updated the code

1 files changed, 25 insertions, 25 deletions

diff --git a/608/CH21/EX21.22/21_22.sce b/608/CH21/EX21.22/21_22.sce
index e6fc14636..45c50cd08 100755
--- a/608/CH21/EX21.22/21_22.sce
+++ b/608/CH21/EX21.22/21_22.sce
@@ -1,26 +1,26 @@
-//Problem 21.22: A 220 V, d.c. shunt-wound motor runs at 800 rev/min and the armature current is 30 A. The armature circuit resistance is 0.4 ohm. Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and (b) the steady state value of the armature current at the new value of flux, assuming the shaft torque of the motor remains constant.

-

-//initializing the variables:

-Ia1 = 30; // in Amperes

-Ra = 0.4; // in ohm

-n = 800/60; // in Rev/sec

-V = 220; // in Volts

-x= 0.1;

-

-//calculation:

-//For a d.c. shunt-wound motor, E = V - (Ia*Ra),Hence initial generated e.m.f.,

-E1 =  V - (Ia1*Ra)

-//The generated e.m.f. is also such that E proportional to (Phi*n) so at the instant the flux is reduced, the speed has not had time to change, and

-E = E1*(1-x)

-//Hence, the voltage drop due to the armature resistance is

-Vd = V - E

-//The instantaneous value of the current is

-Ia = Vd/Ra

-//T proportional to (Phi*Ia), since the torque is constant,

-//Phi1*Ia1 = Phi2*Ia2,  The flux 8 is reduced by 10%, hence

-Phi2 = (1-x)*Phi1

-Ia2 = Phi1*Ia1/Phi2

-

-printf("\n\n Result \n\n")

-printf("\n (a)instantaneous value of the current %.0f A ",Ia)

+//Problem 21.22: A 220 V, d.c. shunt-wound motor runs at 800 rev/min and the armature current is 30 A. The armature circuit resistance is 0.4 ohm. Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and (b) the steady state value of the armature current at the new value of flux, assuming the shaft torque of the motor remains constant.
+
+//initializing the variables:
+Ia1 = 30; // in Amperes
+Ra = 0.4; // in ohm
+n = 800/60; // in Rev/sec
+V = 220; // in Volts
+x= 0.1;
+
+//calculation:
+//For a d.c. shunt-wound motor, E = V - (Ia*Ra),Hence initial generated e.m.f.,
+E1 =  V - (Ia1*Ra)
+//The generated e.m.f. is also such that E proportional to (Phi*n) so at the instant the flux is reduced, the speed has not had time to change, and
+E = E1*(1-x)
+//Hence, the voltage drop due to the armature resistance is
+Vd = V - E
+//The instantaneous value of the current is
+Ia = Vd/Ra
+//T proportional to (Phi*Ia), since the torque is constant,
+//Phi1*Ia1 = Phi2*Ia2,  The flux 8 is reduced by 10%, hence
+//Phi2 = (1-x)*Phi1
+Ia2 = Ia1/0.9
+
+printf("\n\n Result \n\n")
+printf("\n (a)instantaneous value of the current %.0f A ",Ia)
 printf("\n (b)steady state value of armature current, %.2f A ",Ia2)
\ No newline at end of file