IGEE 402 Power System Analysis. FINAL EXAMINATION Fall 2004

Transcription

1 IGEE 40 Power System Analysis FINAL EXAMINATION Fall 004 Special instructions: - Duration: 150 minutes. - Material allowed: a crib sheet (double sided 8.5 x 11), calculator. - Attempt 4 out of 7 questions. - Make any reasonable assumption. - All questions and sub-questions carry equal weight. - Return the completed answer sheet attached. QUESTION 1 (0 points) A single-phase 600 V, 60 Hz feeder, with a series reactance of 0. pu, supplies a motor load. The motor draws 8 times base current at a power factor of 0. (lagging) at start-up under 600 V. The motor can be represented by a series R-L combination. Assume a base power of 5 kva. Give results in real quantities (Ω, V, A). (a) Find the equivalent series R-L combination of the motor in Ω. Taking into account the feeder impedance, compute the line current and the voltage across the motor. (b) A series capacitor is used to cancel 75 % of the feeder inductance. Compute the voltage across the motor and the line current. (c) A capacitor, of 3 pu reactance, is connected in parallel with the motor. Compute the voltage across the motor and the line current. Compare with the previous case and explain the difference. QUESTION (0 points) A three-phase 13.8 kv/138 kv, 00 MVA, 60 Hz transformer, with a series equivalent impedance of j0.08 pu (on the transformer base) is connected to a transmission line of impedance + j0 Ω. The line feeds a 00 MVA load, with a power factor of (a) Give the transformer and line equivalent circuits, in real and pu values. Draw the single line diagram of the system, showing impedances and voltages in pu and in Ω, V. (b) The sending-end voltage is adjusted so that the load voltage is 138 kv. Compute the current flowing in the transmission line. Use pu values. Find the amplitude in V and angle of the voltage on the low voltage side of the transformer. Compute the voltage regulation. (c) Series compensation, equal to 75 % of the line reactance, is added to the transmission line. The sending-end voltage is adjusted so that the load voltage is 138 kv. Find the amplitude and angle of the voltage on the low voltage side of the transformer. Compute the voltage regulation and compare to the uncompensated case.

2 IGEE 40 Power System Analysis FINAL EXAMINATION Joós, G. QUESTION 3 (0 points) A three-phase 600 V, 60 Hz feeder supplies two three-phase loads: (a) a Y connected inductor, of per phase impedance of 0 Ω and (b) a Δ connected capacitor, of per phase impedance of 0 Ω. (a) Compute the current in each capacitor and inductor. Compute the total current drawn by the loads. Draw the phasor diagram showing the phase voltage and inductor, capacitor and total current per leg, based on a Y equivalent. (b) Compute the power associated with each load, real and reactive component. Find the total power. (c) One of the capacitors becomes an open circuit. Find the total line current flowing in each of the three lines. QUESTION 4 (0 points) A 400 km, 765 kv, 60 Hz transmission line has the following distributed parameters: x = 0.30 Ω/km, y = j5.0 x 10-6 S/km. Losses are neglected. (a) Compute the nominal π equivalent circuit parameters and draw the circuit. Compute the corresponding ABCD parameters. Find the surge impedance and surge impedance loading. (b) The line delivers 000 MW at 735 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. (c) Series compensation of 70 % is introduced. Using the short line equivalent, compute the new load angle and maximum steady state power transfer capability. Compare with the uncompensated line. QUESTION 5 (0 points) A two-bus transmission system consists of the following: (i) a voltage regulated generator bus (1.0 pu), the swing bus; (ii) a voltage regulated bus (1.0 pu), with a generator supplying real power (0.5 pu) and reactive power, and a load of power (1.0 + j0.5) pu. The transmission line impedance is (j0.5). (a) Draw the single line diagram. Identify the type of buses. Compute the Y bus matrix. Write the current equation and the power equation. (b) Formulate the Newton Raphson solution. Compute the first iteration. (c) Indicate the changes required to the formulation above if a load bus replaces the voltage controlled bus. Compute the first iteration. QUESTION 6 (0 points) A three-phase generator has a Y connected winding, with a neutral grounded through an impedance of j0. pu. The generator positive (transient), negative and zero sequence components are respectively, j0.4, j0.5, j0.0. The transmission line impedance is j0.4 pu, with a zero sequence component of 0. The generator prefault voltage is assumed to be regulated at 1 pu. Ignore all prefault currents. (a) A three-phase symmetrical short circuit occurs at the end of the transmission line, with a 0.1 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. (b) A single line-to-ground fault occurs on phase a with an impedance of 0.1 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. (c) 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. Page of 4

