12. UNBALANCED DELTA CONNECTED LOAD

When the load impedance in the three phases are not equal in magnitude or phase or both, the load is said to be unbalanced. If three unequal loads are connected to form a delta and connected across a 3-phase supply the currents in the three loads will not be equal in magnitude and/or phase. The three-phase currents and the line currents will also be unbalanced.

Example 40. A 440 V, 50 Hz, 3-phase supply has delta-connected load having 50 Ω between R and y, 159 mH between Y and Band 15.9 µF between B and R. Find;

(i) The line current for the sequence RYB.

(ii) The value of star-connected balanced resistors for the same power.

Solution. Phase voltage (Eph) = line voltage (EL) = 440 V
The potential difference across three-phases are given by
ERY = 440(1 + j0) = 440V

EYB = 440( – 0.5 – j0.866) or 440 120° V
EBR = 440(- 0.5 + j0.866) or 440 120° V

Impedance ,  ZRY = (50 + j0) = 50 0° Ω

Impedance ,  ZYB = (0 + j2rπfL) = (0 + j2π × 50 × 159 × 10-3)
= (0 + j50) or 50 90° Ω

Example 41. A 440 V, 50 Hz, 3-phase supply has delta-connected load having 50 Ω between R and y, 159 mH between Y and Band 15.9 µF between B and R. Find;

(i) The line current for the sequence RYB.

(ii) The value of star-connected balanced resistors for the same power.

Solution.

Phase voltage(Eph) = line voltage,  (EL) = 400 V
The potential differences across three-phases are given by:

ERY = 400 0°
EYB = 400 – 120°
EBR  = 400 120°

The phase impedances are:

ZRY  = 31 + j59 = 66.6 62.3°
ZYB  = 30 – j40 = 50 – 53.1°
ZBR  = 80 + j60 = 100 36.9°

Example 42. For the unbalanced delta-connected load of Fig 52, find the phase currents, line currents and total power consumed by the load, when phase sequence is:

(i) RYB
(ii) RBY

(AMIE Advanced Electrical M/c Winter, 1998)

Solution.

(i) Phase sequence RYB :

The potential differences across three phases are:

ERY = 200
EYB = 200-120°
EBR = 200 120°

Impedance,   ZRY = 12 + j16 = 20 53.13°
Impedance,   ZYB = 16 + j12 = 20 36.8°
Impedance,   ZBR = 8 – j6 = 10 – 36.8°

Phase sequence RBY:

Here ERY = 200
EYB = 200
EBR = 200

Example 43. A 3-phase, 400 V, 50 Hz system has the following load connected in delta:
Between R and Y lines: a non-reactive resistor of 50 Ω
Between Y and B lines: a coil having a resistance of 10 Ω and reactance of 30 Ω
Between B and R lines: a loss free capacitor of 15 µ F.
Calculate: (i) The phase currents (ii) The line currents.
Assume phase sequence to be RYB.

Solution.

Eph = EL = 400 V

The potential difference across the three phases are:

ERY = 400 V
EYB = 400 120° V
EBR = 400 120° V

Line Current

IR =IRY – IBR = 8 0° – 1.886 210°

= 8 – 1.886 (- 0.866 – j0.5) = 8 + 1.633 + j0.943
= 9.633 + j0.943 = 9.68 5.6° A. (Ans.)

IY = IYB – IRY = 12.65 – 191.56° – 8 0°

= 12.65(- 0.979 + j0.2) – 8 = – 12.38 + j2.53 – 8
= – 20.38 + j2.53 = 20.53 7.07° A. (Ans.)

IB = IBR – IYB = 1.886 210° – 12.65 – 191.56°
= 1.886 (- 0.866 – j0.5) – 12.65(- 0.979 + j0.2)
= – 1.633 – j 0.943 + 12.38 – j2.53

= 10.747 -j3.473 = 11.29 17.4°. (Ans.)

Example 45. A 400 V, 3-phase 4 wire system has the loads 6 – j8, 6 + j0 and 4 + j3 Ω connected
in star. Find:
(i)                 The line currents,
(ii)               Neutral current, and
(iii)             Total power.

Solution. The connections are shown in Fig. 56.

= – [23.09 53.13° + 38.48-120°+ 46.18 83.1°

= – [23.09(0.6 + j0.8) + 36.18 (- 0.5 -j0.866) – 46.18 (0.12 + j0.99)] = – [13.85 + j18.47 – 18.24 – j31.59 + 5.54 +.j45.72]

= – (1.15 + j32.6) = -1.15. -j32.6 = 32.62 -92° A. (Ans.)

