A network is composed of the utility having Ssc-250MVA, and an 200OKVA transformer, rated at 20KV/400V (no load), whose voltage impedance is 796, and whose load losses are 15750watts. the short circuit at the secondary of the transformer using the impedance method is: Select one: O None of these O 25.8KA O 27.01KA O 30KA
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- A two-winding single-phase transformer rated 60kVA,240/1200V,60Hz, has an efficiency of 0.96 when operated at rated load, 0.8 power factor lagging. This transformer is to be utilized as a 1440/1200-V step-down autotransformer in a power distribution system. (a) Find the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the ratings as a two-winding transformer. Assume an ideal transformer. (b) Determine the efficiency of the autotransformer with the kVA loading of part (a) and 0.8 power factor leading.An infinite bus, which is a constant voltage source, is connected to the primary of the three-winding transformer of Problem 3.53. A 7.5-MVA,13.2-kV synchronous motor with a sub transient reactance of 0.2 per unit is connected to the transformer secondary. A5-MW,2.3-kV three-phase resistive load is connected to the tertiary Choosing a base of 66 kV and 15 MVA in the primary, draw the impedance diagram of the system showing per-unit impedances. Neglect transformer exciting current, phase shifts, and all resistances except the resistive load.The transformer of Problem 3.16 is supplying a rated load of 50 kVA at a rated secondary voltage of 240 V and at 0.8 posr factor lagging. Neglect the transformer exciting current. (a) Determine the input terminal voltage of the transformer on the high-voltage side. (b) Sketch the corresponding phasor diagram. (c) If the transformer is used as a step-down transformer at the load end of a feeder whose impedance is 0.5+j2.0, find the voltage VS and the power factor at the sending end of the feeder.
- A network is composed of the utility having Ssc=250MVA, and an 2000KVA transformer, rated at 20KV/400V (no load), whose voltage impedance is 7%, and whose load losses are 15750watts. the short circuit at the secondary of the transformer using the impedance method is: a)27.01KA b)None of these c)25.8KA d)30KAA transformer with a 10:1 ratio and rated at 60kVA, 2200/220V, 60Hz is used to step down the voltage of a distribution system. The low tension voltage is to kept constant at 220V. (a) What load Impedance connected to the low-tension side will be loading the transformer fully at 0.8 power factor (lag) ? (b) What is the value of this impedance referred to the high tension side?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.
- A network is composed of the utility having Ssc=500MVA, and an 800KVA transformer, rated at 20KV/410V (no load), whose voltage impedance is 5%, and whose load losses are 5400watts. the short circuit at the secondary of the transformer using the impedance method is: Select one: a)30.5KA b)None of these c)45.8KA d)21.83KAA single-phase 100-kVA, 2,400/240 V, 60-Hz distribution transformer is used as a step-down transformer. The load, which is connected to the 240 V secondary winding, absorbs 80 kVA at 0.8 power factor lagging and is at 230 V. Assuming an ideal transformer, calculate the following:(a) primary voltage;(b) load impedance;(c) load impedance referred to the primary;(d) the active and reactive power supplied to the primary winding.1) A single-phase 100-kVA, 2,400/240 V, 60-Hz distribution transformer is used as a step-down transformer. The load, which is connected to the 240 V secondary winding, absorbs 80 kVA at 0.8 power factor lagging and is at 230 V. Assuming an ideal transformer, calculate the following: (a) primary voltage;(b) load impedance;(c) load impedance referred to the primary;(d) the active and reactive power supplied to the primary winding.
- A two-winding transformer rated at 100 kVA, is connected as an autotransformer to provide 10kV from a 15kV supply. Calculate the new rated kVA output. consider the power transformer mode and calculate the related currents Where, Vp = 10000VIp = S/(√3Vp) Is = S/(√3Vs) When find the new currents and voltages proceed to find the primary and secondary powers of the autotransformer. And conclude the rating of the autotransformer.630kVA 31500/400 DY 50Hz transformer has no load and short circuit tests power measurements are given as 1kW and 20kW respectively. Under the fourth of the nominal load with 0,85 capacitive power factor, calculate the efficiency of the transformer at nominal voltage as %?A 400-turns auto transformer, operating in the step-down mode with 25 percent tap supplies a 4.8 kVA, 0.85 lagging power factor load. The input to the transformer is 2400 V, 60 Hz. Neglect the small losses and leakage effects. What will be the apparent power(in kVA) conducted and transformed respectively?