Question 1 |
A 20 MVA, 11.2 kV, 4-pole, 50 Hz alternator has an inertia constant of
15 MJ/MVA. If the input and output powers of the alternator are 15 MW and
10 MW, respectively, the angular acceleration in mechanical degree/s^2 is
__________. (round off to nearest integer)
25 | |
50 | |
75 | |
100 |
Question 1 Explanation:
We have, swing equation
\frac{2HS}{\omega _s}\frac{d^2\delta }{dt^2}=P_a
Put the values,
\begin{aligned} \frac{2 \times 15 \times 20}{360 \times 50}\frac{d^2\delta }{dt^2}&=15-10\\ \frac{d^2\delta }{dt^2}&=\frac{5 \times 360 \times 50}{2 \times 15 \times 20}\\ &=150 \text{ electric degree per} sec^2\\ \frac{d^2\delta }{dt^2}&=\alpha =\frac{2}{4} \times 150 \text{ Mech. degree/}sec^2\\ &=75\text{ Mech. degree/}sec^2 \end{aligned}
\frac{2HS}{\omega _s}\frac{d^2\delta }{dt^2}=P_a
Put the values,
\begin{aligned} \frac{2 \times 15 \times 20}{360 \times 50}\frac{d^2\delta }{dt^2}&=15-10\\ \frac{d^2\delta }{dt^2}&=\frac{5 \times 360 \times 50}{2 \times 15 \times 20}\\ &=150 \text{ electric degree per} sec^2\\ \frac{d^2\delta }{dt^2}&=\alpha =\frac{2}{4} \times 150 \text{ Mech. degree/}sec^2\\ &=75\text{ Mech. degree/}sec^2 \end{aligned}
Question 2 |
Two generating units rated for 250 MW and 400 MW have governor speed
regulations of 6% and 6.4%, respectively, from no load to full load. Both the
generating units are operating in parallel to share a load of 500 MW. Assuming free
governor action, the load shared in MW, by the 250 MW generating unit is
_________. (round off to nearest integer)
100 | |
150 | |
200 | |
250 |
Question 2 Explanation:
Let no-load frequency is 50 Hz.
Draw the curve :
From the curve,
\begin{aligned} \frac{50-f}{3}&=\frac{P_1}{250} \;\;\;..(1) \\ \frac{50-f}{3.2}&=\frac{P_2}{400} \;\;\;..(2) \end{aligned}
From eq. (1) & (2),
\begin{aligned} \frac{3P_1}{250}&= \frac{3.2P_2}{400} \\ 3P_1&=2P_2\;\;\;...(3) \end{aligned}
Given: P_1+P_2=500
From eq. (3),
P_1+1.5P_1=500
P_1=200MW
Draw the curve :
From the curve,
\begin{aligned} \frac{50-f}{3}&=\frac{P_1}{250} \;\;\;..(1) \\ \frac{50-f}{3.2}&=\frac{P_2}{400} \;\;\;..(2) \end{aligned}
From eq. (1) & (2),
\begin{aligned} \frac{3P_1}{250}&= \frac{3.2P_2}{400} \\ 3P_1&=2P_2\;\;\;...(3) \end{aligned}
Given: P_1+P_2=500
From eq. (3),
P_1+1.5P_1=500
P_1=200MW
Question 3 |
In the figure shown, self-impedances of the two transmission lines are 1.5j\:p.u each, and Z_{m}=0.5j \:p.u is the mutual impedance. Bus voltages shown in the figure are in p.u. Given that \delta>0, the maximum steady-state real power that can be transferred in p.u from Bus-1 to Bus-2 is
\left | E \right |\left | V \right | | |
\frac{\left | E \right |\left | V \right |}{2} | |
2\left | E \right |\left | V \right | | |
\frac{3\left | E \right |\left | V \right |}{2} |
Question 3 Explanation:
\begin{aligned} L_{e q}&=\frac{L_{1} L_{2}-M^{2}}{\left(L_{1}+L_{2}-2 M\right)} \\ X_{\mathrm{eq}}&=\frac{1.5 \times 1.5-0.5^{2}}{1.5+1.5-2 \times 0.5}=1 \mathrm{p.u.} \end{aligned}
\begin{aligned} P_{\max }&=\frac{|E||V|}{1}\\ P_{\max }&=|E||V| \end{aligned}
Question 4 |
In the single machine infinite bus system shown below, the generator is delivering the real power of 0.8pu at 0.8 power factor lagging to the infinite bus. The power angle of the generator in degrees (round off to one decimal place) is _________
12.8 | |
28.4 | |
20.5 | |
32.6 |
Question 4 Explanation:
\begin{aligned} X &=0.25+0.2+0.4||0.4 \\ &=0.45+0.2=0.65pu \\ P&=V_{pu} \times I_{pu} \cos \phi \\ 0.8 &=1 \times I_{pu} \times 0.8 \\ I_{pu} &= 1pu\\ E &= V+jI_aX_s\\ &=1+1\angle -36.86^{\circ} \times j0.65 \\ &= 1.484 \angle 20.51^{\circ}pu\\ \delta &= 20.51^{\circ} \end{aligned}
Question 5 |
Consider a lossy transmission line with V_{1} \; and \; V_{2} as the sending and receiving end voltages, respectively. Z and X are the series impedance and reactance of the line, respectively. The steady-state stability limit for the transmission line will be
greter than |\frac{V_{1}V_{2}}{X}| | |
less than |\frac{V_{1}V_{2}}{X}| | |
equal to |\frac{V_{1}V_{2}}{X}| | |
equal to |\frac{V_{1}V_{2}}{Z}| |
Question 5 Explanation:
With only x:
P_{max}=\left | \frac{V_1V_2}{x} \right |
With Lossy Tr, Line
P=\left | \frac{V_1V_2}{z} \right | \cos (\beta -\delta )-\left | \frac{AV_2^2}{z} \right |\cos (\beta -\delta )
Therefore, with Lossy Line P_{max} \lt \left | \frac{V_1V_2}{x} \right |
P_{max}=\left | \frac{V_1V_2}{x} \right |
With Lossy Tr, Line
P=\left | \frac{V_1V_2}{z} \right | \cos (\beta -\delta )-\left | \frac{AV_2^2}{z} \right |\cos (\beta -\delta )
Therefore, with Lossy Line P_{max} \lt \left | \frac{V_1V_2}{x} \right |
There are 5 questions to complete.