电机学变压器涉及公式学习笔记(待补全)

变压器涉及公式学习笔记

绪论

铁磁材料及其特性

  • 铁耗(磁滞损耗与涡流损耗之和): p f e = P 1 50 ( f 50 ) β B m 2 G , β = 1.2 1.6 , f p_{fe}=P_{\frac{1}{50}}(\frac{f}{50})^\beta B_m^2G,\beta=1.2\sim1.6,f 为频率。

磁路相关的基本物理定律

  • 磁通连续性原理: ϕ = 0 \sum\phi=0

  • 全电流定律: H l = N i , Hl=Ni, 定义 N i Ni 为磁路磁动势,则磁路 K V L KVL 方程 F = N i = H l F=Ni=Hl 。当磁路存在气隙 δ \delta F = N i = H f e l f e + H δ δ F=Ni=H_{fe}l_{fe}+H_{\delta}\delta

  • 磁路欧姆定律与电感:

    • { F = ϕ R m N i = F = H l ϕ = B S B = μ H \begin{cases}F=\phi R_m\\Ni=F=Hl\\\phi=BS\\B=\mu H \end{cases} \Rightarrow { R m = l μ S Λ m = μ S l , \begin{cases} R_m=\frac{l}{\mu S}\\ \Lambda_m=\frac{\mu S}{l}\end{cases}, 铁磁性物质中 μ \mu μ f e \mu_{fe} 替代
    • ψ = N ϕ = N F Λ m = N 2 i Λ m = L i     L = N 2 Λ m \psi=N\phi=NF\Lambda_m=N^2i\Lambda_m=Li \ \ \ \Rightarrow L=N^2\Lambda_m
    • X = ω L = ω N 2 Λ m = ω N 2 μ S l X=\omega L=\omega N^2\Lambda_m=\omega N^2\frac{\mu S}{l}
  • 电磁感应定律: e = d ψ d t = N d ϕ d t e=-\frac{d\psi}{dt}=-N\frac{d\phi}{dt}

  • 能量转换效率: η = P 2 P 1 = ( 1 p P 2 + p ) , \eta=\frac{P_2}{P_1}=(1-\frac{\sum p}{P_2+\sum p}), 式中 P 1 P_1 为输入功率 P 2 P_2 为输出功率 p \sum p 为总损耗

变压器的基本工作原理和结构

额定值

  • 单相: S N = U 1 N I 1 N = U 2 N I 2 N S_N=U_{1N}I_{1N}=U_{2N}I_{2N}
  • 三相: S N = 3 U 1 N I 1 N = 3 U 2 N I 2 N S_N=\sqrt{3}U_{1N}I_{1N}=\sqrt{3}U_{2N}I_{2N}

变压器的运行分析

空载运行

  • 磁场分析与电动势分析:

    • U ˙ 1 I ˙ 0 F ˙ 0 = N 1 I ˙ 0 { ϕ ˙ { e 2 = N 2 d ϕ d t e 1 = N 1 d ϕ d t ϕ ˙ σ 1 e σ 1 = N 1 d ϕ σ 1 d t \dot U_1\rightarrow\dot I_0\rightarrow\dot F_0=N_1\dot I_0\Rightarrow\begin{cases}\dot\phi\rightarrow\begin{cases}e_2=-N_2\frac{d\phi}{dt}\\e_1=-N_1\frac{d\phi}{dt} \end{cases}\\\dot{\phi}_{\sigma_1}\rightarrow e_{\sigma_1}=-N_1\frac{d\phi_{\sigma_1}}{dt} \end{cases}
    • 主磁通感应电势: u 0 i 0 r 1 = e 1 + e σ 1 u_0-i_0r_1=e_1+e_{\sigma_1}
      • { ϕ = ϕ m s i n ω t e 1 = N 1 d ϕ d t e 1 = N 1 ϕ m ω c o s ω t = ω N 1 ϕ m s i n ( ω t π 2 ) = E 1 m s i n ( ω t π 2 ) \begin{cases} \phi=\phi_msin\omega t\\e_1=-N_1\frac{d\phi}{dt}\end{cases}\Rightarrow e_1=-N_1\phi_m\omega cos\omega t=\omega N_1\phi_msin(\omega t-\frac{\pi}{2})=E_{1m}sin(\omega t-\frac{\pi}{2})
      • 相量形式: E 1 ˙ = E 1 m ˙ 2 = j ω N 1 ϕ ˙ m 2 = j 4.44 f N 1 ϕ ˙ m \dot{E_1}=\frac{\dot{E_{1m}}}{\sqrt{2}}=-j\frac{\omega N_1\dot\phi_m}{\sqrt{2}}=-j4.44fN_1\dot\phi_m
    • 漏电动势分析: ϕ σ 1 = ϕ σ 1 m s i n ω t \phi_{\sigma_1}=\phi_{\sigma_{1m}}sin\omega t
      • e σ 1 = N 1 d ϕ σ 1 d t = ω N 1 ϕ σ 1 m s i n ( ω t π 2 ) = E 1 m s i n ( ω t π 2 ) ) e_{\sigma_1}=-N_1\frac{d\phi_{\sigma_1}}{dt}=\omega N_1\phi_{\sigma_{1m}}sin(\omega t-\frac{\pi}{2})=E_{1m}sin(\omega t-\frac{\pi}{2}))
      • 相量形式: E ˙ σ 1 = E ˙ σ 1 m 2 = j ω N 1 ϕ ˙ σ 1 m 2 = j 4.44 f N 1 ϕ ˙ σ 1 m \dot E_{\sigma_1}=\frac{\dot E_{\sigma_{1m}}}{\sqrt{2}}=-j\frac{\omega N_1\dot\phi_{\sigma_{1m}}}{\sqrt{2}}=-j4.44fN_1\dot\phi_{\sigma_{1m}}
      • 漏电抗: E ˙ σ 1 = j ω N 1 ϕ ˙ σ 1 m 2 I 0 ˙ I 0 ˙ = j ω N 1 ϕ σ 1 m 2 I 0 = j ω L σ 1 I 0 ˙ = j x 1 I 0 ˙ \dot E_{\sigma_1}=-j\frac{\omega N_1\dot\phi_{\sigma_{1m}}}{\sqrt{2}}\cdot \frac{\dot{I_0}}{\dot{I_0}}=-j\frac{\omega N_1\phi_{\sigma_{1m}}}{\sqrt{2}I_0}=-j\omega L_{\sigma_1}\dot{I_0}=-jx_1\dot{I_0} 其中 L σ 1 = N 1 ϕ σ 1 m 2 I 0 L_{\sigma_1}=\frac{N_1\phi_{\sigma_{1m}}}{\sqrt{2}I_0} 称为原绕组的漏电感, x 1 = ω L σ 1 = ω N 1 2 Λ σ 1 x_1=\omega L_{\sigma_1}=\omega N_1^2\Lambda_{\sigma_1} 称为原绕组的漏电抗,其大小不随电流大小变化。
    • 电动势平衡方程式:
      • 原边: U ˙ 1 = E ˙ 1 E ˙ σ 1 + I ˙ 0 r 1 = E ˙ 1 + I ˙ 0 ( r 1 + j x 1 ) = E 1 ˙ + I 0 ˙ z 1 , z 1 \dot U_1=-\dot E_1-\dot E_{\sigma_1}+\dot I_0r_1=-\dot E_1+\dot I_0(r_1+jx_1)=-\dot{E_1}+\dot{I_0}z_1,z_1 称为原绕组的漏阻抗。
      • 副边:因为空载, U ˙ 20 = E ˙ 2 \dot U_{20}=\dot E_2
      • 空载电流很小所以可认为 { U ˙ 1 E ˙ 1 U 1 E 1 = 4.44 f N 1 ϕ m \begin{cases} \dot U_1\approx-\dot E_1\\ U_1\approx E_1=4.44fN_1\phi_m\end{cases}
    • 变压器变比:定义为原边电动势与副边电动势之比 k = E 1 E 2 = 4.44 f N 1 ϕ m 4.44 f N 2 ϕ m = N 1 N 2 k=\frac{E_1}{E_2}=\frac{4.44fN_1\phi_m}{4.44fN_2\phi_m}=\frac{N_1}{N_2}
    • 空载电流分析
      • 大小: I 0 = E 1 r m 2 + x m 2 U 1 r m 2 + x m 2 I_0=\frac{E_1}{\sqrt{r_m^2+x_m^2}}\approx\frac{U_1}{\sqrt{r_m^2+x_m^2}}
      • 相位: I ˙ 0 = E ˙ 1 r m + j x m = E ˙ 1 Z m ψ 0 \dot I_0=\frac{-\dot E_1}{r_m+jx_m}=\frac{-\dot E_1}{|Z_m|}\angle-\psi_0 ,由于 x m > > r m x_m>>r_m ,所以 ψ 0 = t g 1 x m r m \psi_0=tg^{-1}\frac{x_m}{r_m} 接近 9 0 90^。
    • 空载运行方程式: { U ˙ 1 = E ˙ 1 + I ˙ 0 Z 1 E ˙ 1 = j 4.44 f N 1 ϕ ˙ m I ˙ 0 = E ˙ 1 Z m E ˙ 2 = j 4.44 f N 2 ϕ ˙ m U ˙ 20 = E ˙ 2 \begin{cases} \dot U_1=-\dot E_1+\dot I_0Z_1\\ \dot E_1=-j4.44fN_1\dot\phi_m\\ \dot I_0=\frac{-\dot E_1}{Z_m}\\ \dot E_2=-j4.44fN_2\dot\phi_m\\ \dot U_{20}=\dot E_2 \end{cases}