Power:

Power taken by phase R,

Example 46. A 440/254 volt, 3 phase, 4-core supplies an unbalanced load represented by the following impedance in ohms connected between R, Y and B phases respectively and the neutral 16 + j12; 14 – j21 and 25. The phase sequence is RYE. Calculate the current in each conductor of the cable and the readings on each of the three wattmeters connected in each line and neutral.

Solution. The e.m.fs between the three lines and neutral are given by :

ER = 254(1 + j0)

EY= 254(- 0.5 – j0.866)

EB = 254(- 0.5 + j0.866) taking ER as the reference vector

Fig.57 (a) Circuit Diagram

Fig.57 (b) Phasor Diagram

Example 47. A 3-phase, 4-wire system having a 440 V line to line has the following loads connected between the respective lines and neutral:
ZR = 5 0° Ω, ZY = 5 37° Ω, ZB = 5 -53° Ω.
Calculate the current in neutral wire and power taken by each load when phase sequence is:
(i) RYB
(ii) RBY.

Solution. (i) Phase sequence RYB (Figs. 58 and 59).

The potential differences across the three phases are:

Current in the neutral:

IN = – (IR + IY + IB)

= – (50.8 + 50.8 157° + 50.8 173°)
= – [50.8 + 50.8 (- 0.92 – j0.39) + 50.8(- 0.99 + j0.122)]
= – [50.8 – 46.74 – j19.8150.3 + j6.2]}

= – (- 46.24 – j13.61) = 46.24 + j13.61

= 48.2 16.4°. (Ans.)

Now RR = 5 Ω, RY = 5 cos 37° = 4 Ω,

RB = 5 cos (- 53°) = 3 Ω
Power taken by phase R,

PR = 50.82 × 5 = 12903 W. (Ans.)
PY = 50.82 × 4 = 10322 W. (Ans.)
PB = 50.82 × 3 = 7742 W. (Ans.)

Phase Sequence RBY (Fig. 60):

ERN = 254 0°, EYN = 254 120°
EBN = 254 -120°

Line currents:

Current in the neutral:

IN=-(IR + IY + IB)

= – (50.8 0° + 50.8 83° + 50.8 – 67°)

= – [50.8 + 50.8(0.122 + j0.992) + 50.8(0.39 – j0.92)] = – (50.8 + 6.197 + j50.4 + 19.81- j46.74)

= – (76.81 + j3.66) = – 76.81 – j3.66

= 76.89 -177.3° A. (Ans.)

Power would remain the same because magnitude of branch currents is unaltered.

Example 48. In a 3-phase, 4-wire system, the line voltage is 400 V and non-inductive loads of
5, 4 and 2.5 kW Ware connected between the three lines conductors and the neutral, as shown in Fig. 61.Calculate
i. The current in lines; and
ii. The current in the neutral conductor.

Solution. Refer Fig. 61.

Phase voltage across each load,

These currents are mutually 120o out of phase as the circuit is purely resistive. Taking IR as the reference vector, line currents may be given as

IR = 21.64 0° = 21.64 (1 + j0)
IY = 17.32 -120° = 17.32 (- 0.5 – j0.866)
IB = 10.82 120° = 10.82 (- 0.5 + j0.866)

(ii) Current in the neutral wire (Refer Fig. 62):

IN =- (IR + IY + IB)

= – [21.64(1- j0) + 17.32 (- 0.5 – j0.866) + 10.82(- 0.5 + j0.866)]

= – [21.64 – 8.66 – j15 – 5.41 + j9.371 = – (7.57 – j5.63)

= -7.57 + j5.63 = 9.434 143.4°. (Ans.)

Example 49. A 3-phase star-connected system with 230 V between each phase and neutral has resistances 8 Ω, 10 Ω and 12 Ω respectively in the three phase. Calculate:

i. Current in each phase,
ii. Current in neutral wire, and
iii. Total power absorbed.

Solution. The circuit and vector diagrams are shown in Figs. 63 and 64 respectively.

Potential difference across each phase = 230 V

(i) Phase currents:

Since the circuit is purely resistive, so currents will be in phase with their respective voltages and will be mutually 1200 apart. Taking IR as the reference vector (Fig. 64) the line currents are given as:

IR = 28.75  = 28.75(1 + j0)

IY = 23 -120° = 23 (- 0.5 – j0.866)

IB = 19.17 120° = 19.17 (- 0.5 + j0.866).
(ii) Current in neutral wire,

IN = – (IR + IY + IB) = – [28.75(1 + j0) + 23(- 0.5 – j0.866) + 19.17(- 0.5 + j0.866)
= – [28.75 – 11.5 – j19.92 – 9.58 + j16.6)

= – [7.67 – j3.32) = – 7.67 + j3.32 = 8.357 156.6° A. (Ans.)
(iii) Total power absorbed

= 6612.5 + 5290 + 4409.8 = 16312.3 W. (Ans.)

