate the energy storage in the moment of inertia of the rotor and in the inductances of the armature and field windings, respectively.

Also, it’s clear that --- are nonlinear because of the products of the state variables in these equations. As a result, the Laplace transform approach would not be appropriate to get closed-form solutions for oJt) and i,W. However, a simplifying assumption can be made to linearize these equations.

In an electric motor, the time constant of the electric circuit is much smaller than the time constant dependent upon its mechanical members. Therefore, it can be considered that the time constant of the field circuit is much smaller than the mechanical time constant of the motor. In other words, the variation of the motor speed and thereby the back emf will be much slower than the variation of the field current with time. Thus, we can assume that the field current has already achieved its new steady-state value (constant current) before the armature (rotor) responds to the changes in the field current. In this case --- become linear equations as ....where , is the steady-state value of field current.

…after the change has taken pce in the… Now, to obtain the dynamic response of the motor, we can use the Laplace transform ...

The following example shows the use of these equations to predict the dynamic response of a dc motor when the flux is adjusted.

EXAMPLE 4: A 240-V, 12-hp, separately excited dc motor operating on a load of 15 N-m in the linear region of its magnetization characteristic has the following parameters R, = 0.28 a, La = 2.81 mH, Rf = 320 a, Lf = 2 H, J = 0.087 kgm2, D = 0.02 N-ms, and KT = 1.03. Determine the variation of the motor speed, armature current, and field current as a function of time when the field voltage is suddenly reduced from 240 V to 192 V a t = 0.

SOLUTION: Since the motor has been operating at steady state on a load of TL = 15 N.m before the field voltage is suddenly changed, we have to evaluate first the initial conditions on o,(t), z,(t), and Yt) as applied at steady state for t < 0,

Simultaneous solution of the above equations yields:

o,(O) = 300.79 rad/s, i,(O) = 27.2 A, and \$(O) = 0.75 A

When the field voltage is reduced to 192 V suddenly, the field current will drop from 0.75 A to a steady-state value of I = 192/320 = 0.6 A after a short transform of the angular velocity, armature current, and field current as duration. Using equations, we can determine the Laplace ...

....in partial fraction expansion form. Taking the inverse Laplace transform yields ...

....as the variation of the angular velocity, armature current, and field current, respectively. The waveforms. It’s clear that the field current attains its steady state at about 30 ms whereas it takes about 300 ms for the speed and thereby the armature current to do so.

This is in accordance with our assumption that the mechanical response is sluggish in comparison with the electrical response that resulted in the change of states. Another important fact that can be observed from this problem is that the armature current increases to a peak, which is well over the motor rating. Again this is mainly due to the mechanical time constant of the motor that does not allow a rapid change in the back emf of the motor. Therefore, it’s recommended that the field current be gradually varied so that high currents won’t take place in the armature circuit. Lastly, you can notice that the final value of the armature current is higher than what we had before the field current was reduced. The reason is that the mechanical losses increased significantly with increasing motor speed leading to a higher torque demand. Consequently, the increase in developed torque caused an increase in armature current.

====7 The field current as a function of time.

====8 The variations in (a) the motor speed and (b) armature current as a function of time.

Exercises:

A 24-V, 3000-rpm, 50-W, 70% efficient, separately excited dc motor operates at its rated conditions. If the speed is reduced to 1500 rpm at constant load and the time allowed to restore the steady state is 2 ms, determine the moment of inertia for this motor. The resistance of the armature winding is 5 n. The armature inductance and the effect of saturation are negligible.

Assume a viscous friction coefficient of 0.001 N-m-s.

To run the motor given in Exercise 3 at 4000 rpm at no load, calculate (a) the total resistance in the field circuit and (b) the time taken to reach the steady state. The rated field voltage is 24 V, and the field-winding resistance and inductance are 250 R and 100 mH, respectively. The flux per pole is given as O, = O.l\$.

====9 Equivalent circuit of a separately excited dc generator.

Dynamics of DC Generators:

A separately excited dc generator delivering power to a static load. Once again, we assume that the generator operates in the linear region.

To determine the dynamic behavior of the generator with respect to the changes in the armature and field currents, a considerable simplification can be achieved by making an assumption that the shaft speed is practically constant.

During the transient state, the field voltage is ... and the generated voltage is ... However, the generated voltage can be expressed as .... where Ke = K,Kfo, is a constant of proportionality.

