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+//Book Name:Fundamentals of Electrical Engineering
+//Author:Rajendra Prasad
+//Publisher: PHI Learning Private Limited
+//Edition:Third ,2014
+
+//Ex8_16.sce.
+
+clc;
+clear;
+P_in_HP=20;
+P=(P_in_HP)*736;
+N=450;
+Ra=0.18;
+Rf=0.12;
+R=8.7+Ra+Rf;
+omega=(2*%pi*N)/60;
+Tf=P/omega;
+
+//The voltage developed for 450 rpm is 289 volt which is taken from the curve
+E=289;
+P_not=(E*E)/R;
+Pi=(2*%pi*N*Tf)/60;
+
+//The mechanical input is greater than electrical output , so  the motor speed increases
+//The voltage developed for 550 rpm is 403 volt which is taken from the curve
+N=550;
+E=403;
+P_not=(E*E)/R;
+Pi=(2*%pi*N*Tf)/60;
+
+printf("\n Electrical input=%5.2f W \n",P_not)
+printf("\n Mechanical input=%5.2f W \n",Pi)
+if Pi<P_not then
+    N1=540;
+else
+    N1>N
+end
+printf("\n Desired speed=%d rpm \n",N1)
+//Answer vary dueto roundoff error
+//since mechanical input is less than electrical output the motor cannot attain a speed as 550 rpm 
+//So the speed is 540 rpm which is obtained using trial and error method