《电力拖动基础》魏炳贵著课后习题答案.pdf
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第 一 章 1- 2 生 产 机械 切 削 力 或 物 重 力 ( N ) 切 削 速 度 或 升 降 速 度 ( m / s ) 电 动 机 转 速 ( r / m i n ) 传 动 效率 负载转矩 (Nm) 传动损耗 转矩(N m) 电磁转矩 (Nm) 刨床 3400 0.42 975 0.80 17.48 3.50 17.48 起 重 机 9800 提 升 1.4 1200 0.75 145.57 36.39 145.57 下 降 1.4 1200 0.67 72.79 36.39 72.79 电梯 15000 提 升 1.0 950 0.42 359.0 208.2 359.0 下 降 1.0 950 - 0.38 - 157.4 208.2 - 57.4 1- 3 解: 旋转部分飞轮矩 2 2 2 2 2 2 2 2 2 3 4 5 6 0 1 2 2 2 2 2 2 2 1 2 1 4 3 2 1 4 3 6 5 ( / ) ( / ) ( / ) ( / ) ( / ) ( / ) a G D G D G D G D G D G D G D G D Z Z Z Z Z Z Z Z Z Z Z Z + + = + + + + = 2 2 2 2 2 2 40.2 19.6 56.8 37.3 137.20 230 8.25 ( 55 / 20) ( 55 / 20) ( 64 / 38 ) ( 55 / 20) ( 64 / 38 ) ( 78 / 3 0) + + + + + + = 2 251.49 N m i 工作台和工作总重量 1 2 14715 9810 24525 G G G N = + = + = 切削速度 4 3 / m i n 0 . 7 2 / v m m s = = 齿轮 6 的转速 6 6 6 43 27.56 / m i n 0.02 78 k v n r t Z = = = 电动机转速 6 4 2 6 5 3 1 331.88 / m i n Z Z Z n n r Z Z Z = = 直线运动部分飞轮矩 2 2 2 2 2 2 24525 0.72 365 365 42.13 331.88 b G v G D N m n = = = 总飞轮矩 2 2 2 2 251.49 42.13 293.62 a b G D G D G D N m = + = + = 工作台及工作与导轨的摩擦力 1 2 ( ) ( 147.15 9810) 0.1 2452.5 f G G N = + = + = 折算到电机轴上的负载转矩 ( ) ( 9810 2452.5 ) 0.72 9.55 9.55 317.52 0.8 331.88 F F f v T N m n + + = = = 切削时电动机输出功率 2 2 317.57 331.88 11.03 60 F P T K W = = = 空载时电动机轴上的总飞轮矩为 2 2 2 14715 0.72 251.49 365 276.77 331.88 G D N m = + = 1 78 64 55 15.44 / ( m i n ) 0.02 78 30 38 20 dn dv r s dt dt = = 14715 0.1 0.72 9.55 38.11 0.8 331.88 F T N m = = 49.50 375 F F G D dn T T N m dt = + = 1- 4 解: 旋转部分飞轮矩 2 2 2 2 2 2 2 2 2 3 4 5 6 7 8 1 2 2 2 2 2 2 2 4 2 4 2 6 3 4 2 6 3 7 5 ( / ) ( / ) ( / ) ( / ) ( / ) ( / ) a G D G D G D G D G D G D G D G D G D Z Z Z Z Z Z Z Z Z Z D D + + + = + + + + = 2 2 2 2 2 2 2.94 17.05 98.10 294 3.92 3.92 5.59 0.98 ( 30 / 2) ( 30 / 2) ( 65 / 15 ) ( 30 / 2) ( 65 / 15 ) ( 0.15 / 0. 5 ) + + + + + + + = 5.59 0.98 0.089 0.093 0.021 + + + + = 2 6.773 N m 重物、吊钩的总重量为 9 10 490 19620 20110 G G G N = + = + = 提升速度 12 / m i n 0.2 / k v m m s = = 绳索的速度 2 24 / m i n k v v m = = = 卷筒外圆线速度 卷筒转速 5 5 24 15.3 / m i n 0.5 v n r D = = = 电动机转速 6 4 5 3 2 65 30 15.3 993 / m i n 15 2 Z Z n n r Z Z = = = 于是得直线部分飞轮矩为 2 2 2 2 2 2 20110 0.2 365 365 0.298 993 k b G v G D N m n = = = 折算到电动机轴上的系统总飞轮矩为 2 2 2 2 6.773 0.298 7.071 a b G D G D G D N m = + = + = 重物吊起时阻转矩为 20110 0.2 9.55 9.55 55.25 0.7 993 k L c G v T N m n = = = 重物及吊钩转矩折算值为 20110 0.2 9.55 9.55 38.68 993 k L G v T N m n = = = 所以传动机构损耗转矩为 55.25 38.68 16.57 