3 IGEE 40 Power System Analysis FINAL EXAMINATION Joós, G. QUESTION 7 (0 points) A three-phase. 60 Hz, round rotor hydroelectric generator has an H constant of 4 s. It delivers P m = 1.0 pu power at a power factor of 0.85 lagging to an infinite bus through a transmission line of reactance X = 0.7. The voltage at the infinite bus is 1.0 pu with an angle of 0 deg. The machine transient reactance X d = 0.35 pu. (a) Compute the transient machine internal voltage and angle δ. Give the corresponding equation for the electric power. Write the pu swing equation. (b) A three-phase bolted short occurs midway along the transmission line. Determine the power angle 5 cycles after the initiation of the short circuit. Assume the mechanical input power remains constant at the initial value. Draw the pre-fault and post-fault P- δ curves. Indicate the initial conditions, and the fault clearing point. Explain the application of the equal area criterion to this case. (c) The fault is cleared after 5 cycles. Find the maximum δ angle. Indicate if the system recovers from the fault. APPENDIX A SYMMETRICAL COMPONENT TRANSFORMATION MATRICES A = 1 a a 1 a a A -1 = 1/3 1 a a 1 a a Page 3 of 4

4 QUESTION 1 (0 points) A single-phase 600 V, 60 Hz feeder, with a series reactance of 0. pu, supplies a motor load. The motor draws 8 times base current at a power factor of 0. (lagging) at start-up under 600 V. The motor can be represented by a series R-L combination. Assume a base power of 5 kva. Give results in real quantities (Ω, V, A). Vbase=600; Sbase=5e3; xf=0.; xc=3; pf=0.; (a) Find the equivalent series R-L combination of the motor in Ω. Taking into account the feeder impedance, compute the line current and the voltage across the motor. Ibase=Sbase/Vbase; %Find the phasor current teta=acos(pf); Im=8*Ibase*exp(-j*teta); %Ohms law to find the impedance zm=vbase/im; zbase=vbase^/sbase; %Voltage divider to find the voltage across the impedance vm=vbase*zm/(zm+j*xf*zbase); %Ohms law to find the line current Iline=Vbase/(zm+j*xf*zbase); R= 1.8 X= 8.81 Iline= 5.76A Vm= 31.8V (b) A series capacitor is used to cancel 75 % of the feeder inductance. Compute the voltage across the motor and the line current. vm_series_compensation=vbase*zm/(zm+j*(xf-0.75*xf)); Iline_series_compensation=Vbase/(zm+j*(xf-0.75*xf)); Iline= 47.81A Vm= 430V (c)

5 A capacitor, of 3 pu reactance, is connected in parallel with the motor. Compute the voltage across the motor and the line current. Compare with the previous case and explain the difference. zm_parallel_xc=zm*(-j*xc)/(zm-j*xc); vm_parallel_compensation=vbase*zm_parallel_xc/(zm_parallel_xc+j*xf); Iline_parallel_compensation=Vbase/(zm_parallel_xc+j*xf); Iline= 5.36A Vm= 37.9V The system seems to be adequately compensated in series. However, the high current and the over voltages in the parallel shows that the system is overcompensated with the parallel capacitor. QUESTION (0 points) A three-phase 13.8 kv/138 kv, 00 MVA, 60 Hz transformer, with a series equivalent impedance of j0.08 pu (on the transformer base) is connected to a transmission line of impedance + j0 Ω. The line feeds a 00 MVA load, with a power factor of Vt_high=138; Vt_low=13.8; St=00; zt=0.0+j*0.08; zl_ohms=+j*0; Pl=00; pf=0.85; (a) Give the transformer and line equivalent circuits, in real and pu values. Draw the single line diagram of the system, showing impedances and voltages in pu and in Ω, V. Vbase=Vt_high; Sbase=St; Zbase=Vbase^/Sbase; zt_high_ohms=zt*zbase; zl_pu=zl_ohms/zbase; Rtr(pu)= 0.0 Xtr(pu)= 0.08 Rtr(ohm)= 1.9 Xtr(ohm)= 7.61 Rl(pu)= 0.01 Xl(pu)= 0.1 Rl(ohm)= Xl(ohm)= /138kV 00MVA 0.0+j0.08 p.u 1.9+j7.61ohm from the high voltage side Line: +j0homs 0.01+j0.1pu Load 00MVA pf 0.85 lagging

