Transmission & distribution formulas

$\rho =1000{\text{kg/m}}^3$, $g=9.81{\text{m/s}}^2$, $Q$ 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.