IGEE 402 Power System Analysis. FINAL EXAMINATION - SAMPLE Fall 2004

Transcription

1 IGEE 402 Power System Analysis FINAL EXAMINATION - SAMPLE Fall 2004 Special instructions: - Duration: 80 minutes. - Material allowed: a crib sheet (double sided 8.5 x ), calculator. - Attempt 5 out of 7 questions. - Make any reasonable assumption. - All questions and sub-questions carry equal weight. QUESTION (20 points) A single phase 550 V, 60 Hz feeder supplies 2 loads in parallel: (a) a 20 kw heating load; (b) a 50 kw motor load, with an efficiency of 0.90 and a power factor of 0.80 (lagging). (a) Draw the instantaneous current and voltage waveforms for the total load, indicating peak and rms values. Give the equation for the total instantaneous power and draw the waveform. Indicate the value of the total real power. (b) Draw the V-I vector diagram and the power diagram. Compute total reactive power and power factor. (b) A capacitor is connected in parallel with the loads to increase the power factor to.0. Compute the reactive power required from the capacitor. Redraw the power diagram. (c) Explain the benefits in having a unity power factor from (i) the utility point of view; (ii) the consumer point of view. QUESTION 2 (20 points) A single phase 24 kv/2.4 kv, 60 Hz transformer, with a series equivalent impedance of j.0 Ω referred to the low-voltage side, is connected at the end of a transmission line of impedance 50 + j400 Ω. The transformer feeds a load drawing 200 kw, at a 0.8 power factor lagging, and 2300 V. (a) The feeder voltage is adjusted so that the load voltage is 2300 V. Assume the load voltage has an angle of 0 deg. Find the amplitude and angle of the voltage at the sending end of the transmission line. (Hint: refer all quantities to the low voltage side). (b) Compute the current in the transmission line, the losses in the transmission line and transformer, and the overall system efficiency. (c) Calculate the no-load voltage and the voltage regulation. (d) Explain how the voltage regulation can be reduced. (Hint: consider shunt compensation). QUESTION 3 (20 points) A three-phase feeder supplies three identical single-phase 35 kv/6.9 kv, 60 Hz, 00 MVA transformers, each with a leakage reactance of 0.5 pu on the transformer base, and connected in a

2 IGEE 402 Power System Analysis FINAL EXAMINATION SAMPLE Joós, G. -Y connection. The load is a three-phase Y-connected 2 kv load, with per phase parameters of 60 MW, 0.85 power factor (lagging). A capacitor bank of three elements of impedance (-j5 Ω) connected in is used for power factor correction. The secondary voltage is assumed to be 2 kv. (a) Compute the total power and total reactive power drawn from the feeder. (b) Draw the single line pu circuit, using a 300 MVA (three-phase), 2 kv base. (c) Compute the total line current, the total power and total reactive power drawn from the feeder, in pu. Compute the voltage at the feeder. Compare with results in (a). (d) Phase c of the load becomes an open circuit. In the original V-A three-phase circuit, compute the current in the lines feeding the load, when (i) the load neutral is solidly grounded; (ii) the load neutral is floating. QUESTION 4 (20 points) A 400 km, 38 kv, 60 Hz transmission line has the following distributed parameters: r = 0.06 Ω/km, x = Ω/km, y = j3.36 x 0-6 S/km. Losses are neglected. (a) Compute the nominal π equivalent circuit parameters and draw the circuit. Compute the corresponding ABCD parameters. (b) Find the surge impedance and surge impedance loading. (c) The line delivers 40 MW at 32 kv with a power factor of 0.95 lagging. Using the ABCD parameters, compute the sending end voltage, current and δ angle. Confirm using the nominal π equivalent circuit, and the short line equivalent. (d) Draw the approximate voltage profile of this line for the following power delivered: (i) 0 MW, 20 MW, 50 MW, and surge impedance loading. Indicate the methods available to maintain the voltages within the range of 0.95 and.05. QUESTION 5 (20 points) A two-bus transmission system consists of the following: (i) a voltage regulated bus (.0 pu), the swing bus; (ii) a load bus, with a load of power (0.3 = j.0), and a fixed capacitor, supplying a reactive power of (j.); (iii) a transmission line of impedance (j0.4). (a) Draw the single line diagram. Assuming the load bus voltage magnitude is pu, compute the load angle δ. Compute the Y bus matrix. (b) Formulate the Gauss Seidel solution. Indicate the unknowns. Compute the first iteration. (c) Formulate the Newton Raphson solution. Compute the first iteration. (d) Comment on the features of the 2 methods described above. Indicate approaches to accelerate the convergence to the solution. Indicate the changes required to the formulation above if a voltage-controlled bus replaces the load bus. QUESTION 6 (20 points) A three-phase power system is fed from the secondary of a -Y connected transformer, with a neutral grounded through an impedance of j pu. The transformer series impedance is j0.5 pu. The transmission line impedance is j0.2 pu, with no zero sequence component. The sending end voltage is assumed to be pu. The load is a Y-connected impedance of pu (resistive). (a) Draw the sequence component networks, indicating all the impedance values, and voltage sources. (b) A three-phase symmetrical short circuit occurs at the load end, with a 0. pu impedance. Give the sequence impedance matrix for the fault current. Draw the sequence networks corresponding to the fault. Compute the sequence components of the fault currents. Convert into phase quantities. (c) A single line-to-ground fault occurs on phase a with an impedance of 0. pu. Calculate the sequence components of the fault current. Draw the sequence networks corresponding to the Page 2 of 3