Example 50. A 440 V (line-to-line), 3-phase, 4-core supplies an unbalanced load represented by the following impedances in ohms connected between R, Y and B phases respectively and the neutral 8 + j6, 7 – j10.5 and 12.5. Calculate:

(i)  Current in each conductor of the cable,
(ii) Current in the neutral wire, and
(iii) Readings of the three wattmeters connected in each line and neutral.

The phase sequence is RYB.

Solution.The potential differences across the three phases are:

EYN = 254 -120°
EBN = 254 120°.

Impedances:

ZR = 8 + j6 = 10 36.9°

ZY = 7 – j10.5 = 12.62 – 56.3°
ZB = 12.5 = 12.5 0°.

(i) Phase currents:

(ii) Current in the neutral wire:

IN = – (IR + IY + IB)

= – [25.4 -36.9° + 20.12 – 63.7° + 20.32 120°]

= – [25.4(+ 0.8 – j0.6) + 20.12(0.44 – j0.896) + 20.32(- 0.5 + j0.866)]

= – (20.32 – j15.24 + 8.85 – j18.03 – 10.16 + j17.6)

= – (19.01 – j15.67) = – 19.01 + j15.67 = 24.64 39.5° A. (An s.)

(iii) Readings of wattmeters:

KIRCHHOFF’S LAWS (Article 11)

Example 51. Calculate the voltage drop across each of the following three star-connected impedances when they are connected to a 400 V, 3-phase system:
ZR = j40Ω, ZY = j2Ω, ZB = – j2Ω.
Also find the potential of the star-point above ground.
The phase sequence is RYB.

Solution. Refer Fig. 67.

Considering ERY as a reference vector. The vector expressions for the three line voltages are:

ERY = 400(1 + j0) V = (400 + j0) V

EYB = 400(- 0.5 – j0.866) = (- 200 – j346.4) V

EBR = 400(- 0.5 + j0.866) = (- 200 + j346.4) V

Impedances:

ZR = (0 + j40),

ZY= (0 + j2)

ZB = (0 – j2)

Now for finding IR, IY and IB, using the relations:

Voltage drops across impedances:

Voltage drop,    ER = IRZR = (173.2 – j100) × j40 = (4000 + j6928) V. (Ans.)

Voltage drop,    EY = IYZY = (3464 – j1800)(j2) = (3600 + j6928) V. (Ans.)

Voltage drop,    EB = IBZB = (- 3637.2 + j1900) (- j2) = (3800 + j7274.4) V. (Ans.)

Potential of star point above ground:

Example 52. An unbalanced star-connected load shown in Fig. 68, ZR == j40 Ω, ZY = – j40 Ω, ZB = 40 Ω is connected across a balanced 400 V, 3-phase supply. Find the reading of the wattmeter connected in Y-phase.

The phase sequence is R Y B.

Solution. Refer Fig. 68.

Taking ERY as reference vector

ERy = 400 0° = 400 (1 + j0)
EYB = 400 -1200 == 400 (- 0.5 – j0.866)
EBR == 400 120° == 400(- 0.5 + j0.866)
Impedances are
ZR == (0 + j40), ZY = (0 – j40), ZB = (40 + j0)

Voltage drop,   ER = IRZR = (1.34 – j5) × j40 = (200 + j53.6) V

Voltage drop,   EY = IYZY = (- 1.34 – j5) × (- j40) = (- 200 + j53.6) V

Voltage drop,   EB = IBZB = j10 × 40 = (0 + j400) V

Voltage cross potential coil of wattmeter

= voltage between supply neutral and load neutral

Current through wattmeter current coil x conjugate of voltage across wattmeter potential coil
= IY × (- j169) = (- 1.34 – j5) × (- j169) = (- 845 + j226.5)

STAR-DELTA CONVERSION

Example 53. An unbalanced star-connected load has branch impedances of ZR = 10 30° Ω, ZY = 20 45° Ω and ZB = 20 60° Ω and is connected across a balanced 3-phase, 3-wire supply of 440 V. Find:

i. Line currents, and
ii.Voltage across each impedance
Use star=delta conversion method.

