电机与拖动基础
第一章 电机的基本原理
第一节
P-10
\(B =\frac{\Phi}{S}\) (1-3)
\(B =\mu_{0}H\)
可以推得 \(H=\frac{\Phi}{\mu S}\)
公式(1-7)有\(F_{m}=Ni=H_{c}l_{c}=\frac{\Phi}{\mu S_{c}} \cdot l_{c} = \frac{l_{c}}{\mu S_{c}} \cdot \Phi\)
P-12
\(R_{mg} \gg R_{mc}\)
可以根据在电场中,电阻值大于导体电阻,因此数值上可以忽略电阻/铁芯。
P-13
图1-9,教材右端,正负极方向标反。
P-17
右手定则,大拇指为导体运动方向。
左右有力,右手有电。
第二节 机电能量转换
P-20
公式(1-17) \(\Psi = N\Phi\)
结合(1-14) \(e = -N \frac{d\Phi}{dt}\)
每一层线圈,可以看做一个磁通横截面,则N层线圈对应\(N\cdot \Phi\). 其中\(\Phi\) 为每一层的磁通量.
P-23
式(1-24)\(P=-ie=i\frac{d\Psi}{dt}\rightarrow Pdt = id\Psi\)
则 \(W_{e}=\int_{0}^{t}Pdt=\int_{0}^{\Psi}id\Psi\)
第三节 电机的基本结构与工作原理
P-32
励磁绕组直观图
P-33,教材p14,图1-14.
点击查看代码
t = -1.5:0.01:1.5;
f = zeros(size(t));
for i=1:length(t)
if t(i)>=-1.5&t(i)<=-0.5
f(i)=-0.3;
elseif t(i)>=-0.5&t(i)<=0.5
f(i)=0.3;
elseif t(i)>=0.5&t(i)<=1.5
f(i)=-0.3;
end
end
f1 = zeros(size(t));
f2 = zeros(size(t));
f3 = zeros(size(t));
f4 = zeros(size(t));
t1 = +0.2;
for i=1:length(t)
if t(i)>=-1.5 + t1&t(i)<=-0.5 + t1
f1(i)=-0.3;
elseif t(i)>=-0.5+ t1&t(i)<=0.5+ t1
f1(i)=0.3;
elseif t(i)>=0.5+ t1&t(i)<=1.5+ t1
f1(i)=-0.3;
end
end
t1 = +0.4;
for i=1:length(t)
if t(i)>=-1.5 + t1&t(i)<=-0.5 + t1
f2(i)=-0.3;
elseif t(i)>=-0.5+ t1&t(i)<=0.5+ t1
f2(i)=0.3;
elseif t(i)>=0.5+ t1&t(i)<=1.5+ t1
f2(i)=-0.3;
end
end
t1 = -0.2;
for i=1:length(t)
if t(i)>=-1.5 + t1&t(i)<=-0.5 + t1
f3(i)=-0.3;
elseif t(i)>=-0.5+ t1&t(i)<=0.5+ t1
f3(i)=0.3;
elseif t(i)>=0.5+ t1&t(i)<=1.5+ t1
f3(i)=-0.3;
end
end
t1 = -0.4;
for i=1:length(t)
if t(i)>=-1.5 + t1&t(i)<=-0.5 + t1
f4(i)=-0.3;
elseif t(i)>=-0.5+ t1&t(i)<=0.5+ t1
f4(i)=0.3;
elseif t(i)>=0.5+ t1&t(i)<=1.5+ t1
f4(i)=-0.3;
end
end
figure
subplot(5,1,1)
plot(t,f); grid on
subplot(5,1,2)
plot(t,f1); grid on
subplot(5,1,3)
plot(t,f2); grid on
subplot(5,1,4)
plot(t,f3); grid on
subplot(5,1,5)
plot(t,f4)
grid on
hold off
three_in_one = + f1+ f2
five_in_one = f + f1+ f2 + f3+ + f4;
figure
subplot(2,1,1)
plot(t,three_in_one); grid on
subplot(2,1,2)
plot(t,five_in_one)
P-39 教材P-14倒数第二段
正弦波有效值的平方是其峰值二次方的一半,注意,是有效值。
推导:
对于任意一个正弦波,
周期为\(T\)
有效值\(I_{e}\)
峰值(最大值)\(I_{m}\)
则其瞬时电流值为\(I=I_{m}\sin (\omega t)\)
瞬时热功率为\(dP=I^{2}R=I_{m}^{2}\sin^{2}(\omega t)R\)
\(\begin{aligned}
Q &=\int dP dt \\
& =I_{m}^{2}R \int \sin^{2}(\omega t)dt \\
