Chapter 20 Three Phase Systems

CHAPTER 20 THREE PHASE SYSTEMS

Exercise 112, Page 327

1. Three loads, each of resistance 50  are connected in star to a 400 V, 3-phase supply. Determine

(a) the phase voltage, (b) the phase current and (c) the line current.

400 V, 3-phase supply means that 400 V is the line voltage.

(a) For a star connection,

Hence, phase voltage, = 231 V

(b)Phase current, = 4.62 A

Hence, line current, = 4.62 A

2. A star-connected load consists of three identical coils, each ofinductance 159.2 mH and resistance

50 . If the supply frequency is 50 Hz and the line current is 3 A determine (a) the phase voltage

and (b) the line voltage.

Inductive reactance, XL = 2πfL = 2π(50)(159.2 × 10-3) = 50 

Impedance of each phase, Zp = = = 70.71

For a star connection,IL = Ip =

Hence,phase voltage,Vp = IpZp = (3)(70.71) = 212 V

Line voltage,VL = Vp = (212) = 367 V

3. Three identical capacitors are connected in star to a 400 V, 50 Hz, 3-phase supply. If the line

current is 12 A determine the capacitance of each of the capacitors.

For a star connection,

hence, = 231 V

thus,

andcapacitance, C = = 165.4 F

4. Three coils each having resistance 6  and inductance L H are connected in star to a 415 V, 50 Hz,

3-phase supply. If the line current is 30 A, find the value of L.

For a star connection,

hence, = 239.6 V

thus,

from which, = 5.272 Ω

Hence, 5.272 = 2π f L

and inductance, L = = 16.78 mH

5. A 400 V, 3-phase, 4 wire, star-connected system supplies three resistive loads of 15 kW, 20 kW

and 25 kW in the red, yellow and blue phases respectively. Determine the current flowing in each

of the four conductors.

For a star connected system, from which, = 230.94 V

Power, P = VI for a resistive load, hence

Thus, = 64.95 A, = 86.60 A

and = 108.25 A

The phasor diagram of the three currents is shown in (i) below. Adding phasorially gives diagram (ii) below, where is the neutral current.

(i) (ii)

Total horizontal component = 64.95 cos 90 + 108.25 cos 210 + 86.60 cos 330 = - 18.75

Total vertical component = 64.95 sin 90 + 108.25 sin 210 + 86.60 sin 330 = - 32.475

Hence, magnitude ofneutral current, = = 37.50 A

Exercise 113, Page 329

1. Three loads, each of resistance 50  are connected in delta to a 400 V, 3-phase supply.

Determine (a) the phase voltage, (b) the phase current and (c) the line current.

(a) For a delta connection,

Since = 400 V, then phase voltage, = 400 V

(b) Phase current, = 8 A

2. Three inductive loads each of resistance 75  and inductance 318.4 mH areconnected in delta to a

415 V, 50 Hz, 3-phase supply. Determine (a) the phase voltage, (b) the phase current, and (c) the

line current.

(a) For a delta connection,

Since = 410 V, then phase voltage, = 415 V

(b) Phase impedance,

Phase current, = 3.32 A

3. Three identical capacitors are connected in delta to a 400 V, 50 Hz, 3-phase supply. If the line

current is 12 A determine the capacitance of each of the capacitors.

For a delta connection, hence

= 400 V

thus,

andcapacitance, C = = 55.13F

4. Three coils each having resistance 6  and inductance L H are connectedin delta, to a 415 V,

50 Hz, 3-phase supply.If the line current is 30 A, find the value of L.

For a delta connection, hence

= 415 V

thus, 23.96 =

from which, i.e. 2π f L = 23.197 Ω

Hence, inductance, L = = 73.84mH

5. A 3-phase, star-connected alternator delivers a line current of 65 A to a balanced delta-connected

load at a line voltage of 380 V. Calculate (a) the phase voltage of the alternator, (b) the alternator

phase current and (c) the load phase current.

(a) In star, from which, phase voltageof alternator, = 219.4 V

(b) In star, hence, alternator phase current = 65 A

6. Three 24 F capacitors are connected in star across a 400 V, 50 Hz, 3-phase supply. What value

of capacitance must be connected in delta in order to take the same line current?

In star, hence, = 230.94 V

= 132.63

Hence, = 1.741 A = line current for star connection.

In delta, if hence, = 1.00517 A

= 397.94  i.e. 397.94 =

from which, capacitance in delta, C = = 8 F

Exercise 114, Page 331

1. Determine the total power dissipated by three 20  resistors when connected (a) in star and

(b) in delta to a 440 V, 3-phase supply.

(a) In star, hence,

and

Hence in star, power, P = = 9.68 kW

or P = = 9.68 kW

(b) In delta, and

= 38.11 A

Hence in delta, power, P = = 29.04 kW

or P = = 1350 W = 29.04 kW

2. Determine the power dissipated in the circuit of Problem 2, Exercise 112,page 327.

Circuit phase angle, ϕ =

Power, P = = 1348 W = 1.35 kW

or P = = 1.35 kW

3. A balanced delta-connected load has a line voltage of 400 V, a linecurrent of 8 A and a lagging

power factor of 0.94. Draw a complete phasordiagram of the load. What is the total power

dissipated by the load ?

If the power factor is 0.94, then 0.94 = cos ϕ from which, ϕ = = 19.95º

Line voltage = phase voltage = 400 V

Line current = 8 A Phase current = = 4.62 A

From the diagram above, , and

The complete phasor diagram is shown below, with the line current lagging the line voltage by 19.95º

Power, P = since power factor = cos ϕ

= 5210 W = 5.21 kW

4. Three inductive loads, each of resistance 4  and reactance 9  are connected in delta. When

connected to a 3-phase supply the loads consume 1.2 kW. Calculate (a) the power factor of the

load, (b) the phase current, (c) the line current and (d) the supply voltage.

