电力系统分析（第二版）Hadi Saadat matlab 第五章 输电线路模型及其特性(教材搬运)

思维导图

例5.1

 VRLL=220; VR = VRLL/sqrt(3); 
Z = (0.15+j*2*pi*60*1.3263e-3)*40; 
disp('(a)') 
SR=304.8+j*228.6; 
IR = conj(SR)/(3*conj(VR)); IS = IR; 
VS = VR + Z*IR; 
VSLL = sqrt(3)*abs(VS) 
SS = 3*VS*conj(IS) 
REG = (VSLL - VRLL)/VRLL*100 
Eff = real(SR)/real(SS)*100 
disp('(b)') 
SR=304.8-j*228.6; 
IR = conj(SR)/(3*conj(VR)); IS = IR; 
VS = VR + Z*IR; 
VSLL = sqrt(3)*abs(VS) 
SS = 3*VS*conj(IS) 
REG = (VSLL - VRLL)/VRLL*100 
Eff = real(SR)/real(SS)*100

运行结果

(a)

VSLL =

  250.0186


SS =

   3.2280e+02 + 2.8858e+02i


REG =

   13.6448


Eff =

   94.4252

(b)

VSLL =

  210.2884


SS =

   3.2280e+02 - 1.6862e+02i


REG =

   -4.4144


Eff =

   94.4252

例5.2

r = 0.036; g = 0; f = 60; 
L = 0.8; % 毫亨
C = 0.0112; % 微法
Length = 130; VR3ph = 325; 
VR = VR3ph/sqrt(3) + j*0; % kV (末端相电压) 
[Z, Y, ABCD] = rlc2abcd(r, L, C, g, f, Length) 
AR = acos(0.8); 
SR = 270*(cos(AR) + j*sin(AR)); % MVA (末端功率) 
IR = conj(SR)/(3*conj(VR)); % kA (末端电流) 
VsIs = ABCD* [VR; IR]; %列向量 [Vs; Is] 
Vs = VsIs(1); 
Vs3ph = sqrt(3)*abs(Vs); % kV(始端线电压) 
Is = VsIs(2); Ism = 1000*abs(Is); %A (始端电流) 
pfs= cos(angle(Vs)- angle(Is)); % (始端功率因数) 
Ss = 3*Vs*conj(Is); %MVA (始端功率) 
REG = (Vs3ph/abs(ABCD(1,1)) - VR3ph)/VR3ph *100; 
%空载时 IR  0 ，由式(5.10) VR NL ( )  VS 和式(5.9)可得 A
fprintf(' Is = %g A', Ism), fprintf(' pf = %g\n', pfs) 
fprintf(' Vs = %g L-L kV\n', Vs3ph) 
fprintf(' Ps = %g MW', real(Ss)), 
fprintf(' Qs = %g Mvar\n', imag(Ss)) 
fprintf(' Percent voltage Reg. = %g\n', REG)

运行结果

Enter 1 for Medium line or 2 for long line --> 1
 
 Nominal pi model
 ----------------

 Z = 4.68 + j 39.2071 ohms
 Y = 0 + j 0.000548899 Siemens
 
         0.98924     + j 0.0012844     4.68        + j 39.207       
 ABCD =                                                             
         -3.5251e-07 + j 0.00054595    0.98924     + j 0.0012844    
 

Z =

   4.6800 +39.2071i


Y =

   0.0000e+00 + 5.4890e-04i


ABCD =

   0.9892 + 0.0013i   4.6800 +39.2071i
  -0.0000 + 0.0005i   0.9892 + 0.0013i

 Is = 421.132 A pf = 0.869657
 Vs = 345.002 L-L kV
 Ps = 218.851 MW Qs = 124.23 Mvar
 Percent voltage Reg. = 7.30913

