Q1.A Balanced V3φ, Y-Connected Generator with Positive Sequence Has an Impedance Of

Tutorial Sheet 4

Q1.A balanced V3φ, Y-connected generator with positive sequence has an impedance of

0.2 + 0.5j Ω/ф and an internal voltage of 120 V/ф. The generator feeds a balanced V3φ, Y-connected load having an impedance of 39 + 28j Ω/ф. The impedance of the line connecting the generator to the load is 0.8 + 1.5j Ω/ф. The a-phase internal voltage of the generator is the reference phasor.

a)Construct the a-phase equivalent circuit of the system

Single phase equivalent circuit is given below.

b)Calculate the three line currents: IaA, IbB, and IcC

IbB will be same as IaA but will lag by 120 deg.

Hence,

Similarly,

c)Calculate the 3 phase voltages at the load: VAN, VBN, and VCN

VBN will lag VAN by 120 deg.

Hence,

Similarly,

d)Calculate the 3 line voltages at the terminals of the load: VAB, VBC, and VCA

Line voltage is √3 times phase voltage and leads phase voltage by 30 deg.

Hence,

VBC will lag VAB by 120 deg.

Similarly

e)Calculate the 3 phase voltages at the terminals of the generator: Van, Vbn, and Vcn

Vbn will lag Van by 120 lag.

Similarly,

f)Calculate the 3 line voltages at the terminals of the generator: Vab, Vbc, and Vca

Line voltage will √3 times phase voltage and will lead phase voltage by 30 deg.

Hence,

Vbc will be same as Vab but will lag Vab by 120 deg.

Hence,

Similarly,

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Q2. The Y-connected source in Question (1) feeds a Δ-connected load having an impedance of 118 + 85.8j Ω/Φ. The impedance of the line connecting the generator to the load is 0.3 + 0.9j Ω/Φ. The a-phase internal voltage of the generator is specified as the reference.

a)Construct a single phase equivalent circuit of the 3-Φ system

We will convert delta into equivalent star connected load.

Equivalent star connected load per phase = (118+j85.8)/3 = 39.33+j28.6

Single phase equivalent circuit is given below.

b)Calculate the line currents: IaA, IbB, and IcC

IbB will be same as IaA but will lag by 120 deg.

Hence,

Similarly,

c)Calculate the 3 phase voltages at the load

VBN will lag VAN by 120 deg.

Hence,

Similarly,

d)Calculate the 3 line voltages at the terminals of the generator

Vbn will lag Van by 120 lag.

Similarly,

Line voltage will √3 times phase voltage and will lead phase voltage by 30 deg.

Hence,

Vbc will be same as Vab but will lag Vab by 120 deg.

Hence,

Similarly,

Q1 and 2 shows equivalence of star and delta connected loads.

Q3. A V3φ Δ-connected source has an internal impedance of 0.9 + 9j Ω/Φ. The terminal voltage is equal to 13,200 Volts when no load is connected. The generator feeds a Δconnected load with a per-phase impedance of 645 + 171j Ω/Φ. The impedance of the transmission line connecting the generator to the load is 0.7 + 3j Ω/Φ. The a-phase internal voltage of the generator has been specified as the reference.

a)Construct a single phase equivalent circuit of the V3φ system

We will convert delta into equivalent star for single phase representation.

Equivalent star connected per phase load = (645+j171)/3 = 215+j57

Similarly phase voltage at sending end = 13200/√3 = 7621.023V

Equivalent star connected per source impedance = (0.9+j9/3 = 0.3+j3

We can draw the following equivalent circuit for single-phase.

b)Calculate the magnitude of the line current (33.87 Amps)

Line current =

c)Calculate the magnitude of the line voltage at the terminals of the load (13048

Volts)

Line voltage at load end = √3*(215+j57)*33.87=13048.23V

d)Calculate the magnitude of the line voltage at the terminals of the source

(13178.13 Volts)

Line voltage at load end = √3*(215.7+j60)*33.87=13134.41V

e)Calculate the magnitude of the phase current in the load (19.55 Amps)

Phase current in load = 33.87/√3 = 19.55A

f)Calculate the magnitude of the phase current in the source (19.55 Amps)

Phase current in source = 33.87/√3 = 19.55A

Q4. A balanced V3φ load required 480 kW at a lagging power factor of 0.8. The line impedance is 0.005 + 0.025 Ω/Φ. The line voltage at the terminals of the load is 600 Volts.

a)Construct a single-phase equivalent circuit

Equivalent circuit is given below.

b)Calculate the magnitude of the line current (577.35 Amps)

We have

577.35A

c)Calculate the magnitude of the line voltage at the sending end (619.23 Volts)

Supply phase voltage

Hence, line voltage at sending end = √3*357.513=619.23V

d)

Pf at sending end = cos(36.869+1.573) =0.783 lagging

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