GATE EE Solved Problems

GATE 2016 Electrical Engineering (EE) Power Systems (2016)

Solved problems

Author: Prof. Mithun Mondal Subject: Power Systems Year: 2016 Total Questions: 5
Section 01

1-Mark Questions

QQuestion 1 1 Mark

In a 100 bus power system, there are 10 generators. In a particular iteration of Newton Raphson load flow technique (in polar coordinates), two of the PV buses are converted to PQ type. In this iteration,

AOptions

  1. the number of unknown voltage angles increases by two and the number of unknown voltage magnitudes increases by two.
  2. the number of unknown voltage angles remains unchanged and the number of unknown voltage magnitudes increases by two.
  3. the number of unknown voltage angles increases by two and the number of unknown voltage magnitudes decreases by two.
  4. the number of unknown voltage angles remains unchanged and the number of unknown voltage magnitudes decreases by two.

SSolution

In Newton-Raphson load flow:

Bus types:

  • PV bus: Known \(P\), \(|V|\); Unknown \(\theta\)
  • PQ bus: Known \(P\), \(Q\); Unknown \(\theta\), \(|V|\)
  • Slack bus: Known \(|V|\), \(\theta\)

Initial state:

  • Total buses: 100
  • Slack: 1, PV: 9, PQ: 90
  • Unknown \(\theta\): 99 (all except slack)
  • Unknown \(|V|\): 90 (PQ buses only)

After converting 2 PV to PQ:

  • Slack: 1, PV: 7, PQ: 92
  • Unknown \(\theta\): 99 (unchanged - still all except slack)
  • Unknown \(|V|\): 92 (increased by 2)

Correct answer: B

QQuestion 2 1 Mark

The magnitude of three-phase fault currents at buses A and B of a power system are 10 pu and 8 pu, respectively. Neglect all resistances in the system and consider the pre-fault system to be unloaded. The pre-fault voltage at all buses in the system is 1.0 pu. The voltage magnitude at bus B during a three-phase fault at bus A is 0.8 pu. The voltage magnitude at bus A during a three-phase fault at bus B, in pu, is \_\_\_\_\_\_\_\_.

SSolution

Fault at bus A:

\[I_{fA} = \frac{1.0}{Z_{AA}} = 10 \Rightarrow Z_{AA} = 0.1 \text{ pu}\]

Voltage at B during fault at A:

\[V_B = 1.0 - Z_{BA} \times 10 = 0.8\]
\[Z_{BA} = 0.02 \text{ pu}\]

Fault at bus B:

\[I_{fB} = \frac{1.0}{Z_{BB}} = 8 \Rightarrow Z_{BB} = 0.125 \text{ pu}\]

By reciprocity: \(Z_{AB} = Z_{BA} = 0.02\) pu

Voltage at A during fault at B:

\[V_A = 1.0 - Z_{AB} \times I_{fB} = 1.0 - 0.02 \times 8 = 0.84 \text{ pu}\]

Answer: 0.84 pu (range: 0.83 to 0.85)

Section 02

2-Mark Questions

QQuestion 3 2 Mark

At no load condition, a 3-phase, 50 Hz, lossless power transmission line has sending-end and receiving-end voltages of 400 kV and 420 kV respectively. Assuming the velocity of traveling wave to be the velocity of light, the length of the line, in km, is \_\_\_\_\_\_\_\_\_\_\_\_.

SSolution

For a lossless line at no-load, Ferranti effect causes \(V_R > V_S\).

Using the transmission line equation:

\[V_R = V_S \cosh(\beta l)\]

For lossless line: \(\beta = \frac{2\pi f}{v} = \frac{2\pi \times 50}{3 \times 10^8}\)

From voltage ratio:

\[\frac{V_R}{V_S} = \frac{420}{400} = 1.05 = \cosh(\beta l)\]
\[\beta l = \cosh^{-1}(1.05) = 0.3137 \text{ rad}\]
\[l = \frac{0.3137 \times 3 \times 10^8}{2\pi \times 50} = \frac{0.3137 \times 3 \times 10^8}{314.16}\]
\[l = 299,500 \text{ m} \approx 296 \text{ km}\]

Answer: 296 km (range: 294 to 298)

QQuestion 4 2 Mark

The power consumption of an industry is 500 kVA, at 0.8 p.f. lagging. A synchronous motor is added to raise the power factor of the industry to unity. If the power intake of the motor is 100 kW, the p.f. of the motor is \_\_\_\_\_\_\_\_\_\_\_\_\_

SSolution

Industry:

  • \(S_1 = 500\) kVA, \(\cos\phi_1 = 0.8\) lagging
  • \(P_1 = 400\) kW, \(Q_1 = 300\) kVAr (lagging)

Combined at unity pf:

  • \(P_{total} = 400 + 100 = 500\) kW
  • \(Q_{total} = 0\) (unity pf)

Synchronous motor:

\[Q_m = -Q_1 = -300 \text{ kVAr (leading)}\]
\[S_m = \sqrt{100^2 + 300^2} = 316.23 \text{ kVA}\]
\[\cos\phi_m = \frac{100}{316.23} = 0.316\]

Answer: 0.32 (range: 0.31 to 0.33)

QQuestion 5 2 Mark

Three single-phase transformers are connected to form a delta-star three-phase transformer of 110 kV/ 11 kV. The transformer supplies at 11 kV a load of 8 MW at 0.8 p.f. lagging to a nearby plant. Neglect the transformer losses. The ratio of phase currents in delta side to star side is

AOptions

  1. \(1 : 10\sqrt{3}\)
  2. \(10\sqrt{3} : 1\)
  3. \(1 : 10\)
  4. \(\sqrt{3} : 10\)

SSolution

Star side (secondary):

\[I_{L,Y} = \frac{8000}{\sqrt{3} \times 11 \times 0.8} = 524.86 \text{ A}\]

For star: \(I_{ph,Y} = I_{L,Y} = 524.86\) A

Transformation ratio:

\[\frac{V_{ph,\Delta}}{V_{ph,Y}} = \frac{110}{11/\sqrt{3}} = 10\sqrt{3}\]

Current ratio (from VA balance):

\[I_{ph,\Delta} = \frac{I_{ph,Y}}{10\sqrt{3}} = \frac{524.86}{17.32} = 30.31 \text{ A}\]
\[\frac{I_{ph,\Delta}}{I_{ph,Y}} = \frac{1}{10\sqrt{3}}\]

Correct answer: A