6 (b) The sending-end voltage is adjusted so that the load voltage is 138 kv. Compute the current flowing in the transmission line. Use pu values. Find the amplitude in V and angle of the voltage on the low voltage side of the transformer. Compute the voltage regulation. %Compute the current Iline=conj(pf+j*sin(acos(pf))); %voltage in the secondary of the transformer Vtransf_pu=1+Iline*(zl_pu+zt); Vtransf_volts=Vtransf_pu*Vt_low; Vtransf_volts_amplitude=abs(Vtransf_volts)*sqrt(); delta=angle(vtransf_pu)*180/pi; %voltage regulation VR_no=(abs(Vtransf_pu)-1)*100; Iline(pu)= 1 Vt_low= 16.7kV Delta= 10.7 VR= 0.87 (c) Series compensation, equal to 75 % of the line reactance, is added to the transmission line. The sending-end voltage is adjusted so that the load voltage is 138 kv. Find the amplitude and angle of the voltage on the low voltage side of the transformer. Compute the voltage regulation and compare to the uncompensated case. Vtransf_comp=1+Iline*(zl_pu-0.75*zl_pu+zt); Vtransf_comp_amplitude=abs(Vtransf_comp)*Vt_low*sqrt(); delta_compensated=angle(vtransf_comp)*180/pi; VR=(abs(Vtransf_comp)-1)*100; Vt_low= 15.11kV Delta= 5. VR= 9.58 Series compensation improves voltage regulation since reduces the reactance and therefore the voltage drop across the line. QUESTION 3 (0 points) A three-phase 600 V, 60 Hz feeder supplies two three-phase loads: (a) a Y connected inductor, of per phase impedance of 0 Ω and (b) a Δ connected capacitor, of per phase impedance of 0 Ω. V=600; xi=0; xc=0; (a) Compute the current in each capacitor and inductor. Compute the total current drawn by the loads. Draw the phasor diagram showing the phase voltage and inductor, capacitor and total current per leg, based on a Y equivalent. %current in the individual elements Iind=V/sqrt(3)/(j*xi); Icap=V/(-j*xc); %total current Itot=Iind+Icap*sqrt(3);

7 Ic= 30 Iind= 17.3 Iload= Icapacitor phase=j51.96a Itotal=34.64A Vphase=346V Iinductor phase=-j17.3a (b) Compute the power associated with each load, real and reactive component. Find the total power. Sind=3*(Iind)^*j*xi; Scap=3*(Icap)^*(-j*xc); Scap 54000Var Sind Var Stot 36000Var (c) One of the capacitors becomes an open circuit. Find the total line current flowing in each of the three lines. Ic Iind Icap Phase c Iind Iind Icap Ib Phase b Ia Phase a %the voltage is not affected then, one of the currents is the same Va=600/sqrt(3); Vb=Va*exp(-j**pi/3); Vc=Va*exp(j**pi/3);

