diff options
Diffstat (limited to '1319/CH11')
-rw-r--r-- | 1319/CH11/EX11.1/11_1.jpg | bin | 0 -> 135967 bytes | |||
-rw-r--r-- | 1319/CH11/EX11.1/11_1.sce | 24 | ||||
-rw-r--r-- | 1319/CH11/EX11.2/11_2.jpg | bin | 0 -> 128870 bytes | |||
-rw-r--r-- | 1319/CH11/EX11.2/11_2.sce | 26 | ||||
-rw-r--r-- | 1319/CH11/EX11.3/11_3.jpg | bin | 0 -> 128870 bytes | |||
-rw-r--r-- | 1319/CH11/EX11.3/11_3.sce | 21 |
6 files changed, 71 insertions, 0 deletions
diff --git a/1319/CH11/EX11.1/11_1.jpg b/1319/CH11/EX11.1/11_1.jpg Binary files differnew file mode 100644 index 000000000..6a00b2096 --- /dev/null +++ b/1319/CH11/EX11.1/11_1.jpg diff --git a/1319/CH11/EX11.1/11_1.sce b/1319/CH11/EX11.1/11_1.sce new file mode 100644 index 000000000..38f9a48e6 --- /dev/null +++ b/1319/CH11/EX11.1/11_1.sce @@ -0,0 +1,24 @@ +// Determine the size of the conductor for power and lighting circuit
+
+clc;
+clear;
+
+P5=3*800; // Load wattage in 5A circuit
+P15=2*1500;// Load wattage in 15 A circuit
+
+Pt=P5+P15; // Total Load
+
+// Assume a average of 0.8 pf, common for distribution systems
+
+pf=0.8;
+
+V=230; // Supply voltage
+
+I=Pt/(V*pf); // Current at 230 supply
+
+Isc=1.5*I; // Short Circuit Current
+
+printf('The Current is %g A and the short circuit current is %g A\n\n',I,Isc)
+printf('From the result sheet provided along with this code,\n for aluminium wire the size of the conductor comes out to be 25 mm^2.\nIn fact for 43 A it is 16 mm^2 but we should always go for one higher size of the conductor\n and hence we select conductor of size 25 mm^2 or 7/2.24 mm.')
+printf('\n \n Refer the table in the result sheet \n')
+
diff --git a/1319/CH11/EX11.2/11_2.jpg b/1319/CH11/EX11.2/11_2.jpg Binary files differnew file mode 100644 index 000000000..a032d2e23 --- /dev/null +++ b/1319/CH11/EX11.2/11_2.jpg diff --git a/1319/CH11/EX11.2/11_2.sce b/1319/CH11/EX11.2/11_2.sce new file mode 100644 index 000000000..f43a05e30 --- /dev/null +++ b/1319/CH11/EX11.2/11_2.sce @@ -0,0 +1,26 @@ +//Determine the size of the conductor at 25 m distance
+
+clc;
+clear;
+
+V=230; // Supply Voltage
+d=25; // Distance between mains and residence
+I=5; // Supply current
+
+pvd=1+((2/100)*V); // Permissible Voltage drop
+
+// From the table given in the result sheet along with this code, Minimum size of wire for 10A
+
+A=1.5*(10^-6);
+
+dm=2.3;
+Vd=d/dm; // Voltage drop at 10A
+
+Vd5=Vd/2; // Voltage drop at 5A
+
+//According to the table (Refer below) Permissible drop is 5.6 V
+
+printf('The pemissible voltage drop = %g V\n',pvd)
+printf('The voltage drop at 5 A = %g V\n',Vd5)
+printf('As the permissible drop is 5.6 volts \nand the conductor with 1.5 mm^2 section gives \nvoltage drop of 5.4 volts hence the suitable size is 1/1.40 mm.')
+printf('\n \n Refer the table in the result sheet \n')
diff --git a/1319/CH11/EX11.3/11_3.jpg b/1319/CH11/EX11.3/11_3.jpg Binary files differnew file mode 100644 index 000000000..a032d2e23 --- /dev/null +++ b/1319/CH11/EX11.3/11_3.jpg diff --git a/1319/CH11/EX11.3/11_3.sce b/1319/CH11/EX11.3/11_3.sce new file mode 100644 index 000000000..3c1bf8e8a --- /dev/null +++ b/1319/CH11/EX11.3/11_3.sce @@ -0,0 +1,21 @@ +// Size of conductor to be used for wiring a 10 kW 400V 3 Phase induction motor
+
+clc;
+clear;
+
+P=10*(10^3); // Power Rating
+V=400; // Voltage Rating
+f=50;// Frequency of operation
+
+// Assumptions Made
+eff=85/100; // Efficiency
+pf=0.8// Power Factor
+
+I=P/(sqrt(3)*V*eff*pf); // Current flowing in the conductor
+
+Is=2*I; //At the time of starting the induction motor may take 2 times the rated current
+
+printf('The Rated Current = %g A\n\n',I)
+printf(' At the time of starting the induction motor may take\n 2 times the rated current and hence starting current = %g A \n\n',Is)
+
+printf(' From the table corresponding to 42.4 A that is 43 A\n the wire used is 7 /1. 7 mm or 16 mm^2.\n')
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