A 15 KVA, 3.81 KV/381-V, 60 Hz, Y-Y 3-phase transformer is supplying 15 KVA of power to a load at 0.85 Lagging power factor, a line load voltage of 381 Volts. Its per-phase equivalent circuit is given below. Calculate the magnitude of the line input voltage of this transformer 10 2440 12 KV. 0.808 +14.04 02 1700.392 Answer The magnitude of the line input voltage is: 4

A 15 KVA, 3.81 KV/381-V, 60 Hz, Y-Y 3-phase transformer is supplying 15 KVA of power to a load at 0.85 Lagging power factor, a line load voltage of 381 Volts. Its per-phase equivalent circuit is given below. Calculate the magnitude of the line input voltage of this transformer 10 2440 12 KV. 0.808 +14.04 02 1700.392 Answer The magnitude of the line input voltage is: 4

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.34P: Three single-phase, two-winding transformers, each rated 450MVA,20kV/288.7kV, with leakage reactance...

See similar textbooks

Similar questions

Consider the oneline diagram shown in Figure 3.40. The three-phase transformer bank is made up of three identical single-phase transformers, each specified by X1=0.24 (on the low-voltage side), negligible resistance and magnetizing current, and turns ratio =N2/N1=10. The transformer bank is delivering 100 MW at 0.8 p.f. lagging to a substation bus whose voltage is 230 kV. (a) Determine the primary current magnitude, primary voltage (line-to-line) magnitude, and the three-phase complex power supplied by the generator. Choose the line-to-neutral voltage at the bus, Va as the reference Account for the phase shift, and assume positive-sequence operation. (b) Find the phase shift between the primary and secondary voltages.
In developing per-unit circuits of systems such as the one shown in Figure 3.10. when moving across a transformer, the voltage base is changed in proportion to the transformer voltage ratings. (a) True (b) False
The per-unit equivalent circuit of two transformers Ta and Tb connected in parallel, with the same nominal voltage ratio and the same reactan of 0.1 per unit on the same base, is shown in Figure 3.43. Transformer Tb has a voltage-magnitude step-up toward the load of 1.05 times that of Ta (that is, the tap on the secondary winding of Tb is set to 1.05). The load is represented by 0.8+j0.6 per unit at a voltage V2=1.0/0 per unit. Determine the complex power in per unit transmitted to the load through each transformer, comment on how the transformers share the real and reactive powers.
Consider Figure 3.4. For an ideal phase-shifting transformer, the imda nce is unchanged when it is referred from one side to the other. (a) True (b) False
Three single-phase two-winding transformers, each rated 25MVA,54.2/5.42kV, are connected to form a three-phase Y- bank with a balanced Y-connected resistive load of 0.6 per phase on the low-voltage side. By choosing a base of 75 MVA (three phase) and 94 kV (line-to-line) for the high-voltage side of the transformer bank, specify the base quantities for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-voltage side. Then determine the load resistance RL in ohms referred to the high-voltage side and the per-unit value of this load resistance on the chosen base.
Three single-phase, two-winding transformers, each rated 450MVA,20kV/288.7kV, with leakage reactance Xeq=0.10perunit, are connected to form a three-phase bank. The high-voltage windings are connected in Y with a solidly grounded neutral. Draw the per-unit equivalent circuit if the low-voltage windings are connected (a) in with American standard phase shift or (b) in Y with an open neutral. Use the transformer ratings as base quantities. Winding resistances and exciting current are neglected.
4. An Open-Wye/Open-Delta transformer, as shown in Figure, is formed by two identical single-phase transformers. The primary line-to-line voltage is 12.47 kV, and the secondary line-to-line voltage is 208 V. Take
6. A 5,000-kVA, 3-phase transformer, 6.6/33-kV, A/Y, has a no-load loss of 15 kW and a full-load loss of 50 kW. The impedance drop at full-load is 7%. Calculate the primary voltage when a load of 3,200 kW at 0.8 p.f. is delivered at 33 kV.
Here is a single phase transformer, 15 kVA, 15000/380 V, 50 Hz has Ze-0.02+j0.05 perunit, R. 50 perunit, and Xm =30 perunit, (per unit = actual value/base value), for scaling of the problem. (pu). A) Compute and draw the equivalent circuit components in Qs referred to the low-voltage side, B) Low voltage side is connected to a load of 4L 90° (lagging) and HV side is supplied by 15KV, compute voltage drop on the load V2 and voltage regulation VR.
No-load operation and short circuit tests were performed on three-phase, 50 Hz 480/220 delta / Y connected transformer. In these testsThe line voltage, line current and three-phase total power have been measured from the high voltage windings and the test results are below.pictures. 100 A DC current when 1.08 V DC voltage from non-transferable software is applied to the low voltage windingshas attracted.Find the single phase equivalent circuit parameters of this transformer.idle test:Voltage: 480V Power:114w current: 0.71Ashort circiut test:V: 10v P: 78w c: 18.3a
Consider a single phase transformer of rating 150 KVA, 3600/360, 50 Hz. Let the series resistance due to windings be 0.2 ohm and 0.002 ohm on the HV and LV sides, respectively. Let the series reactance due to leakage flux be j0.45 ohm and j0.0045 ohm on the HV and LV sides, respectively. While conducting Short Circuit test, the HV side should be excited with Volt Input power factor while conducting Short Circuit test is lagging
A 50-KVA 2400:240-V 60-Hz distribution transformer has a leakage impedance of 0.72 + j0.922 in the high-voltage winding and 0.0070+ j0.0090 $2 in the low-voltage winding. At rated voltage and frequency, the impedance Z, of the shunt branch (equal to the impedance of R. and jXm in parallel) accounting for the exciting current is 6.32 + j43.7 52 when viewed from the low-voltage side. Draw the equivalent circuit referred to (a) the high-voltage side and (b) the low-voltage side, and label the impedances numerically.