UNIT 4

Poly-Phase Circuits

4.1   Generation of three phase voltages

• Poly phase one which produces many phase simultaneously
• Instead of saying poly phase we use 3 ɸ supply there for 3 ɸ system
• Generation of electrical supply is 3 ɸ only using  alternator (AC generator ) by 3 separate winding placed 1200 a part from each other one winding for I phase

3 windings

3 phase =   = 1290   apart each winding

Here R is the reference phase globally in generation of 3 Ø ac

Ø = 0

Y is the 2nd  phase generated and placed apart from R phase Ø = -1200

B is the 3rd  phase generated and placed apart from R phase Ø = -2400

• Phasor  Diagram :

Equation

VR = Vm Sin wt

VY = Vm Sin (wt-1200)

VB = sin (wt – 2400)

Or VB = Vm sin (wt + 1200)

• Advantages of 3 Ø system over single phase

1. More output : for same size the output of 3Ø machine is always higher than single 1 Ø phase machine.
2. Smaller size : for producing same output the size of 3 phase machine is always smaller than of single phase machine
3. 3 phase motor are self starting as the 3 Ø ac supply is capable of producing a rotating magnetic file when applied are self starting 1 Ø motor need additional starter winding
4. More power is transmitted : in the transmitted system it is possible to transmitted more power using 3 Ø system rather than 1 Ø system, by using conductor of same cross sectional .
5. Smaller cross sectional area of conductor

ɡȴ same amount of power is to transmitted than cross sectional area of conductor used for 3 Ø system is small as compared to that for single Ø system.

4.2   Relations of voltage and current in star and delta connections for balanced systems

• Symmetrical or balanced system

3Ø system in which 3 voltages are of identical magnitude and frequency and are displaced by 1200 from each other called as Symmetrical system.

• Phase sequence :

The sequence in which the 3 phase reach their maximum +ve values Sequence is R-Y-B 3 colours used to denoted 3 phase are red’, yellow, blue.

The direction of rotation of 3 Ø machines depends on phase sequence. ɡȴ the phase sequence is changed ie R-B-Y than the direction of rotation will be reversed.

• Types of loads

1. Star connection of load
2. Delta connection of load

• Balanced load:  balanced load is that in which magnitude of all impedance connected in the load are equal and the phase angle of them are also equal

Ie  Z1  Z2 Z3  then it is unbalanced load

• Line values and phase Values :

1. Line Values: ɡȴ RYB are supply lines then the voltage measured between any 2 Line is called as line voltage and current measured in the supply line is called as line current.

2.     Phase Value: the voltage measured across a single winding or phase is called as phase voltage and the current measured on the single winding or phase is called as line current.

Derive

Relation between line value and phase value of voltage and current for balanced (ʎ) star connected load (load can be resistive, Inductive or capacitive)

For capacitive load

Consider a 3 Ø balanced star connected balanced load capacitive

• Line Value

Line voltage = VRY = VYB = VBR = VL

Line current = IR = IY = IB = IL

Phase value

Phase voltage = VRN = VYN = VBN = Vph

Phase current = VRN = VYN = VBN = Vph

Since for a balanced star connected load the line current is the same current flowing in the phase the line current = phase current IR = IY = IB = IRN = IBN = IYN

dor star connected load IN = Ipn

• Since the line voltage differ from phase voltage we can relate the line value of voltage with phase value of voltage

Referring current diagram

= +

Bar indicates vector addition

=

= -         …..①

Instead of writing or we can write VR and VY for practical purpose

Similarly other line voltage can be writing as follows.

(Resultant)

= -         …..②

= - 

• Phasor Diagram :

Consider equation   …..①

NOTE : We are getting resultant line voltage by subtracting phase voltage   take phase voltage at reference phase as shown

Cos 300 =

=

VL =

• phase for ʎ connected capacitive balance load

• Relation between line value and phase value of voltage and current for a balance () delta connected inductive load

Consider a 3 Ø balance delta connected inductive load

• Line values

Line voltage = VRY = VYB = VBR = VL

Line current = IR = IY = IB = IL

Phase value

Phase voltage = VRN = VYN = VBN = Vph

Phase current = VRN = VYN = VBN = Vph

• Since for a balance delta connected load the voltage measured in line and phase is same because their measuring points are same

for balance delta connected load VL =  Vph

VRV = VYB = VBR = VR = VY = VB = VL = VPh

• Since the line current differ from phase current we can relate the line and phase values of current as follows
• Apply KCL at node R

IR + IRY= IRY

IR     =   IRY - IRY … ….①

Line    phase

Similarly apply  KCL at node  Y

IY  +  IYB  =  IRY … ….②

Apply KCL at node  B

IB  +  IBR  =  IYB … ….③

• Phasor Diagram

Consider equation  ①

Note : we are getting resultant line current IR      by subtracting  2 phase currents IRY and IBR   take phase currents at reference as shown

Cos 300  =

=

• Complete phases diagram for delta connected balanced Inductive load.

Phase current IYB  lags behind VYB  which is phase voltage as the load is inductive

• Power relation for delta load star power consumed per phase

PPh =VPhIPh Cos Ø

For 3 Ø  total power is

PT= 3 VPhIPh Cos Ø …….①

For star

VL and IL = IPh  (replace in ①)

PT  = 3    IL  Cos Ø

PT  = 3   VL IL  Cos Ø – watts

For delta

VL  =VPhand IL =      (replace in ①)

PT = 3VL    Cos Ø

PT VLIL  Cos Ø – watts

Total average power

P = VLIL  Cos Ø – for ʎ and load

K (watts)

Total reactive power

Q = VLIL  Sin Ø – for star delta load

K (VAR)

Total Apparent power

S = VL IL   – for star delta load

K (VA)

• Power triangle

• Relation between power

In star and power in delta

Consider a star connected balance load with per phase impedance ZPh

We know that for

VL = VPhandVL= VPh 

Now  IPh  =

VL = =

And VPh=

IL  =    ……①

Pʎ = VL IL Cos Ø ……②

Replacing ① in ② value of IL

Pʎ = VL IL   Cos Ø

Pʎ = ….A

• Now for delta

IPh  =

IPh = =

And IL  = IPh

IL  = X    …..①

P = VL IL Cos Ø ……②

Replacing ② in ① value of IL

P = Cos Ø

P = …..B

Pʎ from …A

…..C

    =   P

We can conclude that power in delta is 3 time power in star from …C

Or

Power in star is time power in delta from ….D

• Step to solve numerical

1. Calculate VPh from the given value of VL by relation

For star VPh =

For delta VPh= VL

2.     Calculate IPh using formula

IPh =

3.     Calculate IL using relation

IL = IPh - for star

IL  = IPh  - for delta

4.     Calculate P by formula (active power)

P = VL IL Cos Ø – watts

5.     Calculate Q by formula (reactive power)

Q = VL IL Sin Ø – VAR

6.     Calculate S by formula (Apparent power)

S = VL IL– VA

References:

1. Electrical Technology (Volume I & 2), B L Thereja, 22nd edition, S Chand & Company Ltd

2. Basic Electrical Engineering, V K Mehta, Revised edition, S Chand & Company Ltd

3. Basic Electronics Solid State, B L Thereja, Revised edition, S Chand & Company Ltd

4. Digital Principles and Applications, Albert Malvino , Donald Leach, Tata McGraw Hills Publication

5. Principles of Electronic Devices and Circuits (Analog and Digital), B. L. Theraja , R. S. Sedha , S. Chand publication