1

Review of Chapter 2, Plus Matlab Examples (Part II) 2.7 Balanced Three-Phase Circuits Let 1 120a = ∠ o . Clearly then 2 *a a= and 3 1a = . The triple (1, a ,

2a ) form a set of balanced phasors, thus 21 0a a+ + = . This triple represents the three cube roots of 1. Y-connected generator, positive sequence: 0An pE E= ∠

2120Bn p AnE E a E= ∠− =

240Cn p AnE E aE= ∠− = Here pE is the "phase" voltage. Note that if we know AnE we can

find the other two voltages in terms of AnE . It is noted that line-to-line voltage LV is larger than the phase voltage by a factor of 3

and is ahead by 30o , i.e. 3 30L L pE E− = ∠ o . However, line current and phase current

are identical, i.e. L pI I= . Y-connected generator, negative sequence: 0An pE E= ∠

120Bn p AnE E aE= ∠ =

2120Cn p AnE E a E= ∠ − = The negative sequence connection is obtained by "interchanging" any two phases of the positive sequence connection. Power systems use only positive sequence connected generators. Also, every effort is made to maintain balanced voltages and loads.

1

a

a2

EAn

EBn

ECn ωt

EAn

ECn

EBn

ωt

2

Van

n

Vbn

Vcn

Vab

Vbc

Vca

2.8 Y-Connected Loads 0An pV V= ∠

2120Bn p AnV V a V= ∠− =

240Cn p AnV V aV= ∠− = The line-to-line voltages are found as follows: ( )1 0 1 120 3 30ab an bn p pV V V V V= − = ∠ − ∠− = ∠ Similarly we find: 3 90bc pV V= ∠ −

3 150ca pV V= ∠

Note: all these results apply provided the reference phasor is as shown. This is arbitrarily chosen, for example above, it is assumed that Van is the reference vector (at zero degrees). If a different reference is chosen, the diagram stays the same though it will be rotated.

2.9 Delta Connected Loads In a similar way we see that L pV V= (line voltage and phase voltage are the same for delta connected loads) and 3 30a pI I= ∠ −

3 150b pI I= ∠ −

3 90c pI I= ∠ and in general: 3 30L pI I= ∠ −

Van

Vbn

VcnVab

Vbc

V ca

3

2.10 ∆-Y Transformation for Balanced Loads 3 /3Y YZ Z Z Z∆ ∆= = 2.11 Per-Phase Analysis 1. It is assumed that everything is "balanced". 2. Change all delta connected loads/sources to Y connections, this provides a neutral

point. 3. All the neutral points are at the same potential, hence all the neutral points may be

connected 4. This breaks up the circuit into three entirely separate circuits, one for each phase. 5. Solve the single-phase circuit for phase "a". The other phases are the same after a

phase shift of 120± o . 2.12 Balanced Three-Phase Power Recall that single phase instantaneous power was sinusoidal and had a frequency twice that of the line. Its average value was called the real power. Here is the single phase power equation:

( ) ( ) ( )1 cos2 sin 2v vp t P t Q tω θ ω θ = + + + + **Single phase** The three-phase instantaneous power equation will be derived below. Suppose the voltages of a three-phase supply are given by: ( )2 cosan p vv V tω θ= +

( )2 cos 120bn p vv V tω θ= + − o

( )2 cos 240cn p vv V tω θ= + − o Assume the currents are given by: ( )2 cosa p ii I tω θ= +

( )2 cos 120b p ii I tω θ= + − o

( )2 cos 240c p ii I tω θ= + − o The instantaneous power is then found by adding the powers for the three phases:

4

( )3 an a bn b cn cp t v i v i v iφ = + + Thus we have:

( ) ( ) ( )( ) ( )( ) ( )

3 2 cos cos

2 cos 120 cos 120

2 cos 240 cos 240

p p v i

p p v i

p p v i

p t V I t t

V I t t

V I t t

φ ω θ ω θ

ω θ ω θ

ω θ ω θ

= + +

+ + − + −

+ + − + −

Using the identity ( ) ( )1 1cos cos cos cos

2 2A B A B A B= − + + we have:

( ) ( ) ( )( ) ( )( ) ( )

3 cos cos 2

cos cos 2 240

cos cos 2 480

p p v i v i

p p v i v i

p p v i v i

p t V I t

V I t

V I t

φ θ θ ω θ θ

θ θ ω θ θ

θ θ ω θ θ

= − + + +

+ − + + + −

+ − + + + −

Note that the terms involving time are a balanced set hence add to zero and we have: ( ) ( )3 3 cosp p v ip t V Iφ θ θ= − If we let v iθ θ θ= − (the impedance angle) then: ( )3 3 3 cosp pp t P V Iφ φ θ= = It is interesting to note that the RHS is independent of time! Each phase by itself has a pulsating power, but when the three are added, the result is a constant power… compare with single phase! The concept of complex power is now extended to three phase, thus we define 3 3 sinp pQ V Iφ θ= 3 3 3S P jQφ φ φ= + In terms of phase voltages and currents, we also have: *

