ELEC130 - Electrical Engineering 1

1 / 10

# ELEC130 - Electrical Engineering 1 - PowerPoint PPT Presentation

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 &amp; calculating Power Factor Correction Module 5 - Transients in a.c. Circuits

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' ELEC130 - Electrical Engineering 1' - sopoline-knowles

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### 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
• (Will cover in next lecture)

Lecture 8

• 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)
• 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)

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)
• 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

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)
• 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
• How do we calculate how much correction to add?
• Why is power factor correction equipment usually added in parallel instead of in series?

Lecture 8

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
• 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