1000 KVA 11kV/415V RTR = 1.%, XTR -6.0% 200 KVAR 16 -C VLL = 11 kV, 50 Hz MVASc = 200 X/R = 2.4 Linear and nonlinear plant loads Figure Q3(c)
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- Consider the three single-phase two-winding transformers shown in Figure 3.37. The high-voltage windings are connected in Y. (a) For the low-voltage side, connect the windings in , place the polarity marks, and label the terminals a, b, and c in accordance with the American standard. (b) Relabel the terminals a, b, and c such that VAN is 90 out of phase with Va for positive sequence.Now that a 1:15 step-up transformer is installed at the output of the generator and a 15: 1transformer is installed at the end of the transmission line, as depicted in Figure Q3(b). (i)Determine the new load voltage/generated voltage ratioand power loss in the transmission line. (ii)By assuming the transformer is ideal, evaluate your answers with those obtained in 3(a)A 1000-VA, 230/115-V traformer has been tested to determine its equivalent circuit. The results are shown below: Open-Circuit Test results from the low voltage side: VOC = 115 V, IOC = 0.1 A, POC = 5 W; Short-Circuit Test results from the high voltage side: VSC = 20 V, ISC = 10 A, PSC = 50 W; Find the transformer’s voltage regulation, in percent, at 50% loading conditions and 0.8 PF lagging. Note: (If %1, the you should write as '1', not as '0,01')
- Mechanical Engineering Course, Electrical subject questions! (refer attached pic) A 20kVA 8000/277V distribution transformer has the following resistances and reactances:Re, = 32 ohmRs0.050 ohmXp45 ohmXs0.060 ohmR. = 250k ohmXM30k ohmThe excitation branch impedances aregiven referred to the high-voltageside of thetransformer. Assume that this transformer is supplying rated load at 277 V and 0.85 pflagging, determine the following:=:Draw and label the equivalent circuit of transformer referred to primary sidein)Determine its voltage regulation and efficiency.A short-circuit test was performed on a 10kVA, 2000/400 V single phase transformer. The instruments indicated 100V, 5A and 100 W. Determine the secondary voltage when operating at rated load at 80% pf lagging at rated primary voltage. Provide values in order the space provided below: Re, Xe, I1, I2, and E1, E2A 10-kVA, 2300/230-V transformer is to be tested to determine its excitation branch components, its series impedances and its voltage regulation. The following test data have been taken from the primary side of the transformer: [CLO-1, PLO-2] (Marks-15) Open- circuit test Short- circuit test Voc = 2300 V Vsc = 48 V Ioc = 0.20 A Isc = 5.0 A Poc = 40 W Psc = 180 W The data have been taken by using the connections of open-circuit test and short-circuit test: Calculate the full-load voltage regulation at 0.8 lagging power factor and 0.8 leading power factor What is the efficiency of the transformer at full load with a power factor of 0.8 lagging?
- Q3- A 10 KVA, 230/ 110 V transformer is to be used as an auto-transformer. What will be the voltage ratio and output power rating of auto-transformer Q4-A single-phase, 10-kVA, 440/110-V, two- winding transformer is connected as an autotransformer to supply a load at 550 V from a 440 V supply as shown below. Calculate the following. 1. kVA rating as an autotransformer 2. apparent power transferred by conduction 3. apparent power transferred by electromagnetic inductionHow to generate 480V three phase 400Hz ac from 230V three phase 60Hz ac source? (Please specify the power components you pick, such as transformers, dc-dc converter, dc-ac converter and/or ac-dc converter, how these components are connected and the key parameters of these components, such as the turns ratio of transformer and the duty cycle functions for the converters.)The efficiency of a 20 kVA 15.4 / 0.4 kV three-phase power transformer is 79.1% for a load with unit power factor at rated power. Transformer Relative short circuit voltage% Uk = 30.0 and relative idle current% I0 = 30. Since the transformer is connected star-delta; Note: In 3-phase electrical machines, the current-voltage information given on the label is always line values. Phase value When given, it must be specified separately. Therefore, all current-voltage information given in the question are line values. 1-Draw the primary reduced equivalent circuit. 2-What is the power value measured in the short circuit test of this transformer? -The value of iron losses of this transformer what?
- As an Engineer you have been contracted by a power distribution company to investigate the causes of frequent transformer failures in its network. In the course of your investigation you came across the following: The Transformer installed in urban area failed within the warrant period of 3 years; the diagnostic results showed failure of one phase of the High Voltage (HV), It was further observed that this failure was causes by Overheating and overloading of that phase which caused burning of both HV and Low Voltage (LV). Explain what could be the cause(s) of this overload and the implications it could have on other equipment in the TransformerConsider the following open circuit transformer as shown in figure below which operate at: the voltage sources at no load (Voc)=170 V no load current (Ioc) = 85 m A iron losses power (Poc) = 14 W power factor = cos∅= 0.968 Compute the No load transformer parameter magnetizing (Im) and copper losses current (Ic) , (R0) and (X0) ?A 120:480-V, 10-kVA step-up two-winding transformer is to be used as an autotransformer to supply a 120-V circuit from a 600-V source (a step-down autotransformer). When it is tested as a two-winding transformer at rated load, unity power factor, its efficiency is 97.9 %. 1.Show the connection diagram of the transformer converted as an autotransformer and identify which is the series winding Nse and the common winding Nc.