Draw a radial distribution network feeder (11 kV), with 5 busbars/distribution substations, explain in what case there is a voltage rise problem: i) with the aid of a diagram showing load flow direction and voltage profile with simple calculations (ignore the power losses on the feeder); ii) list three possible solutions to reduce voltage rise issues, and explain using the voltage drop equation.

Draw a radial distribution network feeder (11 kV), with 5 busbars/distribution substations, explain in what case there is a voltage rise problem: i) with the aid of a diagram showing load flow direction and voltage profile with simple calculations (ignore the power losses on the feeder); ii) list three possible solutions to reduce voltage rise issues, and explain using the voltage drop equation.

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.47P: Consider the oneline diagram shown in Figure 3.40. The three-phase transformer bank is made up of...

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b) A fault occurs at bus 3 of the network shown in Figure Q4. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, transformer and load are given in Figure Q4. V₁ = 120° p.u. V₂ = 120° p.u. V₂ = 1/0° p.u. V₂= 120° p.u. jXj0.1 p.u. JX2) 0.1 p.u. jX0j0.15 p.u. jXn-j0.2 p.u. 1 JX(2)-j0.2 p.u. 2 jX)=j0.25 p.u. JX20-10.15 p.u. jXa(z)-j0.2 p.u. 4 jX2(0)=j0.2 p.u. jXT(1) j0.1 p.u. jXT(2)=j0.15 p.u. jXT(0)=j0.1 p.u. Figure Q4. Circuit for problem 4b). = jXj0.1 p.u. j0.1 p.u. - JX(2) JXL(0) 10.1 p.u. = (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 3. (ii) Determine the positive sequence fault current for the case when a three- phase-to-ground fault occurs at bus 3 of the network. (iii) Determine the short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus…
A power generation plant connected to bulk load of high voltage industries shown in Figure Q3 has specifications given in Table Q3. Each equipment reactance is expressed in per unit on a common base of 100 MVA. a) Find the fault level for each bus bar in the generation plant. GI G2 TI Figure Q3: Single line diagram of a power generation plant Table Q3: Specifications of power system components Power System Component Voltage Rating X MVA Rating (kV) (pu) GI 50 33 0.2 G2 40 33 0.18 50 33/11 0.1 T2 40 33/11 0.1
b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. V₁ = 120° p.u. V₂ = 120° p.u. jX(1) j0.1Y p.u. jX2)= j0.1Y p.u. jXko) j0.1X p.u. - 0 jX(1) = j0.2 p.u. 1JX(2) = 0.2 p.u. 2 jX1(0) = j0.25 p.u. jX2(1) j0.2 p.u. V₁=1/0° p.u. jX(2(2) = j0.2Y p.u. jX2(0) = j0.3X p.u. = V₂ = 120° p.u. jXT(1) j0.1X p.u. jXT(2) j0.1X p.u. JX3(1) j0.1Y p.u. JX3(2)=j0.1Y p.u. jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u. 0- = 3 = Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXa(n) = j0.13 p. u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u. 4 (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 4. (ii)…
b) A fault occurs at bus 2 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. JX20 /0.1X p.u. jXa2) 0.1X p.u. JX20 j0.2Y p.u. V,= 120° p.u. V, 120° p.u. V, 120° p.u. jX4-70.2X p.u. jX2 j0.2X p.u. jX o 0.2Y p.u. jXncay J0.25 p.u. jXna J0.25 p.u. 3 jXno0.3 p.u. jXTu) /0.2Y p.u. jXra j0.2Y p.u. - j0.2Y p.u. Xp-10.1X p.u. jXa j0.1X p.u. jXp0)- j0.05 p.u. 0 Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXac1) = j0.22 p.u., jXac2) = j0.22 p.u., and jXaco) = j0.23 p. u. X-2 Y=8 (iv) Determine the short-circuit fault current for the case when a phase-to- phase fault occurs at bus 2.
Q2 A three-phase power system is shown in Figure Q2. The data for this system is given in Table Q2. The pre-fault load current and A-Y transformer phase shift are neglected. (a) Draw all the sequence network using a 250 MYA, 275 kV based in the transmission line zone. Write the values of all the parameters on the sequence network. (b) Compute the subtransient fault current at Bus 4 for a line to line to ground fault (DLG). (c) Discuss the answer in part (b) if the neutral of the generator is not grounded. Bus 2 TI Bus I Line I (GI Bus 3 T2 Line 2 AY Line 3 G2 Тз Bus 4 YA Line 4 G3 AY Figure Q2 Table Q2 Voltage Ground impedance Network |Component Generator, G1 Generator, G2 Generator, G3 Transformer TI MVA X2 X, rating rating (kV) (pu) (pu) (pu) ZN (pu) 250 13.8 0.40 0.40 0.10 j0.023 200 20 0.40 0.15 0.10 j0.26 250 13.8 0.40 0.40 0.10 j0.31 250 13.8A/275Y 0.10 0.10 0.10 Transformer T2 Transformer T3 300 20Δ/275Y 0.08 0.08 0.08 350 13.8A/275Y 0.10 0.10 0.10 Each line X, = X; = X, = 200 2