3 IGEE 402 Power System Analysis FINAL EXAMINATION SAMPLE Joós, G. fault. Compute the value of the sequence components of the fault currents. Convert into phase quantities. (d) For the symmetrical three-phase fault and the single line-to-ground fault, compute the symmetrical components of the voltage at the fault, and convert to abc values. Draw conclusions as to the severity and impact of those two types of faults on the voltages at the load.. QUESTION 7 (20 points) A three-phase. 60 Hz, round rotor hydroelectric generator has an H constant of 5 s. It delivers P m =.0 pu power at a power factor of 0.90 lagging to an infinite bus through a transmission line of reactance X = 0.6. The voltage at the infinite bus is.0 pu with an angle of 0 deg. The machine transient reactance X d = 0.3 pu. (a) Compute the transient machine internal voltage and angle δ. Give the corresponding equation for the electric power. (b) Write the pu swing equation. A three-phase bolted short occurs midway along the transmission line. Determine the power angle 4 cycles after the initiation of the short circuit. Assume the mechanical input power remains constant at the initial value. (c) Draw the P- δ curve for the above operating conditions. Indicate the initial conditions, and the fault clearing point. Explain the application of the equal area criterion to this case. (d) Discuss methods to ensure that the machine does not loose synchronism (breaker operation, series compensation, ) APPENDIX A SYMMETRICAL COMPONENT TRANSFORMATION MATRICES A = a 2 a a a 2 A - = a a 2 a 2 a Page 3 of 3

4 ECSE Power Systems Analysis Final Exam - Solutions QUESTION (8 points) A single phase 550 V, 60 Hz feeder supplies 2 loads in parallel: (a) a 20 kw heating load; (b) a 50 kw motor load, with an efficiency of 0.90 and a power factor of 0.80 (lagging).. Draw the instantaneous current and voltage waveforms for the total load, indicating peak and rms values. Give the equation for the total instantaneous power and draw the waveform. Indicate the value of the total real power. 2. Draw the V-I vector diagram and the power diagram. Compute total reactive power and power factor. 3. A capacitor is connected in parallel with the loads to increase the power factor to.0. Compute the reactive power required from the capacitor. Redraw the power diagram. 4. Explain the benefits in having a unity power factor from (i) the utility point of view; (ii) the consumer point of view. Vl := 550 Pr := Pm := pf_m := 0.80 ω := 2 π 60. Instantaneous voltage, current, and power waveforms Heating Load: Ir := Pr Vl Motor Load: Sm := Pm pf_m Imotor_mag := Sm Vl φ := acos( pf_m) Imotor := Imotor_mag cos φ Combined Load: ( ( ) j sin( φ) ) Is := Ir + Imotor mag( Is) = a_deg( Is) = 28.79

