[PSA] Chapter 2 - Basic Principles

2.1 Phasor Representation

\[\begin{equation} e^{\pm j\theta} = \cos\theta \pm j\sin\theta \tag{2.1} \end{equation}\]

• Remark Euler’s identity

\[\begin{equation} \begin{aligned} v(t) &= V_{\max}\cos(\omega t+\theta_v) \\ &= V_{\max}\operatorname{Re} \!\left[e^{j(\omega t+\theta_v)}\right] \\ &= V_{\max}\operatorname{Re} \!\left[e^{j\theta_v}e^{j\omega t}\right] \end{aligned} \tag{2.4} \end{equation}\]

• Conventional to use the cosine function in analyzing the sinusoidal steady state

Effective Phasor representation

\[\begin{equation} V = \frac{V_{\max}e^{j\theta_v}}{\sqrt{2}} \tag{2.6} \end{equation}\]

• Complex number that carres the amplitude and angle

\[\begin{equation} \begin{aligned} V &= \frac{ V_{\max}\cos\theta_v +jV_{\max}\sin\theta_v }{\sqrt{2}} \\ &= |V|\cos\theta_v +j|V|\sin\theta_v \end{aligned} \tag{2.7} \end{equation}\]

2.2 Complex power supplied to a one-port

\[\begin{equation} v(t) = V_{\max}\cos(\omega t+\theta_v) \tag{2.8} \end{equation}\] \[\begin{equation} i(t) = I_{\max}\cos(\omega t+\theta_i) \tag{2.9} \end{equation}\]

• sometimes we represent phase angle of current and voltage by $\theta_V = \phase{V}, \theta_I = \phase{I}$
• Function of power over time $p(t) = v(t)i(t)$

\[\begin{equation}\begin{aligned}p(t)&= V_{\max} I_{\max}\cos(\omega t+\theta_v)\cos(\omega t+\theta_i) \\&= \frac{1}{2}V_{\max}I_{\max}\left[\cos(\theta_v-\theta_i)+\cos(2\omega t+\theta_v+\theta_i)\right]\end{aligned}\tag{2.10}\end{equation}\]

• Averaged term + sinusoidal compnent of frequency $2\omega$
• Twice as many zero crossings of $p(t)$ as of $v(t) $ or $i(t)$

Power factor angle / Average power over one period

\[\begin{equation}\phi \triangleq \theta_v - \theta_i\tag{2.11}\end{equation}\] \[\begin{equation}\begin{aligned}P&= \frac{1}{T}\int_{0}^{T} p(t)\,dt \\&= \frac{1}{2}V_{\max}I_{\max}\cos\phi\end{aligned}\tag{2.12}\end{equation}\]

• Using effective phasors

\[\begin{equation}v(t)=V_{\max}\cos(\omega t+\theta_v)\quad\Longleftrightarrow\quad V=\frac{V_{\max}}{\sqrt{2}}e^{j\theta_v}\tag{2.13}\end{equation}\] \[\begin{equation}v(t)=\operatorname{Re}\left\{\sqrt{2}\,V e^{j\omega t}\right\}\tag{2.14}\end{equation}\]

→ using effective phasors(where $V$ is RMS value of $v(t)$)

• Average power dissipated in a resistor with resistance $R$ connected with effective voltage $V$ = Same equation with DC

\[\begin{equation}\begin{aligned}P&=\frac{1}{2}V_{\max}I_{\max}\cos\phi \\&=|V||I|\cos\phi \\&=\operatorname{Re}\left[|V|e^{j\angle V}|I|e^{-j\angle I}\right] \\&=\operatorname{Re}\left(VI^{*}\right)\end{aligned}\tag{2.15}\end{equation}\] \[\cos \phi = \Re(e^{j\phase{V}}e^{j\phase{I}})\]

Power Factor(PF)

\[\begin{equation}\mathrm{PF}\triangleq\cos\phi\tag{2.16}\end{equation}\]

\[\begin{equation}\begin{aligned}S &\triangleq VI^{*} \\Q &\triangleq \Im\left(VI^{*}\right)\end{aligned}\tag{2.17}\end{equation}\]

Terminology and descriptive units

• $S$ : Complex power. $[VA, kVA, MVA]$
• $\vert S \vert$ : Apparent power. $[VA, kVA, MVA]$
• $P$ : Average or real or active power. $[W, kW, MW]$
• $Q$ : Reactive power. $[VAr, kVAr, MVAr]$ (r : reactive)

2.4 Balanced Three-phase

• Positive sequence (order of a, b and c) : 0, -120, 120 degree
• Negative sequence (a, c, b) or 0, 120, -120 degree

Wye and delta source

• In case of Case II, assume that $E_{ca}+E_{bc}+E_{ab}=0$ → series resistance is zero, so if not zero, circulating current would be infinite → so make sum of Voltage is zero, current is inderminate(0/0) → assume its circulating current 0

Symmetric three-phase network and neutral voltage

• For every symmetric three-phase network, neutral voltage $V_{nn’}=0$
• when impedence $Z_n$ is connected between two neutrals, $V_{nn’}=0$

Delta-Wye transformation (Symmetrical case)

• For load transformation, $Z_{\lambda} = Z_\Delta /3$ : Use KVL, and KCL
• For source transformation

• For positive-sequence,

\[V_{an} = {1\over \sqrt 3} e^{-j\pi / 6} V_{ab}\]

2.5 Per phase analysis

• Per-phase analysis reduces balanced three-phase network to one.
• Assumption
  1. balanced three-phase (connected) system
  2. all loads and sources are wye-connected
  3. no mutual inductance between phases
• We can say
  1. all the neutrals have same potential
  2. phases are completely decoupled (i.e. can be interpreted independently)
  3. all corresponding network variables occur in balanced sets of same sequence of source (if positive, postive .. i.e.)
• Method of per-phase analysis
  1. convert all $\Delta$-connection to Y connection
  2. analyze only phase a circuit
  3. Subtract 120, 240 degree to get b, c
  4. Get back to original circuit if necessary

2.6 Balanced Three-phase power

• Instantaneous power delived to a load is constant

\[\begin{equation}S_{3\phi}=V_a I_a^{*}+V_b I_b^{*}+V_c I_c^{*}\tag{2.34}\end{equation}\]

• Sum of three phase complex power

\[\begin{equation}\begin{aligned}S_{3\phi}&=V_a I_a^{*}+V_a e^{-j2\pi/3} I_a^{*} e^{j2\pi/3}+V_a e^{j2\pi/3} I_a^{*} e^{-j2\pi/3}\end{aligned}\tag{2.35}\end{equation}\] \[\begin{equation}S_{3\phi}=3V_a I_a^{*}=3S\tag{2.36}\end{equation}\]

• $S_{3\phi}$ is three times of per-phase complex power

\[\begin{equation}\begin{aligned}p_{3\phi}(t)&=v_a(t)i_a(t)+v_b(t)i_b(t)+v_c(t)i_c(t) \\&=|V||I|\left[\cos\phi+\cos\left(2\omega t+\angle V+\angle I\right)\right] \\&\quad+|V||I|\left[\cos\phi+\cos\left(2\omega t+\angle V+\angle I-\frac{4\pi}{3}\right)\right] \\&\quad+|V||I|\left[\cos\phi+\cos\left(2\omega t+\angle V+\angle I+\frac{4\pi}{3}\right)\right] \\&=3|V||I|\cos\phi \\&=3P\end{aligned}\tag{2.37}\end{equation}\]