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Problem 1. Equipment ratings and per-unit reactances for the system shown in figure 1. are given...
Q3: (15 Points) Equi below are given as follows: pment ratings and per-unit reactances for the sy...
Q3: (15 Points) Equi below are given as follows: pment ratings and per-unit reactances for the system shown in circuit 2 4 T1 TL12 T2b G1 G2 TL13 TL23 0.03 j0.03 Synchronous Generators: G1 100 MVA 25 kV G2 100 MVA 3.8 kV Transformers: T1 100 MVA 25/230 kV T2 100 MVA 13.8 /230 kV X1=X2= 0.2 X0-0.05 X1-X2-0.2 X0 0.05 X1 = X2-X0 = 0.05 Transmission lines: TL12 100 MVA TL13 100 MVA TL23 100 MVA X0 0.3 XO...
Ransformer TZXı = Χ, = 0.10 pu, X,-00 Transformer T, x,-X,-Xo = 0.50 pu Transformer Tl: x, = x, =...
ransformer TZXı = Χ, = 0.10 pu, X,-00 Transformer T, x,-X,-Xo = 0.50 pu Transformer Tl: x, = x, = 0.30 pu, X.-0。 Transmission line TL23-X-X2 = 0.15 pu, Xo = 0.50 pu Transmission line TL35: X1-X2 = 0.30 pu, Xo = 1.00 pu Transmission line TL57: Xi = X2 = 0.30 pu, X,-1.00 pu Draw the corresponding positive-sequence network Draw the corresponding negative-sequence network Draw the corresponding zero-sequence network (e) (b) (c) Ti TL23 T2 18.75 MVA 1 △Y1...
A single line diagram of a power system is shown in Fig. 2. The system data with equipment ratings and assumed sequence reactances are given the following table. The neutrals of the generator and A-Y...
A single line diagram of a power system is shown in Fig. 2. The system data with equipment ratings and assumed sequence reactances are given the following table. The neutrals of the generator and A-Y transformers are solidly grounded. The motor neutral is grounded through a reactance Xn 0.05 per unit on the motor base. Assume that Pre-fault voltage is takin as VF-1.0 ,0° per unit and Pre- fault load current and Δ-Y transformer phase shift are neglected In the...
Power System
A simple three-phase power system is shown in Figure 2. Assume that the ratings of the various devices in this system are as follows: • Generators G1, G2: 40 MVA, 13.2 kV, = 0.15 pu, = 0.15 pu, = 0.08 • Generator G3: 60 MVA, 13.8 kV, = 0.20 pu, 0.20 pu, - 0.08 • Transformers T1, T2, T3, T4: 40 MVA, 13.8/138 kV, X1 = X2 = 0.10 pu, XO 0.08 pu Transformers T5, T6: 30 MVA, 13.8/138 kV, X1 = X2...
3.13 A single-line diagram of a three-phase power system is shown in Fig. 3.51. The ratings...
3.13 A single-line diagram of a three-phase power system is shown in Fig. 3.51. The ratings of the equipment are shown below Generator G: 100 MVA, 11 kV, Xi -X2-0.20 pu, Xo -0.05 pu Generator G2 : 100 MVA, 20 kV, Xi=X2=0.25 pu, Xo=0.03 pu, X,,-0.05 pu Transformer T: 100 MVA, 11/66 kV, Xi -X2-Xo 0.06 pu Transformer T2: 100 MVA, 11/66 kV, Xi-X2 = Xo 0.06 pu Line: 100 MVA, X,-X2 = 0.15 pu, Xo = 0.65 pu A...
1) Consider the power system shown in Fig. 1. Use a power base of 500 MVA...
1) Consider the power system shown in Fig. 1. Use a power base of 500 MVA to calculate the fault current in amperes for a double line-to-ground fault at bus B. G: 500 MVA, 13.8 kv, xa = 0.2 p.M., X2 = 0.2 p.j. and x = 0.1 p.u. G2:600 MVA, 26 kv, xa = 0.15 p.u., X2 = 0.15 p.u. and X, = 0.1 p.u. G3:400 MVA, 13.8 kv, x, = 0.2 p.u., x2 = 0.2 p.u. and x...