5 Plot Waveforms: t := 0, ( ) it ():= 2mag ( Is) cos ω t + arg( Is) vt ():= 2Vl Ip := 2 mag( Is) Vlp := 2 Vl cos( ω t) 000 it () vt () Vlp = Vl = 550 Ip = mag( Is) = t Instantaneous Power: pt ():= vt ()it Total Real Power: Ptotal := Pm + Pr pt () Ptotal Ptotal = t

6 2. VI vector diagram, total reactive power, power factor, and power triangle ( ) Qt := Sm sin φ Stotal := ( Pm + Pr) + j Qt 80 cos( pf ) = π pf = 0.88 pf := ( Pm + Pr) mag( Stotal) Qt = mag( Stotal) = Calculate the reactive power required from the capacitor The reactive power required from the capacitor is simply the negative of the total reactive power from part 2. This way the load power factor becomes unity and the feeder apparent power has only a real component. The new power triangle looks as follow: 4. Explain the benefits of having unity power factor from (i) the utility point of view and (ii) the customer's point of view. The utility benefits from unity power factor since in this way the reactive power flows on the line become zero. Although this doesn't directly contribute to losses, reactive power flows contribute to increased currents and the associated losses involved. In addition, the increased current may also play a role in reducing the lifespan of the lines and other equipment. Also, providing reactive compensation at the load can help to improve voltage stability In terms of the customer, unity power factor is desirable since it indirectly affects the power quality through the regulation of the voltage. Large inductive loads can cause undervoltages which can be undesirable for sensitive loads. Also, the customer will likely inherit some of the cost due to the losses mentioned above and therefore saving from the point of view of the utility also result in savings for the customer as well.

7 QUESTION 2 (8 points) A single phase 24 kv/2.4 kv, 60 Hz transformer, with a series equivalent impedance of j.0 Ω referred to the low-voltage side, is connected at the end of a transmission line of impedance 50 + j400. The transformer feeds a load drawing 200 kw, at a 0.8 power factor lagging, and 2300 V The feeder voltage is adjusted so that the load voltage is 2300 V. Assume the load voltage has an angle of 0 deg. Find the amplitude and angle of the voltage at the sending end of the transmission line. (Hint: refer all quantities to the low voltage side). Compute the current in the transmission line, the losses in the transmission line and transformer, and the overall system efficiency. Calculate the no-load voltage and the voltage regulation. Explain how the voltage regulation can be reduced. (Hint: consider shunt compensation).. Calculate the feeder voltage and angle Define Load: Pload := pf load := 0.8 Sload := Pload pf load Vload := 2300 Vload 2 Zload mag := Sload ( ( ( ))) Zload := Zload mag pf load + j sin acos pf load Zload = i Calculate Feeder Voltage: mag( Zload) = 2.6 a_deg( Zload) = a := Ztrans := j.0 Zline := ( 50 + j 400) a 2 Zline = i Ztrans Zline Vs := Vload ( + + Zload) Zload mag( Vs) = a_deg( Vs) = 8.24

