Taming Alternator Voltage: Solved Problems for Stable Power Generation

Problem-1

A 3-phase, star-connected alternator supplies a load of 10 MW at p.f. 0.85 lagging and at 11 kV (terminal voltage). Its resistance is 0.1 ohm per phase and synchronous reactance 0.66 ohm per phase. Calculate the line value of e.m.f. generated.

\[\begin{aligned} I_\mathrm{F.L}&=\frac{10 \times 10^6}{\sqrt{3} \times 11,000 \times 0.85}=618 \mathrm{~A} \\ I R_a \text { drop }&=618 \times 0.1=61.8 \mathrm{~V} \\ I X_S \text { drop }&=618 \times 0.66=408 \mathrm{~V} \\ V_t/\text{phase}&=11,000 / \sqrt{3}=6,350 \mathrm{~V}\\ \phi&=\cos ^{-1}(0.85)=31.8^{\circ} \\ \sin \phi&=0.527 \end{aligned}\]

\[\begin{aligned} E_0&=\sqrt{\left(V \cos \phi+I R_a\right)^2+\left(V \sin \phi+I X_S\right)^2} \\ & =\sqrt{(6350 \times 0.85+61.8)^2+(6350 \times 0.527+408)^2} \\ & =6,625 \mathrm{~V} \\ \text { Line e.m.f. }&=\sqrt{3} \times 6,625=11,486 \text { volt } \\ \end{aligned}\]

Problem-2

In a 50-kVA, star-connected, 440-V, 3-phase, 50-Hz alternator, the effective armature resistance is \(0.25~\Omega\) per phase. The synchronous reactance is \(3.2 ~\Omega\) per phase and leakage reactance is \(0.5~\Omega\) per phase. Determine at rated load and unity power factor :
1. Internal e.m.f. \(E_a\)
2. no-load e.m.f. \(E_0\)
3. percentage regulation on full-load
4. value of synchronous reactance which replaces armature reaction.

\[\begin{aligned} V&=440 / \sqrt{3}=254 \mathrm{~V}\\ I_\mathrm{FL}~\text{at UPF} & =\dfrac{50,000} { \sqrt{3} \times 440}=65.6 \mathrm{~A}\\ IR_a&=65.6 \times 0.25=16.4 \mathrm{~V}\\ I X_L&=65.6 \times 0.5=32.8 \mathrm{~V} \end{aligned}\]

\[\begin{aligned} E_a & =\sqrt{\left(V+I R_a\right)^2+\left(I X_L\right)^2} \\ &=\sqrt{(254+16.4)^2+32.8^2}=272 \text { volt } \\ \text { Line value } & =\sqrt{3} \times 272=471 \text { volt. } \\ E_0 & =\sqrt{\left(V+I R_a\right)^2+\left(I X_S\right)^2}\\ &=\sqrt{(254+16.4)^2+(65.6 \times 3.2)^2}=342 \text { volt } \\ \text { Line value } & =\sqrt{3} \times 342=\mathbf{5 9 2} \text { volt } \end{aligned}\]

\[\begin{aligned} \%~\text{Regulation}& =\frac{E_0-V}{V} \times 100\\ &=\frac{342-254}{254} \times 100=\mathbf{34.65\%} \\ X_a & =X_S-X_L\\ &=3.2-0.5= \mathbf{2.7}~\Omega \end{aligned}\]

Problem-3

A 1000 kVA, 3300-V, 3-phase, star-connected alternator delivers full-load current at rated voltage at 0.80 p.f. Lagging. The resistance and synchronous reactance of the machine per phase are 0.5 ohm and 5 ohms respectively. Estimate the generated voltage.

\[\begin{aligned} V_{p h} & =\frac{3300}{\sqrt{3}}=1905 \text { volts } \\ I_{p h} & =\frac{1000 \times 1000}{\sqrt{3} \times 3300}=175 \mathrm{amp} \end{aligned}\] \[\begin{aligned} OA & = 1905,~AB = I\cdot R ,~ BC = I\cdot X_s \end{aligned}\]

\[\begin{aligned} \text { E along Ref } ~O D&=O A+A B \cos \phi+B C \sin \phi \\ & =1905+(87.5 \times 0.80)+(875 \times 0.60)\\ &=2500~\mathrm{V} \\ \text { E along} \perp ~ C D& =-A B \sin \phi+B C \cos \phi \\ & =87.5 \times 0.6+875 \times 0.80 =647.5 ~\mathrm{V}\\ E = OC &= \sqrt{OD^2+DC^2} = \sqrt{2500^2+647.5^2}=\color{magenta}{2582.5~\mathrm{V}}\\ \delta_1 & = \sin^{-1}\dfrac{CD}{OC} = \sin^{-1}\left(647.5/2582.5\right)=14.52^{\circ} \end{aligned}\]

