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。
空载转速和满载转速
n₀ = 60·24/(2π·0.08) = 2865 rpm
nFL = 60·(24 − 4·0.5)/(2π·0.08) = 2626 rpm
转速调整率
SR = (2865 − 2626)/2626·100 = 9.091 %
答案
SR = 9.091 %
2.直流电机
答案: Rs = 11.53 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 = 1/0.5 = 2 A
E = 0.5·41.89 = 20.94 V
串联电阻
Ra + Rs = (48 − 20.94)/2 = 13.53 Ω
Rs = 11.53 Ω
答案
Rs = 11.53 Ω
该电阻消耗 46.11 W;斩波器调速的损耗要小得多。
3.齿轮与齿轮系
答案: Ft = 72.44 N
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.