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systems analysis design

Questions and Answers of Systems Analysis Design

Reconsider Problem 8.8 and choosing \(V_{b c}\) as the reference, show that\[V_{b c, 0}=0 ; \quad V_{b c, 1}=-j \sqrt{3} V_{a, 1} ; \quad V_{b c, 2}=j \sqrt{3} V_{a, 2}\]Problem 8.8Let an unbalanced,
Given the line-to-ground voltages \(V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}\), and \(V_{c g}=290 \angle 130^{\circ}\) volts, calculate (a) the sequence components of the
A balanced \(\Delta\)-connected load is fed by a three-phase supply for which phase \(\mathrm{C}\) is open and phase \(\mathrm{A}\) is carrying a current of \(10 \angle 0^{\circ} \mathrm{A}\). Find
A Y-connected load bank with a three-phase rating of \(500 \mathrm{kVA}\) and \(2300 \mathrm{~V}\) consists of three identical resistors of \(10.58 \Omega\). The load bank has the following applied
The currents in a \(\Delta\) load are \(I_{a b}=10 \angle 0^{\circ}, I_{b c}=12 \angle-90^{\circ}\), and \(I_{c a}=15 \angle 90^{\circ} \mathrm{A}\). Calculate (a) the sequence components of the
The voltages given in Problem 8.10 are applied to a balanced-Y load consisting of \((12+j 16)\) ohms per phase. The load neutral is solidly grounded. Draw the sequence networks and calculate \(I_{0},
Repeat Problem 8.14 with the load neutral open.Problem 8.14Given the line-to-ground voltages \(V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}\), and \(V_{c g}=290 \angle 130^{\circ}\)
Repeat Problem 8.14 for a balanced- \(\Delta\) load consisting of \((12+j 16)\) ohms per phase.Problem 8.14Given the line-to-ground voltages \(V_{a g}=280 \angle 0^{\circ}, V_{b g}=250
Repeat Problem 8.14 for the load shown in Example 8.4.Problem 8.14Given the line-to-ground voltages \(V_{a g}=280 \angle 0^{\circ}, V_{b g}=250 \angle-110^{\circ}\), and \(V_{c g}=290 \angle
Perform the indicated matrix multiplications in (8.2.21) and verify the sequence impedances given by (8.2.22) through (8.2.27).Eq. (8.2.21)Eq. (8.2.22)Eq. (8.2.27) Zo Zab Zac 1 1 a Zbb Zbc 1 3 Z Z
The following unbalanced line-to-ground voltages are applied to the balanced-Y load shown in Figure 3.3: \(V_{a g}=100 \angle 0^{\circ}, V_{b g}=75 \angle 180^{\circ}\), and \(V_{c g}=50 \angle
(a) Consider three equal impedances of (j27) \(\Omega\) connected in \(\Delta\). Obtain the sequence networks.(b) Now, with a mutual impedance of (j6) \(\Omega\) between each pair of adjacent
The three-phase impedance load shown in Figure 8.7 has the following phase impedance matrix:Determine the sequence impedance matrix \(Z_{\mathrm{S}}\) for this load. Is the load symmetrical?Figure
The three-phase impedance load shown in Figure 8.7 has the following sequence impedance matrix:Determine the phase impedance matrix \(Z_{\mathrm{P}}\) for this load. Is the load symmetrical?Figure
Consider a three-phase balanced Y-connected load with self and mutual impedances as shown in Figure 8.23. Let the load neutral be grounded through an impedance \(Z_{n}\). Using Kirchhoff's laws,
A three-phase balanced voltage source is applied to a balanced Y-connected load with ungrounded neutral. The Y-connected load consists of three mutually coupled reactances, where the reactance of
A three-phase balanced Y-connected load with series impedances of \((6+j 24) \Omega\) per phase and mutual impedance between any two phases of \(j 3 \Omega\) is supplied by a three-phase unbalanced
Repeat Problem 8.14 but include balanced three-phase line impedances of \((3+j 4)\) ohms per phase between the source and load.Problem 8.14The voltages given in Problem 8.10 are applied to a
Consider the flow of unbalanced currents in the symmetrical three-phase line section with neutral conductor as shown in Figure 8.24. (a) Express the voltage drops across the line conductors given by
Let the terminal voltages at the two ends of the line section shown in Figure 8.24 be given byThe line impedances are given by:\[Z_{\mathrm{aa}}=j 60 \Omega \quad Z_{\mathrm{ab}}=j 20 \Omega \quad
A completely transposed three-phase transmission line of \(200 \mathrm{~km}\) in length has the following symmetrical sequence impedances and sequence admittances:\[\begin{aligned}& Z_{1}=Z_{2}=j 0.5
