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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

ELEC130 - Electrical Engineering 1

Week 8

Module 4

Applications in Power Systems


This lecture
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


Administration
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


Rms root mean square revision
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


Power in an ac circuit revision
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


Power terms revision
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


Power triangle revision

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


Power factor correction revision
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


Example 1
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


Example 2
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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