The whole Power Systems formula sheet, to print

$\rho =1000{\text{kg/m}}^3$, $g=9.81{\text{m/s}}^2$, $Q$ in m${}^3$/s, $H$ net head in m, $\eta$ overall efficiency; result in W.

Both over the same period; each is at most 1.

Individual maximum demands occur at different times, so their sum is at least the group's maximum demand. A higher diversity factor means less plant capacity is needed to supply the same consumers.

$D_m$ is the GMD between phases, $D_s$ the GMR of one phase conductor (self GMD). For a three-phase line, $D_m=\sqrt[3]{D_{ab}{D}_{bc}{D}_{ca}}$. For a bundled phase, $D_s$ is the bundle GMR from the table below.

Transposed line, ground effect neglected; $D_m$ is the same GMD as for inductance. The capacitance uses the actual radius $r$ (or the bundle's equivalent radius built with $r$, not $r^{\prime }$), because it depends on the surface charge, not on internal flux.

The loop inductance is twice the inductance of one conductor. The line-to-line capacitance is half the capacitance of either conductor to neutral, $C_n=2\pi {\epsilon }_0/\ln (D/r)$. Valid for $D\gg r$.

$AD-BC=1$ holds for any passive, linear, bilateral two-port, such as a transmission line. $A=D$ holds in addition when the two-port is symmetrical; the short, nominal-$\pi$, nominal-T and long-line models are all symmetrical.

$z$ and $y$ are the series impedance and shunt admittance per unit length.

$V_S$ held constant; magnitudes per phase (or line-to-line throughout). Multiply by 100 for per cent. For a short line $|V_{R,\text{NL}}|=|V_S|$. Regulation can be negative, for example with a leading-power-factor load. Efficiency is $P_R/P_S$ at the same load.

$A=|A|\angle {\theta }_A$, $B=|B|\angle {\theta }_B$, $V_R=|V_R|\angle 0^{\circ }$, $V_S=|V_S|\angle \delta$. With line-to-line voltages the result is three-phase power; with line-to-neutral voltages it is power per phase. For a lossless short line ($A=1$, $B=jX$) it reduces to $P=|V_S||V_R|\sin \delta /X$.

$L$, $C$ per unit length; $V_L$ is the line-to-line voltage; SIL is the total three-phase power at which the line draws zero net reactive power. At SIL the voltage magnitude is the same all along the line. Below SIL the line produces net vars and the receiving voltage tends to rise; above SIL it absorbs vars and the voltage sags.

The receiving-end voltage exceeds the sending-end voltage because the line charging current flows through the series inductance. The approximation holds for small $\beta l$; it equals $V_S(1+{\omega }^{2}LC{l}^2/2)$ with $L$, $C$ per unit length, so the rise grows roughly with the square of the length.

$X$ is the series reactance of the line; $\delta$ is the angle by which $V_S$ leads $V_R$. Maximum power at $\delta =90^{\circ }$ is $|V_S||V_R|/X$ (steady-state limit). Use line-to-line voltages for three-phase power and line-to-neutral voltages for power per phase. For a long lossless line, replace $X$ by $|B|=Z_c\sin \beta l$.

The insulation resistance is radial, so it is inversely proportional to the cable length $l$.

$V$ is the voltage between conductor and sheath (use rms for rms stress), $r\le x\le R$. Stress is greatest at the conductor surface and least at the sheath.

$C_s$ is the core-to-sheath and $C_c$ the core-to-core capacitance. $C_1$: all three cores bunched, measured against the sheath. $C_2$: two cores bunched with the sheath, measured against the third core. $C_n$ is the capacitance of each core to neutral; the charging current per phase is $\omega C_n{V}_{ph}$.

$k$ is the degree of series compensation. Then $P_{\max }=|V_S||V_R|/(X_L(1-k))$, raised by the factor $1/(1-k)$. Voltage drop and regulation also improve.

