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+//CHAPTER 2,ILLUSTRATION 11 PAGE 65

+//TITLE:TRANSMISSION OF MOTION AND POWER BY BELTS AND PULLEYS

+clc

+clear

+//INPUT

+C=4.5//      CENTRE DISTANCE IN metres

+D1=1.35//    DIAMETER OF LARGER PULLEY IN metres

+D2=.9//      DIAMETER OF SMALLER PULLEY IN metres

+To=2100//    INITIAL TENSION IN newtons

+b=12//       WIDTH OF BELT IN cm

+t=12//       THICKNESS OF BELT IN mm

+d=1//        DENSITY IN gm/cm^3

+U=.3//       COEFFICIENT OF FRICTION

+L=1//        length in metres

+PI=3.141

+e=2.71

+

+//CALCULATION

+M=b*t*d*L*10^-2//               mass of belt per metre length in KG

+V=(To/3/M)^.5//                 VELOCITY OF FOR MAX POWER TO BE TRANSMITTED IN m/s

+Tc=M*V^2//                      CENTRIFUGAL TENSION IN newtons

+//                              LET (T1+T2)=X

+X=2*To-2*Tc   //                THE VALUE OF (T1+T2)

+F=(D1-D2)/(2*C)

+ALPHA=asind(F)

+THETA=(180-(2*ALPHA))*PI/180//   ANGLE OF CONTACT IN radians

+//                               LET T1/T2=Y

+Y=e^(U*THETA)//                  THE VALUE OF T1/T2

+T1=X*Y/(Y+1)//                   BY SOLVING X AND Y WE WILL GET THIS EQN

+T2=X-T1

+P=(T1-T2)*V/1000//                 MAX POWER TRANSMITTED IN kilowatts

+N1=V*60/(PI*D1)//                   SPEED OF LARGER PULLEY IN rpm

+N2=V*60/(PI*D2)//                   SPEED OF SMALLER PULLEY IN rpm

+//OUTPUT

+printf('\n MAX POWER TO BE TRANSMITTED =%f KW',P)

+printf('\n SPEED OF THE LARGER PULLEY =%f rpