Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN: 9781305632134
Author: J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher: Cengage Learning
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- 3. The single-line diagram of a three-phase power system is as shown in Figure Al-3. Impedances are marked in per unit on a 100-MVA, 400-kV base. The load at bus 2 is S2 = 15.93 MW – j 33.4 Mvar, and at bus 3 is S3 = 77 MW +jl4 Mvar. It is required to hold the voltage at bus 3 at 40020° kV. Working in per unit, determine the voltage at buses 2 and 1. V2 V1 V3 j0.5 pu j0.4 pu S2 S3 Figure A1-3arrow_forward2. The one-line diagram of a three-phase power system is as shown in the figure below. Impedances are marked in per unit on a 100-MVA, 400-kV base. The load at bus 2 is S2 = 15.93 MW-j33.4 Mvar, and at bus 3 is S3 = 77 MW +j14 Mvar. It is required to hold the voltage at bus 3 at 4006 0 kV. Working in per unit, determine the voltage at buses 2 and 1. V3 j0.5 pu V₂ S₂ j0.4 pu S3arrow_forwardQ1 Figure 1 shows the online diagram of a three-phase power system. By selecting a common base of 100 MVA and 22 kV on the generator side, draw an impedance diagram showing all impedancesincluding the load impedance in per-unit. The data are given as follows: G: 100 MVA 22 kV x= 0.18 per unitT1: 60 MVA 22/220 kVT2: 40 MVA 220/11 kVT3: 40 MVA 22/110 kVT4: 40 MVA 110/11 kVM: 66.5 MVA 10.45 kVLines 1 and 2 have series reactance’s of 48.4 and 65.43 V, respectively. Atbus 4, the three-phase load absorbs 57 MVA at 10.45 kV and 0.6 power factor lagging.Figure 1x =0.10 per unit x = 0.06 per unitx =0.064 per unit x= 0.08 per unit x =0.185 per unitarrow_forward
- Draw an impedance diagram for the electric power system shown in Figure 3.32 showing all impedances in per unit on a 100-MVA base. Choose 20-kV as the voltage base for generator. The three-phase power and line-line ratings are given below. G1; 90 MVA 20 kV T1: 80 MVA T2: 80 MVA 90 MVA G2: Line: Load: G₁ T₁ 38 1 20/200 kV 200/20 kV 18 kV 200 kV 200 kV Load FIGURE 3.32 One-line diagram for Problem 3.13 Line X = 9% X = 16% X = 20% X = 9% X = 120 Ω S = 48 MW + j64 Mvar T₂ 38 2 G₂arrow_forwardQ2) A 13.2-kV single-phase generator supplies power to a load through a transmission line. The load's impedance is Zload = 500 236.87° ohm, and the transmission line's impedance is Zline = 60 253.1° ohm. To reduce transmission line losses to 0.0103 of its losses without using the transformers design and use two transformers T1 between the generator and the transmission line and T2 between the transmission line and the load.arrow_forwardFigure below shows the one line diagram of a 3-o system. By selecting a common base of 100 MVA and 22 KV on the generator side, draw an impedance diagram showing all impedances including the load impedance in per unit. The data are given as follows: G: 100 MVA, 22 KV, X= 0.18 pu 22/220 KV, X= 0.1 pu TR1: 50 MVA, TR2: 40 MVA, 220/11 KV, X=0.06 pu Load 1: 50 MVA, 0.8 pf lag Load 2: 50 MVA, 0.8 pf lead If volt at bus 4 equal 11 KV constant value : calculate i) volt at bus 1 ii) EMF (Eg) of generator ii) Transmission line current TR1 TR2 ZTL=j80 2 G1 3. 4 Load 1 Load 2arrow_forward
- The one-line diagram of a three-phase power system is as shown in the figure attached. Impedances are marked in per unit on a 100-MVA, 400-kv base. The load at bus 2 is S(sub 2) = 15.93 MW - j33.4 Mvar, and bus 3 is S(sub 3) = 77MW + j14 Mvar. I tis required to hold the voltage at bus 3 at 400 angle 0 degrees kV. Working in per unit, determine the voltage at buses 2 and 1.arrow_forwardRefer to the given answer so that i will know how.arrow_forwardQ2) A 13.2-kV single-phase generator supplies power to a load through a transmission line. The load's impedance is Ztoad 500 236.87° ohm , and the transmission line's impedance is Zine = 60 253.1° ohm. To reduce transmission line losses to 0.0103 of its losses without using the transformers design and use two transformers T1 between the generator and the transmission line and T2 between the transmission line and the load.arrow_forward
- The three-phase power and line-line ratings of the electric power system shown in Figure 30 are given below. Vg OBH G T2 2 Vm Line M FIGURE 30 One-line diagram for Problem 3.15 G₁: 60 MVA 20 kV X = 9% T₁: 50 MVA 20/200 kV X = 10% T2: 50 MVA 200/20 kV X = 10% M: 43.2 MVA 18 kV X = 8% Line: 200 kV Z 120+j200 $2 (a) Draw an impedance diagram showing all impedances in per unit on a 100-MVA base. Choose 20 kV as the voltage base for generator. (b) The motor is drawing 45 MVA, 0.80 power factor lagging at a line-to-line ter- minal voltage of 18 kV. Determine the terminal voltage and the internal emf of the generator in per unit and in kV.arrow_forward2. Draw the reactance diagram for the power system shown in Figure 2. Neglect the resistance and use a base of 50 MVA and 13.8 kV on generator G1. G1 : 20 MVA, 13.8 kV, X"= 20% G2 : 30 MVA, 18.0 kV, X"= 20% G3 :30 MVA, 20.0 kV, X"= 20% T1:25 MVA, 220/13.8 kV, X = 10% T2: 3 single phase unit each rated 10 MVA, 127/18 kV, X = 10% T3: 35 MVA, 220/22 kV, X = 10% Determine the new values of per unit reactance of G1, T1, transmission line 1,transmission line 2, T2, G2, T3 and G3. Line I Line 2 j80 2 j100 2 ele T Vele G. Fig.2: One line diagram of the systemarrow_forwardFigure 1 shows a three-phase power system network comprises of synchronous generator, transformers, transmission lines and loads while Table 1 shows the ratings of the components. Using per unit analysis, 100 MVA base and 20 kV as the voltage base at the generator, answer the following questions: Ig T2 TLine Load C Load A Load B Figure 1: A simple power system network. Table 1: The power system component ratings G 90 MVA, 20 kV, XG = 120% T1 80 MVA, 20/132 kV, XTI = 16% T2 80 MVA, 132/11 kV, XT2 = 10% TLine 132 kV, XLine = 18 2 Load A 7 MW +j 3 Mvar Load B 30 MW +j 15 Mvar Load C 22 MVA, 0.89 power factor leading (a) Calculate the impedance per-unit values of all independent components based on given conditions.arrow_forward
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