Q1: Draw the Single Line Diagram for electrical power system which consist of following items: Transmission Line has phase. impedance MVA, 20 per Generator 40 11 kV X-10% Transformer1 (Y/Y), (Y/A), 27 MVA, 11/132 kV X-10% Transformer2 30 MVA, 132/6.6 kV X-15% Motor 29 MVA, 6.6 kV, X-25% Also draw the reactance diagram in per unit value. Assume that the Sh-50MVA, Vb-11KV
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- Consider the oneline diagram shown in Figure 3.40. The three-phase transformer bank is made up of three identical single-phase transformers, each specified by X1=0.24 (on the low-voltage side), negligible resistance and magnetizing current, and turns ratio =N2/N1=10. The transformer bank is delivering 100 MW at 0.8 p.f. lagging to a substation bus whose voltage is 230 kV. (a) Determine the primary current magnitude, primary voltage (line-to-line) magnitude, and the three-phase complex power supplied by the generator. Choose the line-to-neutral voltage at the bus, Va as the reference Account for the phase shift, and assume positive-sequence operation. (b) Find the phase shift between the primary and secondary voltages.7. System Model and Per Unit T1 Line T2 BE 31.25miles 3 Impedance Load R1+jXL-0.1+j0.5 p.u. 30 MVA 30 MVA 13 kV X-0.1 p.u. 14 kV 13/220 kV X-0.2 p.u. 35 MVA * Transformer T2 is composed of "three single-phase" transformers, each rated 40/3 MVA, 132.8:14 kV with a leakage reactance of 0.1 per unit. (a) Fill in the appropriate base values. Gen. G Trans. Ti Line Trans. T: Load New MVA Base New KV Base (b) Calculate the impedance values. Gen G Trans. T1 Line Trans. T2 Load p.u. ImpedanceA 15 MVA, 6.9 KV generator has ( X1=X2= 0.25p.u., Xo= 0.07p.u., and Xn= 0.06p.u.). Determine the line to line voltages when a DL-G fault occurs at the generator terminals and the generator is unloaded and operating at 6 kV. Your answer Determine the line to line voltages when a L-L fault occurs at the generator terminals and the generator is unloaded and operating at 6 kV. Your answer
- A 30 MVA, 11 kV generator has a reactance of 0.10p.u.on its own base. Determine the per-unit reactance when referred to base kVA of 50,000 kVA and base kV of 33 kV.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 7One-line diagram of a power system is shown below. Relevant details are mentioned below. Generator: 100 MVA, 11 kV, 3-phase, reactance 20%; Transformer T1: 100 MVA, 10/132 kV, reactance 6%; Transformer T2: 80 MVA, 132/10 kV, reactance 5%; Line reactance is 100 Q. Motors M1 and M2 are rated at 50 MVA and 40 MVA respectively both at 10 kV and 20% reactance. In the above, reactances are per-phase values. Taking generator rating as base, draw the per- phase impedance diagram in which the elements need to be expressed in per unit. * T1 M, Transmission Line G M2) A
- A 25 MVA, II kV generator has X"d=0.2 p.u. X2 = 0.3p.u. and X0=0.1 p.u. Tht neutral of generator is solidly grounded. Determine the subtransient current in the generator and the line to line voltages for subtransient condition when a Y-B-G fault occurs at the generator terminals. Assume prefault currents and fault resistance to be zero.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.In the single line diagram shown below, generators G₁ and G₂ have a leakage reactance of 0.26 per unit on a 66.5MVA rating at 11kV, and transformers T₁ and T₂ have a voltage ratio of 11/145kV and a leakage reactance of 0.125 per unit on a 75MVA rating. Choosing 100MVA and 132kV (at the lines side) as base quantities, find the reactance of G₁, G₂, T₁ and T2 in per unit to the new base quantities. 66.5MVA 11kV 0.26pu 66.5MVA 11kV 0.26pu (G1 G2 75MVA 11/145kV 0.125pu T1 T2 75MVA 11/145KV 0.125pu 132kV Lines
- The one-line diagram of a simple power system is shown in Figure below. The neutral of each generator is grounded through a current-limiting reactor of 0.25/3 per unit on a 100-MVA base. The system data expressed in per unit on a common 100-MVA base is tabulated below. The generators are running on no-load at their rated voltage and rated frequency with their emfs in phase. G Stark Item Base MVA Voltage Rating X' x² 20 kV 20 kV 20/220 kV 20/220 kV 100 0.05 0.15 0.15 0.10 0.10 220 kV 0.125 0.125 0.30 0.15 0.25 025 0.7125 0.15 100 100 0.15 0.05 0.10 0.10 0.10 100 0.10 100 100 Lu La 220 kV 0.15 220 kV 0.35 100 A balanced three-phase fault at bus 3 through a fault impedance Zf= jo.I per unit. The magnitude of the fault current in amperes in phase b for this fault is: Select one: A. 345.3 B. 820.1 C. 312500 3888888 产产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…Figure below shows a power system. The rating of generators and transformers are:G1:25MVA, 6.6KV, jO.2 p.u. G2:15MVA, 6.6KV, jo.15 p.u. G3:30MVA, 13.2KV, jo.15 p.u. T1: 30MVA, 6.6A- 115Y KV, jo.1 p.u.T2: 15MVA, 6.6A- 115Y KV, jo.1 p.u. T3:single phase units each rated 10MVA, 69/ 6.9 KV, jo.1 p.u. Find te reactances in p.u, take a base of 30MVA, 6.6 KV in the circuit of G1 .=.XG1 G2} 4 A lubu T2 T1 T3 G1 G3 j120 2 j90 N