Date post: | 15-Jun-2015 |

Category: | ## Documents |

View: | 349 times |

Download: | 0 times |

Share this document with a friend

Transcript:

EEE3233 POWER SYSTEMCHAPTER 3: 3-Phase System and Power Conceptnur diyana kamarudin

IntroductionqThe generation, transmission and distribution of electric power is accomplished by means of three-phase circuit. qAt the generating station, three sinusoidal voltage have equal magnitude and an equal 1200phase difference between any two phase. qThis is called a balanced source.

qPositive phase sequence when the voltages reaches their peak value in the sequential ABC qNegative phase sequence when the phase is order is ACB.ECn EBn EAn

EAn

EBnPositive or ABC phase sequence

ECnNegative or ACB phase sequence

Y-Connected Load

A Y-connecter generator supplying a Yconnected load.

Y-Connected Load (cont.) Assume a positive sequence to find the relationship between the line voltage (line-to-line voltage) and phase voltage (line-to-neutral voltage). Line to-neutral voltage of the phase is chosen as a reference.

|Vp|= magnitude of phase voltage

Y-Connected Load (cont.)

The line voltages at the load terminals in terms of the phase voltage are found by application of Kirchhoffs voltage law.

Y-Connected Load (cont.)Comparison of these line-to-line voltages with the line-to-neutral voltages of leads to the following conclusion: In balanced three-phase Y-connected system with positive-sequence source, the line-to-line voltage are 3 times the line-to-neutral voltages and lead by 30. That is; Vab = 3Van Vbc = 3Vbn Vca = 3Vcn +30 +30 +30

Y-Connected Load (cont.)The relationship between the line voltage and phase voltage is shown as below:Vab Vbn Van Vca Vbc Vcn

Phasor diagram showing phase and line voltages

Voltages Triangle

Y-Connected Load (cont.) RMS value of any of the line voltages is denoted by VL, then one of the important characteristic of the Y-connected tree-phase load may be expressed asT h re e - p h a se

cu rre n t a l p o sse ss th re e - p h a se so sym m e try a n d a re g i n b y ve

IL = IP

ExampleCalculate the line voltage and line current of a Y-Y connection

-Connected Loads A balanced -connected load (with equal phase impedances) is shown in below:

Li e n

vo l g e a re th e sa m e a s p h a se ta vo l g e s. ta

-Connected Loads (cont.) The phase current Iab is chosen as referenceI = magnitude of p phase current

T h e

re l ti n sh i b e tw e e n p h a se a n d l n e a o p i cu rre n ts ca n b e o b ta i e d b y a p p l n g n yi K i h o ff s cu rre n t l w a t th e co rn e r o f . rch a

-Connected Loads (cont.) The RMS of any of the line currents and phase is denoted by IL

T h e

m a g n i d e o f l n e cu rre n t i 3 ti e s th e tu i s m m a g n i d e o f th e p h a se cu rre n t, a n d th e l n e tu i cu rre n ts l g s o f p h a se cu rre n ts b y 3 0 . a

-Connected Loads (cont.)The

relationship between the line currents and phase currents is showing in phase diagram.

Three Phase example

Three Phase example(cont.)

-Y TransformationFo r a n a l n g yzi

n e tw o rk p ro b l m s, i i e t s co n ve n i n t to re p l ce th e -co n n e cte d e a ci i w i a n e q u i l n t Y- co n n e cte d rcu t th va e ci i . rcu t C o n si e r th e Y- co n n e cte d ci d rcu i o f Z Y t / phase which is equivalent to a balanced -co n n e cte d ci i o f Z / phase . rcu t

-connection

Y-co n n e cti n o

-Y Transformation ( cont .)Fo r - co n n e cte d

ci i rcu t(1 )

Fro m

th e p h a so r d i g ra m , th e re l ti n sh i a a o p b e tw e e n b a l n ce d p h a se a n d l n e -to - l n e a i i vo l g e ; ta

(2 )

-Y Transformation ( cont .)S u b sti ti g tu n

e q .( 2 ) i to e q .( 1 ), w e g e t n(3 ) (4 )

orFo r Y- co n n e cte d

ci i , w e h a ve rcu t(5 )

S u b sti ti g tu n

e q .( 4 ) i to e q .( 5 ), w e fi d th a t n n(6 )

exa m p l eA balanced, positive-sequence, Yconnected voltage source with E ab = 480 0 volts is applied to a balanced- load with Z = 30 0 . The line impedance between the source and load is Z = 1 85 for each phase. Calculate L the line current, the -load currents, and the

Power in single-phase AC circuitsInstantaneous voltage be, and Instantaneous current be,

Figure show a singlephase sinusoidal voltage supplying a load

The instantaneous power p(t) delivered to the load is the product of voltage v(t) and i(t) given by By using the trigonometric identity

Which results in

RMS value of v(t) is |V|=Vm/2, RMS value of i(t) is |I|=Im/2 . Let =(v i). The above equation, in terms of the rms values, is reduce to

Instantaneous power has been decomposed into two components. The first component is

Since the average value of this sinusoidal function is zero, the average power delivered to the load is given by

Also called Active power or Real power

The second component is

The amplitude of this pulsing power is called reactive powerQ = |V|I| sin = v i > 0 ;Q=+ve = v i < 0; Q=-ve

The Power Factor (pf) is the cosine of the phase difference between voltage and current. It is also the cosine of the angle of load impedance. The power factor may also be regarded as the ratio of the real power dissipated to the apparent power of the load.

