Exam 2: Power Transformers

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Consider an ideal transformer with N1=3000N_{1}=3000 and N2=500N_{2}=500 turns. Let winding 1 be connected to a source whose voltage is e1(t)=100(1t)e_{1}(t)=100(1-|t|) volts for 1t1-1 \leq t \leq 1 and e1(t)=0e_{1}(t)=0 for t>1|t|>1 second. A 3 -farad capacitor is connected across winding 2 . Sketch e1(t),e2(t),i1(t)e_{1}(t), e_{2}(t), i_{1}(t) , and i2(t)i_{2}(t) versus time tt .

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Choosing system bases to be 360/24 kV and 100 MVA, redraw the per-unit equivalent circuit for Test Bank Problem 3.9

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Xpu new =(0.09)(345360)2(100300)=0.02755\quad X_{\text {pu new }}=(0.09)\left(\frac{345}{360}\right)^{2}\left(\frac{100}{300}\right)=0.02755 per unit
\quad X_{\text {pu new }}=(0.09)\left(\frac{345}{360}\right)^{2}\left(\frac{100}{300}\right)=0.02755  per unit      Positive Sequence
Positive Sequence

For the power system in Test Bank Problem 3.11, the synchronous motor absorbs 1200 MW at 0.8 power factor leading with the bus 3 voltage at 18kV18 \mathrm{kV} . Determine the bus 1 and bus 2 voltages in kV\mathrm{kV} . Assume that generators 1 and 2 deliver equal real powers and equal reactive powers. Also assume a balanced three-phase system with positive-sequence sources.

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Vˉ3pu=18200=0.90 per unit Iˉ3=1200(3)(18)(0.8)cos1(.8)=48.1136.87kAIbase X=1003(20)=2.887kAIˉ3pu=48.1136.872.887=16.6736.87 per unit Vˉ1pu=Vˉ2pu=Vˉ3pu+Iˉ3pu(j×T5pu)+12Iˉ3pu(jXLine 25pu+jXT3pu)Vˉ1pu=Vˉ2pu=0.90+16.6736.87(j)(.006666+.01+.013332)=.9+0.30555+126.87=0.7167+j0.2444=0.757218.83puV1=V2=(0.7572)(20)=15.14kV\begin{aligned}\bar{V}_{3 p u} & =\frac{18}{20} \angle 0^{\circ}=0.9 \angle 0^{\circ} \text { per unit } \\\bar{I}_{3} & =\frac{1200}{(\sqrt{3})(18)(0.8)} \angle \cos ^{-1}(.8)=48.11 \angle 36.87^{\circ} \mathrm{kA} \\I_{\text {base } X} & =\frac{100}{\sqrt{3}(20)}=2.887 \mathrm{kA} \\\bar{I}_{3 p u} & =\frac{48.11 \angle 36.87^{\circ}}{2.887}=16.67 \angle 36.87^{\circ} \text { per unit } \\\bar{V}_{1 p u} & =\bar{V}_{2 p u}=\bar{V}_{3 p u}+\bar{I}_{3 p u}\left(j \times_{T 5 p u}\right)+\frac{1}{2} \bar{I}_{3 p u}\left(j X_{\text {Line } 25 p u}+j X_{T 3 p u}\right) \\\bar{V}_{1 p u} & =\bar{V}_{2 p u}=0.9 \angle 0^{\circ}+16.67 \angle 36.87^{\circ}(j)\left(.006666+\frac{.01+.01333}{2}\right) \\& =.9+0.30555+\angle 126.87^{\circ} \\& =0.7167+j 0.2444=0.7572 \angle 18.83^{\circ} \mathrm{pu} \\V_{1} & =V_{2}=(0.7572)(20)=15.14 \mathrm{kV}\end{aligned}

A single-phase 15kVA,2400/24015-\mathrm{kVA}, 2400 / 240 -volt, 60Hz60-\mathrm{Hz} two-winding distribution transformer is connected as an autotransformer to step up the voltage from 2400 to 2640 volts. (a) Draw a schematic diagram of this arrangement, showing all voltages and currents when delivering full load at rated voltage. (b) Find the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the rated values as a two-winding transformer. How much of this kVA rating is transformed by magnetic induction? (c) The following data are obtained from tests carried out on the transformer when it is connected as a two-winding transformer: Open-circuit test with the low-voltage terminals excited: Applied voltage =240 V=240 \mathrm{~V} , Input current =0.68 A=0.68 \mathrm{~A} , Input power =105 W=105 \mathrm{~W} . Short-circuit test with the high-voltage terminals excited: Applied voltage =120 V=120 \mathrm{~V} , Input current =6.25 A=6.25 \mathrm{~A} , Input power =330 W=330 \mathrm{~W} . Based on the data, compute the efficiency of the autotransformer corresponding to full load, rated voltage, and 0.8 power factor lagging. Comment on why the efficiency is higher as an autotransformer than as a two-winding transformer.

