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Diffstat (limited to '2642/CH10/EX10.3')
-rwxr-xr-x | 2642/CH10/EX10.3/Ex10_3.sce | 32 |
1 files changed, 32 insertions, 0 deletions
diff --git a/2642/CH10/EX10.3/Ex10_3.sce b/2642/CH10/EX10.3/Ex10_3.sce new file mode 100755 index 000000000..5af584787 --- /dev/null +++ b/2642/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,32 @@ +// FUNDAMENTALS OF ELECTICAL MACHINES
+// M.A.SALAM
+// NAROSA PUBLISHING HOUSE
+// SECOND EDITION
+
+// Chapter 10 : SYNCHRONOUS MOTOR
+// Example : 10.3
+
+clc;clear; // clears the console and command history
+
+// Given data
+V = 440 // supply voltage in V
+R_a = 1.5 // per phase resistance in ohm
+X_a = 8 // syncronous reactance in ohm
+P = 4 // number of poles
+f = 50 // supply frequency in Hz
+pf = 0.9 // leading power factor
+I_a = 50 // armature current in A
+
+// caclulations
+V_t = V/sqrt(3) // terminal voltage per phase in V
+phi = acosd(pf) // angle in degree
+Z_s = R_a+%i*X_a // impedance per phase ohm
+E_r = I_a*abs(Z_s) // resultant voltage due to impedance in V
+bet = atand(X_a/R_a)
+E_f = sqrt(V_t^2+E_r^2-2*V_t*E_r*cosd(bet+phi)) // excitation voltage per phase in V
+P_dm = (((E_f*V_t)/Z_s)-((E_f^2*R_a)/Z_s^2)) // maximum power per phase in W
+
+// display the result
+disp("Example 10.3 solution");
+printf(" \n Maximum power per phase \n P_dm = %.2f W \n", P_dm );
+printf(" \n In textbook solution they took E_f = 513.5V instead of 533.33V");
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