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Diffstat (limited to 'Working_Examples/83/CH5')
23 files changed, 532 insertions, 0 deletions
diff --git a/Working_Examples/83/CH5/EX5.1/example_5_1.sce b/Working_Examples/83/CH5/EX5.1/example_5_1.sce new file mode 100755 index 0000000..d64c735 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.1/example_5_1.sce @@ -0,0 +1,43 @@ +//Chapter 5 +//Example 5.1 +//page 132 +//To find sending-end voltage and voltage regulation +clc;clear; + +load1=5000; //kW +pf=0.707; +Vr=10000; //receiving end voltage +R=0.0195*20; +X=2*%pi*50*0.63*10^-3*20; + +//to find sending end voltage and voltage regulation +I=load1*1000/(Vr*pf); +Vs=Vr+I*(R*pf+X*sin(acos(pf))); +voltage_regulation=(Vs-Vr)*100/Vr; +printf('\n\nReceiving current =I=%d A\n',I); +printf('Sending end voltage =Vs=%d V\n',Vs); +printf('Voltage Regulation=%0.2f %%',voltage_regulation); + +//to find the value of the capacitor to be connected in parallel to the load +voltage_regulation_desi=voltage_regulation/2; +Vs=(voltage_regulation_desi/100)*Vr+Vr; +//by solving the equations (i) and (ii) +pf=0.911; +Ir=549; +Ic=(Ir*(pf-%i*sin(acos(pf))))-(707*(0.707-%i*0.707)); +Xc=(Vr/imag(Ic)); +c=(2*%pi*50*Xc)^-1; +printf('\n\nCapacitance to be connected across the load so as to reduce voltage regulation by half of the above voltage regulation is given by :\n C = %d uF\n',c*10^6); + +//to find efficiency in both the cases +//case(i) +losses=I*I*R*10^-3; +n=(load1/(load1+losses))*100; +printf('\n Efficiency in : \nCase(i) \t n=%0.1f%%',n); +//caase(ii) +losses=Ir*Ir*R*10^-3; +n=(load1/(load1+losses))*100; +printf('\nCase(ii) \t n=%0.1f%%',n); + + + diff --git a/Working_Examples/83/CH5/EX5.1/result_example_5_1.txt b/Working_Examples/83/CH5/EX5.1/result_example_5_1.txt new file mode 100755 index 0000000..99064b3 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.1/result_example_5_1.txt @@ -0,0 +1,11 @@ + +Receiving current =I=707 A +Sending end voltage =Vs=12174 V +Voltage Regulation=21.75 % + +Capacitance to be connected across the load so as to reduce voltage regulation by half of the above voltage regulation is given by : + C = 87 uF + + Efficiency in : +Case(i) n=96.2% +Case(ii) n=97.7% diff --git a/Working_Examples/83/CH5/EX5.10/example_5_10.sce b/Working_Examples/83/CH5/EX5.10/example_5_10.sce new file mode 100755 index 0000000..af6af96 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.10/example_5_10.sce @@ -0,0 +1,39 @@ +//Chapter 5 +//Example 5.10 +//page 170 +//To determine the MVA rating of the shunt reactor +clear;clc; +v=275; +l=400; +R=0.035*l; +X=2*%pi*50*1.1*l*10^-3; +Z=R+%i*X; +Y=2*%pi*50*0.012*10^-6*l*%i; +A=1+(Y*Z/2); +B=Z; +Vs=275; +Vr=275; +r=(Vs*Vr)/abs(B); +Ce=abs(A/B)*Vr^2; +printf('Radius of the receiving-end circle=%0.1f MVA\n\n',r); +printf('Location of the center of receiving-end circle= %0.1f MVA\n\n',Ce); +printf('From the graph, 55 MVA shunt reactor is required\n\n'); +theta=180+82.5; +x=-75:0.01:450; +a=Ce*cosd(theta); //to draw the circle +b=Ce*sind(theta); +y=sqrt(r^2-(x-a)^2)+b; +x1=a:0.001:0; +y1=tand(theta)*x1; +plot(x,y,x1,y1); +title('Circle diagram for example 5.10'); +xlabel('MW'); +ylabel('MVAR'); +plot(a,b,'markersize',150); +xgrid(2) +set(gca(),"grid",[0,0]) +get("current_axes"); +xstring (-75,25,'55 MVAR'); +xstring(-75,-25,'83.5 deg'); +xstring(-20,-300,'487.6 MVA'); +xstring(300,-100,'544.3 MVA');