As shown in Figure 8.25, a balanced three-phase, positive-sequence source with \(V_{\mathrm{AB}}=480 \angle 0^{\circ}\) volts is applied to an unbalanced \(\Delta\) load. Note that one leg of the
A balanced Y-connected generator with terminal voltage \(V_{b c}=200 \angle 0^{\circ}\) volts is connected to a balanced- \(\Delta\) load whose impedance is \(10 / 40^{\circ} \mathrm{ohms}\) per
In a three-phase system, a synchronous generator supplies power to a 200 -volt synchronous motor through a line having an impedance of \(0.5 ot 80^{\circ}\) ohm per phase. The motor draws \(5
Calculate the source currents in Example 8.6 without using symmetrical components. Compare your solution method with that of Example 8.6. Which method is easier?Example 8.6A Y-connected voltage
A Y-connected synchronous generator rated \(20 \mathrm{MVA}\) at \(13.8 \mathrm{kV}\) has a positive-sequence reactance of \(j 2.38 \Omega\), negative-sequence reactance of \(j 3.33 \Omega\), and
Figure 8.26 shows a single-line diagram of a three-phase, interconnected generator-reactor system, in which the given per-unit reactances are based on the ratings of the individual pieces of
Consider Figures 8.13 and 8.14 of the text with reference to a Y-connected synchronous generator (grounded through a neutral impedance \(Z_{n}\) ) operating at no load. For a line-to-ground fault
Reconsider the synchronous generator of Problem 8.36. Obtain sequencenetwork representations for the following fault conditions.(a) A short-circuit between phases \(b\) and \(c\).(b) A double
Three single-phase, two-winding transformers, each rated 450 MVA, \(20 \mathrm{kV} / 288.7 \mathrm{kV}\), with leakage reactance \(X_{\text {eq }}=0.12\) per unit, are connected to form a three-phase
The leakage reactance of a three-phase, 500-MVA, \(345 \mathrm{Y} / 23 \Delta-\mathrm{kV}\) transformer is 0.09 per unit based on its own ratings. The \(\mathrm{Y}\) winding has a solidly grounded
Choosing system bases to be \(360 / 24 \mathrm{kV}\) and 100 MVA, redraw the sequence networks for Problem 8.39.Problem 8.39The leakage reactance of a three-phase, 500-MVA, \(345 \mathrm{Y} / 23
Draw the zero-sequence reactance diagram for the power system shown in Figure 3.38. The zero-sequence reactance of each generator and of the synchronous motor is 0.05 per unit based on equipment
Three identical Y-connected resistors of \(1.0 \angle 0^{\circ}\) per unit form a load bank that is supplied from the low-voltage Y-side of a Y- \(\Delta\) transformer. The neutral of the load is not
Draw the positive-, negative-, and zero-sequence circuits for the transformers shown in Figure 3.34. Include ideal phase-shifting transformers showing phase shifts. Assume that all windings have the
For Problem 8.14, calculate the real and reactive power delivered to the three-phase load.Problem 8.14Given the line-to-ground voltages \(V_{a g}=280 \angle 0^{\circ}, V_{b g}=250
A three-phase impedance load consists of a balanced- \(\Delta\) load in parallel with a balanced-Y load. The impedance of each leg of the \(\Delta\) load is \(Z_{\Delta}=\) \(6+j 6 \Omega\), and the
For Problem 8.12, compute the power absorbed by the load using symmetrical components. Then verify the answer by computing directly without using symmetrical components.Problem 8.12A Y-connected load
For Problem 8.25, determine the complex power delivered to the load in terms of symmetrical components. Verify the answer by adding up the complex power of each of the three phases.Problem 8.25A
Using the voltages of Problem 8.6(a) and the currents of Problem 8.5, compute the complex power dissipated based on(a) phase components and(b) symmetrical components.Problem 8.6(a)Given the
For power-system fault studies, it is assumed that the system is operating under balanced steady-state conditions prior to the fault, and sequence networks are uncoupled before the fault occurs.(a)
The first step in power-system fault calculations is to develop sequence networks based on the single-line diagram of the system, and then reduce them to their Thévenin equivalents, as viewed from
When calculating symmetrical three-phase fault currents, only _________ sequence network needs to be considered.
