《电机拖动课后题答案汇总.doc》由会员分享，可在线阅读，更多相关《电机拖动课后题答案汇总.doc（18页珍藏版）》请在三一办公上搜索。
1、Chapter 1Problems：1-1 solution : (a) (b) (c) 1-2 solution : Chapter 2Question 2-14: transformer: 480=k60, so k=8When f=50Hz, Load: U1=k * 50=8x50=400v 415v , okProblems：2-1 solution : 2-2 solution : (a)30 x 240 = 7200 va (b)7200va (c)Ip = 7200 / 2400 =3 A2-3 solution :(a)P = kva x power factor = 5 x
2、 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 A2-4 solution : (a) 0.8 power factor, lagging1) With as a reference, the full-load secondary current is 2) The primary voltage seen from the secondary side: 3) (b) 0.8 power factor, leading.1) 2) 3) 2-5solution
3、: 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:02-6 solution:(1) the exciting parameters are:The magnetic resistance is The current of losses is The magn
4、etization 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, (
5、b) using per-unit values2-8 solution：voltage of each turn ，and U1=E1 U2=E2 change cm2 to m2high voltage winding low voltage winding 2-9 solution: the effective section area of the core referred to the primary sidethe effective section area of the core referred to the secondary side2-10 solution: the
6、 two transformer have the same voltage and the same N1and neglect Z1 distribute the voltage in seriesthe primary sidethe secondary side Chapter 3Problems：3-1 solution: When frequency is 50 HzNumber of poles2468101214Field speed 3000 1500 1000 750 600 500428.57When frequency is 60 HzNumber of poles24
7、68101214Field speed 3600 1800 1200 900 720 600514.29When frequency is 400 HzNumber of poles2468101214Field speed2400012000 8000 6000 4800 40003428.63-2 solution:3-3 solution:（1）=Z1/2p=24/4=6（2）=p360/Z1=2360/24=30q=Z1/(2pm)=24/(223)=2（3）（4）（5）The EMF star is:131719234568910111224222120181716151423aU1
8、U2V1W2W1V2（6）The winding connection diagram is:3-4 solution:(1) Pulsate MMF(2) Ellipse MMF3-5 solution:The input power WAIn order to operate at unity power factor 3-6 solution:Form the no-load test we find: From the locked-rotor test we find: Total leakage reactance referred to stator is Total resis
9、tance referred to stator is If we divide from , then Then 3-7 solution:（1）calculate the parameter:=Z1/2p=36/4=9=p360/Z1=2360/36=20q=Z1/(2pm)=36/(223)=3191102822034567891112131415161718363533323130292726252423222134aU1U2V1W2W1V2And the emf star is :（2）the winding connection when U1-(1、2、3) (10、11、12)
10、- (19、20、21 ) (28、29、30) U2V1-(7、8、9) (16、17、18)-(25、26、27 ) (34、35、36)- V2W1-(13、14、15) (22、23、24)-(31、32、33 ) (4、5、6)- W2 And draw the phase U connection U1123456789101112131415161718192021222324252627282930313233343536U2（3） Chapter 44-4 solution：Suppose the system to lift the weight at the speed
11、of the line speed of the reel ，sothe speed of the reel ，sothe speed of the moter ，soChapter 5Problems：5-1 solution：(a) The speed of the magnetic fields is (b) The speed of the rotor is (c) The slip speed of the rotor is (d) The rotor frequency is 5-2 solution：(a) The synchronous speed of this machin
12、e is Therefore, the shaft speed is (b) The output power in watts is 50kW (stated in the problem)(c) The load torque is (d) The induced torque can be found as follows:(e) The rotor frequency is 5-3 solution：(a) The equivalent circuit of this induction motor is shown below:The easiest way to find the
13、line current (or armature current) is to get the equivalent impedance of the rotor circuit in parallel with, and then calculate the current as the phase voltage divided by the sum of the series impedances, as shown below.The equivalent impedance of the rotor circuit in parallel withis:The phase volt
14、age is , so line current is(b) lagging(c) To find the rotor power factor, we must find the impedance angle of the rotorSo lagging(d) The stator copper losses are (e) The air gap power is (f) The power converted from electrical to mechanical form is (g) The induced torque in the motor is(h) (I) (j) 5
15、-4 solution：(a) The slip s will increase.(b) The motor speedwill decrease.(c) The induced voltage in the rotor will increase.(d) The rotor current will increase.(e) The induced torque will adjust to supply the loads torque requirements at the new speed. This will depend on the shape of the loads tor
16、que-speed characteristic. For most loads, the induced torque will decrease.(f) The output power will generally decrease: (g) The rotor copper losses (including the external resistor) will increase.(h) The overall efficiencywill decrease.5-5 solution：（a）（b） （c）overload ability （d）5-6 solution：（a）（b）（
17、c）（d）Chapter 7Problems：7-2 solution：(a) (b) 7-3 solution：(a) , (b) , 7-4 solution：, 7-5 solution：(a) (b) 7-6 solution：(a) (b) (c) (a very dangerous value, which is absolutely not permissible) 7-7 solution：7-8 solution：(a), (b) 1) 2) 3) (c)7-9 solution：(a) (b) The first question:Direct start armature
18、 current: Direct start line current: The second question:Full load armature current: 125 percent load armature current: 7-10 solution：7-11 solution：7-12 solution：7-13 solution：The watts input at full load: =The watts output at full load =Hence the full load efficiency = 7-14 solution：(a) Copper loss
19、es in the field = (b) Armature current = (c) Copper losses in the armature = (d) Total loss = (e) Motor input = (f) Motor output = input losses = 1000 271 = 729 watts(g) Efficiency =output / input =729 / 1000 =72.9%7-15 solution：a) b) When at half load, Ia = 100A,c) 2.5A205A202.5Aa) Full load situation2.5A102.5A100Ab) Half load situation7-16 solutionRc +Ra=6.88Tm=0.444sNw=-3139rpm
