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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /2642/CH5/EX5.7 | |
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initial commit / add all books
Diffstat (limited to '2642/CH5/EX5.7')
-rwxr-xr-x | 2642/CH5/EX5.7/Ex5_7.sce | 31 |
1 files changed, 31 insertions, 0 deletions
diff --git a/2642/CH5/EX5.7/Ex5_7.sce b/2642/CH5/EX5.7/Ex5_7.sce new file mode 100755 index 000000000..053bc52fb --- /dev/null +++ b/2642/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,31 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 5 : DIRECT CURRENT MOTORS
+// Example : 5.7
+
+clc;clear; // clears the console and command history
+
+// Given data
+I_a1 = 65 // supply current to dc series motor in A
+V_t = 230 // supply voltage in V
+N_1 = 900 // speed in rpm
+R_a = 0.2 // armature resistance in ohm
+R_sh = 0.25 // field resistance in ohm
+I_a2 = 15 // supply current to dc series motor in A
+// phi_2 = 0.4*phi_1 value of flux
+
+
+// caclulations
+E_b1 = V_t-I_a1*(R_a+R_sh) // initial back emf in V
+E_b2 = V_t-I_a2*(R_a+R_sh) // final back emf in V
+// phi_2 = 0.4*phi_1 value of flux
+N_2 = (E_b2/E_b1)*(2.5)*N_1 // motor speed when line current is 15A in rpm
+
+// display the result
+disp("Example 5.7 solution");
+printf(" \n motor speed when line current is 15A \n N_2 = %.0f rpm \n", N_2 );
+
+printf(" NOTE: in question they given I_a1=56A , but in solution they took I_a1=65A");
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