In the above set of equations, the armature and field currents are the state variables and Vf is the input variable.

With zero initial conditions expressed in Laplace transform as ....

EXAMPLE 5: A separately excited dc generator operating at 1500 rpm has the following parameters: Rf = 3 0, Lf = 25 mH, and K, = 30 V/A. If a dc voltage of 120 V is suddenly applied to the field winding under no load, determine (a) the field current and the generated voltage as a function of time, (b) the approximate time to reach the steady-state condition, and (c) the steady-state values of the field current and induced voltage.

SOLUTION: Therefore, the field current is ...and the generated voltage is ...

The graphs of if and e, respectively.

For all practical purposes, the field current attains its steady-state value ...after five time constants. Thus, the time required to reach the steady state is ....

The final values of the field current and the induced (no-load) voltage are lj = 40 A and E, = 1200 V, respectively.

====10 The field current as a function of time.

====11 The induced voltage as a function of time.

Exercise:

5. Develop a mathematical model that represents the dynamic state of a dc generator driving a dc motor. Set up the differential equations in the matrix form. Neglect the saturation of the magnetic core, and assume that the motor field current and the generator speed are constant.

Numerical Analysis of DC Motor Dynamics:

Thus far we obtained the closed-form solution for the dynamic response of dc motors using the Laplace transform technique. We resort to a numerical method to obtain the dynamic response when the machine is operating under nonlinear conditions. In this section, our aim is to investigate the dynamic response of dc machines by solving the differential equations numerically employing the fourth -order Runge-Kutta algorithm. Prior to its application, a brief description of the Runge-Kutta algorithm is given.

====12 Flow chart for the Runge-Kutta algorithm.

Fourth-Order Runge-Kutta Algorithm

Consider a set of first-order differential equations given in matrix form as ... where x(t) is a state-variable vector (column matrix) and g(t) is the input-variable vect0r.A and B are constant coefficient matrices that include the parameters of the system. The solution can be given at discrete instants of time as follows:

...where gn and xfl + are the values of the state-variable vector at instants of n and n + 1, respectively. K1, K2, K3, and K4 are constant vectors computed at discrete instants of time as outlined next.

...where h is the step length, which is defined as the time between two discrete instants. The step length is highly dependent on the time constants of the system.

In general, h is chosen less than the smallest time constant of the system in order to maintain the stability of the numerical method. ==== the flow chart of the algorithm, and a sample Fortran program is given.

There are, in all, five sets of inputs to the program matrices A and B, the input variables, and the values for the step length and the total response time.

The Runge-Kutta algorithm is so general that it can be used to solve a set of first-order linear and/or nonlinear differential equations. The following example illustrates the application of the Runge-Kutta method for a linear system.

EXAMPLE 6: The parameters of a 240-V, PM motor are R, = 0.3 a, La = 2 mH, K = 0.8, J = 0.0678 kg-m2. Determine the motor speed and the armature current as a function of time when the motor is subjected to a torque of 100 N-m after 200 ms of starting at no load. Consider a step length of 0.01 s and observe the response for a period of 0.5 s. Neglect the frictional losses and assume that the motor operates in the linear region.

SOLUTION: From Eq. (3), for t < 200 ms, we have ...

====14 The motor speed as a function of time.

====15 The armature current as a function of time.

Exercises

A separately excited dc generator is connected to the armature of a separately excited dc motor. The entire system is at standstill. At t = 0, the field winding of the generator is connected to a 240-V supply as it rotates at 1500 rpm. The motor operates at no load when its field winding cames 1 A. The parameters of the generator are R, = 1.5 R, L, = 11 mH, = 0.07 kgm2, and KO = 360, Rf = 90 R, Lf = 100 mH. The flux per pole in the generator varies as 0.2 9 The motor parameters are R, = 7 R, Lo = 15 mH, I = 0.08 kgm2, K = 5.5, and D = 0.015 N-m-s. Write the computer program to predict the armature current and speed of the motor.

The motor given in Example 6 drives a fan whose speed-torque characteristic can be approximated as TL = 0.5 m,. Write a computer program to determine the motor speed and its armature current.

### Synchronous Generator Dynamics

Whenever there is a sudden change in the torque applied to the rotor or in the current supplied by the generator, a finite time is needed by the generator before it attains its new steady-state condition. The operation of the generator during this finite period is termed transient operation. There are, in fact, two types of transient operations of a synchronous generator, an electrical transient and a mechanical transient.

