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6. STAR OR WYE (Y) CONNECTION

•In this method of inter-connection the similar ends either the ‘start’ or ‘finish’ are joined
together at point N. This common point N [Fig. 5 (a)] is called star point or neutral point
Ordinarily only three wires are carried to the external circuit giving 3-phase, 3-wire start
connected system but sometimes a fourth-wire, known as neutral wire is carried to the
neutral point of the external load circuit giving 3-phase, 4 wire star connected systems.

•The voltage between any line and the neutral point (i.e. voltage across the phase winding
is called the ‘phasevoltage’ (Eph); while the voltage available between any pair of
terminals (or outers) is called the ‘line voltage’ (EL).

6.-STAR-OR-WYE-(Y)-CONNECTION

6.-STAR-OR-WYE-(Y)-CONNECTION

  • In star connection, as is evident in Fig. 5 (a) there are two-phase windings between each
    pair of terminals, but since their similar ends have been joined together, they are in
    opposition. Obviously, the instantaneous value of potential difference between any two
    terminals is the arithmetic difference of the two-phase e.m.fs. concerned. However, the
    r.m.s. value of this potential difference is given by the vector difference of the two-phase
    e.m.fs.

 

  • Fig. 5 (b) shows the vector diagram for phase voltages and currents in a star connection
    where a balanced system has been assumed. [A balanced system is one in which (i) the
    voltages in all phases are equal in magnitude and differ in phase from one another by
    equal angles, in this case, the angle = 360/3 = 120°, (ii).the currents in the three phases
    are equal in magnitude and also differ in phase from one another by equal angles. A
    three-phase balanced load is that in which the loads connected across the three-phases
    are identical]. Thus, we have

ER= EY= EB= Eph(phase e.m.f.)

                                          Line voltage,      ERY(=EL) = Vector difference of ERand EY

=ER-EY

                                         Line voltage,      EYB= EY- EB

Line voltage,      EBR =EB-ER.

 

(a) Relation Between Line Voltages and Phase Voltages. Refer Fig. 6.
The potential difference between outers R any Y is

ERY= EREY                                                        [vector difference]

or                                      ERY= ER+ (-EY)                                                             [vector sum]

 

Hence, ERYis found by compounding ERand EYreversed and its value is given by the diagonal of the parallelogram (Fig.6). Obviously the angle between ER and EY reversed is 60oand the value

ERY (or EL) = …(1)

Similarly                            EYB(= EL) = EY- EB=

and                                   EBR(=EL) =EB-ER =

i.e.,                                    ERY= EYB= EBR= EL=

Hence,                      EL=                                                                                     …(2)

(i.e., Line voltage = phase voltage). 

Relation Between Line Voltages and Phase Voltages. Refer Fig. 6.

(b) Relation between Line Currents and Phase Currents, Since in star-connected system
each line conductor is connected to separate phase, so the current flowing through the line and phase are same.

 

Current in outer (or line)                R = IR

Current in outer                             Y = IY

Current in outer                             B = IB

Since                                              IR= IY= IB= say, Iph-the phase current

Line current,                               IL= Iph                                                                     …(3)

(c) Power. If the phase current has a phase difference of  with phase voltage,

Power per phase                            = Ephlph cos

Total power (true),                         P = 3  power per phase

P = 3  EphIph cos                                                       …[4(a)]

Now                                               Eph=  and Iph= IL.

Hence in terms of line values, the above expression becomes

or                                                                                                          …[4(b)]

(Apparent power = )

 

In a balanced star-connected net work the following points are worthnoting :

  1. Line voltages are times the phase voltages.
  2. Line currents are equal to phase currents.
  3. Line voltages are 120O apart.
  4. Line voltages are 30O ahead of the respective phase voltages.
  5. The angle between line currents and the corresponding line voltages is (30O ± ) + for
    lagging currents – ve for leading currents.
  6. True power = cos , where  is the angle between respective phase current and phase voltage, not between the line current and line voltage.
  7. Apparent power = -
  8. In balance system, the potential of the neutral or star point is zero.
    Potential at neutral (or star) point =ENR+ ENY + ENB = O.

 

Example 1.Three equal impedances each having a resistance of 25 Ω and reactance of 40 Ω
are connected in star to a 400 V, 3-phase, 50 Hz system. Calculate:

(i) The line current                                  (ii) Power factor, and

(iii) Power consumed.

 

Solution. Resistance per phase, Rph= 25 Ω

Reactance per phase,                Xph= 40 Ω

Line voltage,                               EL= 400 V

Line current, IL.:

Power factor, cos

Power consumed, P :

Refer Fig. 7.