8 2. Calculate the transmission line current, all losses, and efficiency Current: Iline_low := ( Vs Vload) ( Zline + Ztrans) Iline_high := Iline_low a mag( Iline_high) = 0.87 a_deg( Iline_high) = Losses: Transmission Line Transformer Pline := mag( Iline_low) 2 Re( Zline) Ptrans := mag( Iline_low) 2 Re( Ztrans) Pline = Ptrans = Efficiency: Eff := Pload 00 Pload + Pline + Ptrans Eff = No-load voltage and voltage regulation Since the line contains no shunt components the no-load voltage will equal the magnitude of the source voltage. Vno_load := Vs Vno_load = VR := ( Vno_load Vload) 00 Vload VR = % 4. Improvement of Voltage Regulation The receiving end voltage varies as a function of the current on line which is dependent on the load connected. We can compensate for the change in voltage by adding shunt compensation at the load. In this way the voltage can be regulated to a given reference, say pu. In this way the receiving end voltage may change by only a very small amount between no-load an full-load. Therefore, a decrease in the VR is possible which will depend on the rating of the reactive power source, in the case where the rating of the compensation is infinite the voltage regulation becomes 0%. VR will also depends upon the impedance of the line and therefore, series compensation or installation of a parallel line could also help to limit VR.

9 QUESTION 3 (8 points) A three-phase feeder supplies three identical single-phase 35 kv / 6.9 kv, 60 Hz, 00 MVA transformers, each with a leakage reactance of 0.5 pu on the transformer base, and connected in a -Y connection. The load is a three-phase Y-connected 2 kv load, with per phase parameters of 60 MW, 0.85 power factor (lagging). A capacitor bank of three elements of impedance (-j 5 ) connected in is used for power factor correction. The secondary voltage is assumed to be 2 kv.. Compute the total power and total reactive power drawn from the feeder. 2. Draw the single line pu circuit, using a 300 MVA (three-phase), 2 kv base. 3. Compute the total line current, the total power and total reactive power drawn from the feeder, in pu. Compute the voltage at the feeder. Compare with results in (). 4. Phase c of the load becomes an open circuit. In the original V-A three-phase circuit, compute the current in the lines feeding the load, when (i) the load neutral is solidly grounded; (ii) the load neutral is floating.. Compute the total real and reactive power drawn from the feeder Sbase := Vbase := 2000 Zbase := Vbase 2 Sbase Zbase = 0.48 Load: pf := 0.85 Pload := Sload := Pload pf Sbase Ibase := Sbase 3 Vbase Zload := pf j sin acos pf Sload ( + ( ( ))) Zload = i Capacitor Bank: Xc := j 5 3 Xcpu := Xc Zbase Xcpu = 3.472i Qc := Xcpu Total Real and reactive power: Ptotal := Pload Ptotal = Qtotal := Sbase ( Sload sin( acos( pf )) Qc ) Qtotal =

10 2. Single Line Diagram 3. Total line current, total real and reactive power and power factor in pu. Feeder voltage. Iline := Zload + Xcpu Iline = a_deg( Iline) = Sline := conj( Iline) Pline := Qline := Re( Sline) Im( Sline) Pline = 0.6 Qline = pf line := Pline Sline pf line = 0.99 Compare with (): Pline Sbase = Ptotal = Qline Sbase = Qtotal = Vfeeder := + Iline 0. Zbase Vfeeder =.029 pu Vf_high := Vfeeder Vf_high = V 4. Currents in the line feeding the load with phase c open circuited (i) neutral is solidly grounded We simply refer to the below figure and solve the two loop equations, we disregard the capacitors since we are interested only in the current feeding the load. Va := π Vb := cos Z := j j 0.75 π + j sin Vab := Va Vb

11 Va ia := Z ia Ibase = Vb ib := Z ib Ibase = io := ia + ib io Ibase = a_deg( ia) = (i) neutral is floating a_deg( ib) = a_deg( io) = In this case, the problem is trivial since I 0 = 0, since there is no path for zero sequence current. Therefore, we can say that I a = -I b and the circuit looks as follows: Ia := Vab 2Z Ib := Ia Ia Ibase = a_deg( Ia) = 6.87 Ib Ibase = a_deg( Ib) = 73.3