L L T T T N m = = = 放下时阻转矩为 2 55.25 2 16.75 22.11 L L T T T N m = = = 空钩吊起时阻转矩为 9 0 490 0.2 9.55 9.55 9.425 0.1 993 k L G v T N m n = = = 空钩其转矩折算值为 9 490 0.2 9.55 9.55 0.9425 993 k L G v T N m n = = = 所以传动机构损耗转矩为 9.425 0.9425 8.48 L L T T T N m = = = 放下时的阻转矩为 2 9.425 2 8.48 7.54 L L T T T N m = = = 在种情况下,由于 L T 和 L T 都大于零,所以电动机是输出机械能; 在种情况下, 由于 L T 大于零, 所以吊起时输出机械能 , 而 0 L T 4 3 4 2 3 1 2 ( 1 ) 0.079 0.133 0.223 0.374 s t a s t s t s t s t s t s t R R R R R R R R = = = = = = = = 、 式中 , 1 s t R , 2 s t R , 3 s t R , 4 s t R 分别是第一级 , 第二级 , 第三级, 第四级切 除阻值。 2- 4 解: 电枢电路不串电阻达稳态时, 0.8 0.8 0.8 126.8 e m L T N a N T N N a N T T T C I T C I I I = = = = = = = 0.20415 N N a N N N a e N N N a e N N U I R U E I R C n I R C n = + = + = = 则电枢回路不串电阻时的转速: 1 220 0.1 126.8 1015.53 / m i n 0.20415 N a a e N U I R n r C = = = 2 2 ( ) N a s N a s N a s a a e N e T N e N e N e N U R R U R R U R R I n T I C C C C C C + + + = = = 220 ( 0.1 0.3 ) 126.8 829.2 / m i n 0.20415 r + = = 1 a a e N a a U E I R C n I R = + = + ( 由 于 机 械 惯 性 , 转 速 n 不 变 ) 将 电 源 电 压 降 至 188V 后的瞬时电枢电流: 188 0.20415 1015.53 193.2 0.1 a a U E I A R = = = 稳态转速: 3 188 0.1 126.8 858.78 / m i n 0.20415 0.20415 a a U I R n r = = = 1 80% N a a e N a a U E I R C n I R = + = + ( 转速 n 不突变 ) 则瞬时电枢电流: 1 80% 220 80% 0.20415 1015.53 541.45 0.1 N e N a a U C n I A R = = = 稳态转速: 4 220 0.1 126.8 1269.4 / m i n 80% 80% 0.20415 N a a e N U I R n r C = = = 5 串电阻调速,设所串电阻为 5 R 5 5 5 ( ) 600 / m i n 220 0.20415 600 0.1 0.669 126.8 N a a e N N e N a a U I R R n r C U C R R I + = = = = = 降压调速 5 5 600 / m i n 135.17 a a e N a a e N U I R n r U C n I R V C = = = + = 2- 5 解: 端电压不变,励磁回路总电阻不变,所以励磁不变。 电枢回路串入电阻、转速不能突变,所以串入瞬间反电势不变。 105.8 110 105.8 6.46 0.15 0.5 1 1 ( ) 9.55 9.55 9.55 0.674 0.674 6.46 4.35 a N N a N a a a e T T T e m T a E U I R V U E I A R R E C n C n C n E C n T C I N m = = = = = + + = = = = = = = = i 1 1 , 3.23 2 2 e m e m a a T T I I A = = = ( ) 107.9 a N a a E U R R I V = + = 0.071 9.55 T e C C = = 107.9 1528.9 / m i n 0.071 a e E n r C = = = 2- 6 解: 0 N N a U I R ,题目数据错误。 2- 7 解: 对电机 A ,拖动恒功率负载,调速前后功率不变,则 A A P P = 2 2 A A A a A a A a N A a A a A a A P E I I R U I I R P = = = 100 A a A a I I A = = 对电机 B 对电机 A A A A A A A P T T P = = = 1650 100 366 450 A T A a A T A a A A a A a A C I C I I I A = = = = 对电机 B ,拖动恒转矩负载,调速前后转矩不变,则 B T B a T B a B T C I C I T = = = 100 B a B a I I A = = 2- 8 解: 最低理想空载转速 01 250 / m i n n r = ,处于降压调速的机械特性上 最高理想空载转速 02 1500 / m i n n r = ,处于弱磁调速的机械特性上 当 01 250 / m i