Poly-phase system costs more than single-phase system, but the former has many advantages over the latter. That is the reason for which at present over 90% of electrical energy used for commercial purposes is generated and transmitted in poly-phase.
Some of the advantages of poly-phase system over single-phase working are as follows:
(i) For a given size of frame, a poly-phase generator or motor has a bigger output than a single-phase machine.
(ii) To transmit a given amount of power at a given voltage over a given distance, a poly-phase transmission line is more economical than a single-phase line.
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(iii) Single-phase motors (except commutator motors) are not self-starting, but poly-phase motors are self-starting.
(iv) While single-phase motors (except commutator motors) have a pulsating torque, poly-phase motors have an absolutely uniform torque.
(v) The pulsating nature of armature reaction in single-phase alternators causes difficulty during parallel operation unless the poles are fitted with unusually heavy dampers. Poly-phase alternators work in parallel without such difficulty.
For the same size of machine, a two-phase system will have 1.414 times and a three-phase system will have 1.57 times more output power than a single-phase system.
Two-Phase Working:
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In a two-phase working the armature has two distinct windings arranged so that there is a phase difference of 90° in the e.m.fs. induced in them. The alternator may be connected either for 4-wire supply or for 3-wire supply, the circuit for 4-wire being the same as two independent single-phase circuits.
Let VL and IL represent line voltage and line current, and VP and IP refer to phase voltage and phase current as shown in fig. 37. In a 4-wire system shown in fig 37(a) VL=VP and IL=IP. The power output of each phase=VPIP cos θ, where θ is the phase angle between VP and IP.
... total output power = 2VPIP cos θ = 2VLIL cos θ.
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In a 3-wire system shown in fig. 37(b) output voltage = √2VL and total output current √2IL, as they have a phase difference of 90° (fig. 38),
Three-Phase Working:
The 3-phase system has the largest application among all poly-phase systems. This is due to the fact that the 3-phase system has the least number of wires of any symmetrical poly-phase system, the line voltages are equal, and with a neutral wire two different values of voltage are available.
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The armature has three similar, separate and distinct windings so that there is a phase difference of 120° in the e.m.fs. induced in them. The armature may be connected in two different ways—Star or Y-connection and Delta (Δ) or Mesh-Connection.
Star-or Y-Connected System:
When three coils or windings, placed 120° apart, are connected together at a common point as shown in fig. 39, they form a star or Y-connected circuit. The three coils oa, ob and oc shown in the diagram represent three phases. They are joined together at the common point o. This point is called the Neutral or Star Point. The potential of this point is zero, and very often it is connected to earth.
The three lines (aa’, bb’, cc’) drawn from the three open terminals (a, b, c) of the three coils are called the main lines. When only three main lines are drawn, it is said to be a 3-phase, 3-wire system. But often a neutral line (oo’) is also drawn from the neutral point. In that case the system is called a 3-phase, 4-wire system.
The current flowing through each coil is called phase current and that flowing through a main line is called line current. It can be seen in fig. 39 that each phase or coil is connected in series with its respective main line. Therefore, in a star-connected system line current is equal to phase current, i.e. the same current flows through a phase and the main line connected in series with it.
In a 3-phase, 4-wire system two different values of voltage are available. Each coil is connected across a main line and the neutral line as shown in fig. 39(a). Therefore, voltage across a coil or phase is equal to voltage between a main line and the neutral. This voltage is called phase voltage. The voltage between any two main lines is called line voltage.
If these two voltages are measured by a voltmeter, it will be found that line voltage is equal to √3 times phase voltage. Thus, if the line voltage is 400 volts, the phase voltage is (400/√3) = 230 volts. That is why in our country motors and other three-phase loads are designed for 400-volt pressure, while lamps, fans and other single-phase loads are supplied at a pressure of 230 volts.
The power in a 3-phase, star-connected system:
Ex. 1:
Show by means of a diagram how you would connect up three 250-volt. 100-watt lamps to 3-phase. 3-wire, 400-volt supply, and determine the power absorbed by each lamp (a) when all the lamps are lighted, and (b) when one lamp is fused.
Solution:
Since the supply available is 3-phase, 3-wire, neutral line has not been drawn. Hence, the lamps may be supplied at a pressure of 400 volts only. But all the lamps are rated for 250-volt supply. Therefore, if each lamp is connected directly across the supply lines, it will be fused immediately.
Lamps rated for 250-volt supply may be connected to 400-volt mains in two different ways:
(i) Two lamps joined in series may be connected across two supply lines. In that case each lamp has a supply pressure of 200 volts, and the lamp cannot burn with usual brightness. Besides, if one lamp is fused, current cannot flow through the other, and both lamps remain dark.
(ii) Three lamps may be connected in star across a 3-phase, 3- wire supply. This is shown in fig. 40. In this case each lamp has a supply pressure of (400/√3) = 231 volts, and the lamps will burn with usual brightness. Moreover, if one lamp is fused, other two lamps will be in series across two lines. Hence, these two lamps will burn with comparatively less brightness, but they will not remain dark.
Lamps are considered to be purely resistive. Therefore power factor of the circuit is unity. Voltage and power ratings of each lamp are 250 volts and 100 watts respectively. Hence, the current rating of each lamp.
I = 100/250 = 0.4 ampere.
Resistance of each lamp,
R = 250/0.4 = 625 ohms.
(i) When all the three lamps connected in star are burning:
Voltage across each lamp,
Current flowing through each lamp,
Io = (VP/ R) = (231/625) = 0.37 ampere.
Power absorbed by each lamp,
P = VPIP = 231 x 0.37 = 85.4 watts.