\ No newline at end of file diff --git a/Working_Examples/83/CH5/EX5.10/result_example_5_10.txt b/Working_Examples/83/CH5/EX5.10/result_example_5_10.txt new file mode 100755 index 0000000..abd9fbc --- /dev/null +++ b/Working_Examples/83/CH5/EX5.10/result_example_5_10.txt @@ -0,0 +1,7 @@ + +Radius of the receiving-end circle=544.3 MVA + +Location of the center of receiving-end circle= 487.6 MVA + +From the graph, 55 MVA shunt reactor is required + diff --git a/Working_Examples/83/CH5/EX5.10/result_fig_example_5_10.jpeg b/Working_Examples/83/CH5/EX5.10/result_fig_example_5_10.jpeg Binary files differnew file mode 100755 index 0000000..522057e --- /dev/null +++ b/Working_Examples/83/CH5/EX5.10/result_fig_example_5_10.jpeg diff --git a/Working_Examples/83/CH5/EX5.11/example_5_11.sce b/Working_Examples/83/CH5/EX5.11/example_5_11.sce new file mode 100755 index 0000000..19bd34f --- /dev/null +++ b/Working_Examples/83/CH5/EX5.11/example_5_11.sce @@ -0,0 +1,19 @@ +//Chapter 5 +//Example 5.11 +//page 172 +//To determine sending-end voltage.maximum power delivered +clear;clc; + +A=0.93*(cosd(1.5)+%i*sind(1.5)); +B=115*(cosd(77)+%i*sind(77)); +Vr=275; +Ce=abs(A/B)*Vr^2; +printf('Centre of the receiving end circle is = %0.1f MVA\n\n',Ce); +CrP=850;Vs=CrP*abs(B)/Vr; +printf('(a) From the diagram,\n\tCrP=%d \n \tSending end voltage|Vs|= %0.1f kV\n\n',CrP,Vs); +Vs=295; //given +r=(Vs*Vr)/abs(B); +Pr_m=556; //from the diagram +printf('(b) Radius of the circle diagram = %0.1f MVA\n\t PR_max=%d MW\n\n',r,Pr_m); +Ps=295; //from the diagram; +printf('(c) Additional MVA to be drawn from the line is = P''S=%d MVAR\n\n',Ps);
\ No newline at end of file diff --git a/Working_Examples/83/CH5/EX5.11/result_example_5_11.txt b/Working_Examples/83/CH5/EX5.11/result_example_5_11.txt new file mode 100755 index 0000000..322c788 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.11/result_example_5_11.txt @@ -0,0 +1,12 @@ + + +Centre of the receiving end circle is = 611.6 MVA + +(a) From the diagram, + CrP=850 + Sending end voltage|Vs|= 355.5 kV + +(b) Radius of the circle diagram = 705.4 MVA + PR_max=556 MW + +(c) Additional MVA to be drawn from the line is = P'S=295 MVAR diff --git a/Working_Examples/83/CH5/EX5.2/example_5_2.sce b/Working_Examples/83/CH5/EX5.2/example_5_2.sce new file mode 100755 index 0000000..925e68d --- /dev/null +++ b/Working_Examples/83/CH5/EX5.2/example_5_2.sce @@ -0,0 +1,28 @@ +//Chapter 5 +//Example 5.2 +//page 134 +//To find voltage at the bus at the power station end +clc;clear; + +base_MVA=5; +base_kV=33; +pf=0.85; +cable_impedance=(8+%i*2.5); +cable_impedance=cable_impedance*base_MVA/(base_kV^2); + +transf_imp_star=(0.06+%i*0.36)/3; //equivalent