n n r = 时: 0.411 N N a e N U I R C n = = 0 1069.6 / m i n N e U n r C = = 60 277 2 N N N N P P T N m n = = = 0 69.6 / m i n N N n n n r = = m i n 01 180.4 / m i n N n n n r = = 1 01 69.6 27.8% 250 N n n = = = 当 02 1500 / m i n n r = 时 02 440 0.29 1500 N e U C n = = = m a x 2 1373 / m i n 9.55 ( ) N a N e e U R T n r C C = = 02 m a x 2 02 8.5% n n n = = 2- 9 解: 0.1478 N N a e N U I R C n = = 0 1488.84 / m i n N e U n r C = = 0 m a x 0 ( ) 1373.7 / m i n a a e I R R n n r C + = = 1 1 5 / m i n n r = 0 n n = 当 20% = 时, 0 115 575.5 / m i n 20% n r = = m i n 0 460.5 / m i n n n n r = = m a x m i n 2.98 n D n = = 当 30% = 时, 0 115 383.3 / m i n 30% n r = = m i n 0 268.3 / m i n n n n r = = m a x m i n 5.1 n D n = = 2- 10 解: 对电动机 220 0.05 305 0.20475 1000 N N a e N N U I R C n = = = 2 ( ) 974.36 / m i n G aG aM G aG aM a e N e T N e N E R R E R R I n T r C C C C + + = = = 0 230 1123.32 / m i n 0.20475 G e N E n r C = = = 0 148.96 / m i n N N n n n r = = 0 13.26% N n n = = m a x 974.36 / m i n n r = 当 30% = 时, m a x 1000 0.3 2.9 ( 1 ) 148.96( 1 0.3 ) N n D n = = = 2- 1 1 解: m a x m a x 0 n n = m a x 0 m a x n n = m a x m a x m i n 0 m a x m a x m a x 0 m a x m a x ( 1 ) N N N N n n n n D n n n n n n n = = = = 2- 12 解: 220 0.26 78.5 0.34 585 N N a e N N U I R C n = = = 0 644.8 / m i n N e U n r C = = 0 59.85 / m i n N N n n n r = = 0 644.8 350 294.82 / m i n n n n r = = = a a N R R n R n + = ( 1 ) 0.207 a N n R R n = = 2- 13 解: 在额定工作点时: 1 220 40 0.5 1000 0.2 / m i n N aN a N e N e N e N e N U I R n C V r C C C = = = 降电压前后: 1 L T a T C I = 为额定值 a I 不变,即 1 40 a I A = 降电压后: 1 1 1 180 40 0.5 800 / m i n 0.2 a a a a e e N U I R U I R n r C C = = = = 由题意知 1 2 2 2 180 180 0.1636 / m i n 220 220 N N N e e N N N U U U C C V r U = = = = = 降电压前后负载转矩为额定值 2 2 2 2 44.44 N L T N aN T a a aN T C I C I I I A = = = = 降电压后 2 2 2 180 44.44 0.5 964.43 / m i n 0.1636 a a e U I R n r C = = = 2- 14 解: 在额定工作点时 1 0.403 / m i n N N a e N N U I R C V r n = = 当磁通减小时 1 3 e e N C C = 同 时 负 载 转 矩 不 变 3 309 N L T N N T a a N N T C I C I I I I A = = = = = 不 可 以 长 期运行 m a x 220 309 0.18 1223.7 / m i n 0.403 / 3 a a e U I R n r n C = = = 所以不能长期运行。 