(a) Phase impedance,

and phase angle,

Hence, power factor of load = cos = cos 66.04 = 0.406

(b) Power, P = i.e.

from which, phase current, = 10 A

(d) Power, P = i.e. 1200 =

from which, supply voltage, = 98.53 V

5. The input voltage, current and power to a motor is measured as 415 V, 16.4 A and 6 kW

respectively, Determine the power factor of the system.

Power, P = i.e. 6000 =

from which, cos = power factor of system = = 0.509

6. A 440 V, 3-phase a.c. motor has a power output of 11.25 kW and operates at a power facror of

0.8 lagging and with an efficiency of 84%. If the motor is delta connected determine (a) the

power input, (b) the line current and (c) the phase current.

(a) Efficiency = i.e. 0.84 =

from which, power input = = 13393 W or 13.39 kW

(b) Power, P = hence, line current, = 21.97 A

Exercise 115, Page 336

1. Two wattmeters are connected to measure the input power to a balanced three-phase load. If the

wattmeter readings are 9.3 kW and 5.4 kW determine (a) the total output power, and (b) the load

power factor

(a)Total input power, P = P1 + P2 = 9.3 + 5.4 = 14.7 kW

(b) tan = = = = 0.459524

Hence,  = tan = 24.68o

Power factor = cos = cos24.68o = 0.909

2. 8 kW is found by the two-wattmeter method to be the power input to a 3-phase motor.

Determine the reading of each wattmeter if the power factor of the system is 0.85

Working in kilowatts, 8 = (1)

If power factor = 0.85, then cos = 0.85 and phase angle,  = = 31.79

and tan  = tan 31.79 = 0.6197

Hence, tan  = 0.6197 =

from which, = 2.862(2)

Adding equations (1) and (2) gives:

from which, = 5.431 kW

and from equation (1), = 2.569 kW

Thus the readings of the two wattmeters are: 5.431 kW and 2.569 kW

3. When the two-wattmeter method is used to measure the input power of a balanced load, the

readings on the wattmeters are 7.5 kW and 2.5 kW, the connections to one of the coils on the meter

reading 2.5 kW having to be reversed. Determine (a) the total input power, and (b) the load power

factor

Since the reversing switch on the wattmeter had to be operated the 2.5 kW reading is taken as

– 2.5 kW

(a) Total input power, P = P1 + P2 = 7.5 + (- 2.5) = 5 kW

(b) tan = = =

Angle  = tan(2) = 73.90o

Power factor = cos = cos 73.90o = 0.277

4. Three similar coils, each having a resistance of 4.0  and an inductive reactance of 3.46  are

connected (a) in star and (b) in delta across a 400 V, 3-phase supply. Calculate for each

connection the readings on each of two wattmeters connected to measure the power by the two-

wattmeter method.

(a) Star connection: and

Hence, = 230.94 V

Phase impedance, = 5.289 

Phase current, = 43.664 A

Total power, P = = 22.879 kW

If the wattmeter readings are and then:

+ = 22.879(1)

Phase angle,  =

and tan 40 86 = from (1)

from which, - =

i.e. - = 11.426(2)

Adding equations (1) and (2) gives: = 22.879 + 11.426 = 34.305

and = = 17.15 kW

Substituting in (1) gives: = 22.879 – 17.15 = 5.73 kW

Hence, in star, the wattmeter readings are 17.15 kW and 5.73 kW

(b) Delta connection: and

Phase current, = 75.629 A

Total power, P = = 68.637 kW

Hence + = 68.637(3)

tan 40 86 = from (3)

from which, - =

i.e. - = 34.278(4)

Adding equations (3) and (4) gives: = 68.637 + 34.278 = 102.915

and = = 51.46 kW

Substituting in (3) gives: = 68.637 – 51.46 = 17.18 kW

Hence, in delta, the wattmeter readings are 51.46 kW and 17.18 kW

5. A 3-phase, star-connected alternator supplies a delta connected load, eachphase of which has a

resistance of 15  and inductive reactance 20 . If the line voltage is 400 V, calculate (a) the

current supplied by the alternator and (b) the output power and kVA rating of the alternator,

neglecting any losses in the line between the alternator and the load.

A circuit diagram of the alternator and load is shown below.

(a) Considering the load: Phase current, Ip =

Vp = VL for a delta connection, hence, Vp = 400 V

Phase impedance, Zp = = = 25

Hence, Ip = = = 16 A

For a delta-connection, line current, IL = Ip = (16) = 27.71 A

Hence,27.71 A is the current supplied by the alternator

(b) Alternator output power is equal to the power dissipated by the load

i.e. P = VL ILcos, where cos = = = 0.6

Hence, P =(400)(27.71)(0.6) = 11518.8 W = 11.52 kW

Alternator output kVA, S = VL IL=(400)(27.71) = 19198 VA =19.20kVA

6. Each phase of a delta-connected load comprises a resistance of 40  and a 40 F capacitor in

series. Determine, when connected to a 415 V, 50 Hz, 3-phase supply (a) the phase current,

(b) the line current, (c) the total power dissipated, and (d) the kVA rating of the load.

(a) Capacitive reactance,

Phase impedance,

Phase current, = 4.66 A (since in delta)

(b) Line current, = 8.07 A

Hence, total power dissipated, P = = 2.605 kW

(d) ThekVA rating of the load, S = = 5.80 kVA