elec130 electrical engineering 1
Download
Skip this Video
Download Presentation
ELEC130 - Electrical Engineering 1

Loading in 2 Seconds...

play fullscreen
1 / 10

ELEC130 - Electrical Engineering 1 - PowerPoint PPT Presentation


  • 118 Views
  • Uploaded on

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

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

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


An Image/Link below is provided (as is) to download presentation

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

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

ad