Power Engineering¶

General¶

Constants¶

1 feet = 12 inches
1 mile = 1.609 km

\[\mu_0=4\pi\times10^{-7}\]

\[\epsilon_0=8.854\times10^{-12}\]

Formulas¶

\[II^* = |I|^2\]

because $(a+jb)(a-jb) = a^2+b^2$

Phasor¶

\[re^{j\theta}=rcos\theta+rjsin\theta=r\angle\theta\]

Hyperbolic functions¶

\[cosh(x)=\dfrac{e^x+e^{-x}}{2}\]

\[sinh(x)=\dfrac{e^x-e^{-x}}{2}\]

\[tanh^{-1}(x)=\dfrac{1}{2}ln(\dfrac{1+x}{1-x})\]

Generator Rotating Speed¶

rotating speed of a generator = 120 x frequency / # of poles

e.g. 60Hz, 2 poles -> 3600 rpm

Ch2 Basic Principles¶

Complex Power¶

\[\phi=\theta_V-\theta_I\]

cos & sin 積起來時被消掉

\[V_{max}=\sqrt{2}|V|\]

\[\begin{align*} v(t) &=V_{max}cos(\omega t+\phi)\\ &=\sqrt{2}|V|cos(\omega t+\phi)\\ \end{align*}\]

I lags V $\phi$ -> phasor = $\phi$

\[\phi = \angle V - \angle I\]

lagging -> $\phi>0$
leading -> $\phi<0$

power factor (PF)

\[PF = cos(\phi)\]

\[S = |S|e^{j\phi} = |V||I|e^{j\phi}\]

\[S = |S|cos(\phi) + j|S|sin(\phi) = P+jQ\]

\[tan(\phi) = \dfrac{Q}{P}\]

\[tan(cos^{-1}(PF))=\dfrac{Q}{P}\]

\[|S| = |I||V|\]

\[|I| = \dfrac{|S|}{|V|}\]

Impedance¶

\[V = IZ\]

\[Z = R+jX\]

\[S = VI^* = ZII^* = |I|^2Z = |I|^2(R+jX) = P+jQ\]

\[P = |I|^2R\]

\[Q = |I|^2X\]

\[S = VI^* = V(\dfrac{V}{Z})^* = \dfrac{|V|^2}{Z^*}\]

Three-Phase Power¶

Each separated by $120^{\circ}$, s.t. the instantaneous power is constant

Balanced Three-Phase Power¶

Each of the neutral points has the same voltage.

pf

Knowing this, we can simplify the circuit

Impedance¶

Delta Wye Transformation

\[Z_\lambda = \dfrac{1}{3}Z_\Delta\]

pf

Voltage¶

We know $V_{an}$, $V_{bn}$, $V_{cn}$ are in sequence with common difference = $-120^\circ$, now we can get the relationship between $V_{an}$ & $V_{ab}$

\[V_{ab}=\sqrt{3}\angle{30^{\circ}}V_{an}\]

Power¶

Power = single phase power x 3

Problems¶

Basic Formula¶

Per phase¶

If 440V is $3\phi$, $I$ should be $\dfrac{S_{1\phi}}{\frac{440}{\sqrt{3}}}=328.04$ A

http://publish.illinois.edu/ece-476-fall-2017/files/2017/08/HW1Sol.pdf

Delta Wye Formula¶

Positive sequence

\[V_{ca}=208\angle{-120^{\circ}}\]

\[V_{bc}=208\angle{0^{\circ}}\]

\[V_{ab}=208\angle{120^{\circ}}\]

\[V_{ab}=\sqrt{3}\angle{30^{\circ}}V_{an}\]

$$V_{an}=\dfrac{V_{ab}}

{\sqrt{3}\angle{30^{\circ}}}=120\angle{90^\circ}$$

\[V_{bn}=\dfrac{V_{bc}}{\sqrt{3}\angle{30^{\circ}}}=120\angle{-30^\circ}\]

\[V_{cn}=\dfrac{V_{ca}}{\sqrt{3}\angle{30^{\circ}}}=120\angle{210^\circ}\]

\[I_a=\dfrac{V_{an}}{Z}=12\angle{105^\circ}\]

\[I_b=\dfrac{V_{bn}}{Z}=12\angle{-15^\circ}\]

\[I_c=\dfrac{V_{bn}}{Z}=12\angle{225^\circ}\]

\[S_{3\phi}=V_{an}I_a^*+V_{bn}I_b^*+V_{cn}I_c^*=\underline{4320\angle{-15^\circ}}\]

Ch3 Transmission Line¶

H = magnetic field intensity
B = magnetic field density

\[B=\mu H\]

Infinite straight wire¶

$H$ outside the condutor

$H$ inside the condutor

Conductor Bundling¶

GMR = geometric mean radius = goemetric mean of $r'$ & the distance from one point to each of the other points

\[r'=re^{-\frac{1}{4}}=0.78r\]

Inductance per meter $l$ $\(l=\dfrac{\mu_0}{2\pi}ln\dfrac{D_m}{R_b}$\)

Inductance per meter¶

Inductance per meter of three-phase tranposed lines

Phase-Neutral capacitance¶

phase-neutral capacitance

\[\bar{c}=\dfrac{2\pi\epsilon}{ln\dfrac{D_m}{R^c_b}}\]

Note that $R_b^c$ uses $r$ instead of $r'=re^{-\frac{1}{4}}$ !!!!

