Q29GATE 2008MCQ

Consider a power system shown below: Given that: $V_{s1}=V_{s2}=1+j0$ p.u; The positive sequence impedance are $Z_{s1}=Z_{s2}=0.001+j0.01$ p.u and $Z_{L}=0.006+j0.06$ p.u 3-phase Base MVA = 100 voltage base = 400 kV(Line to Line) Nominal system frequency= 50 Hz. The reference voltage for phase 'a' is defined as $V(t)=V_m \cos(\omega t)$ . A symmetrical three phase fault occurs at centre of the line, i.e. point 'F' at time ' $t_0$ '. The positive sequence impedance from source S1 to point 'F' equals 0.004+j0.04 p.u. The wave form corresponding to phase 'a' fault current from bus X reveals that decaying d.c. offset current is negative and in magnitude at its maximum initial value, Assume that the negative sequence impedances are equal to postive sequence impedance and the zero sequence impedances are three times positive sequence impedances. Instead of the three phase fault, if a single line to ground fault occurs on phase 'a' at point 'F' with zero fault impedance, then the rms of the ac component of fault current ( $I_x$ ) for phase 'a' will be

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} Z_1&=0.002+j0.02 \;pu\\ Z_2&=Z_1=0.002+j0.02 \;pu\\ Z_0&=3Z_1=0.006+j0.06 \;pu \end{aligned}

\begin{aligned} I_{a0}&=I_{a1}=I_{a2}=\frac{V_s}{Z_1+Z_2+Z_0}\\ &=\frac{V_s}{Z_1+Z_2+3Z_1}\\ &=\frac{V_s}{5Z_1}=\frac{1\angle 0^{\circ}}{5 \times (0.002+j0.02)}\\ &=9.95\angle -84.29\;pu\\ &\text{Fault current}\\ I_f|_{p.u.}&=3|I_{a0}|\\ &=3 \times 9.95=29.85\;p.u.\\ I_x^f&=\frac{|I_f|_{p.u.}}{2}=\frac{29.85}{2}=14.93\;p.u. \end{aligned}

Q30GATE 2007MCQ

A three phase balanced star connected voltage source with frequency $\omega$ rad/s is connected to a star connected balanced load which is purely inductive. The instantaneous line currents and phase to neutral voltages are denoted by ( $i_{a},i_{b},i_{c}$ ) and ( $v_{an},v_{bn},v_{cn}$ ) respectively, and their rms values are denoted by V and I. If

R=[v_{an}\; v_{bn}\; v_{cn}]\begin{bmatrix} 0 & \frac{1}{\sqrt{3}} &-\frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}}& 0&\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 0 \end{bmatrix}

, then the magnitude of R is

🚀 Solve in Practice Mode

📖 Explanation

R=\begin{bmatrix} V_{an} &V_{bn} &V_{cn} \end{bmatrix}\begin{bmatrix} 0 & \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}}\\ -\frac{1}{\sqrt{3}} & 0 & -\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}& -\frac{1}{\sqrt{3}} &0 \end{bmatrix}\begin{bmatrix} i_a\\ i_b\\ i_c \end{bmatrix}\;...(i)

Assuming phase sequence abc, taking $V_{an}$ as the reference,

\begin{aligned} V_{an}&=V\angle 0^{\circ}; \\ V_{bn}&=V\angle -120^{\circ}; \\ V_{cn}&=V\angle 120^{\circ};\;\;...(ii) \end{aligned}

V is the rms value of $V_{an}, V_{bn}$ and $V_{cn}$ I is the rms value of $i_{a}, i_{b}$ and $i_{c}$ From equation (ii), we can say

\begin{aligned} V_{an}&=\sqrt{2}V \sin \omega t;\\ V_{bn}&=\sqrt{2}V \sin (\omega t-120^{\circ}); \\ V_{cn}&=\sqrt{2}V \sin (\omega t+120^{\circ});\;\;...(iii) \end{aligned}

and similarly due to lagging load

\begin{aligned} i_{a}&=\sqrt{2}I \sin (\omega t-90^{\circ});\\ i_{b}&=\sqrt{2}I \sin (\omega t-90-120^{\circ}); \\ i_{c}&=\sqrt{2}I \sin (\omega t-90+120^{\circ}) \end{aligned}

