EEE3405 ELECTRICAL ENGINEERING PRINCIPLES 2 - EXERCISE 1

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1 A 3-phase, 4-wire, 380V, 50Hz power supply with the phase sequence RYB is connected to a balanced star load. Each phase of load consists of a capacitor of 30µF in series with a 100Ω resistor. Determine: (a) the phase voltage; (b) the impedance at each phase; and (c) the line current.

1 Soln (a) V4.219

3

380

3

VV L

P === (b) Ω== − 1.106)1030)(50(2

16x

X C π;

Ω°−∠=−= 7.4680.1451.106100 jZ (c) AI R 5.180.145

4.219 ==

2 Two-wattmeter method is used to measure the total active power taken by a 3-phase motor. The line voltage is 380V and the two wattmeter readings are 5.5kW and –1.5kW respectively. Determine: (a) the total active power; (b) the power factor; and (c) the line current.

2 Soln (a) Total active power = kW4)5.1(5.5 =−+ # (b) 0275.3

)5.1(5.5

)5.1(5.53tan =

−+−−=φ ; ⇒ °=φ 71.71

power factor = cosφ = cos (71.71°) = 0.314 #

(c) φ= cosIV3P LL , A35.19)314.0)(380(3

4000

cosV3

PI

L

L ==φ

= #

3 A 3-phase 4-wire 380V, 50Hz supply feeds power to an unbalanced star load has the following measured data at each phase. Red phase current = 45 A p. f. = 0.8 lagging Yellow phase current = 35 A p. f. = 0.9 lagging Blue phase current = 50 A p. f. = 0.7 lagging Calculate the current in the neutral wire.

3 Soln

Take VRN as reference, V04.219VRN °∠= ; V1204.219VYN °−∠= ; V1204.219VBN °∠=

A9.3645A))8.0(cos0(45I 1R °−∠=−°∠= −

A8.14535A))9.0(cos120(35I 1Y °−∠=−°−∠= − A4.7450A))7.0(cos120(50I 1

R °∠=−°∠= −

A)4.74508.145359.3645(IIII BYRN °∠+°−∠+°−∠=++= A1.45.20 °∠= #

4 An unbalanced three-phase delta load is connected to a 415V, RYB, 3-phase supply as shown in fig. A4. Calculate the RED line current IR.

B

R

Y

415∠0° V Z3

Z2

Z1

Z1 = 10 + j10 Ω

Z2 = 10 + j12 Ω

Z3 = 15 – j5 Ω

IR

Fig. 4 4 Soln Current through Z1 , A4534.29A

10j10

0415I RY °−∠=

+°∠=

Current through Z3 , A4.13825.26A5j15

120415I BR °∠=

−°∠=

A)4.13825.264534.29(III BRRYR °∠−°−∠=−= A4.4357.55 °−∠= # 5 A 3-phase motor of 80% efficiency is connected to a 380V, 50Hz 3-phase supply as shown in fig. B1.

If the motor windings is delta-connected and each phase has a resistance of 8Ω and an inductance of 19mH, determine: (i) the current Im in the motor winding and the current IL fed from the 3-phase supply; (ii) the output power of the motor; and (iii) the power factor of the motor.

5 Soln.

(i) VL = 380V, XL= 2πfL = 2 π (50) (19x10-3) = 5.97 Ω Ω°∠=Ω+=+= 7.3698.997.5j8jXRZM

For the delta-connected motor,

EEE3405 ELECTRICAL ENGINEERING PRINCIPLES 2 - EXERCISE 1

-2-

A7.361.387.3610

0380

Z

VI L

P °−∠=°∠

°∠== A99.65)1.38(3I3I PL ===

the cur∴ rent in motoring winding = IP = 38.1 A# the current from 3-phase supply = IL = 65.99 A#

(ii) MLLM cosIV3P θ= where θM = 36.7° °= 7.36cos)99.65)(380(3 = 34.82kW

the output power = P∴ M x efficiency = (34.82 x 0.8) kW = 27.86 kW # (iii) Power factor = cos θM = cos 36.7° = 0.8 #

6 Three identical coils connected in star are now connected in parallel with the motor to the supply as in part (a). The total power drawn by the coils is found to be 42kW and the overall power factor becomes 0.9. Calculate: (i) the power factor of star-connected load; (ii) the line current taken by the star-connected load; (iii) the resistance and inductance of the coil.

