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ELEC130 - Electrical Engineering 1

ELEC130 - Electrical Engineering 1. Week 8 Module 4 Applications in Power Systems. This Lecture. Module 4 - Applications in Power Systems Revise power triangle Examples using the power triangle & calculating Power Factor Correction Module 5 - Transients in a.c. Circuits

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ELEC130 - Electrical Engineering 1

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  1. ELEC130 - Electrical Engineering 1 Week 8 Module 4 Applications in Power Systems

  2. This Lecture • Module 4 - Applications in Power Systems • Revise power triangle • Examples using the power triangle & calculating Power Factor Correction • Module 5 - Transients in a.c. Circuits • (Will cover in next lecture) Lecture 8

  3. Administration • Laboratory Test Results • Quiz No. 3 • End of this lecture • can pick up quiz from Department Office next week. • Survey - Week 7 • power triangle • complete examples! • TopClass Lecture 8

  4. rms - root-mean-square (revision) • Looking for the value of current (or voltage) which has the effect of delivering the same amount of power whether the source is ac of DC • If the average power delivered by the ac Source is to = the average power delivered by the DC Source then ……………………… • We actually compute the root of the mean(average) of the square of the waveform. Root-mean-square or rms. This is our measure of “average”. • i(t) = Imcos t then Lecture 8

  5. Power in an ac circuit (revision) i(t) + v(t) Z=R- jX - • Instantaneous Power p(t) = v(t).i(t) • Let v(t) = Vmcos t = V2 cos t • then i(t) = Imcos (t+) = I2 cos (t+) • p(t) = [V2 cos t][ I2 cos (t+)] = VI [cos  + cos (2t+) ] • where: VI cos  is a constant term & VI cos (2t+) is a periodic-zero mean Lecture 8

  6. Power Terms (revision) • Average (Real) PowerP = VrmsIrms cos V-I Watts • units of W (Watts) • Case 1, Z = R : =tan-1 X/R = 0°  P = VI Watts • Case 2, Z = X : =tan-1 X/R =  90°  P = VIcos  90° = 0 Watts • Apparent power or complex power S = VI* VA’s • units of VA (volts-amps) • Reactive power is defined as Q = VIsin VAr’s • units are VAr’s (volt-amps-reactive) Lecture 8

  7. R   Z IZ -jX -jIX Power Triangle (revision) Average Power P =I2R IR  Reactive Power Q=I2XC   Complex Power S  |S| = V.I • Multiply Impedance triangle with I to get voltage triangle, then I again to get power triangle…. • S = P - jQ = VI* • Power Factor p.f. = cos  • dimensionless • lies in the range 0  pf  1 • power factor does not discriminate between capacitive (current leads voltage) or inductive (current lags voltage) loads. Hence need to specify about leading or lagging power factors. [e.g. inductive loads have lagging pf] Lecture 8

  8. Power Factor Correction (revision) • As pf decreases, situation gets worse • Distribution Authorities insist on consumers having a power factor of at least 0.8 • To correct power factor, consumers need to add a reactive load of opposite sign to the existing reactive load • Thus for a lagging (inductive) load (the most common type) we add a capacitive load • How do we calculate how much correction to add? • Why is power factor correction equipment usually added in parallel instead of in series? ZLoad ZAdded Lecture 8

  9. Example 1 240 V 50 Hz 100+j100 XA • Find XA such that the new power factor is 0.95 lagging? • Find XA such that the new power factor is 0.95 leading? Lecture 8

  10. Example 2 • A customer’s plant has two parallel loads : • a heating load of 30 kW resistive, and • a set of motors operating at 0.86 lagging power factor. The motors load is 100 kVA. • Power is supplied at 11kV. • Find • (a) total current flowing into the plant • (b) existing pf • (c) additional parallel load to bring pf to 0.95 lagging • (d) new value of current flowing into plant 100 kVA 0.86 lagging 11 kV 30 kW Lecture 8

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