Question 1 |
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 _____.
88.2 | |
108.9 | |
118.8 | |
127.4 |
Question 1 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
\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
Question 2 |
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
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
A | |
B | |
C | |
D |
Question 2 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}
Question 3 |
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
36 | |
1.44 | |
0.72 | |
0.18 |
Question 3 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}
Question 4 |
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
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
4, 0.4 | |
100, 10 | |
4, 10 | |
100, 0.4 |
Question 4 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}
Question 5 |
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
0.578 | |
0.279 | |
0.412 | |
0.44 |
Question 5 Explanation:
(X_d)_{new}=0.4 \times \left ( \frac{100}{75} \right )\left ( \frac{10}{11} \right )^2=0.44077
There are 5 questions to complete.