3) The single-line diagram of a three-phase power system is shown in Fig. 1. Equipment ratings...
3) The single-line diagram of a three-phase power system is shown in Fig. 1. Equipment ratings are given as follows: G1 1,000 MVA, 15.0 kV, 20.18, o 0.07 pu G2 : 1,000 MVA. 15.0 kV, 攻=エ1 =エ2 = 0.20, ro = 0.10 pu G3 : 500 MVA, 13.8 kV. 1" = 띠 z2 = 0.15, zo 0.05 pu G4 : 750 MVA, 13.8 kV. ェd =ェ1 = 0.30, T2 = 0.40 ro = 0.10 pu Ti : 1,000 MVA. 15.0Δ/765Y...
can you calculating this question and explain why? thanks The ratings and sequence reactances of the components for...
can you calculating this question and explain why?
thanks
The ratings and sequence reactances of the components for the power system shown in Figure 2 are given in Table 1. The pre-fault voltage is 1/0° per unit (pu). Bus 8 L3 Bus 1 T1 Bus 4 T2 Bus 2 Bus 5 L1 L2 G1 Bus 3 T3 Bus 7 Bus 6 D.6975 Figure Draw the per unit impedance sequence networks and determine the per unit (a) Thevenin impedances of the...
Assuming there is a FAULT at BUS 3, Determine the thevenin equivalent of each series network...
Assuming there is a FAULT at BUS 3, Determine the thevenin
equivalent of each series network as viewed from the fault bus.
Given:
-Prefault voltage is 1.0 per unit
-Prefault load currents and delta-wye transformer phase shifts
are neglected
Synchronous generators G1 1000 MVA 15 kVX"-X2 0.18, Xo 0.07 per unit G2 1000 MVA 15 kV X: X, = 0.20, X,-0.10 per unit G3 500 MVA 13.8 kV X: X,-0.15, X,-0.05 per unit G4 750 MVA 13.8 kV X,-0.30, X,-0.40,...
Consider the single-line diagram of the three-phase power system shown in Figure 1. Component ratings are...
Consider the single-line diagram of the three-phase power system shown in Figure 1. Component ratings are as follows: Generator G1: 750 MVA, 18 kV, X0.2 per unit Generator G2: 750 MVA, 18 kV, X 0.2 per unit Synchronous Motor M: 1,500 MVA, 20 kV, X-20% A-Y Transformers Ti, T2, T's, & T.: 750 MVA, 500 kV Y/20 kV A, X = 10% Y-Y Transformer T's 1,500 MVA, 500 kV Y/20 kV Y, X-10% ne L:X (a) Using bases of 100...
Q3: (15 Points) Equi below are given as follows: pment ratings and per-unit reactances for the system shown in circuit 2 4 T1 TL12 T2b G1 G2 TL13 TL23 0.03 j0.03 Synchronous Generators: G1 100 MVA 25 kV G2 100 MVA 3.8 kV Transformers: T1 100 MVA 25/230 kV T2 100 MVA 13.8 /230 kV X1=X2= 0.2 X0-0.05 X1-X2-0.2 X0 0.05 X1 = X2-X0 = 0.05 Transmission lines: TL12 100 MVA TL13 100 MVA TL23 100 MVA X0 0.3 XO...
ransformer TZXı = Χ, = 0.10 pu, X,-00 Transformer T, x,-X,-Xo = 0.50 pu Transformer Tl: x, = x, = 0.30 pu, X.-0。 Transmission line TL23-X-X2 = 0.15 pu, Xo = 0.50 pu Transmission line TL35: X1-X2 = 0.30 pu, Xo = 1.00 pu Transmission line TL57: Xi = X2 = 0.30 pu, X,-1.00 pu Draw the corresponding positive-sequence network Draw the corresponding negative-sequence network Draw the corresponding zero-sequence network (e) (b) (c) Ti TL23 T2 18.75 MVA 1 △Y1...
