[PSA] Chapter 7 - Generator modeling II (Circuit Viewpoint)

7.1 Energy conversion

• We can get linear relationship between $\lambda$ and $i$

\[\boldsymbol{\lambda} = {\bf L}(\theta) \bf i\]

• $\bf i, \boldsymbol{\lambda}$ : 4-vectors, $\bf L$ is symmetric $4 \times 4$ matrix

\[\begin{aligned} {\it \boldsymbol{v}} &= {\bf Ri} + {d\boldsymbol{\lambda}\over dt}\bf i \\ &= {\bf Ri} + {\bf L}(\theta){d{\bf i} \over dt}+ {{d{\bf L}(\theta)}\over dt}\bf i \end{aligned}\]

• Where $\it {\boldsymbol{v}} = \text{col}[v_1, v_2, v_3, v_4]$ and ${\bf R } = \text{diag}[R_i]$

Mechanical equation of shaft

\[T = T_M + T_E = J\ddot{\theta}+ D \dot{\theta}\]

• $T_M $ : Mechanical torque, $T_E $ : Electrical torque by magnetic fields

Instantaneous power

\[p = {\bf i}^T {\it \boldsymbol{v}} = {\bf i}^T {\bf Ri}+{\bf i}^T {\bf L}(\theta) {d{\bf i} \over dt }+ {\bf i}^T {d{\bf L}(\theta) \over dt}\bf i\]

• From elementary circuit theory, we get expression for instantaneous stored magnetic energy of coupled coils. Which is :

\[W_{mag} = {1\over 2} {\bf i}^T{\bf L}(\theta) {\bf i} \tag{7.5}\]

• By differentiating, can acquire power supplied to the magnetic field (Used the fact that ${\bf L}(\theta)$ is symmetric matrix

\[\begin{aligned} \frac{dW_{\mathrm{mag}}}{dt} &= \frac{1}{2}\mathbf{i}^{T}\mathbf{L}(\theta)\frac{d\mathbf{i}}{dt} + \frac{1}{2}\frac{d\mathbf{i}^{T}}{dt}\mathbf{L}(\theta)\mathbf{i} + \frac{1}{2}\mathbf{i}^{T}\frac{d\mathbf{L}(\theta)}{dt}\mathbf{i} \\ &= \mathbf{i}^{T}\mathbf{L}(\theta)\frac{d\mathbf{i}}{dt} + \frac{1}{2}\mathbf{i}^{T}\frac{d\mathbf{L}(\theta)}{dt}\mathbf{i} \end{aligned} \tag{7.6}\]

• Combining two results

\[p=\mathbf{i}^{T}\mathbf{R}\mathbf{i}+\frac{dW_{\mathrm{mag}}}{dt}+\frac{1}{2}\mathbf{i}^{T}\frac{d\mathbf{L}(\theta)}{dt}\mathbf{i}\tag{7.7}\]

Interpretation : Electrical power in = Power dissipated in resistances + Power supplied to magnetic fields+Power converted into mechanical form

• By chain rule,

\[P_E=\frac{d\theta}{dt}\frac{1}{2}\mathbf{i}^{T}\frac{d\mathbf{L}(\theta)}{d\theta}\mathbf{i}=\omega T_E\tag{7.8}\]

• Recall fundarmentary physics formula : $P = Fv, P = T\omega$

\[T_E=\frac{1}{2}\mathbf{i}^{T}\frac{d\mathbf{L}(\theta)}{d\theta}\mathbf{i}\tag{7.9}\]

• By plugging in equations above,

\[J\ddot{\theta}+D\dot{\theta}-\frac{1}{2}\mathbf{i}^{T}\frac{d\mathbf{L}(\theta)}{d\theta}\mathbf{i}=T_M\tag{7.10}\]

