The following test results were obtained when these were performed on a 15 MVA, 11 kV, three-phase, 50 Hz star-connected alternator:Find the armature reaction ampere-turns, the leakage reactance and the regulation for full load at 0.8 pf lagging. Neglect resistance.

Question

The following test results were obtained when these were performed on a 15 MVA, 11 kV, three-phase, 50 Hz star-connected alternator:

[latex] \begin{array}{lcccccccccc} ext { Field AT per pole in thousand } & 5 & 10 & 15 & 18 & 25 & 30 & 35 & 40 & 45 & 50 \ ext { Open-circuit line emf in } k V & 2.9 & 5.0 & 7.0 & 8.1 & 10.0 & 11.1 & 11.9 & 12.7 & 13.3 & 13.65 \ ext { Full-load current, zero power } & & & & & & & & & & \ ext { factor test, line pd in kV } & - & - & - & 0 & - & - & - & - & 10.0 & -\end{array}[/latex]

Find the armature reaction ampere-turns, the leakage reactance and the regulation for full load at 0.8 pf lagging. Neglect resistance.

Accepted Answer

From the given data, draw O C C between phase voltage and field current. Full-load zero power factor curve is drawn, taking point A(18,0) and point [latex] B\left(45, \frac{10.2}{\sqrt{3}} ight) [/latex] being known. From the triangle B D F drawn in Fig. 6.68.

Armature reaction ampere turns =B F=15000 AT/pole.

Full-load reactance drop =D F=11.15-10=1.15 kV =1150 volt

Leakage reactance drop per phase, [latex] I X_{L}=\frac{1150}{\sqrt{3}}=664 V[/latex]

[latex] ext { Full-load current }=\frac{15 imes 10^{6}}{\sqrt{3} imes 11000}=787 A[/latex]

Leakage reactance per phase, [latex] X_{L}=\frac{I X_{L}}{I}=\frac{664}{787}=0.844 \Omega[/latex]

The phasor diagram is shown in Fig. 6.69, where

[latex]\begin{array}{l}O V= ext { Terminal phase voltage }=\frac{11000}{\sqrt{3}}=6351 V \V E=1150 ext { volt representing reactance drop } \O E=E=\sqrt{(V \cos \phi+I R)^{2}+(V \sin \phi+I X s)^{2}}\end{array}[/latex]

Since [latex] \cos \phi=0.8 lag: \sin \phi=\sin \cos ^{-1} 0.8=0.6[/latex] and R=0

[latex] E=\sqrt{(6351 imes 0.8)^{2}+(6351 imes 0.6+664)^{2}}=6770 V[/latex]

Excitation corresponding to [latex] 6770 V \left(=\frac{11727}{\sqrt{3}} V ight) [/latex] is 33000 AT

Draw oa equal to 33000 AT, perpendicular to O E.

Draw a b=15000 AT parallel to current vector O I

[latex]\begin{aligned} ext { Total ampere turns, } \quad o b &=\sqrt{\left(o a^{2}+a b^{2}-2 o a imes a b imes \cos \left(90^{\circ}+36.87^{\circ} ight) ight.} \&=\sqrt{(33000)^{2}+(15000)^{2}-2 imes 33000 imes 15000 imes \cos 126.87^{\circ}} \&=43680 AT\end{aligned}[/latex]

[latex]\begin{array}{l} ext { Induced emf corresponding to } 43680 AT =\frac{13200}{\sqrt{3}} V =7621 V \\% R _{ eg }=\frac{E_{0}-V}{V} imes 100=\frac{7621-6351}{6351} imes 100=20 \%\end{array}[/latex]

Question 6.29: The following test results were obtained when these were per...

Related Answered Questions