A shunt capacitor supplies $Q$ and a shunt reactor absorbs it. Use $V$ and $Q$ on a consistent basis: per phase with $V_{ph}$, or three-phase with the line-to-line voltage and star-connected $X_{\text{sh}}$ per phase. A capacitor's output falls as $V^2$, so it gives least support when the voltage is low.

Units numbered from the cross-arm end; the total string voltage is $V_1+V_2+V_3$ (phase-to-earth), and $V_3$ is the largest.

It is at most 100%, reached only if $k=0$, and it falls as $n$ increases. Capacitance grading (units of different capacitance), a guard ring (static shielding) and longer cross-arms (smaller $k$) all improve it.

Lagging load of power factor $\cos \varphi$ and current $I$; $R$ and $X$ are the per-phase resistance and reactance of the section. Valid when the angle between the two end voltages is small. A leading load makes the $X\sin \varphi$ term negative.

Load current per unit length constant along the whole length, at unity power factor (or DC). $I_{\text{tot}}$ is the total load current and $R_{\text{tot}}$ the distributor resistance (the loop resistance of both conductors for a two-wire DC or single-phase distributor). The drop is half that of the same total current concentrated at the far end.

Same symbols and conditions as above. The minimum voltage is at the mid-point.

Three-phase MVA base $S_{base}$ in MVA and line-to-line kV base $V_{base}$ in kV; the impedance is per phase of the equivalent star.

Per-unit voltage, current and power follow in the same way: $V_{pu}=V/V_{base}$, $I_{pu}=I/I_{base}$, $S_{pu}=S/S_{base}$.

Used when an equipment rating (given on its own MVA and kV) must be moved to the system base; the rated MVA and kV are the old base.

Line-to-line rated voltages of the transformer; $S_{base}$ is unchanged across it.

Phasors in per unit; $Y_{ik}$ are elements of the bus admittance matrix.

Here $Y_{ik}=G_{ik}+j{B}_{ik}$ and ${\delta }_{ik}={\theta }_i-{\theta }_k$. The equations are non-linear, so they are solved by iteration.

$y_{i0}$ is the total shunt admittance from bus $i$ to ground; $y_{ik}$ is the series admittance between $i$ and $k$. Valid with no mutual coupling and nominal-ratio transformers.

$P_i$ and $Q_i$ are net injections in per unit. Buses already updated in this iteration use their new values.

Mismatch $\Delta P=P_{spec}-P_{calc}$, $\Delta Q=Q_{spec}-Q_{calc}$; update $\theta \leftarrow \theta +\Delta \theta$, $|V|\leftarrow |V|+\Delta |V|$ and repeat.

$n$ buses, one slack, $m$ PV buses (polar form). $J_1$ is $(n-1)\times (n-1)$, $J_4$ is $(n-1-m)\times (n-1-m)$, $J_2$ is $(n-1)\times (n-1-m)$ and $J_3$ is $(n-1-m)\times (n-1)$.

Assumes $|V|=1$ pu at all buses, $r\ll x$, small angle differences, so $\sin \delta \approx \delta$. Angles in radians. $B^{\prime }$ is the same matrix as in the fast decoupled method (from $1/x$, slack row and column removed), and $\mathbf{P}$ holds the net real-power injections of the non-slack buses. It gives real power flows only, and no losses.

Cost in Rs/h, $P_i$ in MW; incremental cost is in Rs/MWh and is linear in $P_i$.

A unit held at $P_{i,\max }$ has $d{C}_i/d{P}_i\le \lambda$; a unit held at $P_{i,\min }$ has $d{C}_i/d{P}_i\ge \lambda$. $\lambda$ is the system incremental cost in Rs/MWh.

Valid while every unit is inside its limits. The fixed term $a_i$ does not affect the dispatch.

Symmetric $B_{ij}$. For two units: $P_L=B_{11}{P}_1^2+2B_{12}{P}_1{P}_2+B_{22}{P}_2^2$. $\partial P_L/\partial P_i$ is the incremental transmission loss of unit $i$.