Electrical Transients:

The most severe transient that may occur in a synchronous generator is the development of a sudden short circuit across its three terminals. When such a short circuit takes place on a generator that has been operating at the synchronous speed under no load, the current in each phase, is sustained by the generated voltage E, of the machine. It may be surprising that each phase current has a decaying dc component even though only ac voltages are present in the armature circuit. The reason for this is that the machine tries to maintain constant flux linkages for each of the three phases. Another observation is that the waveforms are not symmetric about the zero axis.

However, a symmetric waveform can be obtained by subtracting the dc component. It’s a common practice to examine the waveform by dividing it into three periods-the subtransient period, the transient period, and the steady-state period. During the subtransient period, the current decrement is very rapid and it lasts only few cycles. The transient period, on the other hand, covers a longer time with a slower rate of decrease in the current.

Finally, during the steady-state period the current is determined by the generated voltage and the synchronous reactance of the machine. All three periods of the exponentially decaying envelope of the symmetric short-circuit current are shown, using a semilog graph. We can now determine the equivalent reactances that control the current during the sub-transient and transient periods.

====16 Phase currents in the armature winding after a sudden three-phase short circuit across the terminals of a synchronous generator operating at no load.

====17 Symmetric short-circuit current.

In order to simplify the theoretical development, let us assume that the generator operates in the linear region of its magnetization characteristic and the winding resistances are negligible. Also, we assume that the machine does not carry any damper winding. Just prior to the short circuit, the total flux linkages of the field winding are ...where lf is the field current and Lf is the effective inductance of the field winding.

The equation can also be written as ...after representing the effective inductance has the sum of the leakage inductance Ly of the field winding and the mutual inductance L_af between the field and the armature windings. Since the generator has been operating under no load before the short circuit, the flux linkage due to the armature winding is ...

====18 Semi-log graph of the envelope of the symmetric short-circuit current. 1 (Logarithmic scale) Subtransient --Transient current --(Linear scale).

When a three-phase short circuit occurs across the armature terminals at f = 0, the magnetic axes of the field winding and armature winding, For example phase -a, are considered to be orthogonal. After a short time the rotor attains a certain angular position a with respect to the magnetic axis of phase-a winding and causes currents of i, and Z, + \$ in phase-a and the field winding, respectively, in order to maintain the same total flux linkages. The flux linkages for phase-a can be expressed as ...where L,, is the leakage inductance of phase-a.

====19 The relative positions of the phase and field windings at the inception (t = 0) of a short circuit.

The most severe transient condition takes place when the currents are maximum. From the preceding equations, the currents are maximum when a = 0. Thus, corresponding to the most severe condition, ....

The equation can be used to determine the transient reactance of the synchronous generator that governs its behavior during the transient period. Multiplying both the numerator and the denominator, and after some simplifications, we obtain ....

....where E, = \$daf = If Xaf is the generated voltage prior to the short circuit under no-load condition. The transient reactance can be defined as the ratio of no-load voltage to the short-circuit current during the transient period, or ....and its equivalent-circuit representation is given. Note that E, and i, are the peak values of the generated voltage and the armature current.

In the previous development we assumed no damper winding on the rotor.

If we take the effect of the damper winding into consideration, with the similar approach we used to develop the transient reactance, we obtain ....which is referred to as the subtransient reactance. In this expression, i, is the current that corresponds to the subtransient period, and x/d is the leakage reactance of the damper winding and can be included in the equivalent circuit. It’s evident that Xi is smaller than Xi. Therefore, during the first few cycles after the short circuit has developed, the armature current is very large and we refer to this current as the subtransient current, and its root-mean-square (rms) value is given as ...

====20 Equivalent circuit of the transient reactance.

====21 Equivalent circuit of the subtransient reactance.

The end of the sub-transient period marks the beginning of the transient period, and the rms value of the transient current is ...

The transient period lasts typically for another seven to ten cycles. The transient current is followed by the steady-state current controlled by the synchronous reactance of the machine, and its rms value is ...

It must be borne in mind that our discussion of electrical transients assumes that all three phases of the synchronous generator developed the short circuit at the same time. If the short circuit occurs on only one or two phases, the nature of the response is more complex and is beyond the scope of this article.