Refer Fig-7

 

 

 

Example 2.Three identical coils are connected in star to a 400 V (line voltage), 3-phase A.C.

supply and each coil takes 300 W. If the power factor is 0.8 (lagging). Calculate:

 

(i) The line current,                                 (ii) Impedance, and

(iii) Resistance and inductance of each coil.

 

Solution. Line voltage,     EL = 400 V

Power taken by each coil, Pph= 300 W

Power factor,                 cos  = 0.8 (lagging)

 

IL ; Z ; Zph ; Lph :

 

Example 3.In a 3-phase, 3-wire system with star-connected load the impedance of each phase is (3 + j4) Ω. If the line voltage is 230 V, calculate:

 

 (i) The line current, and                                   (ii) The power absorbed by each phase.

 

Solution. Line voltage,             EL = 230 V

Resistance per phase,                Rph = 3 Ω

Reactance per phase,                Xph = 4 Q

IL; Pph

 

 

Fig-8

Solution by symbolic Notation. In Fig 9 ER, EY and EB are the phase voltages whereas IR, IY and IB are phase currents.

Taking ER as the reference vector, we get

This current lags behind the reference voltage (ER) by 8’ (Fig.9)

 

It lags the reference i.e., ER by 8’ which amounts to lagging behind its phase voltage E by 8’

 

This current leads ER by  which is the same as lagging behind its phase voltage by .

Fig-9

Let us consider R-phase for calculation of power

ER = (133 + j0) ; IR = 26.6 (0.6 – j0.8) = (15.96 – j21.28)

Using method of conjugates, we get

PVA = (133 – jO) (15.96 – j21.28) = 2116 – j2830

Real power absorbed/phase = 2116 W

 

Example 4.A star-connected, 6000 V, 3-phase alternator is supplying 4000 kW at factor of 0.8. Calculate the active and reactive components of the current in each phase.

 

Solution. Line voltage,                         EL = 6000 V

Power supplied,              P = 4000 kW

Power factor,                 cos  = 0.8

Active and reactive components of current:

We know that,             P = ELIL cos

 

4000 x 1000 =   6000  IL.  0.8

Active component                         = Iph  cos  = 481  0.8 = 384.8 A. (Ans.)

 

Reactive component                    = Iph sin = 481  0.6 = 288.6 A. (Ans.)

 

Example 5. In a 3-phase, 4-wire system, two phases have currents of 20 A and 12 Air.
power factors of 0.8 and 0.6 respectively, while the third phase is open-circuited. Calculate the
in the neutral and draw the vector diagram.

Solution. Refer Figs. 10 and 11.

cos 1 = 0.8

1 = cos-1 0.8 = 36° 52′ (lag)

cos 2 = 0.6

2 = cos-1 0.6 = 53° 8′ (lag)

 

Fig-10,11

 

 

Let ER be the reference vector.

Then                           IR = 20 -36° 52′ = 20 (0.8 -j0.6) = (16 – j12)

and                              IY = 12 -173° 8′ = 12 (-1 -j0.12) = (-12 – j1.44)
The current through the neutral

                                             IN=IR+lY

= (16 – j12) + (-12 -j1.44) = 4 -j13.44 = 14 -73° 24′

Hence, current in the neutral = 14 - 73° 24′. (Ans.)

 

Example 6. A balanced 3-phase star connected load of 100 kW takes a leading current of
80 A, when connected across a 3-phase, 1100 V,
50 Hz supply. Find the circuit constants of the load
per phase.                                                                                                       
(AM IE Winter, 1997)
Solution. Given: P = 100 kW; Iph(= IL) = 80 A; EL = 1100 V; f= 50 Hz

Circuit constants of the load per phase, R, C :

As the 3- load is balanced and star connected, line or phase current,

 

Example 7. Three identical star-connected coils take 8 k W at a power factor 0.8 when connected
across a 460 V, 3-phase, 3-wire supply. Find the circuit constants of the load per phase.

Find the line currents if the coil in phase C is short circuited. Supply phase sequence is A-B-C.
(AMIE Summer, 1997)

         Solution. Given:                 P = 8 kW ; V = 460 volts; cos θ = 0.8

Circuit constants of the load:

 

Load resistance, R = Z cos θ = 21.16 × 0.8 = 16.93 Ω. (Ans.)
Load reactance, X = Z sin θ = 21.16 × 0.6 = 12.7 Ω. (Ans.)
Line currents if the coil in phase C is short circuited: 

Fig.12

 

The connection arrangement is shown in Fig. 12 when coil C is shorted.