2- 15 解: 在串励直流电动机中, f a I I = 依题意,可得 a k I = ( k 为系数) 又 a a e U I R n C = ( ) 0.21 ( ) N a a e N N N a N e N N a a N N U I R C I U I R n n C U I R I U I R = = = 0.21 210 / m i n N n n r = = 同时, 2 2 2 a T a T a N N I T T C I C k I T I = = = 1 220 40 0.5 0.2 / m i n 1000 N N e N N U I R C V r n = = = 9.55 0.2 40 76.4 N T N N T C I N m = = = 所以 2 2 1 76.4 19.1 4 a N N I T T N m I = = = 电磁转矩不变 2 T a T C k I = 恒定,所以 20 a a I I A = = ( ) 20 ( 110 20 0.5 ) 10 ( ) 20 ( 220 20 0.5 ) 21 a a a a I U I R n n I U I R = = = 10 210 100 / m i n 21 n r = = 2- 16 解: m i n 900 9 100 N n D n = = = 0 m i n 0 1000 900 10% 1000 N n n n = = = ,因为降电压调速中, N n 不变,所以 m a x m i n 0 m i n 100 50% 100 100 n n n n n = = = = + + 对于降电压调速, 其机械特性硬度是一定的, 只要电压最低的一条人为机械特 性的静差率满足要求时,其他各条机械特性的静差率就都满足要求。 令 m a x m i n 0 m i n 0 100 20% n n n = = = m i n 0 100 500 / m i n 20% n r = = m i n m i n 0 500 100 400 / m i n n n n r = = = 所以转矩下的调速范围是 400 到 900r / m i n 。 2- 17 解: N 220 12.5 0.8 0.175 1200 0.0175 1200 210 2 ( 0) 210 0.8 7.6 2 2 12.5 N aN N a e N N N a N N a e N N aN e e N aB N aB a aB a N U E I R C n I R U I R C n E C n C n V I I I E I R I = + = + = = = = = = = = = = = 能耗制动瞬间 ,所以制动瞬间 要求制动开始后瞬间电流 限制为 , 即 0.9 0.9 0.9 12.5 11.25 R 0 ( ) ( ) 0.175 ( 420) 0.8 5.73 11.25 L N a N a a a e N a a e N a a T T I I A E I R R C n I R R C n R R I = = = = = + + = + + = = = 由于负载为位能性负载,所以n=-420r/min 设电枢回路应串入的电阻为 ,则有 ,由此可得 2- 18 解: 0 m a x 0 0 m a x 220 64 0.25 0.298 685 220 738.255 / m i n 0.298 2 0.298 738.255 0.25 1.47 2 2 64 N aN N a e N N N a N N a e N N N e N a N a e N a a a N U E I R C n I R U I R C n U n r C I E C n R R R I I = + = + = = = = = = = = = = = 空载转速 由此可计算出空载条件下能耗制动停车, 使最大制动电流I 时,电枢应串入的电阻值为 0 0 2 2 2 1 0 1 0 0 ( T 0) 375 30 ( 0.29781 ) 0.493 0.25 1.47 49 49 1 1 0.493 375 375 0.493 0.26 ( | ) 0.265 ( 1 ) 738. e m L L e N t e N e m t N a t N a a t n n n n n N e d G D dn T T J dt dt n C n C C n T C I C n R R R R dn n dt dn dt n t L n L L n U n C = = = = = = = = + + + = = = = = = 空载时 726 / m i n 1 0 0.265 ( 0 738.726) 1.75 1 / m i n n n r L t L s n r = = = , ,当 即认为停车。 0 0 1 0 0 6.6 3.8 6.6 3.8 49 49 1 1 0.493 375 375 0.493 0.26 ( | ) 0.265 ( ) 0.265 1.75 0.298 0.173 0.173 0.25 1.47 t n n n n n n t n t e N a a dn n dt dn dt n t L n L n L n t L n n e C n n I n e R R = = = = = + = = = = = + + 即 n= f ( t ) I a = f ( t ) 2- 19 解: 反接制动时无论是反抗性恒转矩负载还是位能性恒转矩负载, 转速下降到零即 反接制动停车的时间都是一样的。 1 220 31 0.4 0.208 / ( . m i n ) 1000 N N a e N N U I R C V r n = = = 制动前的转速为 0 2 2 220 0.4 49 1010.3 / m i n 9.55 ( ) 0.208 9.55 0.208 N a F L e N e N U R n n T r C C = = = = 反接制动前电动机电枢感应电动势为 0.208 1010.3 210.1 a e N E C n V = = = 反接制动时电枢回路总电阻为 220 210.1 6.94 2 2 31 N a a N U E R R I + = = = 虚稳态点的转速为 2 2 220 6.94 49 1880.7 / m i n 9.55 ( ) 0.208 9.55 0.208 N a L L e N e N U R R n T r C C + = = = 反接制动机电时间常数为 2 2 2 9.8 6.94 0.439 375 9.55 ( ) 375 9.55 0.208 a M e N R R G D T s C + = = = 反接制动停车时间为 0 0 1010.3 ( 1880.7 ) l n 0.439 l n 0.189 ( 1880.7 ) F L M L n n t T s n = = = 分两个过程进行讨论: 从 反 接 开 始 到 转 速 下 降 到 零 这 个 过 程 中 拖 动 反 抗 性 恒 转 矩 负 载 与 位 能 性 恒 转 矩 负载的情况相同,即在这个过程中两者的 ( ) n f t = 、 ( ) a I f t = 相同。 