EXAMPLE 7: The per-unit parameters of a 60-H~~ 71,500-kVA, 13,800-V, cos 8 = 0.8 synchronous generator are X4 = 0.57 Ohm XI, = 0.125 Ohm Xv = 0.239 Ohm and XI, = 0.172 Ohm. Calculate the rms values of the symmetric sub-transient and transient currents if a three-phase short circuit occurs across the armature terminals while the machine has been running at no load.

SOLUTION: To calculate the sub-transient short-circuit current we have to determine the sub -transient reactance as . ...The generated voltage, E, = 1 Ohm occurs at the rated rms voltage right before the short circuit. Thus, ...

The above currents are all normalized with respect to the rated values of the generator. Hence, if we calculate the rated current, we can determine the sub-transient and transient currents in amperes as ...

Mechanical Transients

Let us now examine the mechanical transient in a synchronous generator that may be caused by a major disturbance such as an electrical transient occurring across its armature terminals. Although a mechanical transient develops slowly due to high inertia, it’s one of the most dangerous transients because it can self -destruct the generator.

Let us imagine that a synchronous generator is connected to an infinite bus -bar and is supplying the required power. Under steady-state conditions, its terminal voltage and the frequency are the same as that of the bus-bar. Therefore, it must be rotated at its synchronous speed by a prime mover. Any attempt to increase the speed of the generator translates into electrical power developed by it. Therefore, under steady-state conditions, the power output of the generator is equal to the power input when the losses are neglected.

Let us now consider that there is a sudden change in the power supplied by the generator. A sudden change in power developed requires a sudden change in the power angle 6. However, 6 cannot change suddenly because of rotor inertia. This gives rise to a situation in which the input power is not equal to the output power. The difference between the input and output powers causes a change in the kinetic energy of the rotor, which, in turn, affects its speed. During this transient period, the rotor either slows down or speeds up until the input and the output powers are again equal. When that happens, the rotor regains its synchronous speed. A similar phenomenon of speed adjustments also takes place when a sudden short-circuit develops across generator terminals.

The synchronous speed has been replaced by the rotor speed without introducing any significant error. By doing so, we are assuming that the angular momentum is constant. In terms of the inertia constant of the generator, -- can be expressed as ....

...where p, and pdm are the per-unit powers. Equation --- is a second-order non -linear differential equation and is referred to as the swing equation in terms of the per-unit quantities. It helps us determine the stability of the synchronous generator during the transient state. The swing equation can be solved numerically using the fourth-order Runge-Kutta algorithm described earlier. For the sake of simplicity, the swing equation can be linearized by approximating sin 8,(t) by 6,(t) as long as 6,(t) is a small angle.

====22 Power developed in a round-rotor synchronous generator as a function of power angle.

Equal-Area Criterion:

The equal-area criterion is another method commonly employed to determine the stability of a synchronous generator during a transient state.

The power developed by a round-rotor synchronous generator is shown as a function of power angle. Let us assume that the generator operates at a power p_o corresponding to power angle 6. For a lossless machine, the mechanical power input must be the same as the power output or the power developed under steady state. If a sudden three-phase short circuit occurs, the power developed will become zero as the machine is disconnected from the busbar by the automatic protection system. After the inception of the short circuit has to be modified as ...

...since the prime mover is still supplying mechanical power to the generator. The above equation yields ...

It’s apparent that the shaft speed is higher than the synchronous speed and increases with time, suggesting that the machine can run away. If the short circuit is not cleared within a short period of time (approximately six cycles or so) and the generator is not reconnected to the bus-bar, the generator may self-destruct.

The variation in power angle is ....

....where So is the power angle at the time short-circuit occurs.

If the fault is cleared at an instant t,, the generator will start developing power at a higher level than the pre-fault state, since 6,(t,) is greater than 6, as is evident. This causes the rotor to slow down.

EXAMPLE 8: A 1000-kVA, 4.6-kVI 60-Hz, 4-pole synchronous generator delivers 0.9 Ohm of average power at a power angle of 18" when a three-phase short circuit develops across its terminals. Calculate (a) the per-unit power generated by the generator when the fault is cleared four cycles after its inception and (b) the critical time to clear the fault in order not to lose the stability. The inertia constant is given as 10 J/VA.