8 Ia=Va/(j*xi)+(Va-Vb)/(-j*xc)+(Va-Vc)/(-j*xc); Ib=Vb/(j*xi)+(Vb-Va)/(-j*xc); Ic=Vc/(j*xi)+(Vc-Va)/(-j*xc); I_a= j34.64 I_b= -j17.3 I_c= -j17.3 QUESTION 4 (0 points) A 400 km, 765 kv, 60 Hz transmission line has the following distributed parameters: x = 0.30 Ω/km, y = j5.0 x 10-6 S/km. Losses are neglected. L=400; V=765000; x=0.3; y=5e-6; (a) Compute the nominal π equivalent circuit parameters and draw the circuit. Compute the corresponding ABCD parameters. Find the surge impedance and surge impedance loading. X=x*L*j; Y=y*L*j; Api=1+Y*X/; Bpi=X; Cpi=Y*(1+Y*X/4); Zc=sqrt(x/y); SIL=V^/Zc; beta=sqrt(x*y); Api= 0.88 Cpi= i Zc= 45 Bpi= 10i Dpi= 0.88 SIL= 389 X=j10 Vs Is Ir Vr Y/=j.001 Y/ (b) The line delivers 000 MW at 735 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. P=000e6; Vr=735000; pf=0.95; A=cos(beta*L);

9 B=Zc*sin(beta*L)*j; C=1/Zc*sin(beta*L)*j; Vrl=Vr/sqrt(3); Srec=P/pf; Ir=Srec/3/Vrl*(pf-j*sin(acos(pf))); Vs=A*Vrl+B*Ir; Vspi=Api*Vrl+Bpi*Ir; Vs_short=Vrl+X*Ir; Is=C*Vrl+A*Ir; Ispi=Cpi*Vrl+Api*Ir; Is_short=Ir; Vs= Vs_pi= Vs_short= Is== 143 Is_pi= 144 Is_short= 1653 (c) Series compensation of 70 % is introduced. Using the short line equivalent, compute the new load angle and maximum steady state power transfer capability. Compare with the uncompensated line. Xnew=0.3*X; Vs_new=Vrl+Ir*X; delta_new=asin(abs(xnew)*p/3/(abs(vs_new)*abs(vrl))); Pmax=V^/abs(Xnew)*3; delta_new= 6. Pmax= QUESTION 5 (0 points) A two-bus transmission system consists of the following: (i) a voltage regulated generator bus (1.0 pu), the swing bus; (ii) a voltage regulated bus (1.0 pu), with a generator supplying real power (0.5 pu) and reactive power, and a load of power (1.0 + j0.5) pu. The transmission line impedance is (j0.5). xl=0.5; (a) Draw the single line diagram. Identify the type of buses. Compute the Y bus matrix. Write the current equation and the power equation. Swing bus P=? Q=? V=1 delta=0 j0.5 PV bus P=-0.75 Q=? V=1 delta=? G=0.5 Load=1+j0.5 y=1/(j*xl); Y=[y -y; -y y];

10 (1 δ 1 ) I = xl sin( δ ) P = xl δ =.0 o Y= -i i i -i (b) Formulate the Newton Raphson solution. Compute the first iteration. X = δ ; Y = P X(0)=0; Calculate the Jacobian: P = VV Y sin( δ ) 1 1, P = = cos( ) = = J VV 1 Y1, δ Y1, δ Jacobian= Δy= Δx= x1= (c) Indicate the changes required to the formulation above if a load bus replaces the voltage controlled bus. Compute the first iteration. The number of variables changes and the formulation as follows: X δ P ; Y = ; Q = V Calculate the Jacobian: 0 X(0)= 1

11 P = VV Y sin( δ ) 1 1, Q = VV Y (1 cos( δ )) 1 1, P J = = VV Y = Y = 1 1 1, cos( δ ) 1, δ P J = = V Y δ = Y = 0 1 1, sin( ) 1, V Q J = = VV Y δ = Y = , sin( ) 1, δ Q J = = V Y δ = Y = 4 1 1, (1 cos( )) 1, V Jacobian= Δy= Δx= x1= QUESTION 6 (0 points) A three-phase generator has a Y connected winding, with a neutral grounded through an impedance of j0. pu. The generator positive (transient), negative and zero sequence components are respectively, j0.4, j0.5, j0.0. The transmission line impedance is j0.4 pu, with a zero sequence component of 0. The generator prefault voltage is assumed to be regulated at 1 pu. Ignore all prefault currents. xl=0.4; xg1=0.4; xg=0.5; xg0=0.; xn=0.; xf=0.1; (a) A three-phase symmetrical short circuit occurs at the end of the transmission line, with a 0.1 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. Y11=-j*(1/xg1+1/xl); Y1=j/xl; Y1=Y1; Y=-j*(1/xl); Y=[Y11 Y1; Y1 Y]; Z=inv(Y); Ifault=1/(Z(,)+j*xf); I0=0; I=0; I1=Ifault;