2.28 0 ( ) ( ) 1880.7 2891 M t T t L F L n t n n n e e = + = + 反接制动的瞬间电磁转矩为 9.55 ( 2 ) 123.16 F e N a T C I N m = = 2.28 ( ) ( ) 49 172.16 M t T t L F L T t T T T e e = + = 2.28 ( ) ( ) 24.67 86.67 9.55 t a e N T t I t e C = = 从 转 速 下 降 到 零 到 进 入 新 的 稳 态 这 个 过 程 中 拖 动 反 抗 性 恒 转 矩 负 载 时 稳 态 转 速 为 2 2 220 6.94 49 ( ) 234.64 / m i n 9.55 ( ) 0.208 9.55 0.208 N a D L e N e N U R R n T r C C + = = + = 2.28 ( ) 234.64 234.64 M t T t D D n t n n e e = = + 2.28 ( ) ( ) 49 74.16 M t T t L F L T t T T T e e = + + = 2.28 ( ) ( ) 24.67 37.33 9.55 t a e N T t I t e C = = 拖动位能性恒转矩负载时稳态转速为 1880.7 / m i n D L n n r = = 2.28 ( ) 1880.7 1880.7 M t T t D D n t n n e e = = + 又 49 D T N m = , 123.16 B F T T N m = = 2.28 ( ) ( ) 49 172.16 M t T t D B D T t T T T e e = + = 2.28 ( ) ( ) 24.67 86.67 9.55 t a e N T t I t e C = = 综上所述当拖动反抗性恒转矩负载时 2.28 0 2.28 0 1880.7 2891 , 0 ( ) 234.64 234.64 , t t e t t n t e t t + 2.28 0 2.28 0 24.67 86.67 , 0 ( ) 24.67 37.33 , t a t e t t I t e t t ,所以不能半载启动 1 1.4 0.7 s L N T T T = = 2 2 1 0.7 1 ( ) 1.4 2 S N S N T T W T T W = = = ,得 707 . 0 1 2 = W W 得 0.707 S S I I = , 所以 0.707 0.707 7 20 98.9 S S I I A = = = 7 0 7 . 0 = K 满载起动即要求 N s t T T ,取 N s t T T = ,则有 2 1 1 ) ( 7 5 * N N T N t s U U T K T T s t T = = = ,即 V U U N 16 . 321 * 7 5 1 1 = = 如 用 Y- 起 动 方 法 , 则 要 求 电 机 工 作 时 处 于 接 法 。 题 中 没 有 明 确 表 示 什 么 接法。假设工作时系统为接法。则有 A I K I s t I N I t s 7 . 46 * * 3 1 3 1 = = = N N N T s t s t T T T K T T 5 . 0 447 . 0 * * 3 1 3 1 时,因此无法用降低电压的办法使转速 n= 1 1 00r / m i n 当 2 m s s = 时,为降压调速最低转速 m i n 2 0.9 1350 / m i n N N N n n s n n r = = = 由知,无法通过降压使转速降至 1300r / m i n 3- 9 解: 1 1 600 580 0.033 600 N N n n s n = = = 55 9550 9550 905.6 580 N N N P T N m n = = = m a x 2.3 905.6 2028.88 T N T T N m = = = 2 2 ( 1) 0.33 ( 2.3 2.3 1) 0.1443 m N T T s s = + = + = m a x 2 2 2082.88 0.1443 0.1443 m m T T s s s s s s = = + + 设此时转差率为 1 s ,依题意 0.9 L N T T = m a x 2 2 2 0.9 N m m T T s s s s = + 2 2 2 2 2 2.3 0.1443 0.1443 0 0.9 s s + + = 解得 2 s = + 0.03 或 + 0.71 (舍去) 依题意,下降时电动机运行于回馈制动状态,故 2 1 2 ( 1 ) 600 ( 1 0.03 ) 618 / m i n n n s r = + = + = 2 2 2 0.033 212 0.025 3 3 159 N N N s E r I = = = 设串入 0.4 SA R = 电阻后,转速为 A n , 转差率为 A s 则: 2 2 2 SA A