负载运行

  • 电磁分析及基本方程:

    • 电势分析及电势平衡:

      • U ˙ 1 I ˙ 1 F ˙ 1 = N 1 I ˙ 1 ϕ σ 1 E ˙ σ 1 U ˙ 2 I ˙ 2 F ˙ 2 = N 2 I ˙ 2 ϕ σ 2 E ˙ σ 2 \dot U_1\rightarrow\dot I_1\rightarrow\dot F_1=N_1\dot I_1\rightarrow\phi_{\sigma_1}\rightarrow \dot E_{\sigma_1}\\\dot U_2\rightarrow\dot I_2\rightarrow\dot F_2=N_2\dot I_2\rightarrow\phi_{\sigma_2}\rightarrow \dot E_{\sigma_2}
      • F ˙ m = F ˙ 1 + F ˙ 2 = N 1 I ˙ m ϕ m { E ˙ 1 E ˙ 2 \dot F_m=\dot F_1+\dot F_2=N_1\dot I_m\rightarrow\phi_m\rightarrow\begin{cases}\dot E_1\\ \dot E_2\end{cases}
    • 基本方程:

      • 电动势平衡方程式:

        { U ˙ 1 = ( E ˙ 1 + E ˙ σ 1 ) + I ˙ 1 r 1 = E ˙ 1 + I ˙ 1 r 1 + j I ˙ 1 x 1 = E ˙ 1 + I ˙ 1 z 1 U ˙ 2 = ( E ˙ 2 + E ˙ σ 2 ) I ˙ 2 r 2 = E ˙ 2 I ˙ 2 r 2 j I ˙ 2 x 2 = E ˙ 2 I ˙ 2 z 2 U ˙ 2 = I ˙ 2 z 2 \begin{cases}\dot U_1=-(\dot E_1+\dot E_{\sigma_1})+\dot I_1r_1=-\dot E_1+\dot I_1r_1+j\dot I_1x_1=-\dot E_1+\dot I_1z_1\\ \dot U_2=(\dot E_2+\dot E_{\sigma_2})-\dot I_2r_2=\dot E_2-\dot I_2r_2-j\dot I_2x_2=\dot E_2-\dot I_2z_2\\ \dot U_2=\dot I_2z_2\end{cases}

      • 七个基本方程: { U ˙ 1 = E ˙ 1 + I 1 z 1 U ˙ 2 = E ˙ 2 I ˙ 2 z 2 E ˙ 1 = j 4.44 f N 1 ϕ m E ˙ 2 = E ˙ 1 / k I ˙ 1 = I ˙ m + ( I ˙ 2 / k ) I ˙ m = E ˙ 1 / z m U ˙ 2 = I ˙ 2 Z L \begin{cases} \dot U_1=-\dot E_1+I_1z_1\\ \dot U_2=\dot E_2-\dot I_2z_2\\ \dot E_1=-j4.44fN_1\phi_m\\ \dot E_2=\dot E_1/k\\ \dot I_1=\dot I_m+(-\dot I_2/k)\\ \dot I_m=-\dot E_1/z_m\\ \dot U_2=\dot I_2Z_L\end{cases}

  • 等效归算规律:

