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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /509/CH3 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
| download | Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.gz Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.bz2 Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.zip | |
initial commit / add all books
Diffstat (limited to '509/CH3')
| -rw-r--r-- | 509/CH3/EX3.2/3_2.sci | 27 | ||||
| -rw-r--r-- | 509/CH3/EX3.4/3_4.sci | 54 |
2 files changed, 81 insertions, 0 deletions
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);
+
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