7.2 Application to Synchronous Machine

\[\begin{aligned}\begin{bmatrix}v_{a'a} \\v_{b'b} \\v_{c'c} \\v_{FF'} \\v_{DD'} \\v_{QQ'}\end{bmatrix}&=\begin{bmatrix}r & 0 & 0 & 0 & 0 & 0 \\0 & r & 0 & 0 & 0 & 0 \\0 & 0 & r & 0 & 0 & 0 \\0 & 0 & 0 & r_F & 0 & 0 \\0 & 0 & 0 & 0 & r_D & 0 \\0 & 0 & 0 & 0 & 0 & r_Q\end{bmatrix}\begin{bmatrix}i_a \\i_b \\i_c \\i_F \\i_D \\i_Q\end{bmatrix}+\frac{d}{dt}\begin{bmatrix}\lambda_{aa'} \\\lambda_{bb'} \\\lambda_{cc'} \\\lambda_{FF'} \\\lambda_{DD'} \\\lambda_{QQ'}\end{bmatrix}\\&=\mathbf{R}\mathbf{i}+\frac{d\boldsymbol{\lambda}}{dt}\end{aligned}\tag{7.11}\]

Where Lowercase letters indicate stators quantities, Uppercase indicate rotor quantities

• $i_F$ : Field current, control through excitation control system
• $i_D, i_Q$ are currents in the fictitious rotor coils. (DQ)

• Apply generator convention : $\bf v = -v$ → ${\bf v} = -{\bf Ri}-{d{\bf \lambda} \over dt}$

Self-inductance of stator coils

\[L_{aa}=\frac{\lambda_{aa'}}{i_a}=L_s+L_m\cos 2\theta\qquad L_s > L_m \geq 0\] \[L_{bb}=\frac{\lambda_{bb'}}{i_b}=L_s+L_m\cos\left(2\left(\theta-\frac{2\pi}{3}\right)\right)\] \[L_{cc}=\frac{\lambda_{cc'}}{i_c}=L_s+L_m\cos\left(2\left(\theta+\frac{2\pi}{3}\right)\right)\]

• Because the air gap varies, we must account for varying inductance with $\theta$
• For round rotor, $L_{aa} = L_s$

Shift operator.

• Replace $\theta \leftarrow \theta +{2\over 3}\pi$ : $\mathcal T$ (Shift operator)
• $L_{bb} = {\mathcal T}L_{aa}, L_{cc} = {\mathcal T}^2 L_{aa}$

Mutual inductance between stator coils

\[\begin{aligned}L_{ab}&=\frac{\lambda_{ab'}}{i_b}=-\left[M_s + L_m\cos\left(2\left(\theta+\frac{\pi}{6}\right)\right)\right]\qquad M_s > L_m \geq 0\\L_{bc}&=\frac{\lambda_{bc'}}{i_c}=-\left[M_s + L_m\cos\left(2\left(\theta-\frac{\pi}{2}\right)\right)\right]\\L_{ca}&=\frac{\lambda_{ca'}}{i_a}=-\left[M_s + L_m\cos\left(2\left(\theta+\frac{5\pi}{6}\right)\right)\right]\end{aligned}\]

• Quantitative analysis indicates that $-{\pi \over 6}$ (average value of $-\pi/3, 0$ maximizes.
• $L_{bc} = {\mathcal T}L_{ab}, L_{ca} = {\mathcal T}^2 L_{ab}$

Other inductances

\[\begin{aligned}L_{aF}&=\frac{\lambda_{aF}}{i_F}=M_F\cos\theta\qquad M_F > 0\\L_{FF}&=\frac{\lambda_{FF'}}{i_F}=L_F\qquad L_F > 0\\L_{FQ}&=\frac{\lambda_{FQ'}}{i_Q}=0\end{aligned}\]

• Flux made by $i_F$ is aligned with $d$-axis
• $i_F$ is aligned with $d$-axis, so Flux seen by $i_F$ has no angle. → invarient with $\theta$
• $i_Q$ and $\lambda_{FF’}$ are perpendicular

\[\begin{aligned}\mathbf{L}(\theta)&=\begin{bmatrix}\mathbf{L}_{11}(\theta) & \mathbf{L}_{12}(\theta)\\\mathbf{L}_{21}(\theta) & \mathbf{L}_{22}(\theta)\end{bmatrix}\end{aligned}\tag{7.13}\]

