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Chapter 5—Electrical Network 5–36 I 2 = 57.07/ − 10.73o 0.377/68.2 o = 151.38/ − 78.93 o A The total torque is T m = 90|I 2| 2 R 2 πn s s = 90(151.38)2 (0.14) π(1200)(1) =76.59 N · m/rad This torque is 1.25 times the rated running torque for this particular machine. A majority of induction motors will have a starting torque which is about double the rated running torque. The total three-phase losses will be P loss = 3(57.07)2 120 + 3(155.58) 2 (0.30) + 3(151.38) 2 (0.14) = 31, 500 W This number is about 30 times the loss term for the machine operating at full load. Also, when the machine is not rotating, it is not able to circulate any air for cooling, which makes the temperature rise even more severe. The stator windings will have the most rapid temperature rise for starting conditions because of higher resistance, lower specific heat capacity, and poorer heat conductivity than the rotor windings. This temperature rise may be on the order of 10 o C/saslongastherotorisnotmoving. Obviously, the motor will be damaged if this locked rotor situation continues more than a few seconds. An unloaded motor will typically start in 0.05 s and one with a typical load will usually start in less than 1 s, so this heating is not normally a problem. If a very high inertia load is to be started, such as a large Darrieus, a clutch may be necessary between the motor and the turbine. Very large motors may need special starting techniques even with a clutch. These will be discussed in the next section. Example The induction machine of the previous example is operated as a generator at a slip of -0.025. Terminal voltage is 220 V line to line. Find I 1 , I 2 , input power P m , output power P e , reactive power Q, and efficiency η g . If the rated I 1 is 25 A when operated as a motor, comment on the amount of overload, if any. From the previous example we have V 1 = 127/0 o . The impedance Z 2 is given by Z 2 = 0.14 −0.025 + j0.35 = 5.61/176.42o Ω Z in =0.30 + j0.35 + 13.12/83.72o (5.61/176.42 o ) 13.12/83.72 o +5.61/176.42 o =5.16/147.90o Ω I 1 = − 127/0o 5.16/147.90 o = −24.60/ − 147.90o =24.60/32.10 o A Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
Chapter 5—Electrical Network 5–37 V A = 127 + 24.60/32.10 o (0.30 + j0.35) = 129.16/4.98 o V The total input mechanical power P m is I 2 = −129.16/4.98o 5.61/176.42 o = 129.16/184.98o 5.61/176.42 o =23.02/8.56 o A P m = 3|I 2| 2 R 2 (1 − s) s = 3(23.02)2 (0.14)(1 + 0.025) −0.025 = −9125 W The machine was delivering 9.53 hp to the mechanical load as a motor, but is now requiring 12.23 hp as mechanical shaft power input as a generator. The total output power is The total reactive power Q is P e =3|V 1 ||I 1 | cos θ = 3(127)(24.60) cos(−32.10 o ) = 7940 W Q =3|V 1 ||I 1 | sin θ = 3(127)(24.60) sin(−32.10 o )=−4980 var The negative sign means that the induction generator is supplying negative reactive power to the utility, which is the same as saying it is receiving positive reactive power from the utility. We can compute the efficiency by computing the losses from Eq. 49 and using Eq. 61, or we can merely take the ratio of output to input power. η g = |P e| |P m | = 7940 9125 =0.870 The output current of 24.60 A is slightly under the rated current of 25 A. If the machine is well ventilated, it should operate at this current level for an indefinite period of time. 6 MOTOR STARTING Small induction motors with low to medium inertia loads are normally started by direct connection to a source of rated voltage. Above a rated power of a few kW, the high starting currents usually cause a reduction in line voltage. This reduction may prevent the motor from developing adequate torque to start its load. Electronic equipment and lighting circuits connected to the same source may also be affected by these voltage fluctuations. It is customary, therefore, to start the large induction motors on lowered voltages to limit the starting currents and line voltage fluctuations. This practice is not essential if the supply has sufficient capacity to supply the starting current without objectionable voltage reduction. Motors with ratings up to several thousand Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
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