杭电 电机与拖动基础(总复习)PPT

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T = T TL = 198.153.4 = 242.5 N m
(2)
U IN Ra 150 41×0.376 n= = = 986.7 r/min CeΦN 0.1364
4.5 解: (1)
UN IN Ra 220 68.7×0.224 CeΦN = = = 0.1364 nN 1500 UN 220 n0 = = =1612.9 r/min CeΦN 0.1364
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Umin n01 = CeΦN Umin = CeΦN n01 = 0.1364×376.3 = 51.3V
带额定负载时,有
(3)
(4)
Ia = IN
输入功率
P =Umin Ia = 51.3×68.7 = 3.524kW 1
P = P = Ea Ia = CeΦN nmin IN 2 M = 0.1364×263.4×68.7 = 2.468kW
U EaN 150 204.6 Ia = = = 145.2A Ra 0.376
EaN 204.6 CeΦN = = = 0.1364 nN 1500
T = Ct ΦN Ia = 9.55CeΦN Ia = 9.55×0.1364×(145.2) = 189.1 N m
TL = TN = Ct ΦN IN = 9.55CeΦN IN = 9.55×0.1364×41 = 53.4 N m
电机与拖动基础(总复习)
主讲人:王永忠 杭州电子科技大学自动化学院 2010.5.27
第2章 电力拖动系统动力学
了解: (1)电力拖动系统转动方程式 :什么时候 加速,什么时候减速; (2)负载转矩特性 :恒转矩负载,风机、 泵类负载、恒功率负载, (3)系统稳定运行条件
第3章 直流电机原理
掌握: (1)直流电机的工作原理及结构 (2)直流电机铭牌数据 (3)直流电动机运行原理,基本方程式、功 率关系及工作特性; (4)他励直流电动机的机械特性,固有机械 特性、人为机械特性;
n = n0 n =1100 50 =1050 r/min
0.5TN TL Ia = = = 0.5IN =0.5× 40=20A Ct ΦN Ct ΦN
4.1 解: (1)
UN 220 Is = = =1496.6 A Ra 0.147
(2)
Ts 2TN Is = = = 2IN =2×90=180A Ct ΦN Ct ΦN
I2N
5.2 解: (1)
(2)
U1n 3300 k= = =15 U2N 220 U1n 3 = U1n = 10000 = 25 k= U2N 400 U2N 3
U1n k= U2n 3= 10000 400 3 =14.4
(3)
5.17 解:(1)
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解:
P 17×103 IN = N = = 93.1A UNηN 220×0.83
3 P 17×10 N T2N = 9.55 = 9.55× =108.2 N m nN 1500
17 P= = = 20.5kW 1 ηN 0.83
P N
P 6×103 3.4 解: T2N = 9.55 N = 9.55× = 57.3 N m nN 1000
Ea = CeΦN n = 0.3905×(300) = 117.15V
由于下放的是同一重物,则
Ia = 60A
串入电阻
U Ea 117.15 R= Ra = 0.393 =1.56 Ia 60
电阻上消耗的功率
2 P = RIa =1.56×602 = 5.616kW
(3)
Ea = CeΦN n = 0.3905×(850) = 331.93V
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此时,电枢回路串入的电阻
U Ea 440 + 331.93 R= Ra = 0.393 =12.473 Ia 60
电源输入功率
P =UIa = 440×60 = 26.4kW
串入电阻上消耗的功率
2 P = RIa =12.473×602 = 44.9kW
5.1 解:
I1N
SN 100×103 = = =1.65A 3 1N U 3 ×35000 SN 100×103 = = =144.3A 3 2N U 3 ×400
nmin = n0min nN = 250 69.5 =180.5r/min
最大转差率
nN 69.5 δmax = = = 0.278r/min n0min 250
(2)
弱磁调速,拖动额定恒功率负载时
Ia = IN
UN 440 CeΦ = = = 0.2933 n0max 1500 UN IN Ra 440 76×0.376 nmax = = =1402.7r/min CeΦ 0.2933
(3)
nmax 1402.7 D= = = 7.77 nmin 180.5