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A single-phase 100kVA,2400/240volt,60Hz100-\mathrm{kVA}, 2400 / 240-\mathrm{volt}, 60-\mathrm{Hz} distribution transformer is used as a step-down transformer. The load, which is connected to the 240 -volt secondary winding, absorbs 80kVA80 \mathrm{kVA} at 0.8 power factor lagging and is at 230 volts. Assuming an ideal transformer, calculate the following: (a) primary voltage, (b) load impedance, (c) load impedance referred to the primary, and (d) the real and reactive power supplied to the primary winding.

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The following data are obtained when open-circuit and short-circuit tests are performed on a single-phase, 50kVA,2400/240volt50-\mathrm{kVA}, 2400 / 240-\mathrm{volt} , 60Hz60-\mathrm{Hz} distribution transformer.  The following data are obtained when open-circuit and short-circuit tests are performed on a single-phase,  50-\mathrm{kVA}, 2400 / 240-\mathrm{volt} ,  60-\mathrm{Hz}  distribution transformer.     (a) Neglecting the series impedance, determine the exciting admittance referred to the high-voltage side. (b) Neglecting the exciting admittance, determine the equivalent series impedance referred to the high-voltage side. (c) Assuming equal series impedances for the primary and referred secondary, obtain an equivalent T-circuit referred to the high-voltage side. (a) Neglecting the series impedance, determine the exciting admittance referred to the high-voltage side. (b) Neglecting the exciting admittance, determine the equivalent series impedance referred to the high-voltage side. (c) Assuming equal series impedances for the primary and referred secondary, obtain an equivalent T-circuit referred to the high-voltage side.

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The ratings of a three-phase three-winding transformer are: Primary (1): \quad Y connected, 66kV,2066 \mathrm{kV}, 20 MVA Secondary (2): \quad Y connected, 13.2kV,1513.2 \mathrm{kV}, 15 MVA Tertiary (3): Δ\quad \Delta connected, 2.3kV,52.3 \mathrm{kV}, 5 MVA Neglecting winding resistances and exciting current, the per-unit leakage reactances are: =0.08 on a 20-,66- base =0.10 on a 20-,66- base =0.09 on a 15-,13.2- base (a) Determine the per-unit reactances X1,X2,X3\mathrm{X}_{1}, \mathrm{X}_{2}, \mathrm{X}_{3} of the equivalent circuit on a 20-MVA, 66kV66-\mathrm{kV} base at the primary terminals. (b) Purely resistive loads of 12MW12 \mathrm{MW} at 13.2kV13.2 \mathrm{kV} and 5MW5 \mathrm{MW} at 2.3kV2.3 \mathrm{kV} are connected to the secondary and tertiary sides of the transformer, respectively. Draw the per-unit impedance diagram, showing the per-unit impedances on a 20MVA,66kV20-\mathrm{MVA}, 66-\mathrm{kV} base at the primary terminals.

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A single-phase three-winding transformer has the following parameters: Z1=Z2=Z_{1}=Z_{2}= Z3=0+j0.06,Gc=0Z_{3}=0+j 0.06, \mathrm{G}_{c}=0 , and Bm=0.2\mathrm{B}_{m}=0.2 per unit. Three identical transformers, as described, are connected with their primaries in Y\mathrm{Y} (solidly grounded neutral) and with their secondaries and tertiaries in Δ\Delta . Draw the per-unit sequence networks of this transformer bank.

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Consider a bank of three single-phase two-winding transformers whose high-voltage terminals are connected to a three-phase, 13.8kV13.8-\mathrm{kV} feeder. The low-voltage terminals are connected to a three-phase substation load rated 2.4 MVA and 2.3kV2.3 \mathrm{kV} . Determine the required voltage, current, and MVA ratings of both windings of each transformer, when the high-voltage/low-voltage windings are connected (a) YΔ\mathrm{Y}-\Delta , (b) ΔY\Delta-\mathrm{Y} , (c) YY\mathrm{Y}-\mathrm{Y} , and (d) ΔΔ\Delta-\Delta .

