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# Three Phase Circuits - PowerPoint PPT Presentation

Three Phase Circuits. Introduction: The generator , motor , transformer or rectifier have only one winding is called a single phase system If the current or voltage follows a phase difference 90 0 in a two windings , called two phase systems

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### ThreePhase Circuits

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• The generator , motor , transformer or rectifier have only one winding is called a single phase system

• If the current or voltage follows a phase difference 900 in a two windings , called two phase systems

• If the phase difference is 1200 between voltages or currents in a three winding , called as Three phase systems

• In poly-phase systems , there are more than three windings

• More efficient than single phase system

• Cost is less

• Size is small . Compared to single phase system

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• The amount of conductor material is required less for transmitting same power, over the same distance , under same power loss

• Three phase motors produce uniform torque , where as torque produced by single motor is pulsating

• Three phase generators not produce the harmonics when they are connected in parallel

• Three phase motors are self starting whereas single phase motors are not self starting

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• A three phase supply is said to be balanced, when all the three voltages have the same magnitude but differ in phase by 120° with respect to one another.

• The three phase supply is said to be unbalanced, even if one of the above conditions is not satisfied.

• A three phase load is said to be balanced, when the impedances of all the three phases are exactly the same. Even if one of them is different from the other, then the three phase load is said to be unbalanced

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• In a three phase balanced load, whether star connected or delta connected, the magnitudes of the phase currents are the same but differ in phase by 120o with respected to one another

• But in an unbalanced load, when a three phase balanced supply is given, the magnitudes and phases of all the three phase currents will be different.

Three phase connections:

• There are two types of three phase connections

• Star connection (Y)

• Delta connection (Δ)

Star connection (Y):

• In this method of inter-connection, the similar ends, say, “start” ends of three coils (it could be “finishing” ends also) are joined together at point ‘N’

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• The point ‘N’ is known as star point or delta connected, the magnitudes of the phase currents are the same but differ in phase by 120

neutral point

• If this three-phase voltage is applied across

a balanced symmetrical load, the neutral wire

will be carrying three currents which are exactly

equal in magnitude but are 120o out of phase

with each other. Hence, their vector sum is zero

IR + IY + IB = 0

Voltages and Currents in Y-Connection:

• The voltage induced in each winding is called the

‘phase’ voltage and current in each winding is

known as ‘phase’ current.

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• The vector diagram for phase voltages and currents in a star connection shows that

ER = EY = EB= Eph (phase e.m.f)

• Line voltage VRY between line 1 and line 2 is the vector difference of ER and EY.

• Line voltage VYB between line 2 and line 3 is the vector difference of EY and EB.

• Line voltage VBR between line 3 and line 1 is the vector difference of EB and ER.

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• The p.d. between lines 1 and 2 is connection shows that

VRY = ER - EY (Vector difference)

• VRY is found by compounding ER and EY

reversed and its value is given by the

diagonal of the paral1elogram in figure.

• The angle between ER and EY reversed is 60°.

If ER = EY = EB = Eph the Phase e.m.f then,

C

O

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• Hence, in star connection connection shows that

It will be noted from figure that

• (a) Line voltages are 120° apart.

• (b) Line voltages are 30° ahead of their respective phase voltages.

• (c) The angle between the line currents and the corresponding line voltages is (30 + ɸ) with current lagging.

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Line Currents and Phase Currents: connection shows that

• Current in line 1 = IR

• Current in line 2 = IY

• Current in line 3 = IB

Since IR = IY = IB = say,

Iph - the phase current

Line current IL = Iph

Power:

• The total power in the circuit is the sum of the three phase powers. Hence

• Total Power =3 x phase power=

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Delta ( connection shows thatΔ) or Mesh Connection:

• Phase sequence is R, Y, B

• R leads Y by 120° and Y leads B by 120°.

• The voltage between lines 1 and 2 as VRY

• The voltage between lines 2 and 3 as VYB

• VRY =VYB = VBR = line voltage VL

• Then, it VL = Eph

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Line Currents and Phase Currents: QUESTION PAPERS

• Current in line 1 is I1 = IR – IB

• Current in line 2 is I2 = IY – IR

• Current in line 3 is I3 = IB - IY

• Current in line 1 is found by

compounding IR with IB reversed

and its value is given by the diagonal

of the parallelogram

• The angle between IR and IB reversed (-IB) is 60°.

• If IB = IR = IY = Iph phase current, then current in line 1 is

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Contd.. QUESTION PAPERS

Since all line currents are equal in magnitude i.e., I1= I2 = I3 = IL

From Vector diagram, it should be noted that

• (a) Line currents are 120o apart.

• (b) Line currents are 30o behind the respective phase currents.

• (c) The angle between the line current and the corresponding line voltage is

(30 + ɸ) with the current lagging.

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Power: QUESTION PAPERS

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Measure of power in Three Phase Circuits: QUESTION PAPERS

• Wattmeter is the instrument which

is used to measure power in an electrical circuit.