12 QUESTION 4 (8 points) A 400 km, 38 kv, 60 Hz transmission line has the following distributed parameters: r = 0.06 /km, x = /km, y = j3.36 x 0-6 S/km. Losses are neglected. (a) Compute the nominal equivalent circuit parameters and draw the circuit. Compute the corresponding ABCD parameters. (b) Find the surge impedance and surge impedance loading. (c) The line delivers 40 MW at 32 kv with a power factor of 0.95 lagging. Using the ABCD parameters, compute the sending end voltage, current and angle. Confirm using the nominal equivalent circuit, and the short line equivalent. (d) Draw the approximate voltage profile of this line for the following power delivered: (i) 0 MW, 20 MW, 50 MW, and surge impedance loading. Indicate the methods available to maintain the voltages within the range of 0.95 and.05. (a) Compute the nominal π equivalent circuit parameters, ignore losses rl := 0.06 xl := yl := l := 400 Vl := X := ixl l X = 97.2i Y := iyl l Y =.344i 0 3 Vl 3 = Api := + Y X 2 Api = Bpi := X Bpi = 97.2i Cpi := Y + Y X 4 Cpi =.255i 0 3 (b) Find the surge impedance and the SIL xl ( Vl Vl) Zc := Zc = SIL := SIL = yl Zc β := xl yl β = (c) Calculate sending end voltage and angle using ABCD, nominal pi, and short line equivalent Lossless ABCD A := cos( β l) A = 0.87 ( ) B := i Zc sin β l B = i ( ) C := i sin β l C =.285i 0 3 Zc

13 pf := 0.95 Prec := Srated := Prec pf Vrl := Vrl = Srated Ir := 3 Vrl ( pf + j sin ( acos ( pf ))) Ir = a_deg ( Ir ) = 8.95 Sending end voltage and angle,using ABCD, nominal pi, and short line Vsl := A Vrl + BIr Vsl = i 0 4 Vsl = a_deg( Vsl) = Vsl := Api Vrl + Bpi Ir Vsl = i 0 4 Vsl = a_deg( Vsl) = Vsl := Vrl + X Ir Vsl = i 0 4 Vsl = a_deg( Vsl) = Sending end current using ABCD, nominal pi, and short line Isl := C Vrl + AIr Isl = i Isl = a_deg( Isl) = Isl := Cpi Vrl + Api Ir Isl = i Isl = a_deg( Isl) = Isl := Ir Isl = i Isl = a_deg( Isl) = 8.95 (d) Approximate voltage profile for 0 MW, 20 MW, 50 MW, and SIL (49.7 MW) It is possible to keep the receiving end voltage between 0.95 and.05 using shunt compensation in the form of switched capacitors or inductors. Static compensators which are more elegant can be used as well which provide for better adjustment of the reactive power injected or consumed, and can respond much quicker. Examples include Static var compensators (SVCs) and static shunt compensators (STATCOM).

14 QUESTION 5 (8 points) A two-bus transmission system consists of the following: (i) a voltage regulated bus (.0 pu), the swing bus; (ii) a load bus, with a load of power (0.3 + j.0), and a fixed capacitor, supplying a reactive power of (j.); (iii) a transmission line of impedance (j0.4). (a) Draw the single line diagram. Assuming the load bus voltage magnitude is pu, compute the load angle. Compute the Ybus matrix. (b) Formulate the Gauss Seidel solution. Indicate the unknowns. Compute the first iteration. (c) (d) Formulate the Newton Raphson solution. Compute the first iteration. Comment on the features of the 2 methods described above. Indicate approaches to accelerate the convergence to the solution. Indicate the changes required to the formulation above if a voltage-controlled bus replaces the load bus. (a) Draw the single line diagram, calculate load angle, δ assuming V2 =.0, calculate the admittance matrix. Use: P = V V 2 / X sin(δ -δ 2 ) P := 0.3 X := 0.4 y := X Y bus := j δ 2 := asin( P X) 80 Q 2 := 0. δ 2 = π y y y y 2.5i Y bus = 2.5i 2.5i 2.5i P 2 := P (b) Formulate the Gauss-Siedel problem: state the unknows and calculate the st iteration We apply the Guass-Siedel formula to the voltage at the load bus: V 2 (k+) = /Y 22 [ (P 2 - jq 2 )/V 2 - Y 2 *V (k) ] The unknowns are the voltage and angle at bus 2. Once the solution has converged we are able to obtain the unknown flows which are the real and reactive power at bus.