1. FIGURE 52 shows the one-line diagram of a simple three-bus power system with generation at bus I. The voltage at bus l is V1 = 1.0L0° per unit. The scheduled loads on buses 2 and 3 are marked on the diagram. Line impedances are marked in per unit on a 100 MVA base. For the purpose of hand calculations, line resistances and line charging susceptances are neglected a) Using Gauss-Seidel method and initial estimates of Va 0)-1.0+)0 and V o)- ( 1.0 +j0, determine V2 and V3. Perform two iterations (b) If after several iterations the bus voltages converge to V20.90-j0.10 pu 0.95-70.05 pu determine the line flows and line losses and the slack bus real and reactive power. 2 400 MW 320 Mvar Slack 0.0125 0.05 300 MW 270 Mvar FIGURE 52
The single line diagram of a power system is shown in Figure Q2.1 including generator and transformer winding connection and earthing details. The parameters for this system have been calculated on a common 100 MVA base and are given in Table Q2.1. All resistances and shunt susceptances are neglected. This system experiences a single line to ground fault at a point F on line L1. The point F is at a distance d from Bus 4 along the line L1. The total length ?? of the line L1 is 50 km. Note that the location of ?? is not drawn to scale in Figure Q2.1. The fault current at the fault point F is measured to be 6.106 kA. i) Determine the zero, positive, and negative sequence Thevenin equivalent impedances as seen at the fault point F. These should be evaluated in per unit and shown as a function of d.ii) Use the sequence impedances calculated in part (i) to determine the distance d of the fault (in km) from Bus 4.
1 Example-1: In a single-loop distribution system shown in the figure consider the generator to be of infinite short-circuitcapacity and with a voltage of 1.0 pu. Determine the fault currents flowing in circuit breakers B1, B2 and Bz for a Three phase to ground fault in the midde of line 4-5 when: a) Both transformers, T, and T2 are in service b) Only T1 is in service 1 T1 2 From То Impedance 0.0 + j0.01(T,) 0.0 + j0.01(T2) 0.0 + j0.08 0.02 + j0.05 0.01 + j0.03 0.0 + j0.06 0.01 + j0.09 0.01 + j0.09 1 B. B2 Вз -머 3 4 4 T2 7 6 5 6. 7 7
Fill in the blanks 1. Grounding is connecting …….. of electrical equipment to earth. 2. When a …………………………….… fault occurs on an ungrounded neutral system, the potential of the healthy phases rise to the line voltage. 3. The method of reactance grounding is not used because of ……………………… appearing under fault conditions. 4. The cosine of the angle between the ………………………………………. and the ………………………………...…… is called the sending end power factor. 5. The AC power delivered to a load is equal to ………........
What is a FACTS (Flexible Alternating Current Transmission System) device, and how does it impact power system operation?
b) The single line diagram of a power system is shown in Figure Q2.1 including generator and transformer winding connection and earthing details. The parameters for this system have been calculated on a common 100 MVA base and are given in Table Q2.1. All resistances and shunt susceptances are neglected. This system experiences a single line to ground fault at a point F on line L1. The point F is at a distance d from Bus 4 along the line LI. The total length of the line L1 is 50 km. Note that the location of d is not drawn to scale in Figure Q2.1. The fault current at the fault point F is measured to be 6.106 kA. G1 G1 G2 i) G2 T1 T2 T3 L1 11) 2 Rated voltage (kV) 20 20 20/33 20/33 33/11 11 T1 ΔΕ T2 T3 Figure Q2.1 Table Q2.1 Zero sequence reactance Xo (p.u.) 0.20 0.15 0.12 0.20 0.20 1.20 Positive sequence reactance X₁ (p.u.) 0.30 0.25 0.12 0.20 0.20 0.50 UNIVERSITY OF LIVERPOOL L1 (l-d) 5 Negative sequence reactance X₂ (p.u.) 0.35 0.30 0.12 0.20 0.20 0.50 Determine the zero, positive,…
Q4 (a) Define three (3) advantages of alternating current (AC) for long distance power transmission compare to direct current (DC).