Given that phase sequence is A-B-C.
Hence,                                 EAB = 460

EBC = 460 -120°

ECA = 460 +120°

Also, Z = 21.6 36.8°

 

Line current in line A               = IAB -ICA

= 10.87 36.8° – 21.74 83.2°

= 10.87(0.8 – j0.6) - 21.74(0.118 + j0.993)

= 8.70 – j6.52 - 2.56 – j21.59 = (6.14 – j28.11)A. (Ans.)

Line current in line B               = IBC -IAB

= 21.74 -156.8° -10.87 - 36.8°

= 21.74(- 0.92 – j0.394) - (8.70 – j6.52)

= – 20.0 – j8.56 - 8.70 + j6.52 = (- 28.70 – j2.04)A. (Ans.)

Line current in line C               = ICA - IBC

= (- 2.56 – j21.59) - (- 20 – j8.56) = – 2.56 – j21.59 + 20 +8.56
= (17.44 – j13.03)A. (Ans.)

 

Example 8. In a star-connected load each phase consists of a resistance” of 50 Ω in parallel with a capacitor of capacitance 16 µF. When it is connected to 400 V, 3-phase, 5 Hz supply, calculate:

 

(i)                 The line current                                   (ii) The power factor,

(iii)             The power absorbed, and        (iv) The total kVA

 

Solution. Resistance per phase, R = 50 Ω

Capacitance,               C = 16 µF = 16 × 10-6 F

The circuit is shown in Fig. 13

Fig.13

 

 

Admittance of each phase,

 

 

                                                          lph = Eph.Yph = 231 (0.02 + j0.005) = 4.62 + j1.155 = 4.76 14°

 
  ..                               h =4.76A..                                  h =4.76A

Hence,       line current = 4.76 A.

 

(ii) Power factor
(iii) Now

(Ans.)

 

For a star connection,           Iph = IL

 

IL = 4.76 A

 

Hence,       line current = 4.76 A.  (Ans.)

(ii) Power factor                     = cos 14° = 0.97 (leading). (Ans.)
(iii) Now                                   Eph = (231 + j0); lph = (4.62 + j1.155)
                                                 PYA = (231+ j0) (4.62 – j1.155)

= 231 × 4.62 – j1.155 × 231 = 1067.22 – j266.8

= 1100 14°                                                                       (per phase)

Total power absorbed = 3 × 1067.22 = 3201.66 W = 3.201 kW. (Ans.)

(iv) Total volt-amperes = 3 × 1100 = 3300 VA = 3.3 kVA. (Ans.)

 

Example 9. A 3-phase, star-connected system with 230 V between each phase and neutral has resistance of 8, 10 and 20 Ω respectively in three phases, calculate:

(i) The current flowing in each phase,               (ii) The neutral current, and

(iii) The total power absorbed.

 

Solution. Refer Figs. 14 (a) and (b). 

Fig.14

Fig. 14

Phase voltage,                                Eph = 230 V

ii) The above currents are mutually displaced by 120°. The neutral current IN is the vector sum of these three currents.

IN can be found by splitting up these three-phase currents into their X-components and Y-components and then by combining them together.

ƩX-components             = 23 cos 30° – 11.5 cos 30° =11.5 cos 30° = 9.96 A

ƩY-components             = 28.75 – 23 sin 30° -11.5 sin 30° = 28.75 – 34.5 sin 30° = 11.5 A

Neutral current, IN =  = 15.21 A. (Ans.)

 

(iii) Total power absorbed,

Example 10. In a 3-phase, 4-wire system, there is a balanced 3-phase motor load taking 10 kW
at a power factor of 0.8 lagging while lamps connected between phase conductors and the neutral are
taking
2.5 kW, 2 kW and 5 kW respectively. Voltage between line conductors is 430 V. Calculate:

(i) The current in the neutral wire.          (ii) The current in each conductor.

 

Solution. Motor load                  = 10 kW or 10000 W

Power factor of motor load, cos θ = 0.8

Lamp loads: 2.5 kW, 2 kW and 5 kW

Voltage between line conductors, EL = 430 V.

Refer Figs. 15, and 16.

 

Let us first find the current in the neutral wire due to lamp loads L1 (2.5 kW), L2 (2 kW) and
L3 (5 kW) respectively.

 

Fig.15

 

The neutral current is the vector sum of three lamp currents. Resolving them into their X and Y components, we have

X = 8.06 cos 30° – 20.16 cos 30° = -10.48 A

Y = 10.08 – 8.06 sin 30° – 20.16 sin 30° = – 4.03 A

 

 

Line current drawn by the balanced 3-phase

 

Each conductor carries-two currents:

 

(i) a current of 16.78 A at’O.8 power factor lagging; and

(ii) appropriate lamp current which is in phase with the voltage.
Actual component of load current = 16.78 × 0.8 = 13.42 A
Reactive component = 16.78 × 0.6 = 10.07 A