例5.3

z = 0.036 + j* 0.3; y = j*4.22/1000000; Length = 130; 
Vs3ph = 345; Ism = 0.4; %KA; 
As = -acos(0.95); 
Vs = Vs3ph/sqrt(3) + j*0; % kV (始端相电压) 
Is = Ism*(cos(As) + j*sin(As)); 
[Z,Y, ABCD] = zy2abcd(z, y, Length) 
VrIr = inv(ABCD)* [Vs; Is]; %列向量 [Vr; Ir] 
Vr = VrIr(1); 
Vr3ph = sqrt(3)*abs(Vr); % kV(末端线电压) 
Ir = VrIr(2); Irm = 1000*abs(Ir); % A (末端电流) 
pfr= cos(angle(Vr)- angle(Ir)); % (末端功率因数) 
Sr = 3*Vr*conj(Ir); % MVA (末端功率) 
REG = (Vs3ph/abs(ABCD(1,1)) - Vr3ph)/Vr3ph *100; 
fprintf(' Ir = %g A', Irm), fprintf(' pf = %g\n', pfr) 
fprintf(' Vr = %g L-L kV\n', Vr3ph) 
fprintf(' Pr = %g MW', real(Sr)) 
fprintf(' Qr = %g Mvar\n', imag(Sr)) 
fprintf(' Percent voltage Reg. = %g\n', REG)

运行结果

Enter 1 for Medium line or 2 for long line --> 1
 
 Nominal pi model
 ----------------

 Z = 4.68 + j 39 ohms
 Y = 0 + j 0.0005486 Siemens
 
         0.9893      + j 0.0012837     4.68        + j 39           
 ABCD =                                                             
         -3.5213e-07 + j 0.00054567    0.9893      + j 0.0012837     

Z =

   4.6800 +39.0000i


Y =

   0.0000e+00 + 5.4860e-04i


ABCD =

   0.9893 + 0.0013i   4.6800 +39.0000i
  -0.0000 + 0.0005i   0.9893 + 0.0013i

 Ir = 441.832 A pf = 0.887501
 Vr = 330.68 L-L kV
 Pr = 224.592 MW Qr = 116.612 Mvar
 Percent voltage Reg. = 5.45863

例5.4

z = 0.045 + j*.4; y = j*4.0/1000000; Length = 250; 
gamma = sqrt(z*y); Zc = sqrt(z/y); 
A = cosh(gamma*Length); B = Zc*sinh(gamma*Length); 
C = 1/Zc * sinh(gamma*Length); D = A; 
ABCD = [A B; C D] 
Z = Zc * sinh(gamma*Length) 
Y = 2/Zc * tanh(gamma*Length/2)

运行结果

ABCD =

   0.9504 + 0.0055i  10.8778 +98.3624i
  -0.0000 + 0.0010i   0.9504 + 0.0055i


Z =

  10.8778 +98.3624i


Y =

   0.0000 + 0.0010i

例5.5

L=0.97; C=0.0115; lngth=300; SR=800+j*600; VRLL = 500; 
w=2*pi*60; 
beta=w*sqrt(L*C*1e-9) 
Zc = sqrt(L/C*1e3) 
vel=1/sqrt(L*C*1e-9) 
lambda=vel/60 
betal=beta*lngth; 
VR=VRLL/sqrt(3); 
IR=conj(SR)/(3*conj(VR)); 
VS=cos(betal)*VR+j*Zc*sin(betal)*IR; 
VSLL=sqrt(3)*abs(VS) 
IS = j/Zc*sin(betal)*VR+cos(betal)*IR; 
ISM=abs(IS)*1000 
SS=3*VS*conj(IS) 
A=abs(cos(betal)); 
REG=(VSLL/A - VRLL)/VRLL*100

例5.6

VS=1.0; VR=0.9; lambda=5000; Zc=320; delta=36.87*pi/180; P=700; lngth=315; 
betal=2*pi/lambda*lngth 
SIL=P*sin(betal)/(VS*VR*sin(delta)) 
KVL = sqrt(Zc*SIL) 
disp('(b)') 
Xd=Zc*sin(betal) 
Pmax=KVL*0.9*KVL/Xd

例5.7

Zc=290.43; betal=0.3777; 
VSLL=500; 
VS=500/sqrt(3); 
disp('(a)') 
VRnl = VS/cos(betal); 
VRnlLL=sqrt(3)*VRnl 
disp('(b)') 
XLsh=sin(betal)/(1-cos(betal))*Zc 
QXL = VSLL^2/XLsh