4.9 解:
电机吊住额定负载转矩的重物不动时
n = 0 Ea = 0 Ia = IN
此时需串入的电阻
UN 220 R= Ra = 0.195 = 3.007 IN 68.7
4.12 解: (1)
UN IN Ra 440 76.2×0.393 CeΦN = = = 0.3905 nN 1050
额定转速差
n = n0 nN =1612.9 1500 =112.9 r/min
最低同步转速
n01 =
δmax
nN
112.9 = = 376.3 r/min 0.3
最低转速
nmin = n01 nN = 376.3112.9 = 263.4 r/min
(2)
nmax nN 1500 D= = = = 5.69 nmin nmin 263.4
第3章 直流电机原理
了解: (1)直流电机的励磁方式和空载磁场 ,主 要是励磁回路的饱和现象; (2)直流发电机的运行原理 ,基本方程式 与功率关系; (3)直流电机的换向
第4章 他励直流电动机的运行
掌握 (1)他励直流电动机的起动; (2)他励直流电动机的调速(重点为调压调 速); (3)他励直流电动机的电动和制动 ,包括 功率关系,运行状态,四象限运行;
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3.7 解:
UN IN Ra 220 40×0.5 CeΦN = = = 0.2 nN 1000 UN 220 n0 = = =1100 r/min CeΦN 0.2
额定转速差
nN = n0 nN =1100 1000 =100 r/min
负载时转速差
TL nN = nN =100×0.5 = 50 r/min TN
U Ia Ra 440 60×0.393 n= = = 1187.1 r/min CeΦN 0.3905 TL = T = Ct ΦN Ia = 9.55CeΦN Ia = 9.55×0.3905×60 = 223.8 N m
回馈电源的功率
P =UIa = 440×60 = 26.4kW 1
(2)
3
P 6 N ηN = = = 87% P 6.895 1
P 6.895×103 IN = 1 = = 31.3 A UN 220 pCua 500 Ra = 2 = = 0.51 IN 31.3
3.5 解:
E01 195.9 = = 0.1959 CeΦ1 = n 1000 E02 186.7 CeΦ2 = = = 0.1867 n 1000
当n=1200r/min时
E1 = CeΦ1n = 0.1959×1200 = 235.08V E2 = CeΦ2n = 0.1867×1200 = 224.04V
由于E1>U,所以第1台电机为发电机,由于 E2<U,所以第2台电机为电动机。
U E1 230 235.08 Ia1 = = = 50.8A Ra 0.1 U E2 230 224.04 Ia2 = = = 59.6A Ra 0.1
P 395 0 T0 = 9.55 = 9.55× = 3.8 N m nN 1000
TN = T2N +T0 = 57.3+ 3.8 = 61.1 N m
P P = P + p0 = 6×10 + 395 = 6.395kW TN = 9.55 M 2 M nN P = P + pCua = 6.395×103 + 500 = 6.895kW 1 M
4.7 解: (1)
UN IN Ra 440 76×0.376 CeΦN = = = 0.4114 nN 1000
理想空载转速
UN 440 n0 = = =1069.5r/min CeΦN 0.4114
额定转速差
n = n0 nN =1069.5 1000 = 69.5r/min
额定负载时的最低转速
应串入电阻
UN 220 R= Ra = 0.147 =1.075 Is 180
采用降电压启动,则电压应降到
Us = Is Ra =180×0.147 = 26.46V
4.2 解:(1) 电压降低的瞬间,电枢电动势
EaN =UN IN Ra = 220 41×0.376 = 204.6
电枢电流
第5章 变压器
掌握 (1)变压器的结构、变比及额定数据 ; (2)变压器的折算的原则、折算值的计算(阻抗、 电压、电流) (3)变压器联接组别 了解 (1)变压器的空载运行 (2)变压器的负载运行 (3)变压器运行特性
习题
3.1 某他励直流电动机的额定数据为:PN=17kW, UN=220V,nN=1500r/min,ηN=0.83。计算IN, T2N及额定负载时的P1.
电动机的输入功率
P = UI2 = U(Ia2 + I f ) = 230×(59.6 +1.3) =14.007kW d
发电机的输出功率
P = UI1 = U(Ia1 I f ) = 230×(50.8 1.4) =11.362kW g
∑p = P P =14.007 11.362 = 2.645kW
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