Holds for units inside limits. Here $\lambda$ is the incremental cost of power delivered at the load.

Every unit inside limits has the same product of penalty factor and incremental cost. The approximation holds when $\partial P_L/\partial P_i$ is small. A unit with larger $\partial P_L/\partial P_i$ has a larger $L_i$ and so a lower optimal IC.

$R_i$ in per unit on the unit's own rating; $\Delta P_i$ is the change in generator output in steady state with the speed changer setting fixed.

All units settle at one common frequency. A smaller $R$ or a larger rating takes a larger share of a load change, as long as no unit reaches its limit.

Free governor action, no supplementary control. $\Delta P_D$ is the load increase in MW and $D$ is load damping in MW/Hz (set $D=0$ if not given).

For one isolated area with governor action only (no integral control). $1/R_i$ is the governor response of unit $i$ and $D$ the load damping.

Steady state with no supplementary control; $\Delta P_{12}$ is the change in flow from area 1 to area 2. For a load step in area 1 only, $\Delta P_{12}=-{\beta }_2\Delta P_{D1}/({\beta }_1+{\beta }_2)$: area 2 sends part of the step over the tie line.

$\Delta P_{tie}$ is the tie-line flow deviation (positive out of the area) and $B_f>0$ is the frequency bias in MW/Hz, commonly set equal to the area's $\beta$. The integral controller changes the set point as $\Delta P_{ref}=-K\int \mathrm{ACE}dt$: negative ACE raises generation.

Per phase or per unit consistently; $P$ and $Q$ are the real and lagging reactive power received, $V$ the receiving-end voltage; valid for a small angle across the line. In a mainly inductive line ($X\gg R$) the drop depends mostly on $Q$.

Reactive power $\Delta Q_{inj}$ injected (for example by a shunt capacitor) at a bus whose Thevenin source reactance is $X$; $S_{sc}$ is the three-phase short-circuit level at that bus (per-unit form, $X$ dominant).

Tap-changing transformers at both ends of a line of impedance $R+jX$. $V_1$ and $V_2$ are the nominal voltages at the two ends referred to the line side, $P$ and $Q$ are the power received at end 2, and the angle across the line is small. With $V_1=V_2=V$: $t_s^2=1/\left(1-(RP+XQ)/V^2\right)$.

Total three-phase kVAr for total real power $P$ in kW. The rating is the reactive power the bank delivers at its rated voltage.

$Q_c$ is the three-phase VAr, $V_L$ the line voltage in V, $\omega =2\pi f$, and $C$ comes out in F. Note $C_{\Delta }=C_Y/3$.

At constant $P$ and constant load voltage, the source-side current falls when $\cos {\varphi }_2>\cos {\varphi }_1$, and feeder $I^{2}R$ loss falls with its square.

Constant $P$; demand charge $x$ per kVA of maximum demand per year and capacitor annual cost $y$ per kVAr per year (both linear). It minimises $xP/\cos {\varphi }_2+yP(\tan {\varphi }_1-\tan {\varphi }_2)$.

Multiplying a phasor by $a$ rotates it by $120^{\circ }$ anticlockwise without changing its magnitude. Also $a^2-a=-j\sqrt{3}$.

Same form for currents. Here $V_b=V_{a0}+a^2{V}_{a1}+a{V}_{a2}$ and $V_c=V_{a0}+a{V}_{a1}+a^2{V}_{a2}$.

Same form for currents. The inverse matrix is $\frac{1}{3}$ times the complex conjugate of the synthesis matrix.

Only the zero-sequence component flows in the neutral (or earth) return path. If there is no neutral path, $I_{a0}=0$ in the line currents. So the line currents of a three-wire system (delta or ungrounded star) contain no zero-sequence component, although zero-sequence current can still circulate inside a delta winding.

The factor 3 appears because the transformation used here (with $\frac{1}{3}$ in the analysis equations) is not power-invariant. There are no cross terms between different sequences.

