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Dr. Ali Prf. Dr. Atef Lab:Enge422

JAZAN UNIVERSITY

COLLEGE OF ENGINEERING

Electrical Engineering Department

Experimental Lab of EngE 422

Electrical Power System 1

Prepared By

Prof. Dr Atef El-Emary Prof. Asc. Ali M. Yousef


Contents 1) Power System calculation in per unit using Matlab Software 2) Symmetrical fault analysis using Matlab software 3) Symmetrical fault analysis for large power system using Zbus 4) Symmetrical component analysis using matlab software 5) Un-symmetrical fault analysis 6) Transmission line compensation 7) Power system economical operation


Lab 1

Per unit Calculation

Figure below shows the one line diagram of a 3-ϕ system. By selecting a common base of 100 MVA and 22 KV on the generator side. Write a program using matlab/m-file to convert all impedances to in per unit. Also, draw an impedance diagram showing all impedances including the load impedance in per unit. Also, find the voltage at bus 1 and emf of generator and motor. The data are given as follows:

G: 90 MVA, 22 KV, X = 0.18 pu T1: 50 MVA, 22/220 KV, X = 0.1 pu T2: 40 MVA, 220/11 KV, X = 0.06 pu T3: 40 MVA, 22/110 KV, X = 0.064 pu T4: 40 MVA, 110/11 KV, X = 0.08 pu M: 66.5 MVA, 10.45 KV, X = 0.185 pu

Lines 1 and 2 have series reactances of 48.4 and 65.43 Ω, respectively. At bus 4, the 3-ϕ load absorbs 57 MVA at 10.45 KV and 0.6 pf lagging.


%Lab 1 % per unit calculation updating vb2=22*(220/22); vb4=220*(11/220); vb5=22*(110/22); vb56=vb5; xg=0.18*(100/90) xt1=0.1*(100/50) xt2=0.06*(100/40) xt3=0.064*(100/40) xt4=0.08*(100/40) xm=0.185*(100/66.5)*((10.45/11)^2) %motor in per unit zb2=(220^2)/100; zb5=(110^2)/100; xl1=48.4/zb2 xl2=65.43/zb5; sl=57*(0.6+i*sin(acos(0.6))); zl=10.45^2/conj(sl); zb4=11^2/100; zlpu=zl/zb4 % load impedance in per unit % motor operate at 0.8 pf lead, and v=10.45 kv % find v1 and emf of gen and motor v4=(10.45/11)*(1+i*sin(0)); sm=(66.5*(0.8-i*sin(acos(0.8))))/100 % per unit power of motor im=conj(sm)/conj(v4) il=v4/zlpu i1=im+il xeq=(i*(xt1+xl1+xt2)*(xt3+xl2+xt4))/(xt1+xl1+xt2+xt3+xl2+xt4) v1=v4+xeq*i1 eg=v1+i*xg*i1 em=v4-i*xm*im


Lab 2

Symmetrical fault analysis Using Thevenn's Theory

The above network showed the one line diagram of a simple three lines power system. Each generator is represented by EMF behind the transient reactance. All impedances are represented in per unit on a common 100 MVA base, and for simplicity resistance are neglected.

Write program using matlab software to find the fault current, the bus voltage and the line current during the fault when a balanced three phase fault with a fault impedance Zf = 0.16 pu occurs on bus 3 using Thevenn's theory.


% Lab 2 %fault analysis using Thevenn's therom % Lines & generators data z12 = j*.8; z13 = j*.4; z23 = j*.4; %voltage and current calculation z33 = j*.34; zf = j*.16; v0=[1; 1; 1]; i3f = v0(1)/(z33+zf) ig1 = 0.6/(0.4+0.6)*i3f ig2 = 0.4/(0.4+0.6)*i3f dv1 = 0 - j*0.2*ig1; dv2 = 0 - j*0.4*ig2; dv3 = j*0.16*i3f - v0(3); vf1 = v0(1) + dv1 vf2 = v0(2) + dv2 vf3 = v0(3) + dv3 i12 = (vf1 - vf2)/z12 i13 = (vf1 - vf3)/z13 i23 = (vf2 - vf3)/z23


Lab 3

Symmetrical fault analysis for large power system

Using Z-Bus

The above circuit shown the one line diagram of a simple three lines power system. Each generator is represented by emf behind the transient reactance. All impedances are represented in per unit on a common 100 MVA base, and for simplicity resistance are neglected.