例5.8

PR=800; QR=600; SR=PR+j*QR; VRLL=500; 
Zc=290.43; betal=0.3777; 
disp('(a)') 
Xd=Zc*sin(betal); 
delta=asin((PR*Xd)/(VRLL^2)) 
QR2=VRLL^2/Xd*cos(delta)-VRLL^2/Xd*cos(betal) 
Sc=j*QR2-j*QR 
Xc=VRLL^2/conj(Sc) 
C=1/(2*pi*60*abs(Xc)) 
disp('(b)') 
Xser = .4*Xd; 
Zd = j*(Xd-Xser); B = Zd 
Yd = j*2/Zc*tan(betal/2); A = 1+Zd*Yd/2 
VR = VRLL/sqrt(3); 
IR = conj(SR)/(3*VR) 
VS = A*VR + B*IR 
VSLL = sqrt(3)*abs(VS) 
Reg = (VSLL/abs(A) - VRLL)/VRLL*100

例5.9(习题同理)

lineperf 
%运行结果
 TRANSMISSION LINE MODEL 
 
 Type of parameters for input Select 
 
 Parameters per unit length 
 r(ohms), g(siemens) L(mH) & C (micro F) 1 
 
 Complex z and y per unit length 
 r+j*x (ohms/length), g+j*b (siemens/length) 2 
 
 Nominal pi or Eq. pi model 3 
 
 A, B, C, D constants 4 
 
 Conductor configuration and dimension 5 
 
 To quit 0 
 
Select number of menu --> 1 
 
Enter Line length = 300 
Enter Frequency in Hz = 60 
Enter line resistance/phase in ohms per unit length r = 0 
Enter line inductance/phase in millihenry per unit length L = 0.97 
Enter line capacitance/phase in micro F per unit length C = 0.0115 
Enter line conductance/phase in siemens per unit length g = 0 
Enter 1 for Medium line or 2 for long line --> 2 
Equivalent pi model 
 ------------------- 
 Z' = 0 + j 107.114 ohms 
 Y' = 0 + j 0.00131631 siemens 
 Zc = 290.427 + j 0 ohms 
 alpha l = 0 neper beta l = 0.377735 radian = 21.6426 鳿 n 
 0.9295 + j 0 0 + j 107.11 
 ABCD = 
 0 + j 0.0012699 0.9295 + j 0 
 
 Hit return to continue 
 TRANSMISSION LINE PERFORMANCE 
 ----------Analysis---------- Select 
 To calculate sending end quantities 
 for specified receiving end MW, Mvar 1 
 
 To calculate receiving end quantities 
 for specified sending end MW, Mvar 2 
 
 To calculate sending end quantities 
 when load impedance is specified 3 
 
 Open-end line & inductive compensation 4 
 
 Short-circuited line 5 
 
 Capacitive compensation 6 
 
 Receiving end circle diagram 7 
 
 Loadability curve and voltage profile 8 
 
 To quit 0 
 Select number of menu --> 6 
 CAPACITIVE COMPENSATION 
 Analysis Select 
 -------- ------ 
 Shunt capacitive compensation 1 
 
 Series capacitive compensation 2 
 
 Series & Shunt capacitive compensation 3 
 
 To quit 0 
Select number of menu --> 1 
Enter sending end line-line voltage kV = 500 
Enter desired receiving end line-line voltage kV = 500 
Enter receiving end voltage phase angle?(for Ref. enter 0 ) = 0 
Enter receiving end 3-phase real power MW = 800 
Enter receiving end 3-phase reactive power(+ for lagging & - for leading power factor) Mvar = 
600 
Shunt capacitive compensation 
 ----------------------------- 
 Vs = 500 kV (L-L) at 20.0454? 
 Vr = 500 kV (L-L) at 0? 
 Pload = 800 MW Qload = 600 Mvar 
 Load current = 1154.7 A at -36.8699? PFl = 0.8 lagging 
 Required shunt capcitor: 433.388 ohm, 6.12057 micro F, 576.85 Mvar 
 Shunt capacitor current = 666.089 A at 90? 
 Pr = 800.000 MW Qr = 23.150 Mvar 
 Ir = 924.147 A at -1.65752? PFr = 0.999582 lagging 
 Is = 924.147 A at 21.703? PFs = 0.999582 leading 
 Ps = 800.000 MW Qs = -23.150 Mvar 
 PL = 0.000 MW QL = -46.300 Mvar 
 Percent Voltage Regulation = 7.58444 
 Transmission line efficiency = 100