电机拖动课后题答案汇总.doc

Chapter 1 Problems： solution : (a) (b) (c) solution : Chapter 2 When f=50Hz, Load: U1=k * 50=8x50=400v 415v , ok Problems： 2-1 solution : 2-2 solution : (a)30 x 240 = 7200 va (b)7200va (c)Ip = 7200 / 2400 =3 A 2-3 solution : (a)P = kva x power factor = 5 x 1.0 = 5 kw (b)P = 5 x 0.8 = 4 kw (c)P = 5 x 0.3 = 1.5 kw (d)I = va / E = 5000 / 120 = 41.7 A 2-4 solution : 0.8 power factor, lagging 1) With as a reference, the full-load secondary current is 2) The primary voltage seen from the secondary side: 3) 0.8 power factor, leading. 1) 2) 3) 2-5 solution : A very small amount of iron loss in the transformer is negligible. To get the primary and secondary values separately, the primary and equivalent secondary values are assumed to be equal: thus: 0 2-6 solution: (1) the exciting parameters are: The magnetic resistance is The current of losses is The magnetization current is The magnetization inductance is (2) (a) the equivalent primary impedance (b) the equivalent primary resistance (c) the equivalent primary reactance 2-7 Solution: (a) Power factor for short-circuit condition = Therefore, the phase angle for short-circuit conditions = Therefore, (b) using per-unit values 2-8 solution：voltage of each turn ，and U1=E1 U2=E2 change cm2 to m2 high voltage winding low voltage winding 2-9 solution: the effective section area of the core referred to the primary side the effective section area of the core referred to the secondary side 2-10 solution: the two transformer have the same voltage and the same N1 and neglect Z1 distribute the voltage in series the primary side the secondary side Chapter 3 Problems： 3-1 solution: When frequency is 50 Hz Number of poles 2 4 6 8 10 12 14 Field speed 3000 1500 1000 750 600 500 428.57 When frequency is 60 Hz Number of poles 2 4 6 8 10 12 14 Field speed 3600 1800 1200 900 720 600 514.29 When frequency is 400 Hz Number of poles 2 4 6 8 10 12 14 Field speed 24000 1200