质量 m = 2 kg 的物体挂在弹簧 k = 100 N/m 上,带有黏性阻尼器 c = 8 N·s/m(m·x″ + c·x′ + k·x = F)。在静止平衡状态下,突然施加恒力 F = 100 N。 求位移的百分比超调量 %OS。
m = 2 kgk = 100 N/mc = 8 N·s/mF = 100 N
%OS = %
3溶液
1.直流电机
答案: n0 = 700.3 rpm
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
ω₀ = V/k = 110/1.5 = 73.33 rad/s
n0 = 60·73.33/(2π) = 700.3 rpm
答案
n0 = 700.3 rpm
2.直流电机
答案: n2 = 1021 rpm
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。
变化前后的反电动势
E1 = 120 − 4·1 = 116 V
E2 = 120 − 1.6·1 = 118.4 V
新的转速
n2 = 1000·118.4/116 = 1021 rpm
答案
n2 = 1021 rpm
3.齿轮与齿轮系
答案: Ft = 4025 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.