Exercises: 8. Derive Eq. (42) when the damper winding is present on the rotor and its leakage reactance is given as XI. A 30-MVA, 13-kV, 60-Hz hydro-generator (synchronous generator used in hydro-plants) has X, = 1.0 Ohm, XA = 0.35 Ohm and X: = 0.25 Ohm. The reactances are normalized based on the ratings of the generator. This generator delivers rated power to a load at the rated voltage with a lagging power factor of 0.85. (a) Calculate the current in the armature winding.

(b) Calculate the subtransient and transient currents in the armature winding when a sudden three-phase short circuit occurs across the terminals of the generator.

10. A 60-Hz, 50-MVA, 12.5-kV, 4-pole synchronous generator with an inertia constant of H = 10 J/VA is delivering the rated power at a lagging power factor of 0.8. The load on the generator suddenly reduces to 20% of its rated value owing to a fault in the system. Calculate the accelerating torque after the fault. Neglect the losses and assume constant power input to the generator.

### SUMMARY

When the operating condition of an electric machine changes suddenly, the machine cannot respond to the change instantaneously due to its inertia. Thus, it undergoes a transient (dynamic) state to readjust the energy balance from the time of the inception of the change to the time when the final steady state is achieved.

The transient may be due to a sudden change in load and/or voltage.

The study of transients in electric machines is often complicated by their complex nature. However, some plausible simplifying assumptions reduce the complexity of the study significantly. Here, we assume that the machine is operating within the linear region of its magnetization characteristic. In that case, saturation of the magnetic core is not taken into consideration, which otherwise makes the study a nonlinear problem.

The transient response of a dc machine is exponential in nature and, for all practical purposes, disappears after five time constants. To determine the transient response in a dc machine analytically, the Laplace transform method appears to be a very useful technique. However, the method cannot be used if the operation with saturation is the main concern. Numerical methods, on the other hand, can solve both the linear and nonlinear problems. One of the numerical methods to study electric machine dynamics is based upon the fourth-order Runge-Kutta algorithm.

The most severe transient that may occur in a synchronous generator is the development of a sudden three-phase short circuit across its terminals. In the event of such a short circuit, the current in the armature winding indicates a damped oscillatory nature. The first few cycles of the short-circuit current waveform denote the subtransient period, and the corresponding current is the subtransient current. The second period of the short-circuit current waveform is known as the transient period, and its current is called the transient current.

Finally, the steady-state current occurs after the transient period is over. In synchronous generator models we can use subtransient, transient, and synchronous reactances for the three regimes of short-circuit current.

When the terminal conditions change suddenly on a synchronous generator, the rotor is not able to respond at the same time, and a mechanical transient on the machine occurs. Although mechanical transients on synchronous generators are the slowest, they are the most important because the machine may self -destruct. Following a mechanical transient, the power angle varies as a function of time and is modeled by a second-order differential equation often referred to as the swing equation. If the cause for the mechanical transient is cleared within a short period of time (several cycles), the machine does not lose its stability during the transient state, which otherwise may lead the machine to overspeed and to damage the rotor permanently. Equal-area criterion is a useful tool to determine the transient stability of a synchronous generator.

### QUIZ

1. What is the reason for a transient state in an electric machine?

2. What can cause a transient in an electric machine?

3. How can we model a separately excited dc motor to analyze its transient (dynamic) response?

4. Why does the saturation make the analysis of dc machines more difficult?

5. What are the state variables and input variables in the linear dynamic representation of a separately excited dc motor?

6. Why may the phase currents be asymmetric in a synchronous generator during the initial stage of a three-phase short circuit?

7. What are the subtransient, transient, and steady-state periods after the occurrence of a three-phase short circuit in a synchronous generator?

8. How does the damper winding contribute to the short-circuit current?

9. The excitation of a synchronous generator is accomplished by placing permanent magnets on the rotor. In the presence of the damper winding on the rotor, comment on whether the steady state is achieved more rapidly or more slowly than in a synchronous generator with a conventional field winding when a three-phase short circuit occurs in the machine.

10. Would it be safe to operate a synchronous generator with a power angle of 90" at steady state?

### Example Problems

1. A 250-V, 7.5-hp~ separately excited dc motor is rated at 1500 rpm. The armature resistance and the inductance are 0.8 R and 9 mH, respectively. The constant K = K,O, is 1.6, while the moment of inertia of the rotating system is 0.15 kg.m2. Determine the armature current and the motor speed as a function of time when the armature is subjected to the rated voltage under no load. Neglect the effects of saturation and the frictional losses.