Ch4 Transmission-Line Modeling¶

Terminal¶

$Z_c$ = surge impedance

$P_\mathrm{SIL}$ = surge impedance loading

\[P_\mathrm{SIL}=\dfrac{|V_1|^2}{Z_c}\]

\[Z_c=\sqrt{\dfrac{z}{y}}\]

\[\gamma=\sqrt{zy}\]

$\gamma = \alpha+j\beta$

$\gamma$ = propagation constant
$\beta$ = phase constant

Transmission Matrix¶

Short Line¶

$S_{12}$ = complex power from bus 1 to bus 2

\[S_{12}=P_{12}+jQ_{12}\]

\[\begin{align*} S_{12} &= V_{1}I_{1}^*\\ &= V_1(\dfrac{V_1-V_2}{Z})^*\\ &= \dfrac{|V_1|^2}{Z^*}-\dfrac{V_1V_2^*}{Z^*}\\ &= \dfrac{|V_1|^2}{|Z|}e^{j\angle Z}-\dfrac{|V_1||V_2|}{|Z|}e^{j\angle Z} e^{j\theta_{12}} \end{align*}\]

Power Cycle Diagram¶

Radial Line¶

Voltage at near, complex load at rear

Use Z=jX in the short line $S_{12}$ formula

\[\begin{align*} S_D &=P_D+Q_D\\ &=P_D+j\beta P_D\\ \end{align*}\]

\[\beta=\dfrac{Q_D}{P_D}=tan(\phi)\]

$P_{12}=-P_{21}$

\[\begin{align*} |V_2|^2=\dfrac{|V_1|^2}{2}-\beta P_D X\pm \sqrt{(\dfrac{|V_1|^2}{2})^2-P_DX(P_DX+\beta\dfrac{|V_1|^2}{2})} \end{align*}\]

Ch5 Transformer¶

$a=\dfrac{N_1}{N_2}$
$V_1=aV_2$
$I_1=\dfrac{1}{a}I_2$

Inductance¶

\[R=\dfrac{l}{\mu A}\]

\[L_m'=\dfrac{N^2}{R_m}\]

Autotransformer¶

Redraw into 5.28 (b)
Calculate
- $V_2=V_1+2V_1=3V_1=360V$
- $V_2=I_2(7+j8)=I_2(10.63\angle 48.81^\circ)$
- $I_2=33.87\angle(-48.81^\circ)$
- $I_1=2I_2+I_2=3I_2=101.61\angle(-48.81^\circ)$

Three-Phase Transformer¶

Problem

Wye-Delta connection
$n=58$
$V_{a'n'}=\dfrac{n}{\sqrt{3}}\angle(-30^\circ)V_{an}$
$V_{bn}=V_{an}\angle(-120^\circ)$

Per-Phase Transformer Analysis¶

$K_1$ = gain = $\sqrt{3}n\angle(30^\circ)$

Per Unit Normalization¶

indicated by p.u.

Changing base

Ch6 Generator Modeling 1 - Machine Viewpoint¶

Round-Rotor Machine¶

\[E_a=V_a+I_a(r+jX_s)\]

Problem

\[P=0.5=|V_a||I_a|cos(\phi)=|I_a|cos\phi\]

\[\begin{align*} & E_a=V_a+I_a(j1+j0.5)\\ &= 1+|I_a|\angle(-\phi)(j1.5)\\ &= 1+1.5|I_a|cos(-\phi+90^\circ)+j1.5|I_a|sin(-\phi+90^\circ)\\ &= 1+1.5|I_a|sin(\phi)+j1.5|I_a|cos(\phi)\\ &= 1+1.5|I_a|sin(\phi)+j0.75 \end{align*}\]

\[|E_a|=1.5\]

\[1+1.5|I_a|sin\phi=\sqrt{1.5^2-0.75^2}=1.3\]

\[|I_a|sin\phi=0.2\]

\[|I_a|=\sqrt{0.5^2+0.2^2}=0.539\]

\[\phi=tan^{-1}(\dfrac{0.2}{0.5})=21.8^\circ\]

\[I_a=0.539\angle(-21.8^\circ)\]