380V, 50 Hz 3-phase supply

IL

Motor

R

Im

Y

B

fig. B1

6 Soln

(i) total active power PT = PM + PL= (34.82 + 42) kW = 76.82 kW total reactive power QT = QM + QL where QM = PM tanθM = 34.82 x tan36.7°= 25.95 kVar Since power factor of the circuit = 0.9, θT = cos-1 0.9 = 25.84°

T

LM

T

TT P

)QQ(

P

Qtan

+==θ∴

82.76

)Q95.25(84.25tan L+

=°

⇒ kVAr9.10kVar95.25)82.76)(48.0(QL =−= 26.042

9.10

P

Qtan

L

LL ===θQ

power factor of the star connected load = cos [tan-10.26] = cos 14.57°= 0.97 #

(ii) LLLL cosIV3P θ= A79.65)97.0)(380(3

42000

cosIV3

PI

LLL

LL ==

θ= #

(iii) For a star-connected load,

Ω=== 33.379.65

3380

I

VZ

L

PL RL = ZL cos θL = 3.33 cos14.57° = 3.23Ω #

XL= ZL sin θL = 3.33 sin14.57° = 0.84Ω # mH67.216.314

84.0

f2

XL L ==

π= #

7 a) A star-connected unbalanced 3-phase load is fed by a 3-phase 4-wire system. Explain what will happen if the neutral wire is being open-circuited.

b) A 3-phase, 3-wire, 380V, RYB sequence power system is connected to an unbalanced star load as shown in fig. B2. Take VRN as reference and by using the Millman’s theorem method, determine: (i) the displacement neutral voltage VON; (ii) the phase voltages at the star load; (iii) the three supply line currents; and (iv) the total active power delivered to the load.

VRN ZR= 10 + j5Ω

ZY= 20 + j8Ω

ZB = 5 – j10Ω

R

Y

B

N

IY

IB

VON

IR

fig. B2

O

7 Soln

1. When the neutral wire is disconnected, the star-point voltage of the three-phase load will no longer be zero.

2. A potential will appear at the star point. Its value will depend on the actual unbalanced loading condition. Thus different values of phase voltages will be resulted at each phase of

EEE3405 ELECTRICAL ENGINEERING PRINCIPLES 2 - EXERCISE 1

-3-

the three- phase load. 3. Over-voltage or under-voltage in either one or two phases may appear.

(i) Take VRN as reference, the phase voltages phasors are:

VVRN °∠=°∠= 022003

381 VVYN °−∠= 120220 VVBN °∠= 120220

By Millman’s Therom BYR

BBNYYNRRNON YYY

YVYVYVV

++++=

sjZ

YR

R °−∠=°∠

=+

== 56.26089.056.2618.11

1

510

11

sjZ

YY

Y °−∠=°∠

=+

== 8.21046.08.2154.21

1

820

11

sjZ

YB

B °∠=°−∠

=−

== 43.63089.043.6318.11

1

105

11

43.63089.08.21046.056.26089.0

)43.63089.0)(120220()8.21046.0)(120220()56.26089.0)(0220(VON ∠+−∠+−∠

∠∠+−∠−∠+−∠∠=

#)V48.91j78.75(V64.1298.11897.716.0

67.12101.19 −−°−∠=∠−∠=

(ii) °−∠−°∠=+= 64.1298.1180220VVV NORNRO #)V48.91j8.295(V18.176.309 +°∠=

°−∠−°−∠=+= 64.1298.118120220VVV NOYNYO #)V04.99j21.34(V05.1097.104 −−°−∠=

#NOBNBO )V282j21.34(V9.961.28464.1298.118120220VVV +−°∠=°−∠−°∠=+= (iii) RROR YVI = A)56.26089.0)(18.176.309( °−∠°∠= A38.955.27 °−∠= #