In order of frequency of occurrence of short-circuit faults in three-phase power systems, list those: _________, _________, _________, _________.
For a bolted three-phase-to-ground fault, sequence-fault currents _________ are zero, sequence fault voltages are _________, and line-to-ground voltages are _________.
For a single-line-to-ground fault with a fault-impedance \(Z_{\mathrm{F}}\), the sequence networks are to be connected _________ at the fault terminals through the impedance ; the sequence components
For a line-to-line fault with a fault impedance \(Z_{\mathrm{F}}\), the positive-and negative-sequence networks are to be connected _________ at the fault terminals through the impedance of \(1 / 2 /
For a double line-to-ground fault through a fault impedance \(Z_{\mathrm{F}}\), the sequence networks are to be connected _________ at the fault terminal; additionally, _________ is to be included in
The sequence bus-impedance matrices can also be used to calculate fault currents and voltages for symmetrical as well as unsymmetrical faults by representing each sequence network as a bus-impedance
The rms value of \(v(t)=\mathrm{V}_{\text {max }} \cos (\omega t+\delta)\) is given by(a) \(\mathrm{V}_{\max }\)(b) \(\mathrm{V}_{\max } / \sqrt{2}\)(c) \(2 \mathrm{~V}_{\max }\)(d) \(\sqrt{2}
If the rms phasor of a voltage is given by \(\mathrm{V}=120 / 60^{\circ}\) volts, then the corresponding \(v(t)\) is given by(a) \(120 \sqrt{2} \cos \left(\omega t+60^{\circ}ight)\)(b) \(120 \cos
If a phasor representation of a current is given by \(I=70.7 \angle 45^{\circ} \mathrm{A}\), it is equivalent to(a) \(100 e^{j 45^{\circ}}\)(b) \(100+j 100\)(c) \(50+j 50\)
With sinusoidal-steady-state excitation, for a purely resistive circuit, the voltage and current phasors are(a) In phase(b) Perpendicular with each other with \(V\) leading \(I\) (v) Perpendicular
For a purely inductive circuit, with sinusoidal-steady-state excitation, the voltage and current phasors are(a) In phase(b) Perpendicular to each other with \(V\) leading \(I\)(c) Perpendicular to
For a purely capacitive circuit, with sinusoidal-steady-state excitation, the voltage and current phasors are(a) In phase(b) Perpendicular to each other with \(V\) leading \(I\)(c) Perpendicular to
With sinusoidal-steady-state excitation, the average power in a singlephase ac circuit with a purely resistive load is given by(a) \(\mathrm{I}_{\mathrm{rms}}^{2} R\)(b) \(\mathrm{V}_{\max }^{2} /
The average power in a single-phase ac circuit with a purely inductive load, for sinusoidal-steady-state excitation, is(a) \(\mathrm{I}^{2} X_{\mathrm{L}}\)(b) \(\mathrm{V}_{\max }^{2} /
The average power in a single-phase ac circuit with a purely capacitive load, for sinusoidal-steady-state excitation, is(a) Zero(b) \(\mathrm{V}_{\max }^{2} / X_{\mathrm{C}}\)(c)
The average value of a double-frequency sinusoid, \(\sin 2(\omega t+\delta)\), is given by(a) 1(b) \(\delta\)(c) Zero
The power factor for an inductive circuit ( \(R\) - \(L\) load), in which the current lags the voltage, is said to be(a) Lagging(b) Leading(c) Zero
The power factor for a capacitive circuit ( \(R-C\) load), in which the current leads the voltage, is said to be(a) Lagging(b) Leading(c) One
In a single-phase ac circuit, for a general load composed of RLC elements under sinusoidal-steady-state excitation, the average reactive power is given by(a) Vrms IrmscosϕVrms Irmscos⁡ϕ(b)
