直流电机:电枢是电阻 Ra 与反电动势 E 串联,所以 V = E + Ia·Ra。反电动势随转速增大,E = k·ω;转矩随电流增大,T = k·Ia;在 SI 单位制中两者的电机常数 k 相同(V·s/rad = N·m/A)。转速用 ω = 2π·n/60 换算。电磁功率为 E·Ia = T·ω;输入功率为 V·Ia,差值由铜损 Ia²Ra 和旋转损耗构成。转速随转矩呈直线下降,ω = V/k − Ra·T/k²,从空载转速 V/k 降到堵转转矩 k·V/Ra 处的零。静止时 E = 0,所以起动电流 V/Ra 很大,除非用电阻限流。
以 rad/s 表示的转速
ω = 2π·960/60 = 100.5 rad/s
扭矩
T = P/ω = 11000/100.5 = 109.4 N·m
答案
T = 109.4 N·m
2.直流电机
答案: Rs = 6.611 Ω
直流电机:电枢是电阻 Ra 与反电动势 E 串联,所以 V = E + Ia·Ra。反电动势随转速增大,E = k·ω;转矩随电流增大,T = k·Ia;在 SI 单位制中两者的电机常数 k 相同(V·s/rad = N·m/A)。转速用 ω = 2π·n/60 换算。电磁功率为 E·Ia = T·ω;输入功率为 V·Ia,差值由铜损 Ia²Ra 和旋转损耗构成。转速随转矩呈直线下降,ω = V/k − Ra·T/k²,从空载转速 V/k 降到堵转转矩 k·V/Ra 处的零。静止时 E = 0,所以起动电流 V/Ra 很大,除非用电阻限流。
电流与反电动势
Ia = 1.6/0.4 = 4 A
E = 0.4·41.89 = 16.76 V
串联电阻
Ra + Rs = (48 − 16.76)/4 = 7.811 Ω
Rs = 6.611 Ω
答案
Rs = 6.611 Ω
该电阻消耗 105.8 W;斩波器调速的损耗要小得多。
3.齿轮与齿轮系
答案: Tw = 24.62 N·m
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
蜗轮转速
nw = n·z/Nw = 960·4/36 = 106.7 rpm
输出扭矩
ωw = 2π·106.7/60 = 11.17 rad/s
Tw = η·P/ωw = 0.5·550/11.17 = 24.62 N·m
答案
Tw = 24.62 N·m
4.齿轮与齿轮系
答案: ND = 40
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.