YYOY YVI = A)8.21046.0)(05.1097.104( °−∠°−∠= A5.13082.4 °−∠= #

BBOB YVI = A33.16028.25A)43.63089.0)(A9.961.284( °∠=°∠°∠= #

(iv) Total active power, PT = BYR Z

2

BZ

2

YZ

2

R R.IR.IR.I ++

= W)5()28.25()20()82.4()10()55.27( 222 ++ = 11.25 kW # 8 A 380V 3-phase, 4-wire, 50Hz power

supply of phase sequence RYB is connected to a balanced star load consists of three coils as shown in fig. A1. Each coil has an internal resistance of 5Ω and an inductance of 0.1H. Determine: (a) the supply phase voltage; (b) the power factor of the load; and (c) the line current.

fig. A1

Coil r = 5ΩΩΩΩ; L = 0.1H

8 Soln a) V4.219

3

380

3

VV L

P === # b) Ω=π= 42.31)1.0)(50(2X L ;

157.082.31

5

)42.31(5

5

Z

R.f.p

22==

+== # c) #

PR A9.6

82.31

4.219

Z

VI ===

9 The power input to a 3 kV, 50Hz, 3-phase delta-connected motor is measured by two-wattmeter method. If the two wattmeter readings are 350kW and 200kW respectively, determine: (a) the total input power; and (b) the power factor.

9 Soln

a) W1 = 350 kW, W2 = 200 kW, The total input power = W1+W2 = 550kW

b)[ ] [ ]

472.0550

8.259

200350

)200350(3

WW

)WW(3tan

21

21 ==+

−=+

−=φ #

o

o

9.03.25cos.f.p

3.25

===φ∴

10 A 3-phase 4-wire 380V, 50Hz supply feeds power to an unbalanced star load has the following

measured data at each phase.

EEE3405 ELECTRICAL ENGINEERING PRINCIPLES 2 - EXERCISE 1

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Red phase current = 36 A p. f. = 0.8 lagging Yellow phase current = 45 A p. f. = 0.9 lagging Blue phase current = 60 A p. f. = 0.85 leading, Find the current flowing in the neutral wire.

10 Soln

Take VRN as reference, i.e. V0VV PRN °∠= ; V120VV PYN °−∠= ; V120VV PYN °∠=

A9.3636A))8.0(cos0(36I 1R °−∠=−°∠= −

A8.14545A))9.0(cos120(45I 1Y °−∠=−°−∠= −

A8.15160A))85.0(cos120(60I 1R °∠=+°∠= −

A)8.151608.145459.3636(IIII BYRN °∠+°−∠+°−∠=++= A2.1630.64 °−∠= #

11 A delta connected resistive load is fed by a 380V, 3-phase supply as shown in fig. A5. The power delivered to each load are : Load Z1 = 30kW Load Z2 = 50kW Load Z3 = 80kW Calculate the current IR flowing in the Red line.

fig. A5 B

R

Y

380 V Z3

Z2

Z1

IR

11 Soln

Take VRY as reference,

#oo

31R

BR

BR33BRBR

RY

RY11RYRY

A7.4419.25912053.210095.78III

A53.210V380

W80000

V

PIcosIVP

A95.78V380

W30000

V

PIcosIVP

°−∠=∠−∠=−=

===∴φ=

===∴φ=

.

12 A 415 V, 50 Hz, three-phase 3-wire power supply supplies power to two balanced three-phase loads connected in parallel. They have the following data.

Load 1 : Lighting load of 30kW at 0.5 p.f. lagging Load 2 : Continuous motor load of 40kVA at 0.7 p.f. lagging.

Determine: (a) the line current taken from the three phase supply; (b) the readings of the two wattmeters if the total active power supplied to the two loads is measured by two-wattmeter method; and (c) the value of capacitance per phase of a delta-connected capacitor bank required to improve the overall power factor to 0.95.

