Q2GATE 2014NAT

A three phase star-connected load is drawing power at a voltage of 0.9 pu and 0.8 power factor lagging. The three phase base power and base current are 100 MVA and 437.38 A respectively. The line-to-line load voltage in kV is _____.

🚀 Solve in Practice Mode

📖 Explanation

Given,

\begin{aligned} V&=0.9pu, \cos \phi =0.8\\ S_{base}&=100MVA\\ I_{base}&=437.38A \end{aligned}

For a $3-\phi$ star connected systems,

\begin{aligned} S_{base}&=\sqrt{3}(V_b)_{L-L}(I_b)_L\\ \therefore \; (V_b)_{L-L}&=\frac{S_{base}}{\sqrt{3} \times (I_b)_L}\\ &=\frac{100 \times 10^6}{\sqrt{3} \times (437.38)}=132kV \end{aligned}

$\therefore \;$ Actual line to line load voltage in kV $=0.9 \times 132kV=118.8kV$

Q3GATE 2010MCQ

For the power system shown in the figure below, the specifications of the components are the following : G1: 25 kV, 100 MVA, X = 9% G2: 25 kV, 100 MVA, X = 9% T1: 25 kV/220 kV, 90 MVA, X = 12% T2: 220 kV/25 kV, 90 MVA, X = 12% Line 1: 200 kV, X = 150 ohms Choose 25 kV as the base voltage at the generator G1, and 200 MVA as the MVA base. The impedance diagram is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} Z_{pu(new)}=Z_{pu(old)} \times &\frac{(MVA_B)_{new}}{(MVA)_{old}} \times \frac{(kV_B)_{old}^2}{(kV_B)_{new}^2}\\ (MVA_B)_{new}&=200 MVA\\ (kV_B)_{new}&=25 kV\\ \text{Base impedance,}\\ Z_B&=\frac{(kV_B)^2}{(MVA_B)}\\ (kV_B) \text{ \in line}&=220 kV\\ Z_B&=\frac{220^2}{200}=242\Omega \\ X_{l-pu}&=\frac{X_{l}}{Z_B}=\frac{150}{242}=0.62\;p.u.\\ G_1:25kV,& 100MVA,\\ X_{G1}&=0.09pu\\ X_{G1} &\text{ (0n base 200 MVA )}\\ &=0.09 \times \frac{200}{100}=0.18\;p.u.\\ \{ (kV_B)_{old}&=(kV_B)_{new}\}\\ G_2:2kV,& 100MVA,\\ X_{G2}&=0.09pu\\ X_{G2} &\text{ (0n base 200 MVA )}\\ &=0.09 \times \frac{200}{100}=0.18\;p.u.\\ T_1: 25 kV/220 \;kV, & 90 \;MVA\\ X_{T1}&=0.12\;p.u.\\ X_{T1} &\text{ (0n base 200 MVA )}\\ &=0.12 \times \frac{200}{90}=0.27\;p.u.\\ T_2: 220 kV/25 \;kV, & 90 \;MVA\\ X_{T2}&=0.12\;p.u.\\ X_{T2} &\text{ (0n base 200 MVA )}\\ &=0.12 \times \frac{200}{90}=0.27\;p.u.\\ \end{aligned}

Q4GATE 2006MCQ

A generator is connected through a 20 MVA, 13.8/138 kV step down transformer, to a transmission line. At the receiving end of the line a load is supplied through a step down transformer of 10 MVA, 138/69 kV rating. A 0.72 pu. load, evaluated on load side transformer ratings as base values , is supplied from the above system. For system base values of 10 MVA and 69 kV in load circuit, the value of the load (in per unit) in generator will be

🚀 Solve in Practice Mode

📖 Explanation

Base value on load circuit,

\begin{aligned} (MVA)_B&=10MVA\\ (kV)_B&=69 kV \end{aligned}

Base impedance $=\frac{69^2}{10}=476.1$ Value of load in ohm $=Z_L$ = value of load in p.u. x Base impedance $\Rightarrow \; Z_L=0.72 \times 476.1=342.8\Omega$ Base value in generator circuit,

\begin{aligned} (MVA)_B&=20MVA\\ (kV)_B&=13.8 kV \end{aligned}

Base impedance,

\begin{aligned} Z_B&= \frac{(kV)_B^2}{(MVA)_B}=\frac{13.8^2}{20}\\ Z_B&=9.522\Omega \\ \frac{Z_L}{Z_B} &=\frac{342.8}{9.522}=36\;p.u. \end{aligned}

Q5GATE 2005MCQ

The p.u. parameter for a 500 MVA machine on its own base are: inertia M = 20 p.u.; reactance X = 2 p.u. The p.u. values of inertia and reactance on 100 MVA common base, respectively, are

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📖 Explanation

\begin{aligned} (M_{pu})_{new}&=(M_{pu})_{old} \times \frac{(MVA)_{old}}{(MVA)_{new}} \\ &= 20 \times \frac{500}{100}=100\; p.u.\\ (X_{pu})_{new} &=(X_{pu})_{old} \times \frac{(MVA)_{old}}{(MVA)_{new}} \times \frac{(kV)^2_{old}}{(kV)^2_{new}}\\ &=2 \times \frac{100}{500}=0.4 \;\;[(kV)_{old}=(kV)_{new}] \end{aligned}

Q6GATE 2001MCQ

A 75 MVA, 10 kV synchronous generator has $X_{d}$ = 0.4 p.u. The $X_{d}$ value (in p.u.) is a base of 100 MVA, 11 kV is

🚀 Solve in Practice Mode

📖 Explanation

$(X_d)_{new}=0.4 \times \left ( \frac{100}{75} \right )\left ( \frac{10}{11} \right )^2=0.44077$

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