Write program using matlab software to find the fault current, the bus voltage and the line current during the fault when a balanced three phase fault with a fault impedance Zf = 0.16 pu occurs on using Z-bus.


% Lab 3 symmetrical fault analysis for large power system % using Z-bus % read network data zdata=[0 1 0 0.2 0 2 0 0.4 1 2 0 0.8 1 3 0 0.4 2 3 0 0.4]; % construct bus Admittance Matrix nl=zdata(:,1); nr=zdata(:,2); r=zdata(:,3); x=zdata(:,4); nbr=length(zdata(:,1)); nbus = max(max(nl), max(nr)); z = r + i*x; %branch impedance y= ones(nbr,1)./z; %branch admittance ybus=zeros(nbus,nbus); % initialize Ybus to zero for k = 1:nbr; % formation of the off diagonal elements if nl(k) > 0 & nr(k) > 0 ybus(nl(k),nr(k)) = ybus(nl(k),nr(k)) - y(k) ybus(nr(k),nl(k)) = ybus(nl(k),nr(k)) end end for n = 1:nbus % formation of the diagonal elements for k = 1:nbr if nl(k) == n | nr(k) == n ybus(n,n) = ybus(n,n) + y(k) else, end end end % Z- bus calculation zbus = inv(ybus) % fault calculation zf = j*.16; v0=[1; 1; 1]; i3f = v0(1)/(zbus(3,3)+zf) vf = v0-i3f*zbus(:,3) i12 = (vf(1) - vf(2))/(j*0.8) i13 = (vf(1) - vf(3))/(j*0.4)


Lab 4 Symmetrical component

If Va= 1∟0o, Vb=1∟-30o, Vc=1∟60o Write program to transform the actual values of voltage to symmetrical components Va1, Va2, Va0


% Lab 4 %This program compute symmetrical components zero, %+ve and -ve from actual value of voltage tha=0.; %angle of phase a thb=-30.; %angle of phase b thc=60.; %angle of of phase c vabc=[1 1 1] %magnitude of phase voltages a, b, and c va=vabc(1)*(cos((tha*pi)/180.)+i*sin((tha*pi)/180.)); vb=vabc(2)*(cos((thb*pi)/180.)+i*sin((thb*pi)/180.)); vc=vabc(3)*(cos((thc*pi)/180.)+i*sin((thc*pi)/180.)); %calculation of transformation matrix t a1=cos((120*pi)/180.)+i*sin((120*pi)/180.); a2=cos((240*pi)/180.)+i*sin((240*pi)/180.); t=[1 1 1;1 a2 a1;1 a1 a2] tt=inv(t) vabct=[va;vb;vc] v012=tt*vabct


Lab 5

Unsymmetrical fault

The one line diagram of a simple power system is shown in Figure below. The actual of each generator is grounded through a current limiting reactor of 0.2573 pu on a 100 MVA base. The system data expressed in pu on a common 100 MVA base is tabulated below. The generators are running on no load at their rated voltage and rated frequency with their emfs in phase.

Determine the fault current for the following faults:

a) A balanced three phase fault at bus 3 through a fault impedance Zf = j0.1 pu b) A L-G fault at bus 3 through a fault impedance Zf = j0.1 pu c) A L-L fault at bus 3 through a fault impedance Zf = j0.1 pu d) A 2L-G fault at bus 3 through a fault impedance Zf = j0.1 pu

Item MVA

Base Voltage rating,

KV X1, pu X2, pu Xo, pu

G1 100 20 0.15 0.15 0.05 G2 100 20 0.15 0.15 0.05 T1 100 20/220 0.1 0.1 0.1 T2 100 20/220 0.1 0.1 0.1 L12 100 220 0.125 0.125 0.3 L13 100 220 0.15 0.15 0.35 L23 100 220 0.25 0.25 0.7125