12 a=-0.5+j*sqrt(3)/; A=[1 1 1; 1 a^ a; 1 a a^]; I=A*[I0 I1 I]'; j0.8pu + Zero sequence network V0 - j0.8pu If=I1 + Positive sequence network 1p.u V1 Zf=0.1 - j0.65pu + Negative sequence network V - I 0 =0 I 1 =1.11 I =0 I a = I b = I b = (b) A single line-to-ground fault occurs on phase a with an impedance of 0.1 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.

13 j0.8pu + Zero sequence network V0 - j0.8pu If=I0=I1=I + Positive sequence network 1p.u V1 3xZf=j0.3 - j0.65pu + Negative sequence network V - z0=j*(3*xn+xg0); z1=j*(xg1+xl); z=j*(xg+xl); I0=1/(z0+z1+z+3*j*xf); Ifsingle=A*[I0 I0 I0]'; I 0 =0.39 I 1 =0.39 I =0.39 I a = I b =0 I b =0 (c) 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. V0=-I0*z0; V1=1-I0*z1; V=-I0*z; Vfault=A*[V0 V1 V]'; Three V0= 0 V1= 0.11 V= 0 phase Va= 0.11 Vb= Vc= Single V0= V1= V= phase Va= 0.11 Vb= Vc=.97 13

14 QUESTION 7 (0 points) A three-phase. 60 Hz, round rotor hydroelectric generator has an H constant of 4 s. It delivers P m = 1.0 pu power at a power factor of 0.85 lagging to an infinite bus through a transmission line of reactance X = 0.7. The voltage at the infinite bus is 1.0 pu with an angle of 0 deg. The machine transient reactance X d = 0.35 pu. H=4; xg=0.35; xl=0.7; pf=0.85; f=60; clearing=5; (a) Compute the transient machine internal voltage and angle δ. Give the corresponding equation for the electric power. Write the pu swing equation. I=1/pf*(pf-j*sin(acos(pf))); E=1+j*(xg+xl)*I; ESin( δ ) P = xg+ xl H d ωpu() t δ() t = P m, pu() t Pe, pu() t ωsyn dt Integrating swing equation twice, assuming pm remains constant we obtain ωsync δ () t = t + δ0 4H E= 1.95 δ= 3.46 (b) A three-phase bolted short occurs midway along the transmission line. Determine the power angle 5 cycles after the initiation of the short circuit. Assume the mechanical input power remains constant at the initial value. Draw the pre-fault and post-fault P- δ curves. Indicate the initial conditions, and the fault clearing point. Explain the application of the equal area criterion to this case. wsync=60**pi; delta_new=wsync/(4*h)*(clearing/f)^+angle(e); Pmax=abs(E)/(xg+xl); delta_max=pi-angle(e);

15 1.86 sin(x) Pmax=1.86 pu A A delta= delta=6.8 x delta= delta=3.46 δ_new In order to have transient stability the kinetic energy accumulated in the machine during the fault (A1) has to be less than the energy the generator can return to the system before losing synchronism (A). Then: A1 A δ max δ δ ( Pmax.sin( δ) 1) dδ ; new δ new A1=delta_new-angle(E); A=Pmax*(cos(delta_new)-cos(delta_max))-(delta_max-delta_new); In this case stability is kept (c) The fault is cleared after 5 cycles. Find the maximum δ angle. Indicate if the system recovers from the fault. To find the maximum angle we have to solve the following non-linear equation: A1=A delta_new-angle(e)= Pmax*(cos(delta_new)-cos(delta_max))-(delta_max-delta_new); which is equal to: 1.95 = 1.86 cos( δ max) + δ max Which can be solved graphically as:

16 x δ_max= rad = 6.8degrees