The open-circuit test data of a 500 kVA , 4000 volt, 8 pole, 3 -phase, 50 Hz alternator is: The equivalent armature reaction expressed in ampere-turn per pole is 1.1 × ampere conductors per pole per phase. There are 240 conductors per phase in series. If the inductive voltage drop 8 % on full

Question

The open-circuit test data of a 500 kVA , 4000 volt, 8 pole, 3 -phase, 50 Hz alternator is:

ATs, per pole 2000 3000 3560 5000 6200 7000 8000

Terminal voltage 1990 2900 3400 4000 4400 45900 4800

The equivalent armature reaction expressed in ampere-turn per pole is [latex] 1.1 imes [/latex] ampere conductors per pole per phase. There are 240 conductors per phase in series. If the inductive voltage drop [latex] 8 \%[/latex] on full load and the resistance drop is negligible. Then determine (i) Short circuit characteristic (ii) field excitation and regulation for full load at 0-8 p . f . lagging.

Accepted Answer

[latex] ext { Converting three-phase terminal line voltage into phase values, we have }[/latex]

[latex]\begin{array}{c}\frac{1990}{\sqrt{3}}, \frac{2900}{\sqrt{3}}, \frac{3400}{\sqrt{3}}, \frac{4000}{\sqrt{3}},\frac{4400}{\sqrt{3}},\frac{4800}{\sqrt{3}}=1150,1675,1963,2310,2540,2650,2770 \ ext { Phase voltage, } V=\frac{4000}{\sqrt{3}}=2310 V\end{array}[/latex]

Open-circuit characteristic is drawn by taking ATs per pole along the abscissa and voltage per phase along the ordinate, as shown in Fig. 6.56.

Full load current [latex] I=\frac{k V A imes 1000}{\sqrt{3} imes V_{L}}=\frac{500 imes 1000}{\sqrt{3} imes 4000}=72 A[/latex] .

Armature reaction A T s per pole per phase for full load,

[latex] =1.1 imes ext { ampere conductors per pole per phase }[/latex]

[latex] =1.1 imes I imes \frac{Z}{P}=1.1 imes 72 imes \frac{240}{8}=2376 [/latex]

[latex] ext { Inductive drop } = ext { Leakage reactance drop }=\frac{8}{100} imes \frac{4000}{\sqrt{3}}=185 volt[/latex]

The field A T s or simply field current that is obtained from O C C is used to overcome the effects of armature reaction and leakage reactance. The ATs 2376 are the field A T s for balancing the armature reaction. The field A T s to balance or overcome the leakage reactance can be read off from the O C C graph corresponding to leakage reactance drop of 185 volt and it comes out to be 370 ampere turn.

[latex] herefore \quad ext { Short circuit field } A T s=2376+370=2746 ext { ATs. }[/latex]

So the S C C is drawn with two points, one the origin (0,0) and second point is (2746,72), These two points are joined, hence we get a straight line.

To determine total ampere-turns, proceed as follows:

Draw the phasor diagram as shown in Fig. 6.57. where terminal phase voltage V is taken as reference vector and current lags behind this voltage by an angle [latex] 36.87^{\circ}\left(\phi=\cos ^{-1} 0.8=36.87^{\circ} ight) [/latex]. Here, resistance drop is zero and drop in leakage reactance [latex]IX_{s}[/latex] is 185 V which leads the current vector by [latex] 90^{\circ}[/latex].

[latex]\begin{aligned}E &=\sqrt{(V \cos \phi+I R)^{2}+\left(V \sin \phi+I X_{s} ight)^{2}} \&\left.=\sqrt{(2310 imes 0.8+0)^{2}+(2310 imes 0.6+185)^{2}} \quad ext { (since } R=0 ight) \&=2525 V\end{aligned}[/latex]

From the O C C curve the field A T s corresponding to 2425 volt are 5500 These field A T s, (oa) are drawn at right angle to E as shown in Fig. 6.57. The armature reaction ATs (2376) only are drawn parallel opposition to current I i.e., a b as shown in the Fig. 6.57. The angle between o a and a b, is [latex] (90+\phi) \cdot(\angle \delta[/latex] between E and V is neglected being small). The resultant vector o b is given as below:

[latex]\begin{aligned}o b &=\sqrt{o a^{2}+a b^{2}-2 o a imes a b imes \cos \left(90+36.87^{\prime \prime} ight)} \&=\sqrt{(5500)^{2}+(2376)^{2}+2 imes 5500 imes 2376 imes \cos \left(53.13^{\prime \prime} ight)} \&=\sqrt{(5500)^{2}+(2376)^{2}+2 imes 5500 imes 2376 imes 0.6}=7180 AT ( ext { Ans. })\end{aligned}[/latex]

Corresponding to 7180 AT, the [latex] \operatorname{emf} E_{o}[/latex] from the O C C curve is 2700 volt and lags the o b by [latex] 90^{\circ}[/latex] as shown.

[latex]\% ext { age regulation }=\frac{E_{0}-V}{V} imes 100=\frac{2700-2310}{2310} imes 100=16.88 \% [/latex]

Question 6.25: The open-circuit test data of a 500 kVA , 4000 volt, 8 pol...

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