A single line diagram of a power system is shown in Fig. 2. The system data with equipment ratings and assumed sequence reactances are given the following table. The neutrals of the generator and A-Y transformers are solidly grounded. The motor neutral is grounded through a reactance Xn 0.05 per unit on the motor base. Assume that Pre-fault voltage is takin as VF-1.0 ,0° per unit and Pre- fault load current and Δ-Y transformer phase shift are neglected In the...
3.13 A single-line diagram of a three-phase power system is shown in Fig. 3.51. The ratings of the equipment are shown below Generator G: 100 MVA, 11 kV, Xi -X2-0.20 pu, Xo -0.05 pu Generator G2 : 100 MVA, 20 kV, Xi=X2=0.25 pu, Xo=0.03 pu, X,,-0.05 pu Transformer T: 100 MVA, 11/66 kV, Xi -X2-Xo 0.06 pu Transformer T2: 100 MVA, 11/66 kV, Xi-X2 = Xo 0.06 pu Line: 100 MVA, X,-X2 = 0.15 pu, Xo = 0.65 pu A...
1) Consider the power system shown in Fig. 1. Use a power base of 500 MVA to calculate the fault current in amperes for a double line-to-ground fault at bus B. G: 500 MVA, 13.8 kv, xa = 0.2 p.M., X2 = 0.2 p.j. and x = 0.1 p.u. G2:600 MVA, 26 kv, xa = 0.15 p.u., X2 = 0.15 p.u. and X, = 0.1 p.u. G3:400 MVA, 13.8 kv, x, = 0.2 p.u., x2 = 0.2 p.u. and x...
3) The single-line diagram of a three-phase power system is shown in Fig. 1. Equipment ratings are given as follows: G1 1,000 MVA, 15.0 kV, 20.18, o 0.07 pu G2 : 1,000 MVA. 15.0 kV, 攻=エ1 =エ2 = 0.20, ro = 0.10 pu G3 : 500 MVA, 13.8 kV. 1" = 띠 z2 = 0.15, zo 0.05 pu G4 : 750 MVA, 13.8 kV. ェd =ェ1 = 0.30, T2 = 0.40 ro = 0.10 pu Ti : 1,000 MVA. 15.0Δ/765Y...
can you calculating this question and explain why?
thanks
The ratings and sequence reactances of the components for the power system shown in Figure 2 are given in Table 1. The pre-fault voltage is 1/0° per unit (pu). Bus 8 L3 Bus 1 T1 Bus 4 T2 Bus 2 Bus 5 L1 L2 G1 Bus 3 T3 Bus 7 Bus 6 D.6975 Figure Draw the per unit impedance sequence networks and determine the per unit (a) Thevenin impedances of the...
Assuming there is a FAULT at BUS 3, Determine the thevenin
equivalent of each series network as viewed from the fault bus.
Given:
-Prefault voltage is 1.0 per unit
-Prefault load currents and delta-wye transformer phase shifts
are neglected
Synchronous generators G1 1000 MVA 15 kVX"-X2 0.18, Xo 0.07 per unit G2 1000 MVA 15 kV X: X, = 0.20, X,-0.10 per unit G3 500 MVA 13.8 kV X: X,-0.15, X,-0.05 per unit G4 750 MVA 13.8 kV X,-0.30, X,-0.40,...
Consider the single-line diagram of the three-phase power system shown in Figure 1. Component ratings are as follows: Generator G1: 750 MVA, 18 kV, X0.2 per unit Generator G2: 750 MVA, 18 kV, X 0.2 per unit Synchronous Motor M: 1,500 MVA, 20 kV, X-20% A-Y Transformers Ti, T2, T's, & T.: 750 MVA, 500 kV Y/20 kV A, X = 10% Y-Y Transformer T's 1,500 MVA, 500 kV Y/20 kV Y, X-10% ne L:X (a) Using bases of 100...
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