For a system that is balanced before the fault, with balanced sources, only the positive-sequence network has a source: $E_a$, the pre-fault voltage of the reference phase $a$ at the fault bus. The negative and zero networks are passive. $Z_1,Z_2,Z_0$ are the Thevenin impedances seen from the fault point, and $I_{a1},I_{a2},I_{a0}$ are the sequence currents flowing out of the system into the fault.

A neutral impedance $Z_n$ carries $3I_{a0}$, so it appears as $3Z_n$ in the single-phase zero-sequence network. It does not appear in the positive or negative networks.

$Z_1$ is the positive-sequence Thevenin impedance at the fault bus, $Z_f$ the impedance in each phase of the fault. For a bolted fault $Z_f=0$.

$S_{base}$ in VA for the three-phase rating and $V_{base}$ the line-to-line voltage in volts, giving $I_{base}$ in amperes.

For a bolted three-phase fault with $E=1$ pu. In per unit it equals $1/Z_1$. $S_{base}$ is in MVA, $V_{LL}$ is the base (rated) line-to-line voltage in volts and $I_f$ is the fault current in amperes.

Convert every element to the common MVA base, and to the common voltage base in each voltage zone, before combining.

A voltage $v=V_m\sin (\omega t+\alpha )$ drives a series $R$-$L$ circuit that carries no current at $t=0$. The angle $\alpha$ sets the instant of the fault on the voltage wave. There is no DC offset when $\alpha =\theta$. The DC offset is largest when $\alpha -\theta =\pm 90^{\circ }$. Then, if $R$ is small compared with $\omega L$, the first current peak approaches $2V_m/|Z|$, twice the steady-state peak. This model takes a constant reactance and ignores the machine's change from sub-transient to synchronous reactance.

Fault conditions: $I_b=I_c=0$ and $V_a=Z_f{I}_a$. The fault current through the ground is $I_a$.

$Z_f$ is the impedance connected between $b$ and $c$. $I_a=0$ and $|I_b|=|I_c|=\sqrt{3}|I_{a1}|$. No current flows to ground.

$Z_f$ is the impedance in the common earth path of both phases. $I_a=0$ and the current into the ground is $I_b+I_c=3I_{a0}$.

$V_k(0)$ is the pre-fault voltage at bus $k$, $V_i(F)$ the post-fault voltage at bus $i$, and $Z_{bus}$ the positive-sequence matrix.

Find the sequence fault currents $I_k^{(1)},I_k^{(2)},I_k^{(0)}$ from the formulas of the previous section with $Z_1=Z_{kk}^{(1)}$, $Z_2=Z_{kk}^{(2)}$, $Z_0=Z_{kk}^{(0)}$ and $E=V_k(0)$. Each sequence voltage is updated with its own matrix, then combined into phase voltages with the synthesis matrix. If a star-delta transformer lies between buses $i$ and $k$, shift the positive and negative sequence values by its phase shift before combining.

Apply to every element $Z_{ij}$ of the existing matrix. Both results come from adding a temporary extra row and column and eliminating it.

N_CT = CT ratio and N_VT = VT ratio, each as primary over secondary. Z_pri is the impedance in primary ohms, from the primary voltage and current.

Pickup current = plug setting (as a fraction) times the CT rated secondary current. The relay operates only if PSM > 1.

k and alpha depend on the characteristic (table below). Time is in seconds, and the TMS scales the TMS = 1 curve linearly. If a problem gives its own curve, or a table of operating times against PSM for TMS = 1, read the time from it and multiply by the TMS.

I_1 and I_2 are the CT secondary current phasors at the two ends, both taken in the same sense through the zone. K is the slope (bias) setting, for example 0.2 for 20 %. A small minimum pickup is usually added.

I_L,Delta = line current on the power transformer's delta-winding side, whose CTs are in star. I_L,Y = line current on the star-winding side, whose CTs are in delta. N = ratio of each CT set, primary over secondary. Both sides refer to the same through load.