& = I_{m}^{2}R \int \frac{1}{2} \left ( 1-\cos(2\omega t)\right )dt \\
& = \frac{1}{2} I_{m}^{2}R \int (1-\cos(2\omega t)) dt \\
& = \frac{1}{2} I_{m}^{2}R \left ( t|_{0}^{T}- \frac{1}{2\omega}\sin (2\omega t)|_{0}^{T} \right )\\
& = \frac{1}{2} I_{m}^{2}R \left [ (T-0) - \frac{1}{2\omega}\left ( \sin(2\omega T) - \sin(2\omega 0) \right ) \right ]\\
& = \frac{1}{2} I_{m}^{2}R\left [ T - \frac{1}{2\omega} (0-0) \right ] \\
& = \frac{1}{2} I_{m}^{2}RT
\end{aligned}\)
根据能量守恒定律,该周期\(T\)内,产生的热量相同
\(\begin{aligned}
Q_{equal} &= Q\\
I_{e}^{2}RT &= \frac{1}{2}I_{m}^{2}RT \\
I_{e}^{2} &= \frac{1}{2}I_{m}^{2}
\end{aligned}\)
该句话得证。
P-40
公式(1-38)
根据P-39,余弦定理合成矢量\(F_{sr}\),对角度\(\varphi\)全微分,即可得。
P-41
公式(1-39)下面的关系式,是正弦定理的推论
\(\frac{F_{sr}}{\sin \varphi_{sr}} = \frac{F_{s}}{\sin \varphi_{s}}=\frac{F_{r}}{\sin \varphi_{r}}\). 注:\(\sin \varphi _{sr} = \sin(\pi-\varphi _{sr})\),补角正弦值相等。
P-42
公式(1-42)
\(B_{av}=\frac{2}{\pi}B = \frac{2}{\pi}\mu_{0}H=\frac{2\mu_{0}}{\pi g}F_{sr}\)
\(B_{av}=\frac{2}{\pi}B\)
正弦波的平均值,我们只求\([0 ~ \pi]\)半个周期的平局值,时间长度为\(\pi\),对于幅值为\(B\)的正弦波,其瞬时值为 \(B\sin t\)
\(\begin{aligned}
B_{av} &= \frac{B_{all}}{T} \\
&=\frac{\int_{0}^{\pi}B \sin tdt}{\pi}\\
&=\frac{B}{\pi}\int_{0}^{\pi}\sin tdt \\
&= \frac{B}{\pi} \left( -\cos t\Large|_{0}^{\pi} \right) \\
& = \frac{2B}{\pi}
\end{aligned}\)
\(B_{av}=\frac{2}{\pi}B = \frac{2}{\pi}\mu_{0}H\)
参考公式(1-4)
\(B_{av}=\frac{2}{\pi}B = \frac{2}{\pi}\mu_{0}H=\frac{2\mu_{0}}{\pi g}F_{sr}\)
参考公式(1-6)
第二章
第三节
高等代数里面的常用积分公式表
\(\int \frac{1}{x}dx = \ln |x| +c\)
公式(2-12)
\(\begin{aligned}
\frac{\alpha_{e}-\alpha_{L}}{\frac{GD^{2}}{375}}dt &=\frac{d\Delta n}{\Delta n} ,(K=\frac{375}{GD^{2}}) \\
\frac{\alpha_{e}-\alpha_{L}}{\frac{1}{K}}dt &=\frac{d\Delta n}{\Delta n}\\
K(\alpha_{e}-\alpha_{L}) dt &=\frac{d\Delta n}{\Delta n}\\
\int K(\alpha_{e}-\alpha_{L}) dt &=\int \frac{d\Delta n}{\Delta n}\\
K(\alpha_{e}-\alpha_{L}) \int 1 dt &=\int \frac{1}{\Delta n}d\Delta n \\
K(\alpha_{e}-\alpha_{L}) t &= \ln |\Delta n| + c \\
e^{K(\alpha_{e}-\alpha_{L}) t} &= e^{\ln |\Delta n| + c} \\
e^{K(\alpha_{e}-\alpha_{L}) t} &= e^{\ln |\Delta n|} e^{c} \\
e^{-c} e^{K(\alpha_{e}-\alpha_{L}) t} &= e^{\ln |\Delta n|}, (C=e^{-c})\\
