Module 2.  A.C. series and parallel circuits

Lesson 8

POLYPHASE ALTERNATING CURRENT CIRCUITS

8.1  Introduction

In domestic supply line comprise mostly of single phase electricity. Home appliances like T.V., refrigerator, washing machines are designed to be operated by single phase alternating current. But for individual use where heavy duty machines have to be operated single phase current is not sufficient. There polyphase a.c. is needed to run in plant and machinery. Polyphase means a.c. will have two or more than two phases. Generally 3 phase (also denoted as 3ϕ) alternating systems are employed in the industry.

8.2  Polyphase Systems

8.2.1  Single phase

A generator with one armature winding involves a single phase alternating current. The instantaneous value of emf induced in the eq. is given as

ea1a2 = Em. Sin (ωt)

8.2.2  Two phase

A two phase alternative (generator) has two windings and the angle between them is 90o. As a result the phase difference between the two alternating voltages is 90o. And since the number of turns in the two windings are same the magnitude of e.m.f generated and frequency is same in both the windings. The instantaneous value of e.m.f induced in the two coils is given as:

ea1a2 = Em Sin ωt

ea1a2 = Em sin (ωt – 90o) or Em sin (ωt – π/2)

8.2.3  Three phase

A three phase system has three windings in the alternator. The windings are placed such that the angle between them is 120o (Fig. 8.1). Since the windings are identical the magnitude and frequency of alternating voltage is same for all the three windings.  The instantaneous value of e.m.f. indicates in the three windings are given as:

ea1a2 = Em Sin ωt

eb1b2 = Em Sin (ωt – 120o) or Em Sin (ωt- 2p/3)

ec1c2 = Em Sin (ωt-240o) or Em Sin (ωt- 4p/3)

Fig. 8.1 Wave form of three phase system

8.3  Comparison of Three phase and Single Phase System

Table 8.1 Comparison of three phase and single phase system

 3ϕ System Singleϕ System 1. Transmission line distribution and voltage regulation Better Poor 2. Transmission line conductor (wire) requirement for distribution Less (Only 75% of what is received for single ϕ system) More 3. Vibration due to electrical loads (e.g. motors) Low High 4. Electrical Machine (e.g. motors, generators, transformers etc.) a.       Size Small Bulky b.      Construction Simple Complex c.       Performance Better Average d.      Cost Cheaper Expensive than 3ϕ e.       Efficiency High Low f.       Power factor of motor High (0.7 to 0.8) Low (0.5) g.      Starting of motors Self Starting Not self starting

8.4  Power Sequence

In a poly phase system, the order in which the phases attain the maximum voltages (emf) is called as phase sequence. Maximum voltage or emf is attained in the phase sequence of a1a2®b1b2®c1c2. If the direction notation is reversed then the new phase sequence will be c1c2®b1b2® a1a2.

Note: In 3ϕ induction motor, the direction of rotation depends on phase sequence of the applied voltage. Direction of the rotation can be reversed by interchanging any two lines.

8.5  Numbering of Phases

The phases can be numbered in three methods according to the phase sequence of system.

a)      By numbering 1, 2, 3

b)      By alphabetically a, b, c

c)      Colour code (R, Y, B) where R = Red, Y = Yellow, B = Blue

+ ve phase sequence = RYB

- ve phase sequence = RBY or BYR or YRB

8.6  Double Subscript Notation

Voltages in the 3ϕ systems can be denoted using two subscripts. The two subscripts are the two points between which the voltage or current is being considered. The sequence of subscript gives the direction in which voltage is denoting or current is flowing.

Example

VRY – It denotes the voltage between points RY and the positive direction of voltage from R to Y.

IRY – It denotes the current flowing from point R towards point Y.

8.7  Inter Connection of Three Phases

Let us first understand the construction of a 3ϕ alternator. It comprises of two main parts stator and rotor. Stator is the stationary part of the alternator. Stator have three windings. Winding has two terminal marked on start and finish. A total of six wires will be required to connect the alternator (two for each winding). The type of connection is very expensive and complex. Instead these six terminals are connected by any of the two methods.

1.      Star or wye (Y) connection (Fig 8.2)

2.      Mesh or delta (∆) connection (Fig 8.3)

8.7.1  Star or wye (Y) connection

Here the similar ends (either start or finish) are connected at a point. This point is known as star or neutral point (N). Line conductors are commonly called by colour code R (Red), Y (Yellow) and B (Blue) and are shown in Fig 8.1. as RR¢, YY¢ and BB¢. They can also be sequenced as a,b,c or 123.

