质量 m = 8 kg 的物体挂在弹簧 k = 2000 N/m 上,带有黏性阻尼器 c = 101 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 = 5 N。 求阻尼比 ζ。
m = 8 kgk = 2000 N/mc = 101 N·s/mF = 5 N
ζ =
3溶液
1.直流电机
答案: SR = 14.29 %
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 = 60·48/(2π·0.15) = 3056 rpm
nFL = 60·(48 − 5·1.2)/(2π·0.15) = 2674 rpm
转速调整率
SR = (3056 − 2674)/2674·100 = 14.29 %
答案
SR = 14.29 %
2.直流电机
答案: D = 17.26 %
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。
所需电枢电压
Va = 0.6·52.36 + 10·1 = 41.42 V
占空比
D = 41.42/240 = 17.26 %
答案
D = 17.26 %
3.齿轮与齿轮系
答案: N1 = 18
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.
总齿数
N1 + N2 = 2a/m = 2·63/2 = 63
按传动比分配
N1 = 63/(1 + 2.5) = 18
答案
N1 = 18
4.齿轮与齿轮系
答案: Tout = 104.3 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.