i. Obtain the parameters of the equivalent circuit, Find the full-load copper and iron losses. ii. iii. Calculate the efficiency of 60% of full-load at power factor 0.8 lagging. Find the full-load voltage regulation at power factor 0.8 leading. iv.
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- Consider Figure 3.25 of the text for a transformer with off-nominal turns ratio. (i) The per-unit equivalent circuit shown in part (c) contains an ideal transformer which cannot be accommodated by some computer programs. (a) True (b) False (ii) In the - circuit representation for real c in part (d), the admittance parameters Y11 and Y12 would be unequal. (a) True (b) False (iii) For complex c, can the admittance parameters be synthesized with a passive RLC circuit? (a) Yes (b) NoThe 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.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.
- For the transformer in Problem 3.10. The open-circuit test with 11.5 kV applied results in a power input of 65 kW and a current of 30 A. Compute the values for GcandBm in siemens referred to the high-voltage winding. Compute the efficiency of the transformer for a load of 10 MW at 0.8 p.f. lagging at rated voltage.A single-phase, 50-kVA,2400/240-V,60-Hz distribution transformer has the following parameters: Resistance of the 2400-V winding: R1=0.75 Resistance of the 240-V winding: R2=0.0075 Leakage reactance of the 2400-V winding: X1=1.0 Leakage reactance of the 240-V winding: X2=0.01 Exciting admittance on the 240-V side =0.003j0.02S (a) Draw the equivalent circuit referred to the high-voltage side of the transformer. (b) Draw the equivalent circuit referred to the low-voltage side of the transformer. Show the numerical values of impedances on the equivalent circuits.(a) An ideal single-phase two-winding transformer with turns ratio at=N1/N2 is connected with a series impedance Z2 across winding 2. If one wants to replace Z2, with a series impedance Z1 across winding 1 and keep the terminal behavior of the two circuits to be identical, find Z1 in terms of Z2. (b) Would the above result be true if instead of a series impedance there is a shunt impedance? (c) Can one refer a ladder network on the secondary (2) side to the primary (1) side simply by multiplying every impedance byat2 ?
- On the plate of the one-phase transformer, it is written 20kVA, 2000/200 V, 50 Hz. In the short-circuit test performed on the transformer, the results stated at the end of the problem were obtained. Draw the equivalent circuit of the short-circuit test with the low-voltage side of the transformer short-circuited, by overwriting the circuit element values. open circuit test high voltage side open circuit V= 200 A = 6a W = 200W closed circuit experiment low voltage side short circuit V= 200v A = 10 A W= 400WIn a 100-kVA, 6,000/400-V, Y-Y, 3-ph, 50-Hz transformer, the copper loss is measured at 5000 W during half-load. Apparently, the maximum efficiency also occurs at ¾ full-load, find the efficiencies of the transformer at full-load and at half-load both at 0.8 pf1. A single-phase step-down transformer is rated at 10 kVA, 208V/120 V. a. Compute the rated current of each winding. b. If a 2 Ω load resistance is connected across the 120 V winding, what are the currents in the high-voltage and low-voltage windings? c. What is the equivalent load resistance referred to the 208 V side? d. What is the lowest load resistance that is allowed by the transformer rating?
- 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.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 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')