The instantaneous power absorbed by the load in a single-phase ac circuit, for a general RLC load under sinusoidal-steady-state excitation, is(a) Nonzero constant(b) Zero(c) Containing
With load convention, where the current enters the positive terminal of the circuit element, if \(\Omega\) is positive then positive reactive power is absorbed.(a) True(b) False
With generator convention, where the current leaves the positive terminal of the circuit element, if \(P\) is positive then positive real power is delivered.(a) False(b) True
Consider the load convention that is used for the RLC elements shown in Figure 2.2 of the text.A. If one says that an inductor absorbs zero real power and positive reactive power, is it(a) True(b)
In an ac circuit, power factor improvement is achieved by(a) Connecting a resistor in parallel with the inductive load.(b) Connecting an inductor in parallel with the inductive load.(c) Connecting a
The admittance of the impedance \(-j \frac{1}{2} \Omega\) is given by(a) \(-j 2 S\)(b) \(j 2 S^{2}\)(c) \(-j 4 S\)
Consider Figure 2.9 of the text. Let the nodal equations in matrix form be given by Eq. (2.4.1) of the text.A. The element Y11Y11 is given by(a) 0(b) j13j13(c) −j7−j7B. The element Y31Y31 is
The three-phase source line-to-neutral voltages are given by \(E_{a n}=10 \angle 0^{\circ}\), \(E_{b n}=10 \angle+240^{\circ}\), and \(E_{c n}=10 \angle-240^{\circ}\) volts.Is the source balanced?(a)
In a balanced three-phase Y-connected system with a positive-sequence source, the line-to-line voltages are \(\sqrt{3}\) times the line-to-neutral voltages and lend by \(30^{\circ}\).(a) True(b) False
In a balanced system, the phasor sum of the line-to-line voltages and the phasor sum of the line-to-neutral voltages are always equal to zero.(a) False(b) True
Consider a three-phase Y-connected source feeding a balanced- \(\Delta\) load. The phasor sum of the line currents as well as the neutral current are always zero.(a) True(b) False
For a balanced- \(\Delta\) load supplied by a balanced positive-sequence source, the line currents into the load are \(\sqrt{3}\) times the \(\Delta\)-load currents and lag by \(30^{\circ}\).(a)
A balanced \(\Delta\)-load can be converted to an equivalent balanced- \(Y\) load by dividing the \(\Delta\)-load impedance by(a) \(\sqrt{3}\)(b) 3(c) \(1 / 3\)
When working with balanced three-phase circuits, per-phase analysis is commonly done after converting \(\Delta\) loads to Y loads, thereby solving only one phase of the circuit.(a) True(b) False
The total instantaneous power delivered by a three-phase generator under balanced operating conditions is(a) A function of time(b) A constant
The total instantaneous power absorbed by a three-phase motor (under balanced steady-state conditions) as well as a balanced three-phase impedance load is(a) A constant(b) A function of time
Under balanced operating conditions, consider the three-phase complex power delivered by the three-phase source to the three-phase load. Match the following expressions, those on the left to those on
One advantage of balanced three-phase systems over separate singlephase systems is reduced capital and operating costs of transmission and distribution.(a) True(b) False
While the instantaneous electric power delivered by a single-phase generator under balanced steady-state conditions is a function of time having two components of a constant and a double-frequency
Given the complex numbers \(A_{1}=6 / 30\) and \(A_{2}=4+j 5\), (a) convert \(A_{1}\) to rectangular form; (b) convert \(A_{2}\) to polar and exponential form; (c) calculate
