Question 11.1: A balanced three-phase Y-connected generator with positive s...

A balanced three-phase Y-connected generator with positive sequence has an impedance of 0.2 + j0.5Ω/Φ and an internal voltage of 120 V/Φ .The generator feeds a balanced three-phase Y-connected load having an impedance of 39 + j28 Ω/Φ .The impedance of the line connecting the generator to the load is 0.8+ j 1.5 Ω/Φ .The a-phase internal voltage of the generator is specified as the reference phasor

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

b) Calculate the three line currents \pmb{I}_{aA}, \pmb{I}_{bB}, and \pmb{I}_{cC}.

c) Calculate the three phase voltages at the load, \pmb{V}_{AN}, \pmb{V}_{BN},and \pmb{V}_{CN}.

d) Calculate the line voltages \pmb{V}_{AB}, \pmb{V}_{BC}, and \pmb{V}_{CA} at the terminals of the load.

e) Calculate the phase voltages at the terminals of the generator, \pmb{V}_{an}, \pmb{V}_{bn} ,and \pmb{V}_{cn} .

f) Calculate the line voltages \pmb{V}_{ab}, \pmb{V}_{bc},and \pmb{V}_{ca} at the terminals of the generator.

g) Repeat (a)–(f) for a negative phase sequence.

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a) Figure 11.10 shows the single-phase equivalent circuit.

b) The a-phase line current is

\pmb{I}_{aA}=\frac{120 \angle 0^{\circ}}{(0.2 + 0.8 + 39) + j(0.5 + 1.5 + 28)}

 

=\frac{120 \angle 0^{\circ}}{40+j30}

 

= 2.4 \angle -36.87° A.

For a positive phase sequence,

\pmb{I}_{bB}= 2.4\angle -156.87^{\circ} A,

 

\pmb{I}_{cC}= 2.4\angle 83.13^{\circ} A.

c) The phase voltage at the A terminal of the load is

\pmb{V}_{AN}= \left(39 + j28\right)\left(2.4 \angle -36.87^{\circ}\right)

 

= 115.22 \angle -1.19 ^{\circ}V .

For a positive phase sequence,

\pmb{V}_{BN}= 115.22 \angle -121.19^{\circ} V,

 

\pmb{V}_{CN}= 115.22 \angle 118.81^{\circ} V.

d) For a positive phase sequence, the line voltages lead the phase voltages by thus

\pmb{V}_{AB}= \left(\sqrt{3} \angle 30^{\circ}\right)\pmb{V}_{AN}

 

= 199.58\angle28.81^{\circ} V ,

 

\pmb{V}_{BC}= 199.58 \angle -91.19^{\circ} V,

 

\pmb{V}_{CA}= 199.58 \angle 148.81^{\circ} V.

e) The phase voltage at the a terminal of the source is

\pmb{V}_{an}=120-\left(0.2 + j0.5\right)\left(2.4\angle -36.87^{\circ}\right)

= 120-1.29\angle 31.33 ^{\circ} .

 

= 118.90 – j0.67

 

= 118.90 \angle-0.32 ^{\circ}V .

For a positive phase sequence,

\pmb{V}_{bn}= 118.90 \angle-120.32 ^{\circ}V,

 

\pmb{V}_{cn}= 118.90 \angle 119.68 ^{\circ}V.

f) The line voltages at the source terminals are

\pmb{V}_{ab}=\left(\sqrt{3} \angle 30^{\circ}\right)\pmb{V}_{an}

 

= 205.94\angle 29.68^{\circ} V ,

 

\pmb{V}_{bc}=205.94\angle -90.32^{\circ} V,

 

\pmb{V}_{ca}=205.94\angle 149.68^{\circ} V.

g) Changing the phase sequence has no effect on the single-phase equivalent circuit. The three line currents are

\pmb{I}_{aA}= 2.4\angle-36.87^{\circ}A,

 

\pmb{I}_{bB}= 2.4\angle83.13^{\circ}A,

 

\pmb{I}_{cC}= 2.4\angle -156.87^{\circ}A.

The phase voltages at the load are

\pmb{V}_{AN}= 115.22 \angle -1.19 ^{\circ}V ,

 

\pmb{V}_{BN}= 115.22 \angle 118.81^{\circ} V,

 

\pmb{V}_{CN}= 115.22 \angle -121.19^{\circ} V..

For a negative phase sequence, the line voltages lag the phase voltages by 30° :

\pmb{V}_{AB}= \left(\sqrt{3} \angle -30^{\circ}\right)\pmb{V}_{AN}

 

= 199.58\angle -31.19^{\circ} V ,

 

\pmb{V}_{BC}= 199.58 \angle 88.81^{\circ} V,

 

\pmb{V}_{CA}= 199.58 \angle -151.19^{\circ} V.

The phase voltages at the terminals of the generator are

\pmb{V}_{an}= = 118.90 \angle-0.32 ^{\circ}V ,

 

\pmb{V}_{bn}= 118.90 \angle 119.68 ^{\circ}V,

 

\pmb{V}_{cn}= 118.90 \angle-120.32^{\circ}V.

The line voltages at the terminals of the generator are

\pmb{V}_{ab}=\left(\sqrt{3} \angle -30^{\circ}\right)\pmb{V}_{an}

 

= 205.94\angle -30.32 ^{\circ} V ,

 

\pmb{V}_{bc}=205.94\angle 89.68^{\circ} V,

 

\pmb{V}_{ca}=205.94\angle -150.32^{\circ} V.

11.10

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