15 So, we calculate the first iteration assuming a flat voltage profile as our initial guess. V 2 := Y bus, ( ) P 2 jq 2 Y bus, 0 V 2 =.047 a_deg V 2 ( ) = (c) Formulate the Newton-Raphson problem: state the unknows and calculate the st iteration In order to calculate the first iteration we need to determine the Jacobian, which will define our search direction. The vector of unknows, x is given by the angle and magnitude of the voltage at the load bus: Initial guess: x := 0 The unknown flows again are the real and reactive power at bus. Which are calculated once the voltage at the load bus has been determined. V2 := x Calculate Jacobian: dp2dδ2 := V2 Y bus V, 0 dp2dv2 := 2V2 Y bus cos arg Y bus Y bus, 0 dq2dδ2 := V2 Y bus, 0 V := V2 sin( arg( V2) arg ( V ) arg( Y bus, 0 )) ( (, )) + V cos, ( arg( V2) arg ( V ) arg( Y bus, 0 )) ( ( ) arg ( V ) arg( Y bus, 0 )) ( ( )) ( ( )) V cos arg V2 y := y = dq2dv2 := 2V2 Y bus sin arg Y bus + Y bus V sin arg( V2) arg ( V ) arg Y bus,,, 0, 0 P 2 Q 2 J := dp2dδ2 dq2dδ2 dp2dv2 dq2dv2 J = Calculate the new voltage: x := J y x = x = x = π x := x + x (d) Comments Newton Raphson is more well suited to solve the load flow equations since it is more often used to solve non-linear equations while Gauss-Siedel is more well suited for solving linear equations. Methods to improve speed of convergence are to use fast-decoupled load flow where the Jacobian is only calculated once, and only inverted once as well. Sparse matrices methods are also implemented for larger systems. If bus 2 becomes a PV bus, the formulation is changed slightly in that x becomes the angle δ 2 and whereas y becomes the known flow P 2. In Gauss-Siedel we iterate w.r.t Q 2 and not V 2

16 QUESTION 6 (8 points) A three-phase power system is fed from the secondary of a -Y connected transformer, with a neutral grounded through an impedance of j pu. The transformer series impedance is j0.5 pu. The transmission line impedance is j 0.2 pu, with no zero sequence component. The sending end voltage is assumed to be pu. The load is a Y-connected impedance of pu (resistive). (a) Draw the sequence component networks, indicating all the impedance values, and voltage sources. (b) A three-phase symmetrical short circuit occurs at the load end, with a 0. pu impedance. Give the sequence impedance matrix for the fault current. Draw the sequence networks corresponding to the fault. Compute the sequence components of the fault currents. Convert into phase quantities. (c) A single line-to-ground fault occurs on phase a with an impedance of 0. pu. Calculate the sequence components of the fault current. Draw the sequence networks corresponding to the fault. Compute the value of the sequence components of the fault currents. Convert into phase quantities. (d) For the symmetrical three-phase fault and the single line-to-ground fault, compute the symmetrical components of the voltage at the fault, and convert to abc values. Draw conclusions as to the severity and impact of those two types of faults on the voltages at the load. (a) Draw the sequence component networks (Note: grounding impedance should be 3Zn)

17 (b) Three phase Symmetrical Short Circuit Zo := + j.5 Zo = i Z := + j 0.35 Z = i E := + j 0.35 E = i E = a_deg( E) = 9.29 Z2 := Z Zf := 0. a := j 2 3 I := E Z + Zf I = 2.54 Iph := Aseq I o I I 2 I 2 := 0 a_deg I ( ) = I o := 0 Iph 0 Iph Iph 2 = Aseq := ( ) ( ) ( ) a_deg Iph 0 a_deg Iph a_deg Iph 2 a 2 a = a a