Where those submatrices are

\[\begin{aligned}\mathbf{L}_{11}&=\begin{bmatrix}L_s+L_m\cos2\theta&-M_s-L_m\cos2\left(\theta+\frac{\pi}{6}\right)&-M_s-L_m\cos2\left(\theta+\frac{5\pi}{6}\right)\\[6pt]-M_s-L_m\cos2\left(\theta+\frac{\pi}{6}\right)&L_s+L_m\cos2\left(\theta-\frac{2\pi}{3}\right)&-M_s-L_m\cos2\left(\theta-\frac{\pi}{2}\right)\\[6pt]-M_s-L_m\cos2\left(\theta+\frac{5\pi}{6}\right)&-M_s-L_m\cos2\left(\theta-\frac{\pi}{2}\right)&L_s+L_m\cos2\left(\theta+\frac{2\pi}{3}\right)\end{bmatrix}\end{aligned}\] \[\begin{aligned}\mathbf{L}_{12}=\mathbf{L}_{21}^{T}&=\begin{bmatrix}M_F\cos\theta&M_D\cos\theta&M_Q\sin\theta\\[6pt]M_F\cos\left(\theta-\frac{2\pi}{3}\right)&M_D\cos\left(\theta-\frac{2\pi}{3}\right)&M_Q\sin\left(\theta-\frac{2\pi}{3}\right)\\[6pt]M_F\cos\left(\theta+\frac{2\pi}{3}\right)&M_D\cos\left(\theta+\frac{2\pi}{3}\right)&M_Q\sin\left(\theta+\frac{2\pi}{3}\right)\end{bmatrix}\end{aligned}\] \[\begin{aligned}\mathbf{L}_{22}&=\begin{bmatrix}L_F & M_R & 0\\M_R & L_D & 0\\0 & 0 & L_Q\end{bmatrix}\end{aligned}\]

7.3 The Park Transformation

• Park transformation (also called Blondel transformation, 0dq transformation ..)
• Transformation between abc frame to 0dq perspective

\[\begin{aligned}\begin{bmatrix}i_0\\i_d\\i_q\end{bmatrix}&=\sqrt{\frac{2}{3}}\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\[6pt]\cos\theta&\cos\left(\theta-\frac{2\pi}{3}\right)&\cos\left(\theta+\frac{2\pi}{3}\right)\\[6pt]\sin\theta&\sin\left(\theta-\frac{2\pi}{3}\right)&\sin\left(\theta+\frac{2\pi}{3}\right)\end{bmatrix}\begin{bmatrix}i_a\\i_b\\i_c\end{bmatrix}\end{aligned}\tag{7.16}\]

• In matrix notation,

\[{\bf i}_{0dq} = {\bf P i}_{abc} \tag{7.17}\]

• We can apply same transformation matrix to volatge and flux linkage

\[{\bf v}_{0dq} = {\bf P}{\bf v}_{abc} \tag{7.18}\] \[{\boldsymbol{\lambda}}_{0dq}={\bf P}{\boldsymbol{\lambda}}_{abc}\tag{7.19}\]

• $i_0 $ : zero-sequence current (From eqn 7.16, $i_0 = {1\over \sqrt 3}(i_a+i_b+i_c)$
• Inverse transform, (${\bf P}^T $ is orthonormal matrix)

\[\begin{aligned}{\bf P}^{-1}={\bf P}^{T}&=\sqrt{\frac{2}{3}}\begin{bmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\[6pt]\cos\theta&\cos\left(\theta-\frac{2\pi}{3}\right)&\cos\left(\theta+\frac{2\pi}{3}\right)\\[6pt]\sin\theta&\sin\left(\theta-\frac{2\pi}{3}\right)&\sin\left(\theta+\frac{2\pi}{3}\right)\end{bmatrix}\end{aligned}\tag{7.20}\]

• With original rotor quantities unaffected, constructs Block matrix as follows

\[\begin{aligned}{\bf i}_B\triangleq\begin{bmatrix}i_0\\i_d\\i_q\\i_F\\i_D\\i_Q\end{bmatrix}&=\begin{bmatrix}\begin{array}{c|c}{\bf P} & {\bf 0}\\\hline{\bf 0} & {\bf 1}\end{array}\end{bmatrix}\begin{bmatrix}i_a\\i_b\\i_c\\i_F\\i_D\\i_Q\end{bmatrix}={\bf B}{\bf i}\end{aligned}\tag{7.21}\] \[{\bf v}_B = \bf Bv \tag{7.22}\] \[{\boldsymbol \lambda}_B = \bf B{\boldsymbol{\lambda}} \tag{7.23}\]

