Electrical Engineering

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Three Phase Circuits

P INTRODUCTION

This chapter has been divided into following topics: 1) Basic concepts of Poly phase Circuits and its advantages 2) Generation of three phase supply 3) Star – Delta Connections (Balanced load) 4) Relationship between phase and line currents and voltages 5) Phasor diagrams 6) Measurement of power by two wattmeter method POLYPHASE SYSTEMS 1) Poly phase alternators have two or more windings symmetrically spaced around the armature. Such alternators produce as many independent alternating voltages as the number of windings (Phase). 2) These voltages in the individual windings have the same magnitude and frequency but they have definite phase difference. The amount of phase difference depends upon the number of windings. 3) Thus, by using appropriate poly phase alternator, it is possible to generate two phase and three phase ac and from these to obtain four, six, nine, twelve phase AC depending upon the requirements. These systems are collectively known as poly phase systems. ADVANTAGES OF 3- Φ SYSTEM OVER 1- Φ SYSTEM 1) The output of a 3- φ machine is greater than that of 1- φ machine of the same size. 2) The total power output of a 3- φ machine is not fluctuating as in the case of a 1- φ machine. It has a higher efficiency. 3) A 3- φ transmission needs less conducting material than a 1- φ transmission line. Hence, transmission becomes economical. 4) 3- φ motors are self-starting, whereas 1- φ motors are not self-starting. 5) Power factor of a 1- φ motor is always lower than that of a 3- φ motor of the same output and speed.

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6) 1- φ supply can be obtained from 3- φ supply but 3- φ supply cannot be obtained from 1- φ supply. GENERATION OF THREE PHASE SUPPLY

Waveforms of 3- φ supply Construction 1) The armature of the alternator consists of three single-turn rectangular coils R1R2, Y1Y2 and B1B2 fixed to one another at angles of 120˚. The coils are mounted on a common shaft and have same physical dimensions. 2) The ends of coil are connected to a pair of slip-rings carried on the shaft. The coils are placed in the uniform magnetic field provided by the North and South poles of the magnet. 3) The carbon brushes are pressed against the slip-rings to collect the induced currents in the coils. Operation 1) Suppose the three coils are rotating in an anti-clockwise direction at uniform speed. Because of this each coil will have its own generated e.m.f. and current which will be alternating in nature. 2) As shown in the figure the plane of the coil R1R2 is perpendicular to the magnetic field and hence no e.m.f. is generated in the coil. After 120˚ Y1Y2 will occupy this position and after 120˚ B1B2 will occupy this position. 3) This cycle continues and maximum value attained by every coil has 120˚ phase shift w.r.t e.m.f. in other coil.

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4) If the instantaneous value of the e.m.f. generated in the coil R1R2 is represented by,

eR  Em sin t then, the instantaneous values of the e.m.f.’s generated in the coil Y1Y2 and B1B2 are, eY  E m sin t  120

eB  E m sin t  240

5) Thus, we got three independent alternating voltages from 3- φ alternator which have phase shift of 120˚. The phasor diagram for 3- φ system is as show:

6) If we perform vector addition of these three voltages, it can be observed that, the sum of three voltages at any instant is zero. Mathematically,

eR  eY  eB  Em sin t  Em sin t  120  Em sin t  120   Em sin t  sin t  120   sin t  120 

 Em sin t  sin t.cos120  sin120.cos t  sin t.cos120  sin120.cos t 

 Em sin t  2.sin t.cos120  Em sin t  2.sin t.  1/ 2 

 E m sin t  sin t 

 E m  0

0

eR  eY  eB  0

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Important Definitions related to 3-φ System 1) Symmetrical System:A 3-φ system in which the three voltages are of same magnitude and frequency and displaced by 120¢ª phase angle from each other is defined as ‘Symmetrical System’. 2) Phase Sequence:The sequence in which the voltages in 3-φ reach their maximum positive values is called ‘Phase Sequence’. 3) Balanced System:A 3-φ system is said to be balanced when, a) All phase inpedances are identical. b) Magnitude and phase angle of all phase impedances are identical. c) All phases have same power factor. Connections in Three Phase System In 3- φ system we can have following connections: a) Star or Wie (Y) connection: b) Mesh or Delta (Δ) connection:  Star Connection:-

1) In this connection, like terminals of the coil i.e. either starting terminals R1, Y1, B1 or finishing terminals R2, Y2, B2 are connected together to form a common point ‘N’ which is known as neutral point or star point. 2) This kind of connection is also known as ‘Four wire three phase system’. This connection is shown in figure.