r R s s r + = 即有: 2 2 2 0.025 0.4 0.03 0.51 0.025 SA A r R s s r + + = = = 1 ( 1 ) 600 0.49 294 / m i n A A n n s r = = = 由 3 2 ( 1 ) m SA m s R r s = 可得 1 3 2 ( 1 ) m m R s s r = + 故串入 0.4 电阻时电动机的临界转差率 3 0.4 ( 1 ) 0.1443 2.4531 0.025 m s = + = 依题意,制动前电动机工作在 520r/min 的电动状态,则其转差率 3 600 520 1.87 600 s = = m a x 3 3 3 3 2 2 2.3 2.22 2.4531 1.87 1.87 2.4531 N N m m T T T T s s s s = = = + + 故制动瞬时的电磁转矩 2.22 N T T = 3- 10 解: 1500 1440 0.04 1500 N s = = 2 ( 1) 0.04 ( 2.2 1.96) 0.166 m N T T s s = + = + = 1500 1440 1.96 1500 s + = = 2 4.4 0.37 1.96 0.166 0.166 1.96 N N N m m T T T T s s s s = = = + + 2 0.166 0.166 0.0038 43.98 22 4.48 100 1 ( ) 1 0.1 0.1 m N N T T N N s s T T T T = = = = + + 1 1 0.996 1500 1494 / m i n n n s n r = = = 3- 1 1 解: 1 1 600 577 0.038 600 N N n n s n = = = 2 2 ( 1) 0.038 ( 2.9 2.9 1) 0.215 m N T T s s = + = + = 2 2 2 0.038 253 0.035 3 3 160 N N N s E r I = = = 设该情况下转差率为 1 s , 1 600 35.4 8 0.528 600 s = = , 电动机工作于人为机械特 性上,设其临界转差率为 1 m s 2 1 1 2 ( ) 1 0.528 2.9 2.9 1 2.97 0.095 T N T N m L L T T s s T T = = = 或 (舍去) 1 1 2 2.97 ( 1 ) 0.035 0.45 0.215 m m s R s = = = ( -1)r 起动时 1 s = , 设该情况下其临界转差率为 2 m s , 在人为机械特性上对应的起动 转矩 2 s t T 为 m a x 2 2 2 2 2 2 2 0.4 1 1 T N s t N m m m m T T T T s s s s s s = = = + + 2 2 2 14.5 1 0 m m s s + = 解得 2 14.47 0.07 m s = 或 (舍去) 故 2 2 2 14.43 ( 1 ) ( 1 ) 0.035 2.31 0.215 m m s R r s = = = 由有人为机械特性,设该情况下转差率为 3 s m a x 3 2 2 3 2 0.8 N m m T T T s s s s = = + 3 3 2 2.9 0.8 14.43 14.43 s s = + 解得 3 2.02 102.6 s = 或 (舍去) 因此下放负载时的转速为: 3 1 3 ( 1 ) 612 m i n n n s r = = 设串入 4 0.06 R = 电阻后,转速为 4 n ,转差率为 4 s 则 4 2 4 3 2 2 0.035 0.06 0.04 0.035 2.31 s r R s r R + + = = = + + 4 3 0.04 0.08 s s = = 依题意,运行与回馈制动状态,故 4 1 4 ( 1 ) 600 ( 1 0.08 ) 648 m i n n n s r = + = + = 3- 12 解: 2 2 13 . 1 8 . 9 4 3 . 44 4 m k g g G D J = = = 1 1 = = g S S 0 0267 . 0 3000 2920 3000 2 2 = = = N S S 1 2 2 2 2 1 2 1 2 2 1 1 2 3000 ( 1 ) ( ) 1.13 ( ) ( 1 1.5 ) 139266.85 2 2 60 s t r J S S J r = + = + = 第一级: 0 1500r / m i n 1 1 = S 0 033 . 0 1500 1450 1500 1 2 = = = N S S 2 2 1 2 1500 1.13 ( ) ( 1 1.5 ) 34816.7 2 60 s t J = + = 第二级: 1500 3000r / m i n 5 . 0 3000 1500 3000 1 = = S 0 0267 . 0 2 2 = = N S S 2 2 2 1 2 3000 1.13 ( ) ( 0.5 0) ( 1 1.5 ) 34816.7 2 60 s t J = + = 2 2 2 3 4 8 1 6 . 7 3 4 8 1 6 . 7 6 9 6 3 3 . 4 s t s t s t J = + = + = 总 的 损 耗 为 第 四 章 4- 1 1 60 60 50 1500 / m i n 2 f n r p = = = 1 1 1 1500 500 0.67 1500 N n n s n = = = 1 2 1 1500- 配套讲稿:
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