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- Problem 1 - Series en Parallel AC networks [19] Look at the circuit in Figure 1 and determine the following: (a) Total Admittance. (b) Total Impedance. (c) Total Current (l:). (d) Current (I1) through impedance Z2. (e) Current (12) through impedance Z3. (f) Current (I3) through impedance Z4. (g) Is this an inductive or capacitive circuit? A. B Zs 220V;50HZ Figure 1 (h) Voltage across Z1. (i) Voltage across A and B. G) Voltage across Zs. Z1 = 3 + j5 ohm Z2 = 10 + jo ohm Z3 = 5 + j15 ohm Z4 = 10 – j30 ohm Zs = 20 – j30 ohm Admittance and Impedance in rectangular notation. All currents and voltage in polar notation. Take voltage as reference.Q2) 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.electrical engineering department plzz solve
- Q4/For the three-phase power network shown in Figure. the various components are: GI: 100 MVA, 0.30 pu reactance. G2: 60 MVA, 0.18 pu reactance. Transformers (cach): 50 MVA, 0.10 pu reactance. Inductive reactor X: 0.20 pu on a base of 100 MVA. Lines (each): 80 ohms (reactive); neglect resistance. with the network initially unloaded and a line voltage of 110 kV, a symmetrical short circuit occurs at midpoint E of line 2. Calculate the short circuit MVA to be interrupted by the circuit breakers A and B at the ends of the line. T3 38 L1 L2 G2 Bas 12 T4 BusQ2) 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.5.5 kW, 1450 rpm. L-type equivalent circuit and parameters for a continuous sinusoidal phase of a star connected 3-phase ASM with line voltage 220 V and frequency 50 Hz are given as follows, ignoring core (iron) losses: R1=0.21 Ohm, R2=-0.18 Ohm, X1=X2-0.66 Ohm and Xm = 9.9 Ohm How many times the inrush current will be the rated current when the ASM is started directly? Answer options group 8.0815 2.6455 4.3763 7.1534 1.2558 6.3414 5.2280 3.7473
- 5.5 kW, 1450 rpm. L-type equivalent circuit and parameters for a continuous sinusoidal phase of a star connected 3-phase ASM with line voltage 220 V and frequency 50 Hz are given as follows, ignoring core (iron) losses: R1=0.21 Ohm, R2=-0.18 Ohm, X1=X2-0.66 Ohm and Xm = 9.9 Ohm How many times the inrush current will be the rated current when the ASM is started directly? Answer options group 8.0815 2.6455 4.3763 7.1534 1.2558 6.3414 5.2280 3.7473High voltage engineering subjectIn the system shown in Figure 1, the transformers are connected star-star with both star points grounded and the generator is connected in star with its star points grounded. The per unit impedances of each element on a 40 MVA base are given in Table 1 and the voltage levels are given in Table 2. Z [p.u.] 0 2 tööt Generator 0.01 +j0.08 p.u. L Figure 1: A section of the distribution system Transformer T1 Line V BASE [KV] 0.04 + j 0.03 + j 0.15000000000000002 0.06000000000000000 Generator Table 1: Sequence impedances (p.u. on 40 MVA base) 3 T2 181. Line 3 10 Table 2: Voltage bases (kV) Load 1 Transformer T2 Load Current A load current of 11.316 kA is flowing with a lagging power factor of 90 %. Convert this to a current vector in per-unit. IL = 0.04 + j 0.06000000000000000
- The one-line diagram of a three-phase system is shown in Figure 1. By selecting a common based of 90 MVA and 32.75 kV on the generator bus, all impedances including the load impedance in per-unit are as follows: XG (new) = j0.1421 p.u XT1 (new) = j0.1219 p.u XT2 (new) = j0.1219 p.u ZOH = 0.0016 +j 0.0083 p.u Хм пеw) 3D j0.247 р.u ZLoad = 2.2759+ j1.2306 p.u T1 T2 90 MVA Motor 90 MVA 55 MVA 33/275 kV 275/11kV 10.95 kV XT=12% Xr3=12% Xx=15% Generator 95 MVA M 32.75 kV Xg=15% 2 x OH Line 138 km R = 0.01 ohm/km X = 0.05 ohm/km Load 35 MVA 10.95 kV 0.88 pf lagging Figure 1 Note: Region 1 = Generator bus Region 2 = Transmission Lines busses Region 3 = Load Bus If the motor operates at full load 0.88 power factor lagging at a terminal voltage 10.95 kV, determine the voltage at the generator bus bar.Transkrmers phase with each bank with - adjuste 35. Three single-phase transformers, cach of 100 kVA rating, are conected in a closed-delta arrangement. If one of them is taken out, it would be possible to load the bank in such a manner that cach one is loaded to the extent of: A 86.6 kVA C. 57.7KVA B. 66.7 kVA D. 33.33 kVA when load is at unity power factor. 36. Average power factor at which V-V bank separate is A 86.6% C. 66.67% В. 73.3% D. None of these 37. Scott connections are used for: A. single-phase to three-phase transformation C. three-phase to two-phase transformation B. three-phase to single-phase transformation (D Any of the above 38. The current in low voltage winding of a 25 kVA, 6600/400 V, 50 Hz, 3-phase Y-A transformer, at full load is: A 36 A Č. 65,32 A B. 20.78 A D. 40.2 A 39. The current in high voltage winding of a 20 kVA, 2200/220 V, 50 Hz, 3-phase A-A transformer, at full load is: (A. 3.03 A C. jo.5 A B. 5.25 A D. None of above 40. The current in low voltage…2. 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 S3