258 chapter four fig416 power flow in an induction

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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 energy. 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 phase: ′ ′ 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 short-circuited, find (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 start? 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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