1 files changed, 23 insertions, 23 deletions
diff --git a/608/CH21/EX21.23/21_23.sce b/608/CH21/EX21.23/21_23.sce
index ebf977d2b..0b8538445 100755
--- a/608/CH21/EX21.23/21_23.sce
+++ b/608/CH21/EX21.23/21_23.sce
@@ -1,24 +1,24 @@
-//Problem 21.23: A series motor has an armature resistance of 0.2 ohm and a series field resistance of 0.3 ohm. It is connected to a 240 V supply and at a particular load runs at 24 rev/s when drawing 15 A from the supply. (a) Determine the generated e.m.f. at this load. (b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A. Assume that this causes a doubling of the flux.
-
-//initializing the variables:
-Ia1 = 15; // in Amperes
-Ia2 = 30; // in Amperes
-Rf = 0.3; // in ohms
-Ra = 0.2; // in ohm
-n1 = 24; // in Rev/sec
-V = 240; // in Volts
-x= 2;
-
-//calculation:
-//generated e.m.f., E, at initial load, is given by
-E1 =  V - Ia1*(Ra + Rf)
-//When the current is increased to 30 A, the generated e.m.f. is given by:
-E2 =  V - Ia2*(Ra + Rf)
-//E proportional to (Phi*n)
-//E1/E2 = Phi1*n1/Phi2*n2
-Phi2 = x*Phi1
-n2 = E2*Phi1*n1/(Phi2*E1) 
-
-printf("\n\n Result \n\n")
-printf("\n (a)generated e.m.f., E is %.1f V ",E1)
+//Problem 21.23: A series motor has an armature resistance of 0.2 ohm and a series field resistance of 0.3 ohm. It is connected to a 240 V supply and at a particular load runs at 24 rev/s when drawing 15 A from the supply. (a) Determine the generated e.m.f. at this load. (b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A. Assume that this causes a doubling of the flux.
+
+//initializing the variables:
+Ia1 = 15; // in Amperes
+Ia2 = 30; // in Amperes
+Rf = 0.3; // in ohms
+Ra = 0.2; // in ohm
+n1 = 24; // in Rev/sec
+V = 240; // in Volts
+x= 2;
+
+//calculation:
+//generated e.m.f., E, at initial load, is given by
+E1 =  V - Ia1*(Ra + Rf)
+//When the current is increased to 30 A, the generated e.m.f. is given by:
+E2 =  V - Ia2*(Ra + Rf)
+//E proportional to (Phi*n)
+//E1/E2 = Phi1*n1/Phi2*n2
+//Phi2 = x*Phi1
+n2 = E2*n1/(E1) 
+
+printf("\n\n Result \n\n")
+printf("\n (a)generated e.m.f., E is %.1f V ",E1)
 printf("\n (b)speed of motor, n2, %.2f A ",n2)
 \ No newline at end of file