Determine most economical size
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- Consider three ideal single-phase transformers (with a voltage gain of ) put together as three-phase bank as shown in Figure 3.35. Assuming positive-sequence voltages for Va,Vb, and Vc find Va,Vb, and VC. in terms of Va,Vb, and Vc, respectively. (a) Would such relationships hold for the line voltages as well? (b) Looking into the current relationships, express IaIb and Ic in terms of IaIb and Ic respectively. (C) Let S and S be the per-phase complex power output and input. respectively. Find S in terms of S.Consider the single-Line diagram of a power system shown in Figure 3.42 with equipment ratings given: Generator G1: 50MVA,13.2kV,x=0.15p.u. Generator G2: 20MVA,13.8kV,x=0.15p.u. Three-phase -Y transformer T1: 80MVA,13.2/165YkV,X=0.1p.u. Three-phase Y- transformer T2: 40MVA,165Y/13.8kV,X=0.1p.u. Load: 40MVA,0.8PFlagging,operatingat150kV Choose a base of 100 MVA for the system and 132-kV base in the transmission-line circuit. Let the load be modeled as a parallel combination of resistance and inductance. Neglect transformer phase shifts. Draw a per-phase equivalent circuit of the system showing all impedances in per unit.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.
- 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.Connect three single-phase transformers in Y-Y, with subtractive polarity, so that the line-to-line voltages on the primary are 30 degrees delayed to their respective line-to-line voltages on the secondary. It should show each step that is needed to make this connection (example: the phasor diagram) and should show the final connection of the three transformers including the polarity markings. Assume phase A on the primary is terminal "H1".1. How do you measure three phase line currents by using two current transformers? Draw the connection scheme.
- 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. ImpedanceTranskrmers 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
- Three zones of a single-phase circuit are identified shown in Figure below. The zones are connected by transformers T1 and T2, whose ratings are also shown. Using base values of 33 kVA and 232 volts in zone 1, Find: 1- Draw the per-unit circuit including the per-unit impedances and the per-unit source voltage. 2- Calculate the load current both in per-unit and in amperes (actual or original value). Vs Zone 1 232.940° Vs G. 38 T₁ 30 KVA 240/480 volts Xeq = 0.10 p.u. Zone 2 Xiine = 4 Ω T₂ 20 KVA 460/115 volts Zload = Xea = 0.10 p.u. Zone 3 u 1+j2.2 Ω 2A 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.2) The one-line diagram of a three-phase power system is as shown in Figure. Impedance 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 ...L0 kV. Working in per-unit, determine the voltage at buses 2 and 1. (Please determine the voltage magnitude within the specified limits: 410-450 V. By considering the voltage magnitude value you determined yourself, solve the question.) V₁ j0.5 pu V₂ j0.4 pu S₂ Figure V3 S3