### Frequent Questions

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only risk is using conceptual data offered here in production.

only risk is using conceptual data offered here in production.

the reason why
\sqrt{3}
is common in power analysis equations is due to the relationship between phase voltage and line-line voltage.

V_{LL}=\sqrt{3}V_{\phi}

**example:**three phase apparent power is equal to three times phase voltage times phase current.S=3V_{\phi&space;}I

S=3\frac{V{_{LL}}}{\sqrt{3}}I

S=\sqrt{3}V{_{LL}}I

**consider one phase of a three phase system:**

we know that the average power is:

P=V{_{\phi}}I\cos[\theta_{v}-\theta_{i}]

where:cos[\theta_{v}-\theta_{i}] is known as the power factor.

so with phasor analysis if we used:

\bar{S}=\bar{V{_{\phi}}}\bar{I}=V{_{\phi}}\angle\theta_{v}I\angle\theta_{i}=V_{\phi}I\angle\theta_{v}+\theta_{i}

\bar{S}=V_{\phi}I\cos[\theta_{v}+\theta_{i}]+jV_{\phi}I\sin[\theta_{v}+\theta_{i}]" title="S=V_{\phi }I\cos[\theta _{v}+\theta _{i}]+jV_{\phi }I\sin[\theta _{v}+\theta _{i}]

since the real part is power... this result says that:P=V{_{\phi}}I\cos[\theta_{v}+\theta_{i}]

**← this is not correct !**

now repeat with the complex congugate:

\bar{S}=\bar{V{_{\phi}}}\bar{I}^{*}=V{_{\phi}}\angle\theta_{v}I\angle-\theta_{i}=V_{\phi}I\angle\theta_{v}-\theta_{i}

\bar{S}=V_{\phi}I\cos[\theta_{v}-\theta_{i}]+jV_{\phi}I\sin[\theta_{v}-\theta_{i}]" title="S=V_{\phi }I\cos[\theta _{v}+\theta _{i}]+jV_{\phi }I\sin[\theta _{v}+\theta _{i}]

so now the real power is:P=V{_{\phi}}I\cos[\theta_{v}-\theta_{i}]

**← this is correct !**

reactance comes in two forms:

capacitive reactance and inductive reactance.

capactiors supply reactive power.

inductors (reactors) absorb reactive power.

so if you have and inductive load on a bus...

the system is supplying the reactive power that the load is absorbing.

if you place a shunt capacitor on the bus (in parallel with the load)...

the shunt capacitor is supplying some or all of the reactive power to the load.

therfore, the system is supplying less reactive power (and current) to the load.

this means less voltage drop between the system source and the load bus...

resulting in higher voltage on the load bus.

placing a shunt reactor on a bus has the opposite effect...

will result in a lower voltage on the bus.

capacitive reactance and inductive reactance.

X_{C}=\frac{-j}{\omega C}\hspace{15px}X_{L}=j\omega L

by convention:capactiors supply reactive power.

inductors (reactors) absorb reactive power.

so if you have and inductive load on a bus...

the system is supplying the reactive power that the load is absorbing.

if you place a shunt capacitor on the bus (in parallel with the load)...

the shunt capacitor is supplying some or all of the reactive power to the load.

therfore, the system is supplying less reactive power (and current) to the load.

this means less voltage drop between the system source and the load bus...

resulting in higher voltage on the load bus.

placing a shunt reactor on a bus has the opposite effect...

will result in a lower voltage on the bus.

yes...

PF=0.95\hspace{10px}Q\approx\frac{1}{3}P

PF=0.90\hspace{10px}Q\approx\frac{1}{2}P

Impedance Z is a complex combination of Resistance and Reactance

so B is the Susceptance and is the reactive or imaginary part of Admittance

Y\ne\frac{1}{R}+\frac{1}{jX}

Z=R+jX

Admittance Y is the reciprocal of ImpedanceY=\frac{1}{Z}=G+jB

Admittance is a complex combination of Conductance and Susceptanceso B is the Susceptance and is the reactive or imaginary part of Admittance

Y\ne\frac{1}{R}+\frac{1}{jX}

**← don't fall into this trap !**
this is due to the zero sequence current that circulates in the delta winding

the delta tertiary impedance is effectively connected to neutral in zero sequence

the delta tertiary impedance is effectively connected to neutral in zero sequence

short answer is that an autotransformer is less expensive to construct.

only practical if secondary-primary voltage ratio is less than about 3.

the primary and secondary windings are both magnetically coupled and electrically connected.

therefore, no isolation between the primary and secondary.

the primary and secondary both reference the same "neutral" (or ground) voltage.

only practical if secondary-primary voltage ratio is less than about 3.

the primary and secondary windings are both magnetically coupled and electrically connected.

therefore, no isolation between the primary and secondary.

the primary and secondary both reference the same "neutral" (or ground) voltage.

**as the current increases in a transmission line:**

the line inductance absorbs more and more reactive power.

the reactive power absorbed by the line is supplied by both the system and the line capacitance.

**when the line current is very small:**

the line capacitance supplies the reactive power to the line with any excess supplied to the system.

the line is net capacitive.

**when the line current is very large:**

the line capacitance supplies part of the reactive power to the line and the system supplies the rest.

the line is net inductive

**at some amount of current flow:**

the reactive power supplied by the line capacitance equals the reactive power absorbed by line inductance.

the line is net resistive.

this amount current results in a MW flow known as the

**Surge Impedance Loading (SIL)**.

**another way to say it:**

if a line is terminated with it's surge impedance

**(Z**...

_{o})the line is in a "Balanced" state where the reactive power absorbed by the line inductance is equal to the reactive power supplied by the line capacitance.

Z_{o}=\sqrt{\frac{X}{B}}\hspace{20px} SIL=V^{2}\sqrt{\frac{B}{X}}

**transformer test reports provide a quanitity called %Z, also known as impedance voltage.**

the impedance voltage is a measured value that is a fraction or percent of the full rated phase voltage.

**under test...**

the transformer secondary is shorted.

the primary voltage is increased until full rated current flows.

this voltage divided by the full rated voltage is the impedance voltage or Z

_{pu}

**transformer test reports provide a quanitity called "full-load losses".**

the transformer per-unit resistance is the full-load losses divided by the transformer MVA rating.

R_{pu}=\frac{P_{Loss}}{S_{Base}}

**P**quantifies the

_{ST}**P**erception of voltage flicker (

**S**hort

**T**erm).

P

_{ST}values are measured in real-time with IEEE or IEC compliant flicker-meters.

voltage measurements are gathered then statistically evaluated in successive 10 minute periods.

any P

_{ST}value >= 1.0 is considered to be irritating flicker to humans.

an analytical calculation of P

_{ST}is likely impossible... it requires memory of past events.

a "worst-case" evaluation is usually performed for planning purposes.

the short answer is... because of transformers.

the

there are four (4) quantities to consider...

the other two (2) are calcuated with Ohm's or Watt's law.

with the

all nominal

the

**Per-Unit**method was developed to aid in the analysis of systems containing multiple voltage levels.there are four (4) quantities to consider...

V,\hspace{3px}I,\hspace{3px}Z,\hspace{3px}and\hspace{3px}S

the **Per-Unit**method allows you to choose two (2) of them for "Base" or reference values.the other two (2) are calcuated with Ohm's or Watt's law.

with the

**Per-Unit**method...all nominal

**Per-Unit**voltages = 1.0