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3) Assuming phase sequence R-Y-B all the line quantities and phase quantities of star connection are shown in figure. a) Line Voltage (VL) : The potential difference between any two lines gives line to line voltage called as line voltage. For above fig. Line voltages are VL = VRY = VYB = VBR b) Phase Voltage (Vph) : The potential difference between any line and neutral point is called as phase voltage. For above fig. Phase voltages are Vph = VRN = VYN = VBN c) Line Current (IL): The current flowing through each line is called as line current. For above fig. Line currents are IL = IR = IY = IB d) Phase Current (Iph) : The current flowing through each phase is called as phase current. For above fig. phase currents are Iph = IRN = IYN = IBN  Relation Between Line Quantities and phase Quantities for Star Connected System 1) Consider a 3- φ star connected balanced system as shown in figure. From figure,

IR  IRN , IYN and IB  IBN But,

IL  IR  IY  IB

Hence,

IL  Iph

and

Iph  IRN  IYN  IBN …... (1)

2) To get the relation between line voltage and the phase voltage, plot the vector diagram. Assuming phase sequence to be R-Y-B with lagging power factor. From fig. voltage across line R and line Y is

VL  VRY  VRN  VYN   Similarly, VL  VYB  VYN  VBN  and VL  VBR  VRN  VYN  

…... (2)

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3) Vector Diagram for 3- φ star connected balanced system is shown in below figure. 4) Consider OPQ In OPQ draw PR  OQ From vector diagram,

VRY  l  OQ  and VRN  l  OP  But

l  OQ   2l  OR 

….... (3)

In OPR

l  OR   l  OP  cos300

l  OR   VRN

3 2

5) Substituting this in equation (3),

l  OQ   2VRN

3 2

VRY  3 VRN

Hence,

VL  3 Vph

6) Thus, for a star connected connection we can conclude following points: a) VL  3 Vph and IL  Iph b) Line voltages are 1200 apart. c) Line voltages are 300 ahead of their respective phase voltages. d) The angle between line currents and the corresponding line voltages is  30    with current lagging.

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e) The angle between line currents and the corresponding line voltages is  30    with current leading.  Expression for Power in 3- φ star connected system : 1) Active Power : We know that for a 1-φ system active power is given by, P = VI cos 

……(1)

Hence for a 3- φ system, Total active power = 3 × (Power/phase)

P = 3 Vph Iph cos 

Bus for star connected ckt.

 

P3

Iph  IL and Vph 

VL 3

VL IL cos  3

P  3 VL I L cos 

Where φ is the angle between phase voltage and phase current. 2) Reactive Power : Total reactive power is given by,

Q  3Vph Iph sin 

Q  3 VL IL sin 

3) Apparent Power : Total apparent Power is given by,

S  3Vph Iph

S  3 VL IL

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Delta Connection

1) In this connection finishing terminal of one coil is connected to starting terminal of second coil and finishing of that to starting of next and so on i.e. In Δ connection coils are connected to form a closed loop as shown in fig. (1) 2) This kind of connection is also known as ‘three wire three phase system’. This connection is shown in fig. (1) 3) Assuming phase sequence R-Y-B all the line quantities and phase quantities of delta connection are shown in fig. (1) a) Line Voltage (VL) : The potential difference between any two lines gives line to line voltage called as line voltage. For above fig. Line voltages are VL = VRY = VYB = VBR b) Phase Voltage (Vph) : The potential difference between any line and neutral point is called as phase voltage. For above fig. Phase voltages are Vph = VR = VY = VB c) Line Current (IL): The current flowing through each line is called as line current. For above fig. Line currents are IL = IR = IY = IB d) Phase Current (Iph) : The current flowing through each phase is called as phase current. For above fig. phase currents are Iph = IRY = IYB = IBR

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 Relation Between Line Quantities and phase Quantities for Delta Connected System:1. Consider a 3- φ delta connected balanced system as shown in figure. From figure,

VRY  VYB  VBR  VR  VY  VB But,

VL  Vph

2. To get the relation between line voltage and the phase voltage, plot the vector diagram. Assuming phase sequence to be R-Y-B with lagging power factor. From fig.