2. The motor given in Problem 1 has been operating in steady-state condition at no load. Suddenly it’s subjected to a torque of 5 N-m. Determine the decrease in speed of the motor as a function of time. What is the speed after the motor reaches the steady state?

3. A 380-V, 2-hp, 85% efficient, separately excited dc motor has the following parameters R, = 12 SZ, L, = 110 mH, Rf = 400 R, Lf = 1.5 H, KO = 5, and = 0.098 kg.m2. The characteristic of the mechanical load is given by TL = 1.05~~. Determine the torque developed by the motor as a function of time when the armature circuit is suddenly subjected to the rated voltage. Neglect the saturation and the frictional losses.

4. The motor-load system given in Problem 3 has been operating under a steady-state condition at its rated values. Determine the variations in the field current and the speed if the armature current is kept constant and the field voltage is suddenly reduced to 200 V. Neglect the frictional losses and consider O, = O.Oli,.

5. The speed of a 10-hp, 240-V, separately excited dc motor operating on a load of 8 N-m is to be increased by weakening the field using a series resistor bank in the field circuit. Each resistor in the bank is equipped with a shunt normally-closed switch. Resistors are connected in series with the field winding one by one to achieve the ultimate speed without causing any dangerous increase in the armature current during the transient state. (a) Calculate the switching instants if the resistors are taken into the circuit at the minimum times to the steady state, and (b) determine the variation of the speed, armature current, and field current when all the resistors are connected one by one starting with the smallest resistor. Motor parameters are given as follows R, = 0.3 R, 1, = 3 mH, J = 0.09 kg.m^2, D = 0.015 N-m.s, K, = 1.02, Rf = 300 Q, and Lf = 1.8 H.

6. A 440-V, 10-kW, separately excited dc generator has the following data R, = 3 R, 1, = 80 mH, Rf = 400 R, L -2 H, &I\$= 2.6, and 1 = 0.09 kgm^2. The f -armature connected to a load with a resistance of 75 SZ and an inductance of 20 mH was being driven at a speed of 1500 rpm when the rated voltage was suddenly applied to the field circuit. Determine the variations in the field and armature currents. Neglect the frictional losses and the effect of saturation.

7. A 120-V, I-hp, 200-rpm, PM dc motor is suddenly subjected to its rated voltage and a load torque of 8 N-m. Develop the state equations for the state variables i, and o. The machine parameters are R, = 1 st, L, = 8 mH.

8. Compute the instant at which the motor given in Problem 6 reaches its K = 3, and = 0.3 kg-m i. Neglect the frictional losses and saturation. ---steady state by using the fourth-order Runge-Kutta algorithm.--- The separately excited dc motor for Problem 5.

9. Repeat Problems 7 and 8 if the motor is loaded with a linear load.

10. A 4.6-kV, 1000-kVA, round-rotor synchronous generator with subtransient and transient reactances of 80 Q and 160 0, respectively, experiences a three-phase short circuit across its terminals while operating at no load. (a) Calculate the per-unit reactances. (b) Calculate the subtransient and transient short-circuit currents per unit and in amperes.

11. A 60-Hz, 30-MVA, 7.8-kV, synchronous generator has the following per unit parameters X4 = 0.32 Ohm, Xlf = 0.18 Ohm, and XI, = 0.13 Ohm. The generator experiences a three-phase short circuit at its terminals while operating at no load. The subtransient short-circuit current is measured as 11,150 A. Calculate the leakage reactance of the armature.

12. A 1250-V, 100-kVA, synchronous generator operates a synchronous motor at its rated values of 25 kVA, 1000 V with a leading power factor of 0.9.

The subtransient reactances of the generator and the motor are 0.2 Ohm and 0.15 Ohm, respectively. Calculate the fault currents if a three-phase short circuit occurs across the generator terminals.

13. A 5-MVA, 6.3-kV, 60-Hz, 4-pole synchronous generator experiences a three -phase short circuit at its terminals while delivering an average power of 0.8 Ohm with a power factor of 15". The inertia constant of the synchronous generator is 8 J/VA. (a) Calculate the critical fault-clearing power angle. (b) Calculate the critical fault-clearing time.

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