star impedance of winding of the transformer +Zt=(transf_imp_star*5/(6.6^2))+((0.5+%i*3.75)*5/(33^2)); +total=cable_impedance+2*Zt; + +load_MVA=1; +load_voltage=6/6.6; +load_current=1/load_voltage; + +Vs=load_voltage+load_current*(real(total)*pf+imag(total)*sin(acos(pf))); +Vs=Vs*6.6; +printf('\n\nCable impedance= (%0.3f+j%0.4f) pu\n',real(cable_impedance),imag(cable_impedance)); +printf('\nEquivalent star impedance of 6.6kV winding of the transformer =(%0.2f+j%0.2f) pu\n',real(transf_imp_star),imag(transf_imp_star)); +printf('\nPer unit transformer impedance,Zt=(%0.4f+j%0.3f) pu\n',real(Zt),imag(Zt)); +printf('\nTotal series impedance=(%0.3f+j%0.3f) pu\n',real(total),imag(total)); +printf('\nSending end Voltage =|Vs|=%0.2fkV (line-to-line)',Vs); + diff --git a/Working_Examples/83/CH5/EX5.2/result_example_5_2.txt b/Working_Examples/83/CH5/EX5.2/result_example_5_2.txt new file mode 100755 index 0000000..f9b1d06 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.2/result_example_5_2.txt @@ -0,0 +1,10 @@ + +Cable impedance= (0.037+j0.0115) pu + +Equivalent star impedance of 6.6kV winding of the transformer =(0.02+j0.12) pu + +Per unit transformer impedance,Zt=(0.0046+j0.031) pu + +Total series impedance=(0.046+j0.073) pu + +Sending end Voltage =|Vs|=6.56kV (line-to-line) diff --git a/Working_Examples/83/CH5/EX5.3/example_5_3.sce b/Working_Examples/83/CH5/EX5.3/example_5_3.sce new file mode 100755 index 0000000..195c0b5 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.3/example_5_3.sce @@ -0,0 +1,24 @@ +//Chapter 5 +//Example 5.3 +//page 135 +//problem with mixed end condition +clc;clear; +Vr=3000; //receiving end voltage +pfs=0.8; //sending end power factor +Ps=2000*10^3; //sending end active power +z=0.4+%i*0.4; //series impedance +Ss=Ps/pfs; //sending end VA +Qs=Ss*sqrt(1-pfs^2); //sending end reacive power + +//by substituting all the values to the equation (iii) +deff('[y]=fx(I)',"y=(Vr^2)*(I^2)+2*Vr*(I^2)*(real(z)*((Ps-real(z)*(I^2))/Vr)+imag(z)*((Qs-imag(z)*(I^2))/Vr))+(abs(z))^2*(I^4)-(Ss^2)"); +I=fsolve(100,fx); + +pfR=(Ps-real(z)*(I^2))/(Vr*I); //Cos(phi_r) +Pr=Vr*I*pfR; +Vs=(Ps/(I*pfs)); + +printf('\nLoad Current |I|= %0.2f A',I); +printf('\nLoad Pr=%d W',Pr); +printf('\nReceiving end power factor=%0.2f',pfR); +printf('\nSupply Voltage=%0.2fV',Vs); diff --git a/Working_Examples/83/CH5/EX5.3/result_example_5_3.txt b/Working_Examples/83/CH5/EX5.3/result_example_5_3.txt new file mode 100755 index 0000000..537ce19 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.3/result_example_5_3.txt @@ -0,0 +1,6 @@ + +Load Current |I|= 733.14 A +Load Pr=1785001 W +Receiving end power factor=0.81 +Supply Voltage=3409.99V + diff --git a/Working_Examples/83/CH5/EX5.4/example_5_4.sce b/Working_Examples/83/CH5/EX5.4/example_5_4.sce new file mode 100755 index 0000000..7a4015d --- /dev/null +++ b/Working_Examples/83/CH5/EX5.4/example_5_4.sce @@ -0,0 +1,24 @@ +//Chapter 5 +//Example 5.4 +//page 138 +//to find sending end voltage and voltage regulation of a medium transmission line system +clear;clc; +D=300; +r=0.8; +L=0.461*log10(D/(0.7788*r)); +C=0.0242/(log10(D/r)); +R=0.11*250; +X=2*%pi*50*L*0.001*250; +Z=R+%i*X; +Y=%i*2*%pi*50*C*0.000001*250; +Ir=((25*1000)/(132*sqrt(3)))*(cosd(-36.9)+%i*sind(-36.9)); +Vr=(132/sqrt(3)); +A=(1+(Y*Z/2)); +Vs=A*Vr+Z*Ir*10^(-3); +printf('\n\nVs(per phase)=(%0.2f+%0.2f)kV',real(Vs),imag(Vs)); +Vs=abs(Vs)*sqrt(3); +printf('\n\n|Vs|(line)=%d kV',Vs); +Vr0=Vs/abs(A); +printf('\n\n|Vr0|(line no load)=%0.1fkV',Vr0); +Vol_regu=(Vr0-132)/132; +printf('\n\nVoltage Regulation=%0.1f%%\n\n',Vol_regu*100); diff --git a/Working_Examples/83/CH5/EX5.4/result_example_5_4.txt b/Working_Examples/83/CH5/EX5.4/result_example_5_4.txt new file mode 100755 index 0000000..4a47cad --- /dev/null +++ b/Working_Examples/83/CH5/EX5.4/result_example_5_4.txt @@ -0,0 +1,12 @@ + + +Vs(per phase)=(82.26+7.46)kV + +|Vs|(line)=143 kV + +|Vr0|(line no load)=148.4kV + +Voltage Regulation=12.4% + + + diff --git a/Working_Examples/83/CH5/EX5.5/example_5_5.sce b/Working_Examples/83/CH5/EX5.5/example_5_5.sce new file mode 100755 index 0000000..8c62731 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.5/example_5_5.sce @@ -0,0 +1,29 @@ +//Chapter 5 +//Example 5.5 +//page 147 +//to find maximum permissible length and and frequency +clc;clear; +R=0.125*400; +X=0.4*400; +Y=2.8*(10^-6)*400*%i; +Z=R+X*%i; + +//(i) At no-load +A=1+(Y*Z/2); +C=Y*(1+Y*Z/6); +VR_line=220000/abs(A); +Is=abs(C)*VR_line/sqrt(3); +printf('\n\n |VR|line = %d kV',VR_line/1000); +printf('\n |Is| = %d A',Is); + +//(ii) to find maximum permissible length +//By solving the equations shown in the book,we get +l=sqrt((1-0.936)/(0.56*10^(-6))); +printf('\n\n Maximum permissible length of the line = %d km',l); + +//(iii) to find maximum permissible frequency for the case(i) +//By solving the equations shown in the book,we get +f=sqrt(((1-0.88)*50*50)/(0.5*1.12*10^-3*160)); +printf('\n\n Maximum permissible frequency = %0.1f Hz\n\n',f); + + diff --git a/Working_Examples/83/CH5/EX5.5/result_example_5_5.txt b/Working_Examples/83/CH5/EX5.5/result_example_5_5.txt new file mode 100755 index 0000000..d027f9c --- /dev/null +++ b/Working_Examples/83/CH5/EX5.5/result_example_5_5.txt @@ -0,0 +1,11 @@ + + + + |VR|line = 241 kV + |Is| = 151 A + + Maximum permissible length of the line = 338 km + + Maximum permissible frequency = 57.9 Hz + + diff --git a/Working_Examples/83/CH5/EX5.6/example_5_6.sce b/Working_Examples/83/CH5/EX5.6/example_5_6.sce new file mode 100755 index 0000000..21b98e4 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.6/example_5_6.sce @@ -0,0 +1,35 @@ +//Chapter 5 +//Example 5.6 +//page 149 +//to find incident and reflected voltages +clear;clc; + +R=0.125; +X=0.4; +y=%i*2.8*10^(-6); +z=R+%i*X; + +r=sqrt(y*z); //propogation constant +a=real(r); //attenuation constant +b=imag(r); //phase constant + +//(a) At the receiving-end; +Vr=220000; +Inci_vol=Vr/(sqrt(3)*2); +Refl_vol=Vr/(sqrt(3)*2); +printf('\n\nIncident