• Also, orthonormal property holds

\[\begin{aligned}{\bf B}^{-1}={\bf B}^{T}&=\begin{bmatrix}{\bf P}^{T} & {\bf 0}\\{\bf 0} & {\bf 1}\end{bmatrix}\end{aligned}\tag{7.24}\]

• Using Linear algebra,

\[\begin{aligned}{\boldsymbol{\lambda}}&={\bf L}{\bf i}\\{\bf B}^{-1}{\boldsymbol{\lambda}}_B&={\bf L}{\bf B}^{-1}{\bf i}_B\\{\boldsymbol{\lambda}}_B&={\bf B}{\bf L}{\bf B}^{-1}{\bf i}_B={\bf L}_B{\bf i}_B\end{aligned}\tag{7.25}\]

• Also, we can easily show that

\[\begin{aligned}{\bf L}_B&=\begin{bmatrix}{\bf P} & {\bf 0}\\{\bf 0} & {\bf 1}\end{bmatrix}\begin{bmatrix}{\bf L}_{11} & {\bf L}_{12}\\{\bf L}_{21} & {\bf L}_{22}\end{bmatrix}\begin{bmatrix}{\bf P}^{T} & {\bf 0}\\{\bf 0} & {\bf 1}\end{bmatrix}\\[6pt]&=\begin{bmatrix}{\bf P}{\bf L}_{11}{\bf P}^{T}&{\bf P}{\bf L}_{12}\\{\bf L}_{21}{\bf P}^{T}&{\bf L}_{22}\end{bmatrix}\end{aligned}\tag{7.26}\] \[\begin{aligned}{\bf P}{\bf L}_{11}{\bf P}^{T}&=\begin{bmatrix}L_0 & 0 & 0\\0 & L_d & 0\\0 & 0 & L_q\end{bmatrix}\end{aligned}\tag{7.27}\]

Where

\[\begin{aligned}L_0 &\triangleq L_s-2M_s\\L_d &\triangleq L_s+M_s+\frac{3}{2}L_m\\L_q &\triangleq L_s+M_s-\frac{3}{2}L_m\end{aligned}\]

• If we close look at the matrices, First two rows of $\bf L_{21}$, $\bf P^T$ are aligned (proportional).
• We get

\[\begin{aligned} {\bf L}_{21}{\bf P}^{T} &= \begin{bmatrix} 0 & \sqrt{\frac{3}{2}}M_F & 0\\ 0 & \sqrt{\frac{3}{2}}M_D & 0\\ 0 & 0 & \sqrt{\frac{3}{2}}M_Q \end{bmatrix} \end{aligned} \tag{7.29}\] \[\begin{aligned}{\bf L}_B&=\begin{bmatrix}L_0 & 0 & 0 & 0 & 0 & 0\\0 & L_d & 0 & kM_F & kM_D & 0\\0 & 0 & L_q & 0 & 0 & kM_Q\\0 & kM_F & 0 & L_F & M_R & 0\\0 & kM_D & 0 & M_R & L_D & 0\\0 & 0 & kM_Q & 0 & 0 & L_Q\end{bmatrix}\end{aligned}\tag{7.30}\]

• ${\bf L}_B $ is sparse, symmetric, constant matrix with $k\triangleq \sqrt{3\over 2}$

7.4 Park’s Voltage Equation

• Remark voltage equation with matrix notation

\[{\bf v} = -{\bf Ri}-{d{\bf \lambda} \over dt}\]