Convert the following instantaneous currents to phasors, using \(\cos (\omega t)\) as the reference. Give your answers in both rectangular and polar form.(a) \(i(t)=500 \sqrt{2} \cos (\omega
The instantaneous voltage across a circuit element is \(v(t)=\) \(400 \sin \left(\omega t+30^{\circ}ight)\) volts, and the instantaneous current entering the positive terminal of the circuit element
For the single-phase circuit shown in Figure 2.22, \(I=10 \angle 0^{\circ} \mathrm{A}\). (a) Compute the phasors \(I_{1}, I_{2}\), and \(V\). (b) Draw a phasor diagram showing \(I, I_{1}\),
A \(60-\mathrm{Hz}\), single-phase source with \(V=277 ot 30^{\circ}\) volts is applied to a circuit element.(a) Determine the instantaneous source voltage. Also determine the phasor and
(a) Transform \(v(t)=75 \cos \left(377 t-15^{\circ}ight)\) to phasor form. Comment on whether \(\omega=377\) appears in your answer. (b) Transform \(V=50 / 10^{\circ}\) to instantaneous form. Assume
Let a \(100-\mathrm{V}\) sinusoidal source be connected to a series combination of a \(3-\Omega\) resistor, an \(8-\Omega\) inductor, and a \(4-\Omega\) capacitor.(a) Draw the circuit diagram.(b)
Consider the circuit shown in Figure 2.23 in time domain. Convert the entire circuit into phasor domain. v(t) = 120v2cos(377t-30) V R = 5.76 m2 L = 30.6 H C= : 921 uF R = 59 LL = 5 mH
For the circuit shown in Figure 2.24, compute the voltage across the load terminals. + 0.192 /0.592 120/0 V 1 = 60/0 A VLOAD LOAD
For the circuit element of Problem 2.3, calculate(a) the instantaneous power absorbed,(b) the real power (state whether it is delivered or absorbed),(c) the reactive power (state whether delivered or
Referring to Problem 2.5, determine the instantaneous power, real power, and reactive power absorbed by(a) the \(20-\Omega\) resistor,(b) the \(10-\mathrm{mH}\) inductor,(c) the capacitor with
The voltage \(v(t)=359.3 \cos (\omega t)\) volts is applied to a load consisting of a 10- resistor in parallel with a capacitive reactance \(X_{\mathrm{C}}=25 \Omega\). Calculate(a) the instantaneous
Repeat Problem 2.12 if the resistor and capacitor are connected in series.Problem 2.12The voltage \(v(t)=359.3 \cos (\omega t)\) volts is applied to a load consisting of a 10- resistor in parallel
A single-phase source is applied to a two-terminal, passive circuit with equivalent impedance \(Z=3.0 \angle-45^{\circ} \Omega\), measured from the terminals. The source current is \(i(t)=2 \sqrt{2}
Let a voltage source \(v(t)=4 \cos \left(\omega t+60^{\circ}ight)\) be connected to an impedance \(Z=2 / 30^{\circ} \Omega\). (a) Given the operating frequency to be \(60 \mathrm{~Hz}\), determine
A single-phase, \(120-\mathrm{V}\) (rms), \(60-\mathrm{Hz}\) source supplies power to a series \(\mathrm{R}\) - \(\mathrm{L}\) circuit consisting of \(\mathrm{R}=10 \Omega\) and \(\mathrm{L}=40
Consider a load impedance of \(Z=j \omega L\) connected to a voltage and \(V\) let the current drawn be \(I\).(a) Develop an expression for the reactive power \(\mathrm{Q}\) in terms of \(\omega,
Let a series RLC network be connected to a source voltage \(V\), drawing a current \(I\).(a) In terms of the load impedance \(Z=Z ot Z\), find expressions for \(\mathrm{P}\) and \(\mathrm{Q}\), from
Consider a single-phase load with an applied voltage \(v(t)=150 \cos (\omega t+\) \(\left.10^{\circ}ight)\) volts and load current \(i(t)=5 \cos \left(\omega t-50^{\circ}ight) \mathrm{A}\). (a)

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