Power systems miscellaneous
- Consider the economic dispatch problem for a power plant having two generating units. The fuel costs in Rs/MWh along with the generation limits for the two units are given below:
C1 (P1) = 0.01P1² + 30P1 + 10; 100MW ≤ P1 ≤ 150MW
C2 (P2) = 0.05P2² + 10P2 + 10; 100MW ≤ P2 ≤ 180MW
The incremental cost (in Rs/MWh) of the power plant when it supplies 200 MW is _____.
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NA
Correct Option: D
NA
- The Gauss Seidel load flow method has following disadvantages. Tick the incorrect statement.
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The time required for a single it eration in G-S method is smaller with respect to NewtonRaphson method, but no. of iteration is quite large as it is proportional to the size of system. Also, convergence property also gets affected by the selection of slack bus.
1.= dC1 = 2 × 0.01P1 + 30 = 0.02 P1 + 30 dP2 = dC2 = 2 × 0.05P2 + 10 = 0.01 P2 + 10 dP2 = dC1 = dC2 dP1 dP2
= 0.02P1 + 30 = 30 = 30 = 0.1 P2 + 10
2P1 + 3000 = 10P2 + 1000
2P1 + 2000 = 10P2
P1 + P2 = 200
P2 = 200
P1 = 0= dC1 = 20 Rs/Mwh dP1 Correct Option: D
The time required for a single it eration in G-S method is smaller with respect to NewtonRaphson method, but no. of iteration is quite large as it is proportional to the size of system. Also, convergence property also gets affected by the selection of slack bus.
1.= dC1 = 2 × 0.01P1 + 30 = 0.02 P1 + 30 dP2 = dC2 = 2 × 0.05P2 + 10 = 0.01 P2 + 10 dP2 = dC1 = dC2 dP1 dP2
= 0.02P1 + 30 = 30 = 30 = 0.1 P2 + 10
2P1 + 3000 = 10P2 + 1000
2P1 + 2000 = 10P2
P1 + P2 = 200
P2 = 200
P1 = 0= dC1 = 20 Rs/Mwh dP1
- A generator is connected t hr ough 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 M VA, 138/69 kV rating. A 0.72 puload, 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 circuit will be
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Base impedance on load side, ZBL= (69 kV)² = 476 Ω (10 MVA)
Then, value of load,
ZL = ZL (p.v.) × ZBL
= 0.72 × 476 = 342.8 Ω
Base impedance on generator side,ZBG = (13.8 kV)² = 9.522 Ω (20 MVA)
Then, in p.u. impedenaceZp.u = = ZL = 342.8 = 36p.u ZBG 9.522 Correct Option: A
Base impedance on load side, ZBL= (69 kV)² = 476 Ω (10 MVA)
Then, value of load,
ZL = ZL (p.v.) × ZBL
= 0.72 × 476 = 342.8 Ω
Base impedance on generator side,ZBG = (13.8 kV)² = 9.522 Ω (20 MVA)
Then, in p.u. impedenaceZp.u = = ZL = 342.8 = 36p.u ZBG 9.522
- Three identical star connected resist or s of 1.0 pu are connected to an unbalanced 3 phase supply. The load neutral is isolated. The symmetrical components of the line voltages in pu are:
Vab1 = X ∠1, Vab2 = Y∠2.
If all the pu calculations are with the respective base values, the phase to neutr al sequence voltages are
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Assuming Va as reference phasor,
Van1 = V∠0°
Vbn1 = V∠– 120°
Vab1 = Van1 - Vbn1
= V∠0° – V∠– 120°
= √3Van1 ∠30°
Given, Vab1 = X ∠θ2
Then, Y ∠θ2 = √3 Van2 ∠ - 30°
⇒ Van2 = Y√3 ∠θ2 + 30°Correct Option: C
Assuming Va as reference phasor,
Van1 = V∠0°
Vbn1 = V∠– 120°
Vab1 = Van1 - Vbn1
= V∠0° – V∠– 120°
= √3Van1 ∠30°
Given, Vab1 = X ∠θ2
Then, Y ∠θ2 = √3 Van2 ∠ - 30°
⇒ Van2 = Y√3 ∠θ2 + 30°
- A single phase transmission line and a telephone line are both symmetrically strung one below the other, in horizontal configurations, on a common t ower. The shor t est and l ongest distances between the phase and telephone conductors are 2.5 m and 3 m respectively. The volt age (volt /km) induced in the telephone circuit, due to 50 Hz current of 100 amps in the power circuit is
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Mutual inductance between lower line and telephone line,M = 4 × 10-7 In 
DP1T2 
H/m DP1T1 = 4 × 10-7 In 3 H/m 2.5
= 0.73 × 10–7 H/m
Hence voltage induced by power line into the telephone line,
|VT| = ωMI
= 314 × (0.73 × 10–7) × 100
= 2.29 H/kmCorrect Option: C
Mutual inductance between lower line and telephone line,M = 4 × 10-7 In DP1T2 H/m DP1T1 = 4 × 10-7 In 3 H/m 2.5
= 0.73 × 10–7 H/m
Hence voltage induced by power line into the telephone line,
|VT| = ωMI
= 314 × (0.73 × 10–7) × 100
= 2.29 H/km
