Induction Motor Load Characteristics PDF

## Electrical Machines (9080V LSEGG381A) - Section 9 - Induction Motor Load Characteristics ### 2. MOTOR RATING AND EFFICIENCY Motors convert electrical energy to mechanical energy, as a result motors are always rated in terms of the mechanical output power. The flow of power through a three phase...

## Electrical Machines (9080V LSEGG381A) - Section 9 - Induction Motor Load Characteristics ### 2. MOTOR RATING AND EFFICIENCY Motors convert electrical energy to mechanical energy, as a result motors are always rated in terms of the mechanical output power. The flow of power through a three phase induction motor is shown in figure 2. - **Electrical power input** - Pin = √3 x V₁ x I₁ x λ x P.f - **Mechanical power output** - Pout = 2πnT / 60 The power input to the three phase motor is determined using - where: - V₁ = motor line voltage - I₁ = motor line current - λ = motor power factor The output power from the motor is determined using - where: - n = rotor speed in rpm - T = motor torque in Nm Motor efficiency varies with changes in load conditions and can be determined by taking the ratio of the output power to the input power - - η% = (Pout / Pin) x 100 where: - POUT = motor output power in watts - PIN = motor input power in watts **Note**: output power is always less than the input power due to the losses that occur in the energy conversion process. ### 3. EFFECT OF ROTOR CURRENT ON STATOR CURRENT When operating on no-load, an induction motor produces just enough rotor current to provide the torque for overcoming bearing and windage friction. An increase of load - - retards the rotor, that is, increases the slip - thus causing an increase in rotor emf as the rotor conductors are overtaken more rapidly by the stator flux - consequently, a greater rotor current is provided to meet the demand for greater torque. The currents in the rotor of an induction motor exert a magnetomotive force which would affect the field strength, if the stator or primary currents retained their original values. But, just as in a transformer, any change in the rotor or secondary current is balanced by an opposite change in the stator or primary current. The reason for this is the same as for the corresponding action in a transformer, namely, that the field strength is practically proportional to the applied emf, and so is constant, if the applied emf is kept constant. As in a transformer the stator or primary current of an induction motor is built up of three components - - magnetising current - iron-loss current - current to balance the rotor or secondary mmf ### 4. OPERATION - NO-LOAD When operating on no-load, the energy taken by the motor is that required to - - establish the rotating magnetic field - overcome the motor losses. ### 5. OPERATION - WITH LOAD When operating on load, the energy taken by the motor is that required to - - establish the rotating magnetic field - overcome the motor losses - drive the load. On load, the stator draws a current that is made up of three components - - magnetising current, Iμ - iron-loss current, In - load component of primary current. Effectively the stator current on load is equal to the phasor sum of the three components of current. Figures 6 and 7 show the phasor diagrams for the induction motor under load conditions. Figure 6 half load and figure 7 full load. ### 6. INDUCTION MOTOR LOAD TEST A load test may be applied to an induction motor to determine its performance at various loads . The motor under test may be either direct-coupled to a generator or loaded by some form of brake. During the course of the test - - the the supply voltage to the motor must be kept constant for the various load values - the frequency also should remain constant to obtain reliable test results. If a generator is used as a load, its power output and losses are measured, and the power input obtained by adding the output and the losses. The power output of the motor is equal to the power input to the generator. If a mechanical brake is used, the torque is obtained and the output of the motor calculated from the equation - - Tc = (Pout x 60) / 2πn The quantities usually determined in a load test are as follows - * speed - measured using a tachometer * rotor current - measured by an ammeter or tong tester (slip-ring motors only) * stator current - measured using an ammeter or tong tester * power factor - measured using a power factor meter or determined by calculation * power input - measured by means of the two-wattmeter method * efficiency - determined by calculation * power output - determined by calculation **Note**: Both the stator and rotor currents measured are line values, not phase values. Figure 8 shows the connections required to carry out a load test on a squirrel cage motor. The load performance curves for a 3.75kW slip-ring induction motor are shown in figure 9. These curves are typical performance curves for an induction motor. **Note**: The operation of induction motors at less than full load is one of the chief causes of low power factor in distribution systems. The following table shows representative efficiencies and power factors for moderate voltage, three-phase, 50-cycle squirrel-cage induction motors, at full load. | Synchronous Speed (rpm) | No. of Poles | 7.5kW | Efficiency (%) | 37kW | Efficiency (%) | 75kW | Efficiency (%) | 375kW | Efficiency (%) | 750kW | Efficiency (%) | |---------------------------|---------------|-----------|-------------------|-----------|-------------------|-----------|-------------------|----------|-------------------|----------|-------------------| | 375 | 16 | 75 | 80 | 84 | 87 | 90 | 86 | 91 | 92 | 92 | 86 | | 750 | 8 | 80 | 83 | 87 | 89 | 91 | 92 | 94 | 95 | 95 | 89 | | 1500 | 4 | 84 | 89 | 90 | 92 | 93 | 93 | 95 | 93 | 95 | 93 | ### Read & Study It will be noticed from these curves that the - * **speed falls fairly uniformly as the power output increases.