From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001 From: prashantsinalkar Date: Tue, 10 Oct 2017 12:27:19 +0530 Subject: initial commit / add all books --- 509/CH3/EX3.2/3_2.sci | 27 ++++++++++++++++++++++++++ 509/CH3/EX3.4/3_4.sci | 54 +++++++++++++++++++++++++++++++++++++++++++++++++++ 2 files changed, 81 insertions(+) create mode 100644 509/CH3/EX3.2/3_2.sci create mode 100644 509/CH3/EX3.4/3_4.sci (limited to '509/CH3') diff --git a/509/CH3/EX3.2/3_2.sci b/509/CH3/EX3.2/3_2.sci new file mode 100644 index 000000000..37b922966 --- /dev/null +++ b/509/CH3/EX3.2/3_2.sci @@ -0,0 +1,27 @@ +//Chapter 3 Example 2// +clc +clear +//supply voltage=v,from the figure we can get the required values// +v=220;// in volts// +z1=2+%i*8;// in ohms// +x1=8; +z2=-%i*6; +r=5; +//let the equivalent impedence =zeq// +zeq=(z1)*(z2)/(z1+z2); +printf("\n Equivalent impedence = %.2f%.2fi ohms\n",zeq,imag(zeq)); +//current in the circuit=i// +i=v/(zeq+r); +printf("\n Load current I = %.2f% +.2fi A\n",i,imag(i)); +//power in the 5 ohm resistor=p// +p=abs(i)^2*r; +printf("\n Power in the 5ohm resistor = %.2f W\n",p); +// current in branch ab =i1,current in branch cd=i2// +i1=i*(z2)/(z1+z2); +printf("\n Current in branch ab = %.2f%+.2fi A\n",i1,imag(i1)); +printf("\n = %.2f A\n",abs(i1)); +i2=i*(z1)/(z1+z2); +printf("\n Current in branch cd= %.2f%.2fi A\n",i2,imag(i2)); +printf("\n = %.2f A\n",abs(i2)); + + diff --git a/509/CH3/EX3.4/3_4.sci b/509/CH3/EX3.4/3_4.sci new file mode 100644 index 000000000..73d81dbd0 --- /dev/null +++ b/509/CH3/EX3.4/3_4.sci @@ -0,0 +1,54 @@ +// Chapter 3 Example 4// +clc +clear +//base powers of generators be m1,m2,m3 and base voltages be v1,v2,v3,secondary voltage=v// +v=132; +m1=100;// in MVA// +v1=11; // in kV// +m2=150; +v2=16; +m3=200; +v3=21; +//reactance of generator 1,2,3 are x1,x2,x3 respectively and X1,X2,X3 are the new reactances// +x1=0.25// in p.u// +x2=0.1; +x3=0.15; +X1=x1; +X2=x2*(m1/m2)*(v2/v1)^2; +X3=x3*(m1/m3)*(v3/v1)^2; +printf("\n Reactance of generator 1 = %.2f pu\n",X1); +printf("\n Reactance of generator 2 on the new base of generator1 = %.3f pu\n",X2); +printf("\n Reactance of generator 3 on the new base of generator1 = %.3f pu\n",X3); +// let T1,T2,T3 are the new per unit reactances on new base values and t1,t2,t3 are the old values// +t1=0.05;// in p.u// +t2=0.10; +t3=0.05; +// let the ratings of the tranformers be v1,v2,v3// +tv1=150;// in MVA// +tv2=200; +tv3=250; +T1=t1*(m1/tv1)*(v1/v1)^2; +T2=t2*(m1/tv2)*(v2/v1)^2; +T3=t3*(m1/tv3)*(v3/v1)^2; +printf("\n Per unit reactance of transformer 1 = %.3f pu\n",T1); +printf("\n Per unit reactance of transformer 2 = %.3f pu\n",T2); +printf("\n Per unit reactance of transformer 3 = %.3f pu\n",T3); +// line reactances for 1,2,3 are r1,r2,r3 respectively,base impedence=zb// +r1=100;// in ohms// +r2=50; +r3=80; +zb=(v)^2/(m1); +printf("\n The base impedence Zb = %.2f ohm\n",zb); +// per unit reactances of lines 1,2,3 are pu1,pu2,pu3 respectively// +pu1=r1/zb; +pu2=r2/zb; +pu3=r3/zb; +printf("\n Per unit reactance of line 1 = %.3f pu\n",pu1); +printf("\n Per unit reactance of line 2 = %.3f pu\n",pu2); +printf("\n Per unit reactance of line 3 = %.3f pu\n",pu3); + + + + + + -- cgit