1. Counter-electromotive force (cemf)
Direct-current motors are built the same way as generators are; consequently, a dc machine can operate either as a motor or as a generator. To illustrate, consider a dc generator in which the armature, initially at rest, is connected to a dc source E_{s} by means of a switch (Fig. 5.1). The armature has a resistance R, and the magnetic field is created by a set of permanent magnets.
As soon as the switch is closed, a large current flows in the armature because its resistance is very low. The individual armature conductors are immediately subjected to a force because they are immersed in the magnetic field created by the permanent magnets. These forces add up to produce a powerful torque, causing the armature to rotate.
Figure 5.1 Starting a dc motor across the line.
On the other hand, as soon as the armature begins to turn, a second phenomenon takes place: the generator effect. We know that a voltage E_{o} is induced in the armature conductors as soon as they cut a magnetic field (Fig. 5.2). This is always true, no matter what causes the rotation. The value and polarity of the induced voltage are the same as those obtained when the machine operates as a generator. The induced voltage E_{o} is therefore proportional to the speed of rotation n of the motor and to the flux F per pole, as previously given by Eq. 4.1:
As in the case of a generator, Z is a constant that depends upon the number of turns on the armature and the type of winding. For lap windings Z is equal to the number of armature conductors.
In the case of a motor, the induced voltage E_{o} is called counter-electromotive force (cemf) because its polarity always acts against the source voltage E_{s}. It acts against the voltage in the sense that the net voltage acting in the series circuit of Fig. 5.2 is equal to (E_{s} - _{Eo}) volts and not (E_{s} + E_{o}) volts.
The net voltage acting in the armature circuit in Fig. 5.2 is (E_{s} - E_{o}) volts. The resulting armature current /is limited only by the armature resistance R, and so
When the motor is at rest, the induced voltage E_{o} = 0, and so the starting current is
The starting current may be 20 to 30 times greater than the nominal full-load current of the motor. In practice, this would cause the fuses to blow or the circuit-breakers to trip. However, if they are absent, the large forces acting on the armature conductors produce a powerful starting torque and a consequent rapid acceleration of the armature.
As the speed increases, the counter-emf E_{o} increases, with the result that the value of (E_{s} — E_{o}) diminishes. It follows from Eq. 5.1 that the armature current / drops progressively as the speed increases.
Although the armature current decreases, the motor continues to accelerate until it reaches a definite, maximum speed. At no-load this speed produces a counter-emf E_{o} slightly less than the source voltage E_{s}. In effect, if E_{o} were equal to E_{s} the net voltage (E_{s} — E_{o}) would become zero and so, too, would the current /. The driving forces would cease to act on the armature conductors, and the mechanical drag imposed by the fan and the bearings would immediately cause the motor to slow down. As the speed decreases the net voltage (E_{s} — E_{o}) increases and so does the current /. The speed will cease to fall as soon as the torque developed by the armature current is equal to the load torque. Thus, when a motor runs at no-load, the counter-emf must be slightly less than E_{s} so as to enable a small current to flow, sufficient to produce the required torque.
Example 5-1
The armature of a permanent-magnet dc generator has a resistance of 1 Ω and generates a voltage of 50 V when the speed is 500 r/min. If the armature is connected to a source of 150 V, calculate the following:
a. The starting current
b. The counter-emf when the motor runs at 1000 r/min. At 1460 r/min.
c. The armature current at 1000 r/min. At 1460 r/min.
Figure 5.3 See Example 5.1.
Solution
a. At the moment of start-up, the armature is stationary, so E_{o} = 0 V (Fig. 5.3a). The starting current is limited only by the armature resistance:
b. Because the generator voltage is 50 V at 500 r/min, the cemf of the motor will be 100 V at 1000 r/min and 146 V at 1460 r/min.
c. The net voltage in the armature circuit at 1000 r/min is
The corresponding armature current is
= 50/1 = 50 A (Fig.5.3b)
When the motor speed reaches 1460 r/min, the cemf will be 146 V, almost equal to the source voltage. Under these conditions, the armature current is only
= 4A
and the corresponding motor torque is much smaller than before (Fig. 5.3c).