From equation (iii) by considering line values of voltage

\begin{aligned} V_{ab}&=\sqrt{6}V \sin (\omega t+30^{\circ});\\ V_{bc}&=\sqrt{6}V \sin (\omega t-90^{\circ}); \\ V_{ac}&=\sqrt{6}V \sin (\omega t+150^{\circ}) \end{aligned}

From equation (i), we have

\begin{aligned} &=-\left[ \frac{V_{bn}-V_{cn}}{\sqrt{3}}\frac{V_{cn}-V_{cn}}{\sqrt{3}}\frac{V_{an}-V_{bn}}{\sqrt{3}} \right]\begin{bmatrix} i_a\\ i_b\\ i_c \end{bmatrix}\\ &=-\$[(\sqrt{2}V \sin (\omega t-90^{\circ}))(\sqrt{2}I \sin (\omega t-90^{\circ}))\\ &+(\sqrt{2}V \sin (\omega t+150^{\circ}))(\sqrt{2}I \sin (\omega t+150^{\circ}))\\ &+(\sqrt{2}V \sin (\omega t+30^{\circ}))(\sqrt{2}I \sin (\omega t+30^{\circ}))]\$\\ \because \; &\left ( \sin A \sin B= \frac{\cos (A-B)-\cos (A+B)}{2}\right )\\ &=-\$[VI-VI \cos (2\omega t+180^{\circ})+VI\\ & -VI \cos (2\omega t+300^{\circ})+VI-VI \cos (2\omega t+60^{\circ})+]\$\\ &=-\$[3VI-VI \cos (2\omega t-180^{\circ})\\ &+ \cos (2\omega t+300^{\circ})+ \cos (2\omega t+60^{\circ})]\$\\ &=R=-3VI\\ &=|R|=3VI \end{aligned}

Q31GATE 2007MCQ

Suppose we define a sequence transformation between "a-b-c" and "p-n-o" variables as follows:

\begin{bmatrix} f_{a}\\ f_{b}\\ f_{c} \end{bmatrix}=k\begin{bmatrix} 1 & 1 & 1\\ \alpha ^{2}&\alpha & 1\\ \alpha & \alpha ^{2} & 1 \end{bmatrix} \begin{bmatrix} f_{p}\\ f_{n}\\ f_{o} \end{bmatrix}

where $\alpha =e^{j\frac{2\pi }{3}}$ and k and is a constant. Now, if it is given that :

\begin{bmatrix} V_{p}\\ V_{n}\\ V_{o} \end{bmatrix}= \begin{bmatrix} 0.5 & 0&0 \\ 0 & 0.5 & 0\\ 0 & 0 & 2.0 \end{bmatrix}\begin{bmatrix} i_{p}\\ i_{n}\\ i_{o} \end{bmatrix}

and

\begin{bmatrix} V_{a}\\ V_{b}\\ V_{c} \end{bmatrix}=Z\begin{bmatrix} i_{a}\\ i_{b}\\ i_{c} \end{bmatrix}