12 Soln

Take VRN as reference, V06.239VRN °∠=

For load 1 : 11LL1 cosIV3P φ= ; φ1 = cos-1(0.5) = 60.0∘

A6047.8360)5.0)(415(3

3000060

cosV3

PI

1L

11L °−∠=°−∠=°−∠

φ=

For load 2 : 2LL2 IV3S = ; φ2 = cos-1(0.7) = 45.6∘

A6.4565.556.45)415(3

400006.45

V3

SI

L

22L °−∠=°−∠=°−∠=

Total line current from the 3-phase supply A)6.4565.556047.83(III 2L1LL °−∠+°−∠=+= A2.5407.138 °−∠= #

The magnitude of line current, A07.138I L = and the power angle of the load, φ = 54.2° Therefore the readings shown on the two wattmeters are:

W4.5790)2.5430cos()07.138)(415()30cos(IVW LL1 =°+°=φ+°= #

W6.52263)2.5430cos()07.138)(415()30cos(IVW LL2 =°−°=φ−°= # Total active power, PT = kW58kW)7.0)(40(kW30 =+

Power angle φ before correction = 54.2°, Power angle θ after correction, cos-10.95 = 18.2°

EEE3405 ELECTRICAL ENGINEERING PRINCIPLES 2 - EXERCISE 1

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Capacitive reactive power required for correction is: )2.18tan2.54.(tan58000)tan.(tanPQC °−°=θ−φ=

Var = 61349.6 Var Reactive power rating for each capacitor per phase,

Var8.204493

Var6.61349QC ==

For delta connection, voltage across each capacitor is 415V. ∴the capacitance per phase is:

phase/F0.378)50)(2()415(

8.20449

V

QC

22L

C µ=π

=ω

= #

P=58000 W

54.2°

QC

QT

QL

18.2°T ST

13 a) A star-connected unbalanced 3-phase load

is fed by a 3-phase 4-wire 380V, RYB sequence power system as shown in fig. B2. Determine the current in the neutral wire. (Use VRN as reference)

b) Explain what will happen if the link connected to the neutral wire is being open-circuited.

c) With the link open-circuited, determine: (i) the voltage VON at the star point O; and (ii) the three supply line currents.

Fig. B2

ZR = 5 + j10 Ω

ZY = 12 – j13 Ω

ZB = 6 + j8 Ω

ZB ZY

ZR

O

R

Y

B

fuse

fuse

fuse

link N

380 V

13 Soln

VRN as reference, the phase voltages phasors are:

V04.21903

381VRN °∠=°∠= V1204.219VYN °−∠= V1204.219VBN °∠=

A4.6362.19A10j5

04.219

Z

VI

R

RNR °−∠=

+°∠== A7.7240.12A

13j12

1204.219

Z

VI

Y

YNY °−∠=

−°−∠==

A9.6694.21A8j6

1204.219

Z

VI

B

BNB °∠=

+°∠==

#

BYRN

A6.2301.23

9.6694.217.7240.124.6362.19IIII

°−∠=°∠+°−∠+°−∠=++=

When the neutral wire is disconnected, the star-point voltage of the three-phase load will no longer be zero. Different values of phase voltages will be resulted at each phase of the three- phase load.

By Millman’s Theorem BYR

BBNYYNRRNON YYY

YVYVYVV

++++=

, s43.63089.0

10j5

1

Z

1Y

RR °−∠=

+==

s3.47057.013j12

1

Z

1Y

YY °∠=

−==

, s1.531.0

8j6

1

Z

1Y

BB °−∠=

+==

1.531.03.47057.043.63089.0

)1.531.0)(1204.219()3.47057.0)(1204.219()43.63089.0)(04.219(VON −∠+∠+−∠

−∠∠+∠−∠+−∠∠=

#)V87.36j8.120(V98.1628.1266.4018.0

6.2399.22 −°∠=−∠−∠=

(ii) #

R

ONRN

R

ROR A9.8342.9

10j5

98.1628.12604.219

Z

VV

Z

VI °−∠=

+°∠−°∠=−==

#Y

ONYN

Y

YOY A2.8828.18

13j12

98.1628.1261204.219

Z

VV

Z

VI °−∠=

−°∠−°−∠=−==

#B

ONBN

B

BOB A3.9367.27

8j6

98.1628.1261204.219

Z

VV

Z

VI °∠=

+°∠−°∠=−==