(ii) When one lamp is fused:
When one lamp is fused, other two lamps remain in series across two supply lines. This is shown in fig. 41.
Total resistance of the circuit,
R’ = R + R = 625 + 625 = 1250 ohms.
Current flowing through the circuit,
I = VL/R’ = 400/ 1250 = 0.32 ampere
Total power absorbed by the circuit = VL x I = 400 x 0.32 = 128 watts.
... power absorbed by each lamp = 128/2 = 64 watts.
Ex. 2:
A balanced, star-connected load of resistance 8 ohms and inductive reactance 6 ohms per phase is connected to a 3-phase, 230-volt supply. Find the line current and the power absorbed.
Solution:
The star-connected load circuit is shown in fig. 42. The phase voltage of the circuit,
Impendence of each phase,
Phase current of the circuit,
IP = VP/ Z = 132.8/10 = 13.28 amperes.
Since the load circuit is connected in star, line current and phase current are the same.
... line current IL = IP = 13.28 amperes.
Power factor of each phase,
Cos θ = R/Z = 8.0/10.0= 0.8 lagging.
Power absorbed by whole circuit,
Delta (Δ) or Mesh-Connected Circuit:
In this system three coils or windings, placed 120° apart, are joined together to form a closed circuit or mesh, starting end of one coil being joined to finishing end of the next coil. The connections of three coils are shown in fig. 43. In this diagram three coils oa, ob and oc are placed 120° apart from one another.
The starting end o of coil oa is joined with finishing end c of coil oc, the starting end o of coil ob is joined with finishing end a of coil oa and starting end o of coil oc is joined with finishing end b of coil ob. Thus, the three coils have formed a closed circuit in the form of a triangle as shown in fig. 43(a). Since three coils are not joined together at any common point, in delta connection no neutral point is available. Hence, with this connection only three-phase, three-wire system is possible.
The three main lines (aa’, bb’ & cc’) are drawn from three vertices of the triangle. It can be seen in fig. 43(b) that each coil or phase is connected across two main lines. Therefore in delta connection, line voltage is equal to phase voltage. Two different values of supply voltage cannot be obtained with this connection.
If it becomes imperative to have a neutral point for drawing a neutral line, the same may be obtained by connecting three similar inductive coils in star across three main lines. This is shown in fig. 44. Here the neutral line may be drawn from the star point of three inductive coils. Such star point is called artificial neutral point.
To obtain a neutral point in this way involves considerable expenditure, besides being inconvenient for operation. That is the reason why a star-connected system is usually adopted wherever supply is to be given at two different voltages.
The current flowing through a coil is called phase current and that flowing through a main line is called line current. If these two currents are measured by an ammeter, it is found that the current flowing through the line is √3 times the current flowing through the coil. Therefore, in a delta-connected system line current is equal to √3 times phase current.
The power in a 3-phase, delta-connected system:
Let,
VP = phase voltage,
VL = line voltage = VP,
IP = phase current,
IL = line current = √3 IP, and
cos θ = power factor of each phase, i.e. θ is the phase angle between VP and Ip.
Then,
Power per phase = VP IP cos θ watt.
Hence, the total power for all the three phases is given by,
When the total power is expressed in Kilowatt,
Total Apparent Power of the three phases,
Total Reactive Power of the three phases,
Ex. 3:
Three similar single-phase circuits, each consisting of a fluorescent lamp of resistance 15 ohms connected in series with a choking coil of inductance 0.04 henry and negligible resistance, are connected in ‘star’ to a 3-phase, 50-cycle A.C. supply. 400 volts between lines. Calculate the line current.
If they are now connected in ‘mesh’, calculate the line current and the current taken by each lamp.
Find also the total power absorbed in each case.
Solution:
Since fluorescent lamp is purely resistive, each phase consists of a resistance and an inductance in series.
When three phases are connected in star:
The connections of three phases are shown in fig. 45(a)
Reactance of each phase,
Impendence of each phase,
Voltage per phase,
Current per phase,
Since three phase are connected in star, line current,
IL = IP = 11.8 amperes.
Power factor of each phase,
Cos θ = R/Z = 15.0/19.56 = 0.767 lagging.
Therefore, total power absorbed by the whole circuit,
When three phases are connected in mesh:
The connections of three phases are shown in fig. 45(b).
Resistance, reactance and impedance will remain the same as star connection, but the phase voltage is equal to line voltage in this case, i.e.
VP = VL = 400 volts.
Current per phase,
IP = VP/Z = 400/19.56 = 20.45 amperes.
Phase current flows through each lamp. Therefore,
Current flowing through each fluorescent lamp = IP = 20.45 amperes.
Line current IL = √3Ip = 1.732 x 20.45
= 35.41 amperes.
Power absorbed by the whole circuit,
Note:
Power absorbed by three phases when connected in mesh is three times the power absorbed by three phases when connected in star.
Differences between Star and Mesh Connections:
Differences between star-connected system and mesh-connected system are given below:
Star-Connected System:
1. All the three phases remain connected at a common point, called neutral point or star point.
2. The system may be either 3-phase, 3-wire or 3-phase, 4- wire.
3. Line voltage is √3 times the phase voltage, i.e. VL = √3Vp.
4. Same current flows through the phase and the main line, i.e.
IL = IP
5. The three phases do not form a closed circuit or a mesh.
Mesh-Connected System:
1. The three phases are not connected to any common point. Hence, no neutral point is available.
2. The system can be only 3-phase, 3-wire.
3. Line voltage is equal to phase voltage, i.e.
vL = vp
4. Line current is equal to √3 times phase current, i.e.
IL = √3IP.
5. The three phases form a closed circuit or a mesh, starting ends being joined to finishing ends.