Solution. Taking ERY are reference vector
Then   ERY = 440 0°
EYB = 440 -120°
EBR = 440 120°
Refer Fig.69

Fig. 69 (a)

Fig. 69 (b)

Also      ZR = 10 30°                                                            (given)
ZY= 20 45°                                                             (given)
ZB = 20 60°                                                            (given)

Impedances of equivalent delta-connected load (Refer Fig.69):

(i) Line currents:

IR = IRY – IBR = (9.34 – j6.3) – (3.625 + j10.66)
= 5.715 – j16.96 = 17.89 -71.4° A. (Ans.)
IY = IYB – IRY = (- 5.62 + j0.37) – (9.34 – j6.3)

= -14.96 + j6.67 = 16.4 156° A. (Ans.).

IB = IBR – IYB = (3.625 + j10.66) – (- 5.62 + j0.37) .
= 9.245 + j10.29 = 13.833 48° A. (Ans.)

These are the currents in the phases of the star-connected unbalanced load. Let us find voltage
drop across each star-connected branch impedance.

(ii) Voltage across impedances :

Voltage across  ZR = IR × ZR

= 17.89 -71.4° × 10 30° = 178.9 41.4° V. (Ans.)
Voltage across  ZY = IY × ZY = 16.4 156° × 20 45°

= 328 L210° V. (Ans.)

Voltage across   ZB = IB × ZB = 13.833 48° × 20 60°

= 276.67 108° V. (Ans.)

The phasor diagram is shown in Fig. 70.

Example 54. The branch impedances of an unbalanced star-connected load are: ZR = 20 30° Ω, ZY = 20 45° Ω, ZB = 40 60° Ω. It is connected across a balanced 3-phase, 3-wire supply of 200 V. Find:

(i)   Line currents, and

(ii)               Voltage across each impedance.

(iii)             Use star-delta conversion method.

Solution. The unbalanced star-connected load and its equivalent delta-connected load are shown in Fig. 71.

Fig. 71 (a)

Fig. 71 (b)

ZR = 20 30° Ω, ZY = 20 – 45° Ω, ZB = 40 60° Ω

Now     ZRZY + ZYZB + ZBZR

= (20 300)(20 – 45°) + (20 – 45°)(40 60°) + (40 60°)(20 30°)
= 400 – 15° + 800 15° + 800 90°

= 400(0.966 – j0.26) + 800(0.966 + j0.24) + j800

= 386.4 – j104 + 772.8 + j208 + j800 = 1159.2 + j904 = 1470 38°

Equivalent impedances of delta-connected load are:

(i) Line Currents:

IR = IRY – IBR

= (5.04 + j2.04) – (2.17 + j1.63) = 2.87 + j0.41 = 2.9 8.1 ° A. (Ans.)
IY = IYB – IRY = (- 1.67 – j2.15) – (5.04 + j2.04)

= – 6.71 – j4.19 = 7.91 -148° A. (Ans.)

IB = IBR – IYB = (2.17 + j1.63) – (-1.67 – j2.15)

= 3.84 + j3.78 = 5.39 44.5° A. (Ans.)

These are the currents in the phases of star-connected unbalanced load. Let us find the drop across each star-connected branch impedance.

(ii) Voltage drop across impedances:

Voltage drop across ZR = IRZR = 2.9 8.1° × 20 30° = 58 38.1° V. (Ans.)

Voltage drop across ZY = IYZY = 7.91 -148° × 20 – 45° = 158.2 -193° V. (Ans.)

Voltage across ZB = IBZB = 5.39 44.5° × 40 60° = 215.6 104.5° V. (Ans.)

Example 55. A star-connected load consisting of non-inductive resistors of 30, 12 and 20 ohms connected to the R, Y and B lines respectively is fed by a 300 V (line), 3-phase supply. The phase sequence is RYB. Calculate the voltage across each resistor.

Fig.72

Using star-delta conversion method,

Impedances are :

ZR = 30 Ω,   ZY = 12 Ω   , ZB = 20 Ω

Now ZRZY + ZYZB + ZBR = 30 × 12 + 12 × 20 + 20 × 30 = 360 + 240 + 600 = 1200 Ω

Equivalent impedances of delta-connected load are:

Each current is in phase with its own voltage because the load is purely resistive.

To find line currents for delta-connection using vector algebra and choosing ERY as the reference axis, we get

IRY = 5 + j0

IYB = 7.5(- 0.5 – j0.866) = – 3.75 – j6.5

IBR = 3(- 0.5 + j0.866) = – 1.5 + j2.6

Line currents for delta-connection are:

IR = IRY – IBR

= (5 + j0) – (- 1.5 + j2.6) = (6.5 – j2.6) or 7 A in magnitude

IY = IYB -IRY

= (- 3.75 – j6.5) – (5 + j0) = – 8:75 – j6.5 or 10.9 A in magnitude
IB = IBR =IYB = (- 1.5 + j2.6) – (- 3.75 – j6.5)

= 2.25 – j9.1 or 9.37 A in magnitude

These line currents for delta-connection are the phase currents for star-connection.