Z_r = relay reach (the diameter) along the maximum-torque angle tau. theta = angle of the measured impedance. The reach shrinks as the impedance angle moves away from tau. j is the imaginary unit.

theta = angle by which I lags V, so Z = V/I at angle theta. K_4 (spring restraint) is neglected in the three special cases. Here a negative constant restrains and a positive one operates.

x = distance to a bolted fault, Z_1 and Z_0 = positive- and zero-sequence line impedance per unit length, phasor quantities measured at the relay. The phase pair (b, c) works for any fault involving those two phases. The earth-fault element on phase a needs the residual compensation k_0, or it measures the wrong distance.

Assumes resistance is neglected and the system voltage is constant during the transient (f_n is much higher than the supply frequency). V_m = system voltage across the pole at current zero; for a purely inductive fault circuit this is the peak phase voltage. L and C = inductance and capacitance of the circuit seen from the breaker. The peak occurs at t = pi/omega_n and the maximum RRRV at t = pi/(2 omega_n). The average RRRV up to the peak is 2V_m/(pi/omega_n) = 4 V_m f_n.

For the source inductance L in series, with a resistor R across the contacts in parallel with the capacitance C. If R <= R_crit, the restriking voltage is non-oscillatory.

V_L = rated line voltage in kV. I_b = rated symmetrical (rms) breaking current in kA. Breaking capacity is in MVA. The making current is the peak of the first current loop, about 1.8 x sqrt(2) = 2.55 times the symmetrical rms breaking current. If the current at contact separation has a dc component I_dc, the asymmetrical rms breaking current is sqrt(I_b^2 + I_dc^2).

$\delta$ in electrical radians, $t$ in seconds, $f$ the system frequency in Hz, $P_m$ and $P_e$ in per unit on the same base as $H$; $P_a$ is the accelerating power. Equivalent form: $\frac{2H}{{\omega }_s}\ddot{\delta }=P_m-P_e$ with ${\omega }_s=2\pi f$.

Same equation with $\delta$ in electrical degrees (since $\pi$ rad $=180^{\circ }$). Do not mix the two forms.

Unit: seconds (MJ per MVA). In the second form $J$ is in kg m${}^2$, ${\omega }_{sm}$ the synchronous mechanical speed in rad/s and $S_{\text{machine}}$ in VA. Stored energy is $H{S}_{\text{machine}}$ megajoules when $S$ is in MVA.

$p$ is the number of poles. The swing equation above uses the electrical angle.

The stored energy in MJ is fixed; only the MVA it is divided by changes.

Applies to machines that swing coherently (same $\delta$ and speed), each $H_i$ on its own rating $S_i$. The stored energies in MJ add.

$E$ and $V$ are the voltage magnitudes at the two ends, $X$ the total series reactance between them; per unit values give $P_e$ in per unit.

Of the two angles with this sine, the stable operating point is the one below $90^{\circ }$.

Positive for ${\delta }_0<90^{\circ }$ (stable), zero at $90^{\circ }$, negative beyond. Units: per unit power per electrical radian.

Undamped natural frequency, from linearising the swing equation about ${\delta }_0$ with $\Delta \delta$ in radians: $\frac{H}{\pi f}\Delta \ddot{\delta }=-P_s\Delta \delta$.

Reached when load is increased slowly: beyond $\delta =90^{\circ }$ the coefficient $P_s$ becomes negative and the machine falls out of step.

${\delta }_m$ is the maximum swing angle. Valid for the classical model (constant $P_m$, no damping).

$P_m$ jumps from $P_{m0}$ to $P_{m1}$ with the network unchanged, ${\delta }_0={\sin }^{-1}(P_{m0}/P_{\max })$; angles in radians where they multiply $P_{m1}$. Stable only if ${\delta }_m\le \pi -{\delta }_1$ with ${\delta }_1={\sin }^{-1}(P_{m1}/P_{\max })$.