Vvoltage=%0.2f kV',Inci_vol/1000); +printf('\nReflected Vvoltage=%0.2f kV',Refl_vol/1000); + +//(b) At 200km from the receiving-end +x=200; +Inci_vol=Inci_vol*exp(a*x)*exp(%i*b*x); +Refl_vol=Refl_vol*exp(-a*x)*exp(-%i*b*x); +printf('\n\nIncident voltage=%0.2f @ %0.1f deg kV',abs(Inci_vol)/1000,atand(imag(Inci_vol)/real(Inci_vol))); +printf('\nReflected voltage=%0.2f @ %0.1f deg kV',abs(Refl_vol)/1000,atand(imag(Refl_vol)/real(Refl_vol))); + +//(c) Resultant voltage at 200km from the receiving-end +res=Inci_vol+Refl_vol; +printf('\n\nResultant line-to-line voltage at 200km =%0.2f kV',abs(res)*sqrt(3)/1000); + + + diff --git a/Working_Examples/83/CH5/EX5.6/result_example_5_6.txt b/Working_Examples/83/CH5/EX5.6/result_example_5_6.txt new file mode 100755 index 0000000..cd94d77 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.6/result_example_5_6.txt @@ -0,0 +1,10 @@ + + + +Incident Vvoltage=63.51 kV +Reflected Vvoltage=63.51 kV + +Incident voltage=65.62 @ 12.3 deg kV +Reflected voltage=61.47 @ -12.3 deg kV + +Resultant line-to-line voltage at 200km =215.09 kV diff --git a/Working_Examples/83/CH5/EX5.7/example_5_7.sce b/Working_Examples/83/CH5/EX5.7/example_5_7.sce new file mode 100755 index 0000000..dbd47d2 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.7/example_5_7.sce @@ -0,0 +1,83 @@ +//Chapter 5 +//Example 5.7 +//page 138 +//to tabulate characteristics of a system using different methods +clear;clc; + +Z=40+125*%i; +Y=%i*10^(-3); +Ir=((50*10^6)/(220000*0.8*sqrt(3)))*(cosd(-36.9)+%i*sind(-36.9)); +Vr=220000/sqrt(3); + +//(a) Short line approximation +Vs=Vr+Ir*Z; +Vs_line1=Vs*sqrt(3); +Is1=Ir; +pfs1=cos(atan(imag(Vs)/real(Vs))+acos(0.8)); +Ps1=sqrt(3)*abs(Vs_line1)*abs(Is1)*pfs1; + +//(b) Nominal pi method +A=1+Y*Z/2; +D=A; +B=Z; +C=Y*(1+Y*Z/4); +Vs=A*Vr+B*Ir; +Is2=C*Vr+D*Ir; +Vs_line2=sqrt(3)*Vs; +pfs2=cos(atan(imag(Is2)/real(Is2))-atan(imag(Vs)/real(Vs))); +Ps2=sqrt(3)*abs(Vs_line2)*abs(Is2)*pfs2; + +//(c) Exact transmission line equations +rl=sqrt(Z*Y); //propogation constant +Zc=sqrt(Z/Y); //characteristic impedance +A=cosh(rl); +B=Zc*sinh(rl); +C=sinh(rl)/Zc; +D=cosh(rl); +Vs=A*Vr+B*Ir; +Is3=C*Vr+D*Ir; +Vs_line3=sqrt(3)*Vs; +pfs3=cos(atan(imag(Is3)/real(Is3))-atan(imag(Vs)/real(Vs))); +Ps3=sqrt(3)*abs(Vs_line3)*abs(Is3)*pfs3; + +//(d) Approximation +A=(1+Y*Z/2); +B=Z*(1+Y*Z/6); +C=Y*(1+Y*Z/6); +D=A; +Vs=A*Vr+B*Ir; +Is4=C*Vr+D*Ir; +Vs_line4=sqrt(3)*Vs; +pfs4=cos(atan(imag(Is4)/real(Is4))-atan(imag(Vs)/real(Vs))); +Ps4=sqrt(3)*abs(Vs_line4)*abs(Is4)*pfs4; + +//converting all the values to their standard form before writing it to table + +//voltage to kV +Vs_line1=abs(Vs_line1)/1000; +Vs_line2=abs(Vs_line2)/1000; +Vs_line3=abs(Vs_line3)/1000; +Vs_line4=abs(Vs_line4)/1000; + +//Current to kA +Is1=Is1/1000; +Is2=Is2/1000; +Is3=Is3/1000; +Is4=Is4/1000; + +//power to MW5 +Ps1=Ps1/1000000; +Ps2=Ps2/1000000; +Ps3=Ps3/1000000; +Ps4=Ps4/1000000; + +//preparinf table +printf("\n\n_______________________________________________________________________________________________________________________"); +printf('\n \t\tShort line \t\t Nominal Pi \t\t Exact \t\t Approximation'); +printf("\n_______________________________________________________________________________________________________________________"); +printf('\n|Vs|line\t\t%0.2fkV \t\t %0.2fkV\t\t %0.2fkV \t\t %0.2fkV',Vs_line1,Vs_line2,Vs_line3,Vs_line4); +printf('\nIs \t\t%0.3f@%0.1fdeg kA \t\t%0.2f@%0.1fdeg kA\t\t%0.4f@%0.1fdeg kA\t%0.2f@%0.1fdeg kA',abs(Is1),tand(imag(Is1)/real(Is1)),abs(Is2),tand(imag(Is2)/real(Is2)),abs(Is3),tand(imag(Is3)/real(Is3)),abs(Is4),tand(imag(Is4)/real(Is4))); +printf('\npfs \t\t%0.3f lagging \t\t%0.3f leading \t\t%0.3f leading \t\t%0.3f leading',pfs1,pfs2,pfs3,pfs4); +printf('\nPs \t\t%0.2f MW \t\t%0.2f MW \t\t%0.2f MW \t\t%0.2f MW',Ps1,Ps2,Ps3,Ps4); +printf("\n_______________________________________________________________________________________________________________________\n\n\n"); + diff --git a/Working_Examples/83/CH5/EX5.7/result_example_5_7.txt b/Working_Examples/83/CH5/EX5.7/result_example_5_7.txt new file mode 100755 index 0000000..bb57c69 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.7/result_example_5_7.txt @@ -0,0 +1,14 @@ + + + +_______________________________________________________________________________________________________________________ + Short line Nominal Pi Exact Approximation +_______________________________________________________________________________________________________________________ +|Vs|line 251.34kV 238.08kV 237.40kV 237.28kV +Is 0.164@-0.0deg kA 0.13@0.0deg kA 0.1289@0.0deg kA 0.13@0.0deg kA +pfs 0.746 lagging 0.988 leading 0.987 leading 0.987 leading +Ps 53.23 MW 52.19 MW 52.28 MW 52.24 MW +_______________________________________________________________________________________________________________________ + + + diff --git a/Working_Examples/83/CH5/EX5.8/example_5_8.sce b/Working_Examples/83/CH5/EX5.8/example_5_8.sce new file mode 100755 index 0000000..6479e6a --- /dev/null +++ b/Working_Examples/83/CH5/EX5.8/example_5_8.sce @@ -0,0 +1,56 @@ +//Chapter 5 +//Example 5.8 +//page 162 +//to estimate the torque angle and station powerfactor +clear;clc; +Sd1=15+%i*5; +Sd2=25+%i*15; +//case(a) cable impedance=j0.05pu +r=0; +x=%i*0.05; +PG1=20; +PG2=20; +Ps=5;Pr=5; +V1=1; +V2=1; +d1=asind(Ps*abs(x)/(V1*V2)); //delta1 +V1=V1*(cosd(d1)+%i*sind(d1)); +Qs=((abs(V1)^2)/abs(x))-((abs(V1)*abs(V2))*cosd(d1)/(abs(x))); +Qr=(((abs(V1)*abs(V2))*cosd(d1)/(abs(x)))-(abs(V1)^2)/abs(x)); +Ql=Qs-Qr; +Ss=Ps+%i*Qs; +Sr=Pr+%i*Qr; +Sg1=Sd1+Ss; +Sg2=Sd2-Sr; +pf1=cos(atan(imag(Sg1)/real(Sg1))); +pf2=cos(atan(imag(Sg2)/real(Sg2))); +printf('\n\nCase(a)\nTotal load on station1=%d+j%0.3f pu',real(Sg1),imag(Sg1)); +printf('\nPower factor of station1=%0.3f pu lagging',pf1); +printf('\n\Total load on station2=%d+j%0.3f pu',real(Sg2),imag(Sg2)); +printf('\nPower factor of station2=%0.3f pu lagging',pf2); +//case(b) cable impedance=0.005+j0.05; +r=0.005; +PG1=20; +V1=1;V2=1; +Ps=5; +//from