3 3 p pS V Iφ =

5

Since in the Y-connection we have 3L

p

VV = and p LI I= , then we have:

( )3 3 3 cos

3 cos

p p

L L

p t P V I

V I

φ φ θ

θ

= =

=

3 3 sinL LQ V Iφ θ=

In case of the ∆-connection p LV V= and 3L

p

II = giving rise to the same equations

above. The two equations above are valid for ∆- as well as Y-connected circuits: They involve line-to-line quantities. Very Important: The angle θ in the power equations above is the angle between the phase voltage and phase current (not line voltages nor line currents). Failure to recognize this fact can lead to serious computational errors. Example 2.7 A three phase line has an impedance of 2 4j+ Ω . The line-to-line voltage is 207.85 V. Two three-phase loads are connected to this line in parallel: one is a ∆-load, the other is a Y-load. Each leg of the ∆-load has an impedance of 60 45j− Ω , and each leg of the Y-load has an impedance of 30 40j+ Ω . Taking the phase voltage aV as a reference, determine: (a) The current, real power, and reactive power drawn from the supply. (b) The line voltage at the combined loads. (c) The current per phase in each load. (d) The total real and reactive powers in each load and the line. The Matlab program follows: V1 = 207.85/sqrt(3); ZL = 2 + j*4; Z1 = 30 + j*40; % Note ZL is BETTER than Zl, which looks like Z1 Z2 = (60-j*45)/3; % Z2 is Z∆/3 Z = ZL + Z1*Z2/(Z1+Z2); I=V1/Z, S=3*V1*conj(I) V2 = V1- ZL*I V2ab =sqrt(3)*(cos(pi/6)+j*sin(pi/6))*V2 I1=V2/Z1, I2=V2/Z2 Iab=I2/(sqrt(3)*(cos(pi/6)-j*sin(pi/6))) S1=3*V2*conj(I1), S2=3*V2*conj(I2) SL = 3*ZL*abs(I)^2 Stotal=S1+SL+S2

V1 V2

ZL

ZYZ∆

/3

I I1 I2+ +

- -

6

I = 5.0001 S = 1.8001e+003 V2 = 1.1000e+002 -2.0000e+001i V2ab = 1.8232e+002 +6.5264e+001i I1 = 1.0000 - 2.0000i I2 = 4.0001 + 2.0000i Iab = 1.4227 + 2.1547i S1 = 4.5002e+002 +6.0002e+002i S2 = 1.2000e+003 -9.0003e+002i SL = 1.5001e+002 +3.0001e+002i Stotal = 1.8001e+003 Example 2.8 A three-phase line has impedance of 0.4 2.7j+ per phase. The line feeds two balanced three phase loads that are connected in parallel. The first load is absorbing 560.1 kVA at 0.707 power factor lagging. The second load absorbs 132 kW at unity power factor. The line-to-line voltage at the load end of the line is 3810.5 V. Determine: (a) The magnitude of the line voltage at the source end of the line. (b) Total real and reactive power losses in the line. (c) Real and reactive power supplied at the sending end of the line. The Matlab program follows: V2 = 3810.5/sqrt(3); Zl = 0.4 +j*2.7; S1 = 560.1*(cos(pi/4) + j*sin(pi/4)); S2 = 132; SR = S1+ S2 I = conj(SR)*1000/(3*conj(V2)) V1 = V2 + Zl*I V1L = sqrt(3)*abs(V1) SL = 3*Zl*abs(I)^2/1000 SS = 3*V1*conj(I)/1000 SS = SR + SL SR = 5.2805e+002 +3.9605e+002i I = 80.0079 -60.0078i V1 =

7

2.3940e+003 +1.9202e+002i V1L = 4.1599e+003 SL = 12.0026 +81.0179i SS = 5.4005e+002 +4.7707e+002i SS = 5.4005e+002 +4.7707e+002i Note on units. In reality, P, Q, and S, all have unit of Watt, but to distinguish them apart we assign units as follows:

S=VI*=P+jQ P Q |S|=|V||I| NAME Complex Power Real Power Reactive Power Apparent Power

unit W+jvar or VA at power factor…

W, kW, MW var, kvar, Mvar VA, kVA, MVA