Problem-4

A 100-kVA, 3000-V, 50-Hz 3-phase star-connected alternator has effective armature resistance of 0.2 ohm. The field current of 40 A produces short-circuit current of 200 A and an open-circuit emf of 1040 V (line value). Calculate the full-load voltage regulation at 0.8 p.f. lagging and 0.8 p.f. leading. Draw phasor diagrams.

\[\begin{aligned} Z_S & =\frac{\text { O.C. voltage } / \text { phase }}{\text { S.C. current } / \text { phase }} =\frac{1040 / \sqrt{3}}{200}=3 \Omega \\ X_S & =\sqrt{Z_S^2-R_a^2}=\sqrt{3^2-0.2^2} =2.99 \Omega \\ I_{FL}&=100,000 / \sqrt{3} \times 3000 =19.2 \mathrm{~A} \\ I R_a&=19.2 \times 0.2=3.84 \mathrm{~V} \\ I X_S&=19.2 \times 2.99=57.4 \mathrm{~V} \\ \text { Voltage/phase } & =3000 / \sqrt{3}=1730 \mathrm{~V} \\ & \cos \phi=0.8 \quad \sin \phi=0.6 \end{aligned}\]

\[\begin{aligned} & \color{magenta}{\textbf{0.8 p.f. lagging}} \\ E_0 & =\left[\left(V \cos \phi+I R_a\right)^2+\left(V \sin \phi+I X_S\right)^2\right]^{1 / 2} \\ & \left.=(1730 \times 0.8+3.84)^2+(1730 \times 0.6+57.4)^2\right]^{1 / 2}\\ &=1768 \mathrm{~V} \\ \% \text { regn. } & =\frac{(1768-1730)}{1730} \times 100=2.2 \% \\ & \color{magenta}{\textbf{0.8 p.f. leading}}\\ E_0 & =\left[\left(V \cos \phi+I R_a\right)^2+\left(V \sin \phi-I X_S\right)^2\right]^{1 / 2} \\ & =\left[(1730 \times 0.8+3.84)^2+(1730 \times 0.6-57.4)^2\right]^{1 / 2} \\ & =1699 \mathrm{~V} \\ \% \text { regn. } & =\frac{1699-1730}{1730} \times 100=-1.8 \% \end{aligned}\]

Problem-5

A given 3-MVA, 50 Hz, 11 kV, \(3-\phi\), Y-connected alternator when supplying 100 A at zero p.f. leading has a line-to-line voltage of 12370 V; when the load is removed, the terminal voltage falls down to 11,000 V. Predict the regulation of the alternator when supplying full-load at 0.8 p.f. lag. Assume an effective resistance of 0.4 \(\Omega\) per phase.

\[\begin{aligned} E_0^2 &=\left(V \cos \phi+I R_a\right)^2+\left(V \sin \phi-I X_S\right)^2 \\ E_0 & =11,000 / \sqrt{3}=6350 \mathrm{~V} \\ V & =12370 / \sqrt{3}=7142 \mathrm{~V} \\ \cos \phi&=0, \quad \sin \phi=1 \\ & \therefore \quad 6350^2=(0+100 \times 0.4)^2+\left(7142-100 X_S\right)^2 \\ & \therefore \quad 100 X_S=790 \\ & \Rightarrow~ X_S=7.9 \Omega \end{aligned}\]

\[\begin{aligned} I_{FL} & =\frac{3 \times 10^6}{\sqrt{3} \times 11,000}=157 \mathrm{~A} \\ I R_a & =0.4 \times 157=63 \mathrm{~V} \\ I X_S & =157 \times 7.9=1240 \mathrm{~V} \\ \therefore \quad E_0 & =\left[(6350 \times 0.8+63)^2+\right. (6350 \times 0.6+ \left.1240)^2\right]^{1 / 2}\\ &=7210 \mathrm{~V} / \text { phase } \\ \therefore \quad \% \text { regn } & =\frac{7210-6350}{6350} \times 100=13.5 \% \end{aligned}\]