the eq(i) in the textbook,we can calculate d1 +z=r+x; +theta=atand(imag(z)/real(z)); +z=abs(z); +d1=acosd(z*(V1^2*cosd(theta)/z-Ps)/(V1*V2))-theta; +Qs=(V1^2*sind(theta)/z)-(V1*V2*sind(theta+d1)/z); +Qg1=5+Qs; +Pr=(V1*V2*cosd(theta-d1)/z)-(V1^2*cosd(theta)/z); +Pg2=25-Pr; +Qr=(V1*V2*sind(theta-d1)/z)-(V1^2*sind(theta)/z); +Qg2=15-Qr; +Ss=Ps+%i*Qs; +Sr=Pr+%i*Qr; +Sg1=Sd1+Ss; +Sg2=Sd2-Sr; +pf1=cos(atan(imag(Sg1)/real(Sg1))); +pf2=cos(atan(imag(Sg2)/real(Sg2))); +printf('\n\nCase(b)\nTotal load on station1=%d+j%0.3f pu',real(Sg1),imag(Sg1)); +printf('\nPower factor of station1=%0.3f pu lagging',pf1); +printf('\n\Total load on station2=%d+j%0.3f pu',real(Sg2),imag(Sg2)); +printf('\nPower factor of station2=%0.3f pu lagging\n\n',pf2); diff --git a/Working_Examples/83/CH5/EX5.8/result_example_5_8.txt b/Working_Examples/83/CH5/EX5.8/result_example_5_8.txt new file mode 100755 index 0000000..b7f4dee --- /dev/null +++ b/Working_Examples/83/CH5/EX5.8/result_example_5_8.txt @@ -0,0 +1,15 @@ + + +Case(a) +Total load on station1=20+j5.635 pu +Power factor of station1=0.963 pu lagging +Total load on station2=20+j15.635 pu +Power factor of station2=0.788 pu lagging + +Case(b) +Total load on station1=20+j5.132 pu +Power factor of station1=0.969 pu lagging +Total load on station2=20+j16.119 pu +Power factor of station2=0.781 pu lagging + + diff --git a/Working_Examples/83/CH5/EX5.9/example_5_9.sce b/Working_Examples/83/CH5/EX5.9/example_5_9.sce new file mode 100755 index 0000000..7ca7bf2 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.9/example_5_9.sce @@ -0,0 +1,33 @@ +//Chapter 5 +//Example 5.9 +//page 165 +//to determine power,voltage,compensating equipment rating +clear;clc; +A=0.85; +B=200; + +//case(a) +Vs=275000; +Vr=275000; +a=5;b=75; //alpha and beta +Qr=0; +//from equation 5.62 +d=b-asind((B/(Vs*Vr))*(Qr+(A*Vr^2*sind(b-a)/B))); //delta +Pr=(Vs*Vr*cosd(b-d)/B)-(A*Vr^2*cosd(b-a)/B); +printf('\n\ncase(a)\nPower at unity powerfactor that can be received =%0.1f MW',Pr/10^6); + +//case(b) +Pr=150*10^6; +d=b-acosd((B/(Vs*Vr))*(Pr+(A*Vr^2*cosd(b-a)/B))); //delta +Qr=(Vs*Vr*sind(b-d)/B)-(A*Vr^2*sind(b-a)/B); +Qc=-Qr; +printf('\n\ncase(b)\nRating of the compensating equipment = %0.2f MVAR',Qc/10^6); +printf('\ni.e the compensating equipment must feed positive VARs into the line'); + + +//case(c) +Pr=150*10^6; +Vs=275000; +//by solving the two conditions given as (i) and (ii), we get +Vr=244.9*10^3; +printf('\n\ncase(c)\nReceiving end voltage = %0.1f kV',Vr/1000); diff --git a/Working_Examples/83/CH5/EX5.9/result_example_5_9.txt b/Working_Examples/83/CH5/EX5.9/result_example_5_9.txt new file mode 100755 index 0000000..18b9c60 --- /dev/null +++ b/Working_Examples/83/CH5/EX5.9/result_example_5_9.txt @@ -0,0 +1,11 @@ + + +case(a) +Power at unity powerfactor that can be received =117.6 MW + +case(b) +Rating of the compensating equipment = 27.40 MVAR +i.e the compensating equipment must feed positive VARs into the line + +case(c) +Receiving end voltage = 244.9 kV |