    • 凡是单位为伏特的物理量的归算值等于其原来的 k k 倍,如 U ˙ = k U ˙ \dot U'=k\dot U
    • 电流的归算值等于其原来的 1 k \frac1k 倍,如 I ˙ = 1 k I ˙ \dot I'=\frac1k\dot I
    • 凡是单位为欧姆的物理量的归算值等于其原来的 k 2 k^2 倍,及 Z = k 2 Z Z'=k^2Z
  • 等效归算后的方程: { U ˙ 1 = E ˙ 1 + I 1 z 1 U ˙ 2 = E ˙ 2 I ˙ 2 z 2 E ˙ 1 = j 4.44 f N 1 ϕ m E ˙ 2 = E ˙ 1 I ˙ 1 = I ˙ m + ( I ˙ 2 ) I ˙ m = E ˙ 1 / z m U ˙ 2 = I ˙ 2 Z L \begin{cases} \dot U_1=-\dot E_1+I_1z_1\\ \dot U'_2=\dot E'_2-\dot I'_2z'_2\\ \dot E_1=-j4.44fN_1\phi_m\\ \dot E'_2=\dot E_1\\ \dot I_1=\dot I_m+(-\dot I'_2)\\ \dot I_m=-\dot E_1/z_m\\ \dot U'_2=\dot I'_2Z'_L\end{cases}

  • 功率分析: { P 1 = U 1 I 1 c o s φ 1 = p C u 1 + p f e + P M P 2 = P M p c u 2 = U 2 I 2 c o s φ 2 P M = E 2 I 2 c o s ψ 2 p c u 1 = I 1 2 r 1 p c u 2 = I 2 2 r 2 p f e = I m 2 r m \begin{cases}P_1=U_1I_1cos\varphi_1=p_{Cu1}+p_{fe}+P_M\\P_2=P_M-p_{cu2}=U'_2I'_2cos\varphi_2\\P_M=E'_2I'_2cos\psi_2\\p_{cu1}=I_1^2r_1\\p_{cu2}=I'^2_2r'_2\\p_{fe}=I_m^2r_m\end{cases}

参数测定

  • 空载实验:

    • 参数计算:
      • p f e = p 0 p c u 1 p 0 = I 0 2 r m r m = p 0 I 0 2 p_{fe}=p_0-p_{cu1}\approx p_0=I_0^2r_m\Rightarrow r_m=\frac{p_0}{I_0^2}
      • 空载总阻抗: z 0 = z 1 + z m z m = U 1 N I 0 z_0=z_1+z_m\approx z_m=\frac{U_{1N}}{I_0}
      • 激磁电抗: x m x 0 = z 0 2 r 0 2 x_m\approx x_0=\sqrt{z_0^2-r_0^2}
      • k = = U 20 U 1 N k=\frac{高压边匝数}{低压边匝数}=\frac{高压边电动势}{低压边电动势}\approx\frac{U_{20}}{U_{1N}}
  • 短路实验

    • 参数计算:

      • 负载损耗: p k = p c u 1 + p c u 2 + p f e p c u 1 + p c u 2 p_k=p_{cu1}+p_{cu2}+p_{fe}\approx p_{cu1}+p_{cu2} I 1 = I 1 N I_1=I_{1N} 时,功率损耗为额定的负载损耗 p k N p_{kN} ,所以稳态短路时 p k = p c u 1 + p c u 2 + p f e p c u 1 + p c u 2 = p k N p_k=p_{cu1}+p_{cu2}+p_{fe}\approx p_{cu1}+p_{cu2}=p_{kN}

      • 漏阻抗: z k = U K I K z_k=\frac{U_K}{I_K}

      • 短路电阻: r k = p k I k 2 r_k=\frac{p_k}{I_k^2}

      • 短路阻抗: x k = z k 2 r k 2 x_k=\sqrt{z_k^2-r_k^2}

      • 若要分离原副绕组的阻抗值,则 { r 1 = r 2 = r k 2 x 1 = x 2 = x k 2 \begin{cases} r_1=r'_2=\frac{r_k}2\\ x_1=x'_2=\frac{x_k}2\end{cases}

      • 由于电阻受温度影响,实际油浸电力变压器工作温度为75℃,所以:

        { r k 75 = r k 235 + 75 235 + θ z k 75 = r 75 2 + x k 2 \begin{cases} r_{k75℃}=r_k\frac{235+75}{235+\theta}\\ z_{k75℃}=\sqrt{r^2_{75℃}+x_k^2}\end{cases}