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A single-phase 100kVA,2400/240volt,60Hz100-\mathrm{kVA}, 2400 / 240-\mathrm{volt}, 60-\mathrm{Hz} distribution transformer has a 2ohm2-\mathrm{ohm} equivalent leakage reactance and a 6000 -ohm magnetizing reactance referred to the high-voltage side. If rated voltage is applied to the high-voltage winding, calculate the open-circuit secondary voltage. Neglect I2R\mathrm{I}^{2} \mathrm{R} and Gc2 V\mathrm{G}_{c}^{2} \mathrm{~V} losses. Assume equal series leakage reactances for the primary and referred secondary.

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For a conceptual single-phase, phase-shifting transformer, the primary voltage leads the secondary voltage by 3030^{\circ} . A load connected to the secondary winding absorbs 50kVA50 \mathrm{kVA} at 0.9 power factor leading and at a voltage E2=2770E_{2}=277 \angle 0^{\circ} volts. Determine (a) the primary voltage, (b) primary and secondary currents, (c) load impedance referred to the primary winding, and (d) complex power supplied to the primary winding.

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Consider the single-line diagram of the power system shown in Figure P3.11. Equipment rating are:} generator 1: 750,18,=0.2 per unit generator 2: 750,18,=0.2 synchronous motor 3: 1500,20,=0.2 3-phase \Delta- transformers ,,,: 750,500/20\Delta,=0.1 3-phase Y-Y transformer : 1500,500/20,=0.1 Neglecting resistance, transformer phase shift, and magnetizing reactance, draw the equivalent reactance diagram. Use a base of 100MVA100 \mathrm{MVA} and 500kV500 \mathrm{kV} for the 40 -ohm line. Determine the per-unit reactances.  Consider the single-line diagram of the power system shown in Figure P3.11. Equipment rating are:}  \begin{array}{ll} \text { generator 1: } & 750 \mathrm{MVA}, 18 \mathrm{kV}, \mathrm{X}^{\prime \prime}=0.2 \text { per unit } \\ \text { generator 2: } & 750 \mathrm{MVA}, 18 \mathrm{kV}, \mathrm{X}^{\prime \prime}=0.2 \\ \text { synchronous motor 3: } & 1500 \mathrm{MVA}, 20 \mathrm{kV}, \mathrm{X}^{\prime \prime}=0.2 \\ \begin{array}{l} \text { 3-phase } \Delta-\mathrm{Y} \text { transformers } \\ \mathrm{T}_{1}, \mathrm{~T}_{2}, \mathrm{~T}_{3}, \mathrm{~T}_{4}: \end{array} & 750 \mathrm{MVA}, 500 \mathrm{kV} \mathrm{Y} / 20 \mathrm{kV} \Delta, \mathrm{X}=0.1 \\ \text { 3-phase Y-Y transformer } \mathrm{T}_{5}: & 1500 \mathrm{MVA}, 500 \mathrm{kV} \mathrm{Y} / 20 \mathrm{kV} \mathrm{Y}, \mathrm{X}=0.1 \end{array}  Neglecting resistance, transformer phase shift, and magnetizing reactance, draw the equivalent reactance diagram. Use a base of  100 \mathrm{MVA}  and  500 \mathrm{kV}  for the 40 -ohm line. Determine the per-unit reactances.     FIGURE P3.11 FIGURE P3.11