then,

🚀 Solve in Practice Mode

📖 Explanation

Here,

\begin{aligned} \begin{bmatrix} V_p\\ V_n\\ V_0 \end{bmatrix}&=\begin{bmatrix} 0.5 & 0 & 0\\ 0& 0.5 &0 \\ 0&0 &2 \end{bmatrix}\begin{bmatrix} I_p\\ I_n\\ I_o \end{bmatrix}\\ \text{where, } \begin{bmatrix} V_a\\ V_b\\ V_c \end{bmatrix}&=[Z]\begin{bmatrix} I_a\\ I_b\\ I_c \end{bmatrix}\;\;...(i)\\ \begin{bmatrix} I_a\\ I_b\\ I_c \end{bmatrix}&=K\begin{bmatrix} 1 & 1 & 1\\ \alpha ^2 &\alpha &1 \\ \alpha & \alpha ^2& 1 \end{bmatrix}\begin{bmatrix} I_p\\ I_n\\ I_o \end{bmatrix}\\ &=KA\begin{bmatrix} I_p\\ I_n\\ I_o \end{bmatrix}\\ \text{similarly, }V_{ph}&=k[A]V_{sequence}\\ &=K[A]\begin{bmatrix} 0.5 &0 & 0\\ 0 & 0.5& 0\\ 0& 0 & 2 \end{bmatrix}[I_{sequence}]\\ &=K[A]\begin{bmatrix} 0.5 &0 & 0\\ 0 & 0.5& 0\\ 0& 0 & 2 \end{bmatrix}\frac{[A^{-1}]\$}{K}I_{ph}\;\;...(ii)\\ &\text{comparing eq. (i) with (ii),}\\ \therefore \; Z&=A\begin{bmatrix} 0.5 &0 & 0\\ 0 & 0.5& 0\\ 0& 0 & 2 \end{bmatrix}\$[A^{-1}]\$\\ &=\frac{1}{3}\begin{bmatrix} 1 & 1 & 1\\ \alpha ^2 &\alpha &1 \\ \alpha & \alpha ^2& 1 \end{bmatrix}\begin{bmatrix} 0.5 &0 & 0\\ 0 & 0.5& 0\\ 0& 0 & 2 \end{bmatrix}\begin{bmatrix} 1 & \alpha & \alpha ^2\\ 1 &\alpha ^2 & \alpha \\ 1 & 1& 1 \end{bmatrix}\\ &=\begin{bmatrix} 1 &0.5 & 0.5\\ 0.5 & 1& 0.5\\ 0.5& 0.5 & 1 \end{bmatrix} \end{aligned}

Q39GATE 2003MCQ

A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive sequence, negative sequence and zero sequence impedances of the alternator are j0.1, j0.1 and j0.04 respectively. The neutral of the alternator is connected to ground through an inductive reactor of j0.05 p.u. The per unit positive, negative and zero sequence impedances of transmission line are j0.1, j0.1 and j0.3, respectively. All per unit values are based on the machine ratings. A solid ground fault occurs at one phase of the far end of the transmission line. The voltage of the alternator neutral with respect to ground during the fault is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} &\text{For alternator} \\ Z_{1g}&=j0.1p.u. \\ Z_{2g}&=j0.1p.u.\\ Z_{3g}&=j0.04p.u. \\ &\text{For line} \\ Z_{1l}&=j0.1p.u. \\ Z_{2l}&=j0.1p.u. \\ Z_{0l}&=j0.3p.u. \\ Z_{n}&=j0.05p.u. \\ \text{Equivalent}& \text{sequence impedance:} \\ Z_{1eq}&=Z_{1g}+Z_{1l} \\ &= j0.1+j0.1=j0.2p.u.\\ Z_{2eq}&=Z_{2g}+Z_{2l} \\ &= j0.1+j0.1=j0.2p.u.\\ Z_{0eq}&=Z_{0g}+Z_{0l}+3Z_n \\ &= j0.04+j0.3+3 \times j0.05\\ &=j0.49p.u. \end{aligned}

For single line to ground fault

\begin{aligned} I_{a1}&=I_{a2}=I_{a0}\\ &=\frac{1\angle 0}{Z_{1eq}+Z_{2eq}+Z_{0eq}}\\ &=\frac{1\angle 0}{j(0.2+0.2+0.49)}\\ &=-j1.1236 p.u. \end{aligned}

Fault current $=I_f=3I_{a1}=-j3.3708p.u.$ Voltage of neutral w.r.t. ground

\begin{aligned} V_n&=I_f \times Z_n \\ &= -j3.3708 \times j0.05\\ &= 0.16854p.u. \end{aligned}

Per phase base voltage

\begin{aligned} V_p&=\frac{6.6}{\sqrt{3}}kV=\frac{6.6}{\sqrt{3}}\times 10^3V \\ V_n&=\frac{6.6}{\sqrt{3}} \times 0.16854 \times 10^3 \\ &= 642.2V \end{aligned}