% Lab 5

% Unsymmetrical Short Circuit (L-G, 2L-G, L-L)

z133 = j*0.22; z033 = j*0.35; zf = j*0.1; disp('(a) Balanced three-phase fault at bus 3') ia3F = 1.0/(z133+zf) disp('(b) Single line-to-ground fault at bus 3') i03 = 1.0/(z033 + 3*zf + z133 + z133); i012=[i03; i03; i03] % transformation matrix from 012 to actual abc a =cos(2*pi/3)+j*sin(2*pi/3); tt = [1 1 1; 1 a^2 a; 1 a a^2]; iabc3 = tt*i012 disp('(c) Line-to-line fault at bus 3') i13 = 1.0/(z133 + z133 + zf); i012 = [0; i13; -i13] iabc3 = tt*i012 disp('(b) Double line-to-ground fault at bus 3') i13 = 1/(z133 + z133*(z033+3*zf)/(z133+z033+3*zf)); i23 = -(1.0 - z133*i13)/z133; i03 = -(1.0 - z133*i13)/(z033+3*zf); i012 = [i03; i13; i23] iabc3 = tt*i012; %Fpol = [abs(Frec) angle(Frec)*180/pi]; %Iabc3p = rec2pol(Iabc3) iabc3p= [abs(iabc3) angle(iabc3)*180/pi]


Lab 6 Transmission Line Compensation

If l =200 mile Theta=23 degree Pf= 0.85 lag Draw relation between Vr/Vs with P/Po with and without degree compensation (Kse=0.3)


% Lab 6 %transmission Line Compensation (series compensation) l=200 % miles pf=0.85; delta=0:5:30; delt=delta*pi/180 theta=0.405 phi=31.87*pi/180 %(VR/VS)=V V=(cos(phi+delt))/(pf*cos(theta)) %(P/Po)=P P=(V.*cos(delt)-V.^2*cos(theta))/(tan(phi)*sin(theta)) %figure %plot(P,V) subplot(2, 1, 1), plot(P,V), xlabel('P in pu'), grid title('load ability curve') %Kse1=0.3 %(VR/VS)=V1 at Kse1=0.3 thetay=theta*sqrt(1-0.3) V1=(cos(phi+delt))/(pf*cos(thetay)); %(P/Po)=P1 at Ksh2=0.3 P1=(V1.*cos(delt)-V1.^2*cos(theta))/(tan(phi)*sin(thetay)) subplot(2, 1, 2), plot(P1,V1), xlabel('P in pu'), grid %figure %plot(P1,V1) %Ksh2=0.5 %(VR/VS)=V2 at Kse2=0.5 %thetat=theta*sqrt(1-0.5) %V2=(cos(phi+delt))/(pf*cos(thetat))


Lab 7 Economic operation

The fuel cost functions for three thermal plants in $/h are given by C1 = 500 +5.3P1 + 0.004 P1

2 C2 = 400 + 5.5 P2 + 0.008P2

2 C3 = 200 + 5.8P1 + 0.009P3

2 Where P1, P2, and P3 are in MW. The total load, PD, is 800 MW. Neglecting line losses and generator limits, find the optimal dispatch and the total cost in $/h

a) By analytical method b) By graphical demonstration c) By iterative technique using the gradient method


% Lab 7 %Power System Economical Operation % Iterative solution Using Newton method alpha =[500; 400; 200]; beta = [5.3; 5.5; 5.8]; gama=[.004; .006; .009]; PD=800; DelP = 10; % Error in DelP is set to a high value lambda = input('Enter estimated value of Lambda = '); fprintf('\n ') disp([' Lambda P1 P2 P3 DP'... ' grad Delambda']) iter = 0; % Iteration counter while abs(DelP) >= 0.001 % Test for convergence iter = iter + 1; % No. of iterations P = (lambda - beta)./(2*gama); DelP =PD - sum(P); % Residual J = sum( ones(length(gama), 1)./(2*gama)); % Gradient sum Delambda = DelP/J; % Change in variable disp([lambda, P(1), P(2), P(3), DelP, J, Delambda]) lambda = lambda + Delambda; % Successive solution end totalcost = sum(alpha + beta.*P + gama.*P.^2) %Graphical Demonstration of Example 7.4 axis([0 450 6.5 10.5]); P1=250:10:450; P2 = 150:10:350; P3=100:10:250; IC1= 5.3 + 0.008*P1; IC2= 5.5 + 0.012*P2; IC3= 5.8 + 0.018*P3; Px = 0:100:400; plot(P1, IC1, P2, IC2, P3, IC3, Px, lambda*ones(1, length(Px)),'-m'), xlabel('P, MW'), ylabel(' $/MWh'), grid

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