3. Mechanical power and torque
The power and torque of a dc motor are two of its most important properties. We now derive two simple equations that enable us to calculate them.
1. According to Eq. 4.1 the cemf induced in a lap-wound armature is given by
Referring to Fig. 5.2, the electrical power P_{a} supplied to the armature is equal to the supply voltage E_{s} multiplied by the armature current I:
However, E_{s} is equal to the sum of E_{o} plus the IR drop in the armature:
It follows that
= (E_{o} + IR)I
= E_{o}I + I_{2}R (5.4)
The I_{2}R term represents heat dissipated in the armature, but the very important term E_{o}I is the electrical power that is converted into mechanical power. The mechanical power of the motor is therefore exactly equal to the product of the cemf multiplied by the armature current
where
P = mechanical power developed by the motor [W]
E_{o} = induced voltage in the armature (cemf) [V]
/ = total current supplied to the armature [A]
2. Turning our attention to torque T, we know that the mechanical power P is given by the expression
where n is the speed of rotation.
Combining Eqs. 3.5,4.1, and 5.5, we obtain
= ZnFI/60
and so
The torque developed by a lap-wound motor is therefore given by the expression
where
T = torque [N×m]
Z = number of conductors on the armature
F = effective flux per pole [Wb]*
/ = armature current [A]
6.28 = constant, to take care of units
[exact value = 2p]
Eq. 5.6 shows that we can raise the torque of a motor either by raising the armature current or by raising the flux created by the poles.
Example 5-2
The following details are given on a 225 kW (» 300 hp), 250 V, 1200 r/min dc motor (see Figs. 5.4 and 5.5):
armature coils 243
turns per coil 1
type of winding lap
armature slots 81
commutator segments 243
field poles 6
diameter of armature 559 mm
axial length of armature 235 mm
* The effective flux is given by F = 60 E_{o}/Zn.
Figure 5.4 Bare armature and commutator of a dc motor rated 225 kW, 250 V, 1200 r/min. The armature core has a diameter of 559 mm and an axial length of 235 mm. It is composed of 400 stacked laminations 0.56 mm thick. The armature has 81 slots and the commutator has 243 bars. (H. Roberge)
Figure 5.5
a. Armature of Fig 5.4 in the process of being wound, coil-forming machine gives the coils the desired shape.
b One of the 81 coils ready to be placed in the slots
c Connecting the coil ends to the commutator bars.
d. Commutator connections ready for brazing (H Roberge)
Calculate
a. The rated armature current
b. The number of conductors per slot
c. The flux per pole
Solution
a. We can assume that the induced voltage E_{o} is nearly equal to the applied voltage (250 V).
The rated armature current is
=900A
b. Each coil is made up of 2 conductors, so altogether there are 243 X 2 = 486 conductors on the armature.
Conductors per slot = 486/81 = 6
Coil sides per slot = 6
c. The motor torque is
= 9.55 X 225 000/1200
= 1791N×m
The flux per pole is
When a dc motor drives a load between no-load and full-load, the IR drop due to armature resistance is always small compared to the supply voltage E_{s}. This means that the counter-emf E_{s} is very nearly equal to E_{s}.
On the other hand, we have already seen that Eo may be expressed by the equation
Figure 5.6 Ward-Leonard speed control system.
Replacing E_{o} by E_{s} we obtain
That is,
where
n = speed of rotation [r/min]
E_{s} = armature voltage [V]
Z = total number of armature conductors
This important equation shows that the speed of the motor is directly proportional to the armature supply voltage and inversely proportional to the flux per pole. We will now study how this equation is applied.
5. Armature speed control
According to Eq. 5.7, if the flux per pole F is kept constant (permanent magnet field or field with fixed excitation), the speed depends only upon the armature voltage E_{s}. By raising or lowering E_{s} the motor speed will rise and fall in proportion.