• Resistance is invarient with transformation : ${\bf B R B}^{-1} ={\bf R}$
• Applying transformation,

\[\begin{aligned}{\bf B}^{-1}{\bf v}_B&=-{\bf R}{\bf B}^{-1}{\bf i}_B-\frac{d}{dt}\left({\bf B}^{-1}{\boldsymbol{\lambda}}_B\right)\\[6pt]{\bf v}_B&=-{\bf B}{\bf R}{\bf B}^{-1}{\bf i}_B-{\bf B}\frac{d}{dt}\left({\bf B}^{-1}{\boldsymbol{\lambda}}_B\right)\\[6pt]&=-{\bf R}_B{\bf i}_B-{\bf B}\frac{d{\bf B}^{-1}}{dt}{\boldsymbol{\lambda}}_B-\frac{d{\boldsymbol{\lambda}}_B}{dt}\end{aligned}\]

• We can easily calculate second term, by symmetry,

\[\begin{aligned} {\bf B}\frac{d{\bf B}^{-1}}{d\theta} &= \begin{bmatrix} 0 & 0 & 0 & \vert & \\ 0 & 0 & 1 & \vert & {\bf 0}\\ 0 & -1 & 0 & \vert & \\ \hline & {\bf 0}& & \vert & {\bf 0} \end{bmatrix} \end{aligned} \tag{7.37}\] \[\begin{aligned}{\bf v}_B&=-{\bf R}_B{\bf i}_B-\dot{\theta}\begin{bmatrix}0\\\lambda_q\\-\lambda_d\\0\\0\\0\end{bmatrix}-\frac{d{\boldsymbol{\lambda}}_B}{dt}\end{aligned}\tag{7.38}\]

1. If the shaft rotation is uniform, $\dot \theta =\text{constant}$, the system is linear and time invarient
2. Very simple equation compared to abc with $L = L(\theta)$

7.5 Park’s Mechanical equation

• Applying same formula as above,

\[\begin{aligned}T_E=\begin{bmatrix}i_0 & i_d & i_q & i_F & i_D & i_Q\end{bmatrix}\begin{bmatrix}0\\\lambda_q\\-\lambda_d\\0\\0\\0\end{bmatrix}&=i_d\lambda_q-i_q\lambda_d\end{aligned}\tag{7.43}\]

• We can substitute the result to mechanical equation

\[J\ddot{\theta} + D\dot{\theta} + i_q \lambda_d - i_d\lambda_q = T_M\]

7.6 Circuit model

• Can construct equivalent circuit model with results of Park Transformation

Zero-sequence

\[v_0 = -ri_0 - {d\lambda_0 \over dt} \tag{7.45}\]

Direct-axis

\[\begin{aligned} v_d &= -ri_d - \dot{\theta}\lambda_ q - {d\lambda_d \over dt} \\ v_F &= r_Fi_F + {d\lambda_F \over dt} \\ v_D &= r_Di_D + {d\lambda_D \over dt} = 0 \end{aligned}\tag{7.46}\]

Quadrature-axis

\[\begin{aligned} v_q &= -ri_q + \dot{\theta}\lambda_d - {d\lambda_q \over dt} \\ v_Q &= r_Qi_Q + {d\lambda_Q \over dt} = 0 \end{aligned}\tag{7.47}\]

• Except for $\dot \theta \lambda_q, \dot \theta \lambda_d$ terms, the three groups of equations are completely decoupled

\[\begin{aligned}\lambda_0&=L_0 i_0\end{aligned}\tag{7.48}\] \[\begin{aligned}\begin{bmatrix}\lambda_d\\\lambda_F\\\lambda_D\end{bmatrix}&=\begin{bmatrix}L_d & kM_F & kM_D\\kM_F & L_F & M_R\\kM_D & M_R & L_D\end{bmatrix}\begin{bmatrix}i_d\\i_F\\i_D\end{bmatrix}\end{aligned}\tag{7.49}\] \[\begin{aligned}\begin{bmatrix}\lambda_q\\\lambda_Q\end{bmatrix}&=\begin{bmatrix}L_q & kM_Q\\kM_Q & L_Q\end{bmatrix}\begin{bmatrix}i_q\\i_Q\end{bmatrix}\end{aligned}\tag{7.50}\]

Complete equivalent-circuit model

7.7 Instantaneous Power Output

• Calculate three-phase output

\[p_{3\phi} = {\bf i}^T_{abc} {\bf v}_{abc}\tag{7.51}\] \[p_{3\phi}(t) = {\bf i}^T_{0dq}{\bf PP}^{-1} {\bf v}_{0dq} = i_0v_0 + i_d v_d+i_qv_q\tag{7.51}\]