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Figure P3.15 shows a one-line diagram of a system in which the three-phase generator is rated 300 MVA, 20kV20 \mathrm{kV} with a subtransient reactance of 0.2 per unit and with its neutral grounded through a 0.4Ω0.4-\Omega reactor. The transmission line is 64 km64 \mathrm{~km} long with a series reactance of 0.5Ω/km0.5 \Omega / \mathrm{km} . The three-phase transformer T1T_{1} is rated 350MVA,230/350 \mathrm{MVA}, 230 / 20kV20 \mathrm{kV} with a leakage reactance of 0.1 per unit. Transformer T2T_{2} is composed of three single-phase transformers, each rated 100MVA,127/13.2kV100 \mathrm{MVA}, 127 / 13.2 \mathrm{kV} with a leakage reactance of 0.1 per unit. Two 13.2kV13.2-\mathrm{kV} motors M1M_{1} and M2M_{2} with a subtransient reactance of 0.2 per unit for each motor represent the load. M1M_{1} has a rated input of 200 MVA with its neutral grounded through a 0.4Ω0.4-\Omega current-limiting reactor. M2M_{2} has a rated input of 100 MVA with its neutral not connected to ground. Neglect phase shifts associated with the transformers. Choose the generator rating as base in the generator circuit and draw the positive-sequence reactance diagram showing all reactances in per unit.  Figure P3.15 shows a one-line diagram of a system in which the three-phase generator is rated 300 MVA,  20 \mathrm{kV}  with a subtransient reactance of 0.2 per unit and with its neutral grounded through a  0.4-\Omega  reactor. The transmission line is  64 \mathrm{~km}  long with a series reactance of  0.5 \Omega / \mathrm{km} . The three-phase transformer  T_{1}  is rated  350 \mathrm{MVA}, 230 /   20 \mathrm{kV}  with a leakage reactance of 0.1 per unit. Transformer  T_{2}  is composed of three single-phase transformers, each rated  100 \mathrm{MVA}, 127 / 13.2 \mathrm{kV}  with a leakage reactance of 0.1 per unit. Two  13.2-\mathrm{kV}  motors  M_{1}  and  M_{2}  with a subtransient reactance of 0.2 per unit for each motor represent the load.  M_{1}  has a rated input of 200 MVA with its neutral grounded through a  0.4-\Omega  current-limiting reactor.  M_{2}  has a rated input of 100 MVA with its neutral not connected to ground. Neglect phase shifts associated with the transformers. Choose the generator rating as base in the generator circuit and draw the positive-sequence reactance diagram showing all reactances in per unit.     Figure P3.15 Figure P3.15

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Three single-phase transformers, each rated 10MVA,66.4/12.5kV,60 Hz10 \mathrm{MVA}, 66.4 / 12.5 \mathrm{kV}, 60 \mathrm{~Hz} , with an equivalent series reactance of 0.12 per unit divided equally between primary and secondary, are connected in a three-phase bank. The high-voltage windings are Y connected and their terminals are directly connected to a 115kV115-\mathrm{kV} three-phase bus. The secondary terminals are all shorted together. Find the currents entering the high-voltage terminals and leaving the low-voltage terminals if the low-voltage windings are (a) Y\mathrm{Y} connected, (b) Δ\Delta connected.

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A 100-MVA, 13.2-kV three-phase generator, which has a positive-sequence reactance of 1.2 per unit on the generator base, is connected to a 110 -MVA, 13.2Δ/115YkV13.2 \Delta / 115 \mathrm{Y}-\mathrm{kV} step-up transformer with a series impedance of (0.005+j0.1)(0.005+j 0.1) per unit on its own base. (a) Calculate the per-unit generator reactance on the transformer base. (b) The load at the transformer terminals is 80MW80 \mathrm{MW} at unity power factor and at 115kV115 \mathrm{kV} . Choosing the transformer high-side voltage as the reference phasor, draw a phasor diagram for this condition. (c) For the condition of part (b), find the transformer low-side voltage and the generator internal voltage behind its reactance. Also compute the generator output power and power factor.

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A balanced Y-connected voltage source with Eag=2770E_{a g}=277 \angle 0^{\circ} volts is applied to a balanced- YY load in parallel with a balanced- Δ\Delta load, where ZY=30+j10Z_{Y}=30+j 10 and ZΔ=Z_{\Delta}= 45j2545-j 25 ohms. The Y\mathrm{Y} load is solidly grounded. Using base values of Sbasel ϕ=5kVA\mathrm{S}_{\text {basel } \phi}=5 \mathrm{kVA} and VbaseLN =277\mathrm{V}_{\text {baseLN }}=277 volts, calculate the source current IaI_{a} in per-unit and in amperes.

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Using the transformer ratings as base quantities, work Test Bank Problem 3.5 in per unit.

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The leakage reactance of a three-phase, 500 -MVA, 345Y/23Δ345 \mathrm{Y} / 23 \Delta -kV transformer is 0.09 per unit based on its own ratings. The Y\mathrm{Y} winding has a solidly grounded neutral. Draw the per-unit equivalent circuit. Neglect the exciting admittance and assume American standard phase shift.

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