In practice, we can vary E_{s} by connecting the motor armature M to a separately excited variable-voltage dc generator G (Fig. 5.6). The field excitation of the motor is kept constant, but the generator excitation I_{x} can be varied from zero to maximum and even reversed. The generator output voltage E_{s} can therefore be varied from zero to maximum, with either positive or negative polarity. Consequently, the motor speed can be varied from zero to maximum in either direction. Note that the generator is driven by an ac motor connected to a 3-phase line. This method of speed control, known as the Ward-Leonard system, is found in steel mills, high-rise elevators, mines, and paper mills.
In modem installations the generator is often replaced by a high-power electronic converter that changes the ac power of the electrical utility to dc, by electronic means.
The Ward-Leonard system is more than just a simple way of applying a variable dc voltage to the armature of a dc motor. It can actually force the motor to develop the torque and speed required by the load. For example, suppose E_{s} is adjusted to be slightly higher than the cemf E_{o} of the motor. Current will then flow in the direction shown in Fig. 5.6, and the motor develops a positive torque. The armature of the motor absorbs power because I flows into the positive terminal.
Now, suppose we reduce E_{s} by reducing me generator excitation F_{G}. As soon as E_{s} becomes less than E_{o}, current/reverses. As a result, (1) the motor torque reverses and (2) the armature of the motor delivers power to generator G. In effect, the dc motor suddenly becomes a generator and generator G suddenly becomes a motor. The electric power that the dc motor now delivers to G is derived at the expense of the kinetic energy of the rapidly decelerating armature and its connected mechanical load. Thus, by reducing E_{s}, the motor is suddenly forced to slow down.
What happens to the dc power received by generator G? When G receives electric power, it operates as a motor, driving its own ac motor as an asynchronous generator!* As a result, ac power is fed back into the line that normally feeds the ac motor. The fact that power can be recovered this way makes the Ward-Leonard system very efficient, and constitutes another of its advantages.
* The asynchronous generator is explained in Chapter 14.
Example 5-3
A 2000 kW, 500 V, variable-speed motor is driven by a 2500 kW generator, using a Ward-Leonard control system shown in Fig. 5.6. The total resistance of the motor and generator armature circuit is 10 mΩ. The motor turns at a nominal speed of 300 r/min, when E_{o} is 500 V.
Calculate
a. The motor torque and speed when
b. The motor torque and speed when
Solution
a. The armature current is
= 2000 A
The power to the motor armature is
The motor speed is
The motor torque is
= (9.55 X 760 000)/228
= 31.8kN×m
b. Because E_{o} = 380 V, the motor speed is still 228 r/min.
The armature current is
= -3000 A
The current is negative and so it flows in reverse; consequently, the motor torque also reverses.
Power returned by the motor to the generator and the 10 mΩ resistance:
Braking torque developed by the motor:
= (9.55 X 1 140 000)/228
= 47.8 kN×m
The speed of the motor and its connected mechanical load will rapidly drop under the influence of this electromechanical braking torque.
Rheostat Speed Control
Another way to control the speed of a dc motor is to place a rheostat in series with the armature (Fig. 5.7). The current in the rheostat produces a voltage drop which subtracts from the fixed source voltage E_{s}, yielding a smaller supply voltage across the armature. This method enables us to reduce the speed below its nominal speed. It is only recommended for small motors because a lot of power and heat is wasted in the rheostat, and the overall efficiency is low. Furthermore, the speed regulation is poor, even for a fixed setting of the rheostat. In effect, the IR drop across the rheostat increases as the armature current increases. This produces a substantial drop in speed with increasing mechanical load.
Related Links:
DC Motor Calculations, part 2
DC Motor Calculations, part 3
DC Motor Calculations, part 4
Publication Information | |
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Author: Theodore Wildi | Book: Electrical Machines, Drives, and Power Systems, SixthEdition |
Copyright: 2006 | ISBN: 0-13-177691-6 |
Purchase Electrical Machines, Drives, and Power Systems, SixthEdition
Legal Note
Excerpt from the book published by Prentice Hall Professional (http://www.phptr.com). Copyright Prentice Hall Inc. 2006. All rights reserved.