Ce^{K(\alpha_{e}-\alpha_{L}) t} &= \Delta n \\
\end{aligned}\)
此时求\(C\)值,带入\(t=0\)初始值\(\Delta n =\Delta n_{\rm st}\)
\(\begin{aligned}
\Delta n &= Ce^{K(\alpha_{e}-\alpha_{L}) t},t=0 \\
\Delta n &= Ce^{K(\alpha_{e}-\alpha_{L}) 0} \\
\Delta n &= Ce^{0} \\
\Delta n &= C ,\Delta n = n_{\rm st}\\
& \rightarrow C = n_{\rm st} \\
& \rightarrow \Delta n = Ce^{K(\alpha_{e}-\alpha_{L}) t} \\
& \rightarrow \Delta n = n_{\rm st} e^{K(\alpha_{e}-\alpha_{L}) t}
\end{aligned}\)
第三章 电力拖动系统的动力学基础
第一节 直流电机的基本原理和结构
\(T_{e}=C_{T}\Phi I_{a}\),\(C_{T}\)是与电机有关的常数;\(I_{a}\)是转子电流。
P-10 槽楔(xie):封槽口,压紧绝缘纸及线圈,防止松脱。
第三节
图3-14
\(\int_{-\frac{T}{2}}^{\frac{T}{2}} f(t) e^{iwt}= \int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)\cos(wt)dt + i \int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)\sin(wt)dt\)
该函数为奇函数
This function is an odd function with \(T=2 \pi\). Thus, the \(\cos\) part \(a_{n}\) turns to zero. \(b_{0}=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)dt\)
the \(\sin\) part comes as:
\(\begin{aligned}
b_{n} &=\frac{2}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)\sin(wt)dt \\
&=\frac{2}{T}* 4*\int_{0}^{\frac{T}{4}}f(t)\sin(wt)dt \leftarrow f(t)=\frac{F_{a}}{ \frac{T}{4}} \cdot t \\
&= \frac{2}{T}* 4*\frac{F_{a}}{ \frac{T}{4}} \int_{0}^{\frac{T}{4}}t\sin(wt)dt \\
&=\frac{32}{T^{2}}F_{a} \int_{0}^{\frac{T}{4}}t\sin(wt)dt \leftarrow T=2\pi\\
&=\frac{8}{\pi^{2}}F_{a}\int_{0}^{\frac{T}{4}}t\sin(wt)dt \leftarrow w=\frac{2\pi}{T}=1\\
&=\frac{8}{\pi^{2}} F_{a}\int_{0}^{\frac{T}{4}}t\sin(t)dt \\
&=\frac{8}{\pi^{2}}F_{a} \left [ -t\cos t + \sin t \right ] \Large |_{0}^{\frac{\pi}{2}} \\
& = \frac{8}{\pi^{2}}F_{a}
\end{aligned}\)
with
\(\begin{aligned}
\int t \sin t dt & = - \int t d\cos t \\
& = -t \cos t + \int \cos t dt \\
&= -t \cos t + \sin t + C
\end{aligned}\)
第四章
第一节他励直流电动机的机械特性
P-5 $T_{st} $ 是启动电流
第三节 例题4-2 第二问
拖动电机采用很功率负载,采用弱磁调速时,功率恒定,电枢电流不变。有
\(\begin{aligned}
P_{L}&=P_{x}\\
T_{e}w&=T_{x}w_{x}\rightarrow T_{e}=C_{T}\Phi I_{a},w=\frac{2\pi}{60}n \\
C_{T}\Phi I_{a}*\frac{2\pi}{60}n_{N} &= C_{T}\Phi_{x} I_{a}*\frac{2\pi}{60}n_{x}\\
\Phi n_{N}& = \Phi_{x}n_{x}\rightarrow \Phi_{x}=\frac{1}{3}\Phi\\
\Phi n_{N}& = \frac{1}{3}\Phi n_{x}\\
3 n_{N}& = n_{x}
\end{aligned}\)