7.10 Steady-State Model

• Assume synchronous, positive-sequence, steady-state operation.
• Equations reduced to

\[\begin{aligned}v_d&=-ri_d-\omega_0\lambda_q\\v_q&=-ri_q+\omega_0\lambda_d\\v_F&=r_F i_F\end{aligned}\]

Where

\[\begin{aligned}\lambda_d&=L_d i_d+kM_F i_F\\\lambda_F&=kM_F i_d+L_F i_F\\\lambda_q&=L_q i_q\end{aligned}\]

• By substitution and replacement from lowercase to uppercase (Effective) terms,

\[(V_q+jV_d)e^{j\delta} = e^{j\delta}[-r(I_q+jI_d)+\omega_0 L_dI_d-j\omega_0L_qI_q+{1\over \sqrt 2}\omega_0M_Fi_F]\]

We defined EMF

\[\sqrt 2 E_a = \omega_0 M_Fi_Fe^{j\delta}\]

By rearranging, we get

\[\begin{aligned}E_a&=V_a+rI_a+jX_d I_d e^{j\delta}+jX_q I_q e^{j\delta}\\&=V_a+rI_a+jX_d I_{ad}+jX_q I_{aq}\end{aligned}\tag{7.67}\]

Where $I_{ad} \triangleq jI_de^{j\delta}, I_{aq} \triangleq I_qe^{j\delta}$

Same phasor diagram we considered at chapter 6

7.11 Simplified Dynamic Model

Assume following

1. Positive-sequence, synchronous operation
2. $\theta = \omega_0 t + {(\pi / 2)} + \delta$ with $\vert \dot \delta \vert « \omega_0$
3. $\dot \lambda_d$ and $\dot \lambda_q$ are small compared to $\omega_0 \lambda_q, \omega_0 \lambda_d$
4. Damper circuit is negligible ($i_D = i_Q = 0$) → Non-transient behavior
   • Governing Equations

\[v_d = -ri_d - \dot \theta \lambda_q \tag{7.68}\] \[v_q = -ri_q - \dot \theta \lambda_d \tag{7.69}\] \[v_F = r_Fi_F + {d\lambda_F \over dt} \tag {7.70}\]

• By assumption 2, $\dot \theta \approx \omega_0$
• In transient situation, $E_a $ is not always proportional to $i_F$ but $\lambda_F$ (Flux linkage)

\[v_F = r_Fi_F+{d\lambda_F \over dt}\]

• From park transformation, using equation (7.50), we can gather transient inductance and reactance.

\[L'_d \triangleq L_d-{(kM_F)^2 \over L_F}\]

• Define internal voltage proportional to $\lambda_F$

\[E_a' \triangleq {\omega_0 M_F \over \sqrt 2 L_F}e^{j\delta }\lambda_F\tag{7.71}\]

Where

\[\lambda_F = kM_F i_d + L_Fi_F \tag{7.65}\]

• Multiply by $\omega_0 M_F / \sqrt 2 r_F$ to Field voltage equation

\[E_{fd} \triangleq {\omega_0M_F \over \sqrt 2 r_F} v_F = {\omega_0 M_F \over \sqrt 2 r_F} r_Fi_F + {\omega_0 M_F \over \sqrt 2 r_F} {d\lambda_F \over dt}\]

Define timing constant $T’_{do} = {M_F \over r_F}$

\[E_{fd} = \vert E_a \vert + T'_{do} {d\vert E'_a \vert \over dt} \tag{7.75}\]

• Effective as one unit of $\vert E_a \vert$, often more preferred to use $E_{fd}$ than $v_F$ (same p.u)
• Recall mechanical equation

\[J\ddot \theta + D \dot \theta + i_q \lambda_d - i_d \lambda_q = T_M \tag{7.44}\]

Multiplying by $\dot \theta$

\[{d \over dt} \big({1\over 2 }J \dot \theta^2\big) + D\dot \theta^2 + \dot \theta ( i_q \lambda_d - i_d \lambda_q) = \dot \theta T_M\]

