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Unformatted text preview: uency f 2 of the rotor, these may be
negligible at normal operating speeds, where f 2 is very low. 258 Chapter Four Fig.4.16 Power flow in an induction motor.
The remaining power is converted into mechanical form. Part of
this is lost as windage and friction losses, which are dependent on
speed. The rest is the mechanical output power Pout , which is the
useful power output from the machine.
The efficiency of the induction motor is: η = Pout
Pin (4.71) The efficiency is highly dependent on slip. If all losses are
neglected except those in the resistance of the rotor circuit, Pag = Pin (4.72) P2 = sPag (4.73) Pout = Pmecj = Pag (1 − s ) (4.74) And the ideal efficiency is: ηideal = Pout
= (1 − s )
Pin (4.75) Three-Phase Induction Machine Sometimes 259 ηideal is also called the internal efficiency as it represents the ratio of the power output to the air gap power. It
indicates that an induction machine must operate near its
synchronous speed if high efficiency is desired. This is why the
slip is very low for normal operation of the induction machine.
If other losses are included, the actual efficiency is lower than
the ideal efficiency of Equation (4.75). The full-load efficiency of
a large induction motor may be as high as 95 percent.
4.12 Power Flow In Three Modes Of Operation
The induction machine can be operated in three modes: motoring,
generating, and plugging. The power flow in the machine will
depend on the mode of operation. However, the equations derived
in Section 4.11 for various power relationships hold good for all
modes of operation. If the appropriate sign of the slip s is used in
these expressions, the sign of the power will indicate the actual
power flow. For example, in the generating mode, the slip is
negative. Therefore, from Equation (4.185) the air gap power Pag
is negative (note that the copper loss P2 in the rotor circuit is
always positive). This implies that the actual power flow across the
air gap in the generating mode is from rotor to stator. 260 Chapter Four
The power flow diagram in the three modes of operation is
shown in Fig.4.17. The core losses and the friction and windage
losses are all lumped together as a constant rotational loss.
In the motoring mode, slip s is positive. The air gap power Pag
equation (4.18) and the developed mechanical power Pmech
Equation (4.27) are positive, as shown in Fig.4.17a.
In the generating mode s is negative and therefore both Pag , and Pmech are negative, as shown in Fig.4.17b. In terms of the
equivalent circuit of Fig.4.7e the resistance [(1 − s ) / s]R2 is negative, which indicates that this resistance represents a source of
In the plugging mode, s is greater than one and therefore Pag is
positive but Pmech is negative as shown in Fig.4.17c. In this mode
the rotor rotates opposite to the rotating field and therefore
mechanical energy must be put into the system. Power therefore
flows from both sides, and as a result the loss in the rotor circuit, P2 , is enormously increased. In terms of the equivalent circuit of
Fig.4.7e, the resistance [(1 − s ) / s ]R2 is negative and represents a
source of energy. Three-Phase Induction Machine 261 Fig.4.17 Power flow for various modes of operation of an
induction machine. (a) Motoring mode, 0 < s < 1. (b) Generating
mode, s < 0. (c) Plugging mode, s > 1. 262 Chapter Four
Example 4.4 A three-phase, 460 V, 1740 rpm, 60 Hz, four-pole
wound-rotor induction motor has the following parameters per
R1 = 0.25 Ω, R2 = 0.2 Ω, X 1 = X 2 = 0.5 Ω, X m = 30 Ω
The rotational losses are 1700 watts. With the rotor terminals
(a) (i) Starting current when started direct on full voltage.
(ii) Starting torque. (b) (i) Full-load slip.
(ii) Full-load current.
(iii) Ratio of starting current to full-load current.
(iv) Full-load power factor.
(v) Full-load torque.
(iv) Internal efficiency and motor efficiency at full load. (c) (i) Slip at which maximum torque is developed.
(ii) Maximum torque developed. (d) How much external resistance per phase should be connected in the rotor circuit so that maximum torque occurs at
Solution: Three-Phase Induction Machine 263 264 Chapter Four =163.11 N.m Three-Phase Induction Machine ηmotor = 265 28022.3
*100 = 87.5%
32022.4 ηint ernal = (1 − s ) *100 = (1 − 0.0333) *100 = 96.7%
(c) (i) From Equation (4.60) (i) From Equation (4.61) Note that for parts (a) and (b) it is not necessary to use Thevenin
equivalent circuit. Calculation can be based on the equivalent
circuit of Fig.4.13 as follows: 266 Chapter Four Example 4.5 A three-phase, 460 V, 60 Hz, six-pole wound-rotor
induction motor drives a constant load of 100 N - m at a speed of
1140 rpm when the rotor terminals are short-circuited. It is
required to reduce the speed of the motor to 1000 rpm by inserting
resistances in the rotor circuit. Determine the value of the
resistance if the rotor winding resistance per phase is 0.2 ohms.
Neglect rotational losses. The stator-to-rotor turns ratio is unity.
Solution: Three-Phase Induction Machine 267 From the equivalent circuits, it is obvious that if the value of ′
R2 / s remains the same, the rotor current I 2 an...
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