  • 三相变压器计算激磁阻抗时应用一相的功率,电压,电流计算

变压器运行性能

  • 阻抗电压: u k = I 1 N Z k 75 U 1 N 100 = Z k 75 Z 1 N = Z k 75 u k r = I 1 N r k 75 U 1 N 100 = r k 75 u k x = I 1 N x k U 1 N 100 = x k u_k=\frac{I_{1N}Z_{k75℃}}{U_{1N}}\cdot 100%=\frac{Z_{k75℃}}{Z_{1N}}=Z^*_{k75℃}\\u_{kr}=\frac{I_{1N}r_{k75℃}}{U_{1N}}\cdot 100%=r^*_{k75℃}\\u_{kx}=\frac{I_{1N}x_k}{U_{1N}}\cdot 100%=x^*_k

  • 电压调整率:

    • 定义: Δ U = U 20 U 2 U 2 N 100 = U 2 N U 2 U 2 N 100 = U 2 N U 2 U 2 N 100 = U 1 N U 2 U 1 N 100 \Delta U=\frac{U_{20}-U_2}{U_{2N}}\cdot100%=\frac{U_{2N}-U_2}{U_{2N}}\cdot100%=\frac{U'_{2N}-U'_2}{U'_{2N}}\cdot100%=\frac{U_{1N}-U'_2}{U_{1N}}\cdot100%
    • 由一般变压器 φ 1 φ 2 \varphi_1\approx\varphi_2 可得: Δ U I 1 r k c o s φ 2 + I 1 x k s i n φ 2 U 1 N = β u k r c o s φ 2 + β u k x s i n φ 2 \Delta U\approx\frac{I_1r_kcos\varphi_2+I_1x_ksin\varphi_2}{U_{1N}}=\beta u_{kr}cos\varphi_2+\beta u_{kx}sin\varphi_2 ,其中 β = I 1 I 1 N = I 2 I 2 N \beta=\frac{I_1}{I_{1N}}=\frac{I_2}{I_{2N}}
  • 变压器损耗与效率:

    • 效率: η = P 1 P 2 100 = ( 1 p f e + p c u p 2 + p f e + p c u ) 100 \eta=\frac{P_1}{P_2}\cdot100%=(1-\frac{p_{fe}+p_{cu}}{p_2+p_{fe}+p_{cu}})\cdot100%

    • 假定:

      • 计算 P 2 P_2 时忽略负载时 U 2 U_2 的变化。即:

        P 2 = U 2 I 2 c o s φ 2 U 2 N I 2 N ( I 2 I 2 N ) c o s φ 2 = β S N c o s φ 2 P_2=U_2I_2cos\varphi_2\approx U_{2N}I_{2N}(\frac{I_2}{I_{2N}})cos\varphi_2=\beta S_Ncos\varphi_2

      • 认为空载到负载,主磁通基本不变且忽略空载铜耗的影响:

        p f e = P 0 = p_{fe}=P_0=常数

      • 认为额定负载时的负载损耗等于额定电流时的短路损耗,稳态短路实验室外电压低磁密小,故忽略铁耗:

        p c u = I 1 2 r k = ( I 1 I 1 N ) 2 I 1 N 2 r k = β p k N p_{cu}=I_1^2r_k=(\frac{I_1}{I_{1N}})^2I_{1N}^2r_k=\beta p_{kN}

      • 从而 η = ( 1 p 0 + β 2 p k N β S N c o s φ 2 + p 0 + β 2 p k N ) 100 \eta=(1-\frac{p_0+\beta^2p_{kN}}{\beta S_Ncos\varphi_2+p_0+\beta^2p_{kN}})\cdot100%

      • d η d β = 0 \frac{d\eta}{d\beta}=0

        • 则最大值条件为: p 0 = β 2 p k N β = p 0 p k N p_0=\beta^2p_{kN}或\beta=\sqrt{\frac{p_0}{p_{kN}}}

        • 最大值为:
          η m a x = [ 1 2 p 0 p 0 p k N S N c o s φ 2 + 2 p 0 ] \eta_{max}=[1-\frac{2p_0}{\sqrt{\frac{p_0}{p_{kN}}}S_Ncos\varphi_2+2p_0}]

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