Fig. 8.2 Star or Y connection

·         Neutral connection: - The wire joining neutral point N is known as neutral connection.

·         Phase voltage: - As the name suggests it is the voltage measured between any line and Neutral (N) is known as phase voltage (Vph). Since the coils are held at 1200.

·         Line voltage: - It is the voltage existing between any two phase winding or phases (Vph).

Line voltage = √3 × phase voltage

·         Phase current: - Current flowing through any winding or phase is known as phase current.

·         Line current: - Current flowing through any line is known as line current.

For star connection line current = phase current

Note: Star connection is also known as 3 phase 4 wire system. In case neutral wire in not there, it may be called as 3ϕ 3 wire system.

Power

Output power per phase = Vph Iph Cosϕ

Total output power = 3 Vph Iph Cosϕ

since VL = √3 Vph

Iph = IL

Total output power = √3 VL IL Cosϕ

Where

Vph = Phase voltage

Iph = Phase current

VL = Line voltage

IL = Line current

Cosϕ = Power factor

P = Power

8.7.2  Delta or mesh connection

In delta or Mesh connection the three windings are connected in series pattern (Fig. 8.3). The connections are made in the following way:

Start of winding  a1 → Finish terminal of winding  c2

Start of winding  b1 → Finish terminal of winding a2

Start of winding  c1 → Finish terminal of winding b2

Fig. 8.3 Delta or mesh connection

·         Phase voltage – The voltage measured in windings or phases is known as phase voltage.

·         In Mesh or Delta ∆ connection

Line voltage (VL) = Phase voltage (Vph)

·         Phase current – Is the current flowing through any winding or phase.

·         Line current – Is the current flowing through any line. For ∆ connection:

Line current (IL) = √3 Iph

Where

Iph = phase current

Power

Output power per phase = Vph Iph Cosϕ

Total power P = 3 Vph Iph Cosϕ

Since Vph = VL and IL = √3 Iph

P = √3 VL IL Cosϕ

True power P = √3 VL IL Cosϕ units Watts (W) or kilowatts (kW)

Apparent power Papp = √3 VL I L units VA or kVA

Reactive power Prec = √3 VL IL Sinϕ units VAR or kVAR

8.8  Differences Between Star(Y) and Delta(∆) Connection

Table 8.2 Differences between star(Y) and delta (∆) connection

 Star (Y) Delta (∆) 1.      Structure 1.      Structure 2.      Winding are connected by joining similar ends 2.      Windings are connected by joining dissimilar ends 3.      3ϕ 4 wire system is possible 3.      Only 3ϕ 3 wire system possible 4.      Provision for neutral 4.      No neutral wire 5.      Line voltage VL = √3 Vph Where, Vph = Phase voltage 5.      Line voltage VL = Phase voltage Vph 6.      Line current IL = Phase current Iph 6.      Line current IL = √3 Iph Where, Iph = phase current 7.      Line voltage are at phase difference of 1200 7.      Lines current are at phase difference of 1200. 8.      Alternators are generally designed as 3 ϕ star system because of following reason i. Less insulation requirement ii. Require less number of turns in the winding 8.      Alternators are not generally connected as Delta system.

Numericals

1. Calculate line voltage of Y-system if phase voltage is 340 V.

VL = √3 Vph

VL = √3 X 340

VL = 588.89 V

2. Calculate line current of Δ-system if phase current is 25 A.

IL = √3 Iph

IL = √3 X 25

IL = 43.3 A

3. Calculate total power in a Y-system if line voltage and line current is 440 V and 25 A respectively. Consider power factor as 0.65.

Total output power = √3 VL IL Cosϕ

= √3 X 440 X 25 X 0.65

= 12.38 kW

4. For a star connected system calculate the line voltage and output power if phase voltage = 410 v and line current is 7.5 A. Take power factor as 0.86.

Solution

Given

Vph = 410 v

IL = 7.5 A

Cos Φ = 0.86

= 710.14 v

Total outer power =

= 7933 w

5.  For a data connected system calculate the line current and output power if phase voltage is 440 v and phase current is 5A. Take power factor as 0.75. Also calculate the reactive power.

Solution

Vph = 440 v

IL = 5 A

Cos Φ = 0.76

= 8.66 A

Output power p = 3 Vph Iph cosΦ

=

P = 4950 w

Reactive power prec =

prec = 3 Vph Iph Sin Φ

Cos Φ = 0.75

Φ = 41.40°

Sin Φ  = 0.661

Prec =

= 4362.6 W