** The speed of the motor adjusts itself to such a value that the magnitude of the induced emf in the rotor is just sufficient to exert the required torque. As the load increases the speed must decrease, that is, the slip must increase, so that a greater emf will be induced in the rotor to establish a greater current in the rotor circuit, to develop a larger torque, for the purpose of overcoming the increased load. The induction motor therefore has a drooping speed characteristic. Although the motor, for which the curves under consideration were drawn, was overloaded, it can be seen from figure 8 that the reduction in speed was not excessive. * **rotor current curve is observed to rise almost uniformly over the greater range of loads.** If the supply voltage remains constant, the air gap flux, or rotating magnetic field, remains practically constant from no-load to full-load. The torque of the motor is proportional to the product of the air gap flux and the component of the rotor current in phase with it, therefore as the load increases, the rotor current must increase so that the magnitude of its torque component will be sufficient to develop the required torque to overcome the increased load. * **stator current depends upon the rotor current, therefore it rises almost uniformly over the greater range of loads.** It has been shown in the phasor diagram for the induction motor that the total stator current is the phasor sum of the no-load current and that component of the stator current required to neutralise the mmf of the rotor current. As the rotor current increases, so must the neutralising component of the stator current. Therefore, for every increase of secondary load current there must be a proportional increase in the stator current. * **power-factor also has a rising characteristic.** For any increase in load up to full-load there is an improvement in the power-factor. The power-factor of an induction motor running light is very low, because the major portion of the current supplied to the motor is magnetising current, lagging 90° behind the applied stator voltage; this current remains practically constant at all loads. However, as the load increases the component of the stator current required to neutralise the mmf of the rotor current also increases. This has the effect of bringing the resultant stator current more nearly in phase with the applied voltage, that is, reducing the angle of lag of the stator current, resulting in an improvement of the power factor. * **power-input also has a rising characteristic.** This is to be expected, because the motor is converting energy from one form to another and therefore the rate at which energy is taken in (i.e. power input) must at least equal the rate at which energy is given out (i.e., power output). * **efficiency curve is seen to rise as the load is increased, reaching a maximum at approximately full-load, but it commences to fall when the motor is overloaded.** The losses occurring in the motor under consideration may be grouped, for all practical purposes, into three classes: * the copper (I'R) losses in the stator and rotor; * iron losses in stator and rotor cores; * friction and windage losses. The iron losses, friction and windage losses remain practically constant at all loads, but the copper losses vary as the square of the current. At no-load, the input is required only to supply the losses; the output is zero and consequently the efficiency also is zero. As the load increases, the efficiency increases as the ratio of the output to the input increases. It can be proved that the maximum efficiency occurs at the load which makes the variable I'R copper losses equal the constant losses, and a motor is usually designed so that this occurs at approximately full-load. When a motor is loaded beyond this value, the variable copper losses increase at a greater rate than the output and the efficiency falls accordingly. This is illustrated in figure 8 by the efficiency curve bending downwards after having passed through a maximum value. ### 7. EFFECTS OF REDUCED VOLTAGE There are problems associated with the running of induction motors at reduced voltage. For example, assume that the voltage applied to the motor terminals is reduced to 50 percent of the rated nameplate voltage. The effects of the reduced voltage are - * the strength of the rotating magnetic field of the stator is also reduced to half of the normal value. * the voltages and currents induced in the rotor are similarly reduced to half of their normal values. * the resulting torque output of the motor is reduced to one-fourth of its original value. The torque equation given previously shows why the large reduction in the torque output occurs. - T=K x φ x Ir x s **Not important** Both the stator flux (φ) and the rotor current (Ir) are reduced to half of their original values. This means that the product (torque) of these terms is only one-fourth of its original value. For a given value of slip, the torque varies as the square of the applied voltage. Figure 10 shows that a 50% reduction in voltage causes the torque to decrease to 25 percent of its normal value. The torque developed as a result of reduced voltage, at any particular value of slip, may be determined provided the following values are known - * rated voltage * torque developed at rated voltage * new voltage. **Example: 3** - Tnew = Trated x (Vnew / Vrated)² A 415V squirrel cage induction motor develops a torque of 40Nm. Determine the torque developed by the motor if the supply voltage drops to 380V. - Tnew = Trated x (Vrated / Vnew)² = 40 x (415 / 380)² = 33.5 Nm In addition to the decrease in developed torque the other effects of reduced supply voltage include - * increased supply current * possible stalling of the motor * prolonged acceleration period during start up.