Example

Calculate the power factor seen by the source and the average power supplied by the load

Characteristics of the instantaneous powerPurely resistive v i = 0, Pf = 1 P/S = 1, all power are load (R) consumed Purely reactive load (L or C)

v i = 90o, =0

pf P = 0, no real powerconsumption

Resistive and v i > 0 ;Q=+ve Lagging - inductive load Leading - capacitive load reactive load (R v i < 0; Q=-ve P/S < 1, Part of the apparent and L/C) power is consumed

Complex Power For circuits operating in sinusoidal-steady state, real and reactive power are conveniently calculated from complex power, defined below. Complex power S is product of voltage and the conjugate of the current:

Complex Power(cont.) Thus, the complex power designated by S is given by

The

magnitude of S,

* its unit is volt-amperes and the larger units areThe

kVA or power ,Q : reactive MVA positive : (impedance angle) between V and I is +ve . In this case, load impedances is inductive (I lag V). negative : (impedance angle) between V and I is -ve . In this case, load impedances is capacitive (I leadsV).impedances of the complex power S is given by

The

Power TriangleThe COMPLEX Power is represented by the POWER TRIANGLE similar to IMPEDANCE TRIANGLE. Power triangle has four items: P, Q, S and .

a) Power Triangle

b) Impedance Triangle

Power Triangle

Complex Power(cont.)The Complex Power contains all the information pertaining to the power absorbed by a given load.

Real Power is the actual power dissipated by the load. Reactive Power is a measure of the energy exchange between source and reactive part of the load.

Complex Power(cont.)

Examples

Complex Power(cont.)

Solutio n First method

Solution complex power (cont.)Second method

Third method

Solution complex power (cont.)Current phasor diagram and the complex power vector representation.

Power Factor Correction It is necessary to get PF is close to the unity. If the PF < 1, the larger current I is supplied and additional cost to the utility company. In order to maintain the power factor close to the unity, power companies install banks of capacitor throughout the network as needed. A method of connecting a capacitor in parallel with an inductive load is known as power factor correction. The effect of the capacitor is to increase the power factor of the source that delivers

Power Factor Correction(cont.)Increasing the power factor without altering the voltage or current to the load is called Power Factor CorrectionOriginal Inductive Load Inductive Load with improved power factor correction

Effect of capacitor on total Power triangle of power factor current correction

Power Factor Correction(cont.)Increasing the power factor without altering the voltage or current to the load is called Power Factor Correction.

Q c = Q1 Q 2 = P (tan 1 - tan 2) = CVrms 2

Q1 = S1 sin 1 = P tan 1 P = S1 cos 1 Q2 = P tan 2

Power Factor Correction (Cont.)

Power Factor Correction (Cont.)

Power Factor Correction(cont.)

Balanced Three-Phase Power(cont.)Consider a balanced three-phase source supplying a balanced Y-or -connected load with the following instantaneous voltage and current .

3

Balanced Three-Phase Power(cont.)So the real power of the three-phase is

Three-phase reactive power is

The complex three-phase power is or

Balanced Three-Phase Power(cont.)ExampleA three-phase line has an impedance of 2 + j4 as shown in figure below

The line feeds two balanced three-phase loads that are connected in parallel. The first load is Y-connected and has an impedance of 30 + j40 per phase. The second load is connected and has impedance of 60-j45. The lines is energized at the sending end from a three-phase balanced supply of line voltage 207.85V. Taking the phase voltage Va as reference, determine:

Balanced Three-Phase Power(cont.)a)The current, real power, and reactive power drawn from the supply. b) c) d) The line voltage at the combined loads. The current per phase in each load. The total real and reactive powers in each load and the line.

Solution

Solutio n (cont.)

Unbalanced Three Phase SystemsAn unbalanced system is due to unbalanced voltage sources or unbalanced load. In a unbalanced system the neutral current is NOT zero.

Unbalanced three phase Y connected load.Line currents DO NOT add up to zero.

In= -(Ia+ Ib+ Ic) 0

THANK YOU FOR YOUR ATTENTION !!

Popular Tags:

Click here to load reader

Embed Size (px)

Recommended