IR  IRY  IBR   Similarly, I Y  I YB  I RY  and IB  I BR  I YB 

…… (2)

3. Vector Diagram for 3- φ delta connected balanced system is shown below figure.

4. Consider OPQ In OPQ draw PR  OQ From vector diagram, IRY  l  OP  and IR  l  OQ  But

l  OQ   2l  OR 

…… (3)

In OPR

l  OR   l  OP  cos300

l  OR   I RY

3 2

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5. Substituting this in equation (3),

l  OQ   2 IRY

3 2

IR  3 IRY Hence,

IL  3 Iph

6. Thus, for a delta connected connection we can conclude following points : a) IL  3 Iph and VL  Vph b) Line currents are 120 ˚ apart. c) Line currents are 30˚ ahead of their respective phase currents. d) The angle between line currents and the corresponding line voltages is (30+  ) with current lagging. e) The angle between line currents and the corresponding line voltages is (30-φ) with current leading.  Expression for Power in 3- φ delta connected system : 1) Active Power : P = 3Vph Iph cos  But for delta connected ckt.

Vph  VL and Iph 

IL 3

IL cos  3

P  3VL

P  3VL IL cos 

Where φ is the angle between phase voltage and phase current. 2) Reactive Power : Total reactive power is given by,

Q  3Vph Iph sin 

Q  3 VL IL sin 

3) Apparent Power : Total apparent Power is given by,

S  3Vph Iph

S  3 VL IL

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Relationship between Power drawn by Star and Delta Connected Load

Vph  phase voltage, Iph  phase current

1) Let

VL  Line voltage, I L  Line current cos   Power factor, Zph  Impedance per phase 2) For star connected system, we have,

Vph  But

VL 3

and IL  Iph

Vph  Zph Iph VL  Zph IL 3 IL 

VL 3 Zph

3) Now,

(Power)star  3VL IL cos   3VL

VL cos  3 Zph

(Power)star

VL2 cos   Zph

…..[From (1)]

….(2)

4) For delta connected system, we have, IL  3 Iph and VL  Vph

But

I ph 

Vph Zph

V  IL  3  L   Zph 

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5) Now,

(Power)delta  3VL IL cos 

 3VL   3VL   cos   Zph  (Power)delta 

….[From (3)] 3VL2 cos  Zph

….(4)

6) Thus, from eq n . (2) and (4), we have

(Power)delta  3  (Power)star From this eq n . we can conclude that for the same load, power drawn in delta ckt. is three times more than the power drawn in star connected ckt.

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MEASUREMENT OF POWER IN THREE PHASE SYSTEM Wattmeter 1. It is a device which is used to measure power drawn by single phase circuit.. It consists of two coils (1) Current Coil (CC) and (2) Potential Coil (PC) 2. Current coil senses the current and it is always connected in series with the load. The resistance of the coil is very small and hence its cross sectional area is large and it has less number of turns. 3. Potential coil is also known as ‘Pressure Coil’. This senses the voltage and it is always connected across the supply terminals. The resistance of this coil is very high and hence its cross sectional area is small and it has more number of turns 4. The symbol for wattmeter is as shown in fig. (1)

M  Mains, C  Common, L  Line, V  Voltage

5. The circuit connection for measurement of power of a single phase circuit is shown in fig.(2)

6. Wattmeter reading is directly proportional to a) Current (I) through the coil. b) Voltage (V) across the coil. c) Cosine of the angle between this voltage and current.

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Thus, wattmeter reading is

W = V I cos  Hence, wattmeter gives the direct reading of the power absorbed by the single phase circuit. 7. It is also used for measurement of power in 3-φ system. There are three methods available for measuring power in 3-φ system. They are as follows a) Three wattmeter method b) Two wattmeter method c) One wattmeter method

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TWO WATTMETER METHOD Star Connection

1. The circuit diagram of two wattmeter method for star connected 3-  system is as shown above. 2. Current coil of each wattmeter is connected in any two lines and the potential coils of both the wattmeter are connected to third line. 3. In above fig. current coils of wattmeter W1 and W2 are connected in line-R and line-Y respectively and their potential coils are shorted to line B. 4. For W1 current through current coils is I R and voltage across potential coil is VRB . Hence W1 reading is, W1  VRBIR cos

(VRB IR )

….(1)

5. Similarly, for W2 current through current coil is IY and voltage across potential coil is VYB . 6. Hence W2 reading is,

W2 = W2  VYBIY cos (VYB IY )

….(2)

7. Now consider the phasor diagram for star ckt. Assuming phase sequence R-Y-B and lagging p.f. we have,

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1) From circuit diagram.