18 (c) Single Line to Ground Fault I := E ( Z + Z2 + Zo + 3Zf ) I 2 := I I o := I I = I ph := Aseq I o I I 2 a_deg I ( ) = I ph0 I ph I ph2 = ( ) = a_deg I ph0 (d) Calculate fault voltages Three phase fault: V 2 := 0 V o := 0 Zf V := E Zf + Z V = i V = 0.25 a_deg V ( ) = 75.44

19 V ph := Aseq V o V V 2 V ph0 V ph V ph2 = ( ) = a_deg V ph0 Line to ground fault: V o := I o Zf V := V o V 2 := V V o = 0.06 a_deg V o ( ) = V ph := Aseq V o V V 2 V ph0 V ph V ph2 = ( ) = a_deg V ph0

20 QUESTION 7 (8 points) A three-phase. 60 Hz, round rotor hydroelectric generator has an H constant of 5 s. It delivers Pm =.0 pu power at a power factor of 0.90 lagging to an infinite bus through a transmission line of reactance X = 0.6. The voltage at the infinite bus is.0 pu with an angle of 0 deg. The machine transient reactance X'd = 0.3 pu. (a) Compute the transient machine internal voltage and angle. Give the corresponding equation for the electric power. (b) Write the pu swing equation. A three-phase bolted short occurs midway along the transmission line. Determine the power angle 4 cycles after the initiation of the short circuit. Assume the mechanical input power remains constant at the initial value. (c) Draw the P- curve for the above operating conditions. Indicate the initial conditions, and the fault clearing point. Explain the application of the equal area criterion to this case. (d) Discuss methods to ensure that the machine does not loose synchronism (breaker operation, series compensation, ) (a) Compute the transient machine internal voltage and angle X := 0.6 Xd := 0.3 pf := 0.9 H:= 5 P := I := P ( pf j sin( acos( pf ))) pf I =. a_deg( I) = E:= + j ( X + Xd) I E =.695 a_deg( E) = (b) Provide and use the swing equation ( 2H) ωsyn ω pu() t 2 d 2 dt δ() t = p m,pu (t) -p e,pu (t) ωsyn := 2 π 60 δ := arg( E) Integrating swing equation twice,assuming p m remains constant we obtain δ() t = ωsyn 4H t2 + δ For four cycles following the fault we determine the new δ δ new := ωsyn 4H δ δ new = δ new = π

21 (c) Draw the P-δ curve for the fault operation p m ( δ) = E sin( δ) = ( X + Xd).8 sin( δ) E ( X + Xd) =.883 δ A := p m dδ δ 0 δ δ 0.0 d δ = δ new δ = δ 2 A 2 := δ ( ).8sinδ p m dδ ( ( ) cos( δ 2 )) ( δ 2 δ ).8 cos δ.8 cos( δ 2 ) + δ 2 =.46 = A := 0.68 δ 2 := δ 2 = π We calculate A using the swing equation and the fault duration to determine the angle at which the fault is cleared, δ. The machine continues to accelerate until A2 = A. We can determine whether the machine retains stability determining whether there in fact does exist and angle δ 2 for which A2 does in fact equal A. Following the above calculations we see that this is true for approximately 22 degrees, and therefore stability is maintained.

22 (d) Discuss methods to maintain stability There exist various methods to help maintain stability. Many of the most common are listed here: Smaller equivalent line impedance (parallel lines, different conductors, smaller transformer leakage reactances, series compensation...) Higher voltages FACTS devices Faster clearing of faults Larger machine inertia

23 Define function mag( x) := ( Re( x) ) 2 + ( Im( x) ) 2 conj( x) := Re( x) Im( x) j a_deg( x) := arg( x) 80 π

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