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AC POWER ANALYSIS PART III

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### Transcript of AC POWER ANALYSIS - The Citadel, The Military...

AC POWER ANALYSIS PART III

AC POWER IN THE PHASOR DOMAIN

• Compose a complex quantity denoted by S whose real part is P, the real or average power, and imaginary part is Q, the reactive

power as

jQPS += )sin()cos( ivrmsrmsivrmsrms IVjIV θθθθ −+−=

• Rewriting in polar form, sSS = θ∠

where powerapprent 22 ==+== rmsrmsIVQPSS and anglefactor power the)(cos 1 ==−= − pfivs θθθ

• S is called COMPLEX POWER and its unit of measure is VA (volt-ampere).

• Alternatively, ∗∗ == rmsrms IVIVS21

POWER TRIANGLE

• Contain all the relavant power information in a given load

• Four pieces of information: apparent power (S), real power (P), reactive power (Q), and power factor angle (θpf)

• Given any two, the other two can be determined from the triangle using trigonometry.

ivrms

rms

rms

rmsIV

IV

IVZ θθ −∠===

∗∗

∗∗ ===

ZV

ZVVIVS rmsrms

rmsrmsrms2

ZIIIZIVS rmsrmsrmsrmsrms2=== ∗∗

ZIZ

VIVS rmsrms

rmsrms2

2=== ∗

QjPXjRIZIS rmsrms +=+== )(22

RISP rms2}Re{ ==

XISQ rms2}Im{ ==

Ex. Practice Problem 11.11

For a load, o85110∠=rmsV V, o154.0 ∠=rmsI A. Determine: (a) the complex and apparent powers, (b) the real and reactive powers, (c) the power factor and the load impedance.

Ex. Practice Problem 11.12

A sinusoidal source supplies 20 kVAR reactive power to load .75250 Ω−∠= oZ Determine:

(a) the power factor, (b) the apparent power delivered to the load, and (c) the rms voltage.

PRINCIPLE OF CONSERVATION OF AC POWER

• In a given AC circuit, the instantaneous, real (average), reactive, and complex powers supplied by the sources equal the

respective sums of each type of powers of the individual loads regardless of how the loads are connected.

• In general, for a source connected to N loads,

Reactive power: Nloadloadloadsource QQQQ ,2,1, +++= LL • The principle does not apply to apparent power !!!!!!

Ex. Practice Problem 11.14

Two loads connected in parallel are respectively 2 kW at a leading pf of 0.75 and 4 kW at a lagging pf of 0.95. Calculate the pf of the two loads. Find the complex power supplied by the source.

POWER FACTOR CORRECTION

• The process of increasing the power factor closer to unity without altering the voltage or current to the original load.

• Most loads are inductive (+Q) → reduce the reactive power by adding a capacitor (−Q) in parallel with the load

• With the same supplied voltage, the first circuit draws larger current than the second circuit.

222

}Im{ rmsccrmsc

rmscc CVSQCVj

ZVjQS ωω ==→−==−= ∗

)tan(tan 212 θθω −== PVCQ rmsc → 2

21 )tan(tan

rmsVPC

ωθθ −

=

Ex. Practice Problem 11.15

Find the value of parallel capacitance needed to correct a load of 140 kVAR at 0.85 lagging pf to unity pf. Assume that the load is supplied by a 110-V(rms), 60-Hz line.

Ex. An industrial client is charged a penalty if the plant power factor drops below 0.85. The equivalent loads are as shown. The frequency is 60 Hz. (a) Determine the total complex power and the power factor. (b) What value of capacitance is required to avoid a penalty. (c) Determine generator current before and after correction.