VRB =VRN  VBN  VRN  (VBN ) and

VYB =VYN  VBN  VYN  (VBN )

2) From vector Diagram we have, Angle between VRB and IR is (30- ) and angle betweenVYB and IY is (30- )

Thus eqn . (1) and (2) becomes, W1  VRBIR cos (30- )

….(3)

W2  VYBIY cos (30+ )

….(4)

3) But for star connected balanced system,

VL  VRB  VYB and IL  IR  IY Hence from eq n . (3) and (4), we have

W1 =VL IL cos(30   )

….(5)

W2 =VL IL cos(30   )

….(6)

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4) Adding eq n .’s (5) and (6) we get,

W1 +W2 =VL IL cos(30   )  VL IL cos(30   ) =VL IL [cos(30   )  cos(30   )]   30    30     30    30     =VL I L  2 cos   .cos   2        60   2   =VL I L  2 cos   .cos    2   2  

  3 =VL IL  2. .cos    2 

W1 +W2 = 3VL IL cos 

….(7)

5) But for star connected system, P= 3VL IL cos  Thus,

P=W1 +W2 = 3VL IL cos 

i.e. the total power absorbed is equal to algebraic sum of the two wattmeter readings.

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DELTA CONNECTION

1. The circuit diagram of two wattmeter method for delta connected 3-  system is as shown above. 2. Current coil of each wattmeter is connected in any two lines and the potential coils of both the wattmeter are connected to third line. 3. In above fig. current coils of wattmeter W1 and W2 are connected in line-R and line-Y respectively and their potential coils are shorted to line B. 4. For W1 current through current coil is I R and voltage across potential coil is VRB . Hence W1 reading is,

W1  VRBIR cos (VRBIR )

….(1)

5. Similarly, for W2 current through current coil is I Y and voltage across potential coil is VYB . Hence W1 reading is,

W2  VYBIY cos (VYBIY )

….(2)

6. Now consider the phasor diagram for delta ckt. Assuming phase sequence R-Y-B and lagging p.f. we have,

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7. From circuit diagram. IR =IRY -IBR and

IY =IYB -IRY

8. From vector Diagram we have, Angle between VRB and I R is (30   ) and angle between VYB and IY is (30 + φ) Thus eqn. (1) and (2) becomes, W1 = VRB IR cos (30 - φ) W2 = VYB IY cos (30 + φ) 9. But for delta connected balanced system, VL = VRB = VYB and IL = IR = IY Hence from eqn. (3) and (4), we have

….(3) ….(4)

W1  VLIL cos  30   

….(5)

W2  VLIL cos  30   

….(6)

10. Adding eqn.’s (5) and (6) we get,

W1  W2  VL I L cos  30     VL I L cos  30     VL I L  cos  30     cos  30   

  60   2    VL I L  2cos    cos    2   2    VL I L  2cos 30o  cos  

  3  VL I L  2   cos   2  

W1  W2 

3 VL I L cos 

….(7)

11. But for delta connected system, P  3 VL I L cos  Thus,

P  W1  W2  3 VL I L cos 

i.e. the total power absorbed is equal to algebraic sum of the two wattmeter readings.

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EXPRESSION FOR POWER FACTOR USING TWO WATTMETER METHOD 1. When two wattmeter are used for measurement of power in 3-φ system, the wattmeter readings are given by, 2. Adding (1) and (2) we get,

W1  W2  VL I L cos  30     VL I L cos  30     VL I L  cos  30     cos  30   

  30    30     30    30      VL I L  2cos    cos   2 2        60   2    VL I L  2cos    cos    2   2    VL I L  2cos 30o  cos  

  3  VL I L  2   cos   2  

3 VL I L cos 

W1  W2 

….(3)

3. Subtracting (2) from (1) we get,

W1 W2  VL I L cos  30    VL I L cos  30     VL I L cos  30     cos  30   

  30    30     30    30     VL I L  2cos    sin   2 2       VL I L 2sin 30o  sin  

 1   VL I L  2   sin    2  W1  W2 VL I L sin 

….(4)

4. From eqn. (3) and (4), W1  W2 sin   W1  W2 3 cos 

tan  

3 W1  W2 W1  W2

….(5)

 3 W1  W2    W1  W 

  tan 1 

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5. But p.f  cos 

  3 W1  W2   p.f  cos  tan 1      W1  W2  

….(6)

6. From the above expressions we can conclude following points 1) If   0o , then the readings of two wattmeter are equal. 2) If   60o , then the readings of both the wattmeter are positive. 3) If   60o , then the reading of wattmeter W2 is zero. 4) If   60o , then the reading of wattmeter W1 is negative.

ADVANTAGES OF TWO WATTMETER METHOD 1. Availability of neutral point in star connected system is not compulsory, for power measurement using two wattmeter method. 2. This method can be used for balanced as well as unbalanced systems. 3. It is very easy to determine p.f. of a 3   system using two wattmeter method. 4. Total volt-amperes can be obtained using two wattmeter readings for balanced load. 5. Circuit connections are very simple.

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