68. Choose the correct drawing for a three-phase bank with a 480 V Delta primary and a 240 V Delta secondary. a) H M M
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- Single phase non-ideal transformer. Turn Ratio= 3000/600. R₁ = 0.9Q R₂= 36 mQ Ro= 3000 Q X₁₁ = 0.6Q X12= 0.0024 Q X0= 6000 Q Input voltage-1440 V (also known as rms voltage) Load impedance of secondary winding is z₂= 0.12 + j0.3 Q a) Secondary winding current phasor 12 (NOT to be confused for I'₂) b) Load voltage phasor V2 (NOT to be confused for V'2) c) Primary winding current phasor ī₁ → X11 X12 R₁ R'2 www V₂ V₁ ī₁ R m ×°The direct electrical connection of the windings allows transient over voltages to pass through the auto transfonner more easily, and that is an important disadvantage of the autotransformer. (a) True (b) FalseA single-phase step-down transformer is rated 13MVA,66kV/11.5kV. With the 11.5 kV winding short-circuited, rated current flows when the voltage applied to the primary is 5.5 kV. The power input is read as 100 kW. Determine Req1andXeq1 in ohms referred to the high-voltage winding.
- In order to avoid difficulties with third-harmonic exciting current, which three-phase transformer connection is seldom used for step-up transformers between a generator and a transmission line in power systems. (a) Y- (b) -Y (c) Y-YMatch the following: (i) Hysteresis loss (a) Can be redud by constructing the core with laminated sheets of alloy steel (ii) Eddy current loss (b) Can be reduced by the use of special high grades of alloy stl as core material.A single-phase, 120V(rms),60Hz source supplies power to a series R-L circuit consisting of R=10 and L=40mH. (a) Determine the power factor of the circuit and state whether it is lagging or leading. (b) Determine the real and reactive power absorbed by the load. (c) Calculate the peak magnetic energy Wint stored in the inductor by using the expression Wint=L(Irms)2 and check whether the reactive power Q=Wint is satisfied. (Note: The instantaneous magnetic energy storage fluctuates between zero and the peak energy. This energy must be sent twice each cycle to the load from the source by means of reactive power flows.)
- Draw three transformers for each of the different three-phase connections. That is, draw three transformers and then make the connection, eg star-star, then redraw three transformers and make another three-phase connection, and so on until all four combinations of three-phase connections are made.2) 9.2Three - 25 KVA single-phase transformers are connected in delta – delta. One of the transformers has been defected. What is the maximum real power that the transformer bank can deliver at a power factor of 0.8?From the circuit below, determine secondary winding voltage Vsec if the primary winding voltage Vpri = 120Vrms? %3D Np:Ns 6:1 D. Vp 120 Vac 60 Hz Vout R = 200 Q O 20Vrms O 720Vrms O 120Vrms O Ovrms wi m
- sformer 1200 KVA, 11000/2200 V, 50 Hz e reactance/phase of the primary winding ar ding values for the secondary winding are C primary winding is connected in delta andSingle phase non-ideal transformer. Turn Ratio= 3000/600. R₁ = 0.9Q R2= 36 mQ Ro= 3000 Q X₁₁ = 0.60 X12= 0.0024 02 X= 6000 Input voltage= 1440 V (also known as rms voltage) Load impedance of secondary winding is Z₂= 0.12 + j0.3 a) Copper losses on primary and secondary windings b) Core loss c) Transformer efficiency V₁ Ī₁ Ro ww m Xu R₁ R'2 ww m ww I₂ X12 mDraw the complete circular stator winding for a three phase delta connected AC generator consisting of 4 poles and 24 slots using a parallel connection. Your submission must consist of two drawings as follows: One drawing must show the winding arrangement of the phasegroups in the slots of the stator highlighting the start and finish of each phasegroup The other drawing must show only the end connections of each phase group for a parallel connection of the phasegroups and a delta connection of the phases The use of AutoCad or any other software is encouraged.