Voltage drop across each limb of star-connected load is :

Voltage drop across ZR = IRZR = (6.5 – j2.6) × 30 = 195 – j78 or 210 V. (Ans.)
Voltage drop across ZY = IYZY = (- 8.75 -j6.5) × 12 = – 105 – j78 or 130.8 V. (Ans.)
Voltage drop across ZB = IBZB = (2.25 – j9.1) × 20 = 4.5 – j182 or 182.05 V. (Ans.)

Example 56. A 400 V, 3-phase, 3-wire symmetrical system RYB supplies a star-connected load whose branch circuit impedances are:

Determine the current in each branch.
Take the phase sequence as RYB.

Solution. The circuit is shown in Fig. 73.

Let us solve this question by four possible ways of tackling 3-wire unbalanced star-connected load problems.

Fig.73

1.      By using Kirchhoff’s law :

We know that

Now ZRZY + ZYZB + ZBZR

= (10 30°) (20° 60°) + (20 60°) (5 -90°) + (5 – 90°) (10 30°)

= 200 90° + 100 -30° + 50 -60°

= j200 + (86.6 – j50) + (25 – j43.3)

= 111.6 +j106.7 = 154.4 43.7°

ERYZB – EBRZY = 400 × 5 -90° – 400 120° × 20 60°

= 2000 -90° – 8000 180° = – j2000 – (- 8000)
= 8000 – j2000 = 8246 -14°

EYBZR – ERYZB = 400 -120° × 10 30° – 400 0° × 5-90°
= 4000 -90° – 2000 -90°

=-j4000 + j2000 = – j2000 = 2000 -90°

EBRZY – EYBZR = 400 120o × 20 60° – 400 -120° × 10 30°

=80001800 – 4000 -90° = – 8000 + j4000 = 8944 153.4°
Current in each branch [from (i), (ii) and (iii)]

2. By using Star-delta conversion method: The unbalanced star-connected load and its equivalent delta-connected load are shown in Fig 74.

Fig.74 (a)

Fig.74 (b)

Now,      ZRZY + ZYZB +ZBZR = 154.4 43.7°                            [Already calculated]

Equivalent delta-connected impedances are:

Line currents for delta-connection are:

IR =IRY – IBR

= (- 8.95 – j9.36) – (- 37.15 + j35.78)

= 28.5 – j15.11 = 53.38 57.7° A. (Ans.)

IY =IYB – IRY

= (- 17.89 – j18.72) – (- 8.95 + j9.36)

= -8.94 – j9.36 = 12.94 133.7°

IB=IBR – IYB

= (- 37.45 +j35.78) – (- 17.89 – j18.72) = – 19.56 + j54.5 = 57.9 109.7°

ƩI = IR + IY + IB = 0 ……… as a check.

3. By using Maxwell’s Loop Current Method:

Fig. 75 shows the loop or mesh currents.

Fig. 75

It may be noted that IR = I1
IY = I2 – I1
and                            IB = -I2

Considering drops across R and Y arms, we get
I1ZR + ZY(I1 – I2) = ERY

or               I1(ZR + ZY) – I2ZY = ERY                                          …(i)

Similarly, considering the arms Y and B, we have
(I2 – I1) ZY + I2ZB = EYB

-I1ZY + I2(ZY + ZB) = EYB

IR = I1 = 53.5 – 57.7° A. (Ans.)

IY = I2 – I1 = (19.58 – j54.7) – (28.59 – j45.22)
= – 90 – j9.48 = 13.07 -133.5° A. (Ans.)

IB = -I2 = – 58.08 -70.3° = 58.08 109.7° A. (Ans.)

4. Using Milliman’s Theorem:

Refer Figs. 76 and 77.

According to Milliman’s theorem, the voltage of the load star point 0′ with respect to the star point or neutral 0 of the generator or supply (normally zero potential) is given by

where ERO, EYO and EBO are the phase voltages of the generator or 3-phase supply.

Voltage across each phase of the load is
ERO´ = ERO – EO´O

EYO´ = EYO – EO´O

EBO´ = EBO -EO´O

Now,     IRO´ = (ERO -EO´O)

IYO´ = (EYO – EO´O) YY

IBO´ = (EBO – EO´O) YB