最大盈亏功
出自 MBA智库百科(https://wiki.mbalib.com/)
目录[隐藏] |
什么是最大盈亏功[1]
最大盈亏功是指机械在变速稳定运转的一个周期内,动能的最大值和最小值之间的驱动功与阻抗功之差。
最大盈亏功的确定[2]
计算飞轮转动惯量必须首先确定最大盈亏功。若给出作用在主轴上的驱动力M'和阻力矩M”的变化规律,Amax便可确定如下:
图1(a)所示为某组稳定运转一个周期中,作用在主轴上的驱动力M'和阻力M”随主轴转角变化的曲线。μM为力矩比例尺,实际力矩值可用纵坐标高度乘以μM得到,即M=yμM;




![A_{0a}=\int_0^a(M'-M'')d\varphi= \int_0^a\mu M(y'-y'')dx\mu_\varphi=\mu_M\mu_\varphi[S_1]](/w/images/math/8/6/1/86128ef144e1f9634e6b351a6c152fb9.png)
式中[S1]——0a区间与
曲线之间的面积(mm2);
AOa——区间的盈亏功,以绝对值表示。
由图可见,0a区间阻力矩大于驱动力矩,出现亏功,机器动能减小,故标注负号;而ab区间驱动力矩大于阻力矩,出现盈功,机器动能增加,故标注正号。同理,bc、d0区间为负,cd区间为正。
盈亏功等于机器动能的增减量。设E0为主轴角位置

![E_a=E_0-A_{0a}=E_0-\mu_M\mu_\varphi[S_1]](/w/images/math/2/d/0/2d0b5f02053f9163e03481a4551841bb.png)
![E_b=E_a-A_{ab}=E_a-\mu_M\mu_\varphi[S_2]](/w/images/math/8/f/7/8f7c38fdec1cc2fa18b11cb8b52f558c.png)
![E_0=E_d-A_{d0}=E_d-\mu_M\mu_\varphi[S_5]](/w/images/math/2/9/3/293ef6840c246cb01fe73527c42df03d.png)






【例17.1】某机组作用在主轴上的阻力矩变化曲线如图2(a)所示。已知主轴上的驱动力M'为常数,主轴平均角速度
,机械运转速度不均匀系数δ = 0.02。
(1)求驱动力矩M'。
(2)求最大盈亏功Amax。
(3)求安装在主轴上的飞轮转动惯量J。
(4)若将飞轮安装在转速为主轴3倍的辅助轴上,求飞轮转动惯量f'。
【解】(1)求M'。因给定M'为常数,故为一水平直线。在一个运动循环中驱动力矩所做的功为2πM',它应当等于一个运动循环中阻力矩所做的功,即
2πM'=100×2π+400××2
解上式得M’=200 N·m。由此可作出的水平直线。














做错了吧 Amax算的不对啊