• It consists of (i) a current coil ML’

through which the line current flows

• (ii) a potential coil PV, which is connected

across the circuit.

• The full voltage is applied across the potential coil and it carries a very small current proportional to the applied voltage.

• Three single phase watt-meters may be connected in each phase

• The algebraic sum of their readings gives the total power consumed by the three phase circuit.

• It can be proved that only two watt-meters are sufficient to measure power in a three phase circuit.

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Two Wattmeter Method: QUESTION PAPERS

• The current coils of the two watt meters are inserted in any two lines

• The potential coils joined to the third line.

• Sum of the instantaneous powers indicated by

W1 and W2 gives the instantaneous power

absorbed by the three loads L1, L2 and L3.

• Instantaneous current through W1 = IR

• Instantaneous P.D. Across W1 = VRB = ER - EB

• Power read by W1 = IR (ER – EB)

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Contd.. QUESTION PAPERS

• Instantaneous current through W2 = IY

• Instantaneous P.D. across W2 = VYB = EY - EB

• Power read by W2 = IY (EY – EB)

Therefore, W1 + W2= IR (ER – EB) + IY (EY – EB)

= IR ER +IY EY – EB (IR + IY)

IR + IY + IB = 0

IR + IY = -IB

= IR ER +IY EY – EB (-IB)

= IR ER +IY EY + EB IB

= P1 + P2 + P3

• Where P1 is the power absorbed by load L1, P2 that absorbed by L2 and P3 that absorbed by L3.

W1 + W2 = total power absorbed

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• The load is said to be balanced, when the impedances of the three phases are equal

• The supply is said to be balanced,

if the three voltages are equal and

are displaced by 120o with respect

to one another.

• When a balance supply is given to

a balanced load, the currents flowing

through the three phases will be equal

in magnitude and are displaced by 120o

with respect to each other.

• Two wattmeter connections to measure power in a three phase balanced circuit

as shown above figure

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• Let E QUESTION PAPERSR, EY and EB be the r.m.s. values of the three-phase voltages and IR, IY and IB be the r.m.s. values of the currents

• The currents lagging behind their phase voltages by ‘ɸ’.

• Current through wattmeter W1 = IR

• P.D. across voltage coil of W1 = VRB = ER – EB

• Now reading W1 = VRB IR cos (30° - ɸ)

• Current through wattmeter W2 = IY

• P.D. across voltage coil of W2 = VYB = EY – EB

• W2 = VYB IY cos (30° + ɸ)

• Since the load is balanced, VRB = VYB = line voltage, VL

IR = IY = line current, IL

W1 = VL IL cos (30° - ɸ) --------(1)

W2 = VL IL cos (30° + ɸ)-------(2)

-EB

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Contd.. QUESTION PAPERS

W1 + W2 = VL IL cos (30° - ɸ) + VL IL cos (30° + ɸ)

= VL IL [cos 30° cos ɸ + sin 30° sin ɸ + cos 30° cos ɸ - sin 30° sin ɸ]

= VL IL (2 cos 30° cos ɸ)

• Expression for Power Factor (p.f):

W1 - W2 = VL IL cos (30° - ɸ) - VL IL cos (30° + ɸ)

= VL IL [cos 30° cos ɸ + sin 30° sin ɸ - cos 30° cos ɸ + sin 30° sin ɸ]

= VL IL (2 sin 30° sin ɸ)

= VL IL sin ɸ

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Contd.. QUESTION PAPERS

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• Effect of QUESTION PAPERSp.f. on W1 and W2:

W1 = VL IL cos (30° - ɸ) = VL IL cos 30° (ɸ = 0)

= (√3/2) VL IL

W2 = VL IL cos (30° + ɸ) = VL IL cos 30° (ɸ = 0)

= (√3/2) VL IL

The two wattmeter readings are positive and equal.

W1 = VL IL cos (30° - 60o) = VL IL cos (-30°) (ɸ = 60o)

= (√3/2) VL IL

W2 = VL IL cos (30° + 60o) = VL IL cos 90° (ɸ = 60o)

= 0

One of the watt meters reads zero.

W1 = VL IL cos (30° - 90o) = VL IL cos (-60°) (ɸ = 90o)

= (1/2) VL IL

W2 = VL IL cos (30° + 90o) = VL IL cos (120°) (ɸ = 90o)

= - (1/2) VL IL

One of the watt-meters, reads negative (-ve).

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Observations: QUESTION PAPERS

• One of the watt-meters reads -ve. The pointer of this watt-meter kicks back and hence the reading can not be taken.

• Then, either the current coil connections or potential coil connections are interchanged, pointer moves in the forward direction and the reading can be taken.

• But this reading has to be considered as -ve.

Conclusion:

• For p.f. lying between 0 to 0.5, one of the watt-meters, reads negative (-ve).

• When p.f. = 0.5, one wattmeter reads zero (0).

• When p.f. lies between 0.5 to 1.0, both watt-meter readings are positive (+).

• When p.f. = 1, the readings of both watt-meters are equal.

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