Can write with physical interpretation :

\[{d\over dt} W_{kinetic}+ P_{friction} + \dot \theta (i_q\lambda_d-i_d\lambda_q) = P_M\]

By equation 7.68, 7.69.

\[\begin{aligned}\dot{\theta}(i_q\lambda_d-i_d\lambda_q)&=i_qv_q+i_dv_d+r(i_q^2+i_d^2)\\&=3(I_QV_Q+I_DV_D)+3r(I_Q^2+I_D^2)\\&=3\Re(V_a I_a^*)+3r|I_a|^2\\&=3P_G+3r|I_a|^2\end{aligned}\tag{7.78}\]

Since $I^2 R$ losses are very small, we can neglect and end up with

\[M\ddot \delta + D \dot \delta + P_G = P_M\]

7.12 Generator connected to Infinite Bus (Linear Model)

• As a preliminary, define

\[K_3 = {\tilde X'_d \over \tilde X_d} = {X'_d + X_L \over X_d + X_L} \qquad (0<K_3 <1)\]

• merge transmission line reactance into generator reactance

\[|E_a| = \frac{1}{K_3}|E_a'| + \left(1-\frac{1}{K_3}\right) |V_\infty|\cos\delta\\] \[K_3T_{do}'\frac{d|E_a'|}{dt} + |E_a'| = K_3E_{fd} + (1-K_3)|V_\infty|\cos\delta \tag{7.84}\]

• $E’_a$ is a function of $\delta$
• $P_G$ is also a function of $\delta, \vert E’_a\vert$
• Plugging it into mechanical equation, assume steady-state operation conditions, denote as $^0$
• Also add small perturbations around the operating point

\[K_3T_{do}'\frac{d\Delta|E_a'|}{dt} + \Delta|E_a'| = K_3\Delta E_{fd} - (1-K_3)|V_\infty| \sin\delta^0\,\Delta\delta \tag{7.86}\]

and Let

\[K_4 \triangleq \left(\frac1{K_3}-1\right) |V_\infty|\sin\delta^0\]

Then

\[K_3T_{do}'\frac{d\Delta|E_a'|}{dt} + \Delta|E_a'| = K_3\Delta E_{fd} - K_3K_4\Delta\delta\]

• Perform linearlization

\[T\triangleq \left.{\partial P_G \over \partial \delta} \right|_0 = {\vert V_\infty \vert \over \tilde X'_d} \sin \delta^0\] \[K_2 \triangleq \left. \frac{\partial P_G}{\partial|E_a'|} \right|_0 = {\vert E'_a\vert^0\vert V_\infty \vert \over \tilde X'_d} \cos \delta^0 + \vert V_\infty \vert ^2 \big({1\over \tilde X_q} - {1 \over \tilde X'_d}\big)\cos 2\delta^0\]

Where $T$ : synchronizing power coefficient, slope over $P-\delta$ graph

Then

\[M\Delta\ddot\delta + D\Delta\dot\delta + T\Delta\delta + K_2\Delta|E_a'| = \Delta P_M \tag{7.87}\]

• Applying Laplace transformation into eqn 7.87

\[(K_3T_{do}'s+1)\Delta\widehat{|E_a'|} = K_3\Delta\widehat E_{fd} - K_3K_4\Delta\widehat\delta \tag{7.88}\] \[(Ms^2+Ds+T)\Delta\widehat\delta = \Delta\widehat P_M - K_2\Delta\widehat{|E_a'|} \tag{7.89}\]

\[\frac{\Delta\widehat\delta}{\Delta\widehat P_M}=\frac{G_2(s)}{1-K_2K_4G_1(s)G_2(s)}\tag{7.90}\] \[\frac{\Delta\widehat{|E_a'|}} {\Delta\widehat E_{fd}} = \frac{G_1(s)} {1-K_2K_4G_1(s)G_2(s)} \tag{7.91}\]

• Consider two types of pole

Rotor-angle poles $s_1, s_2$

\[Ms^2 + Ds+T = 0\]

• Indicates slightly damped electromechanical oscillation

Flux-decay pole $s_3$

\[1+K_3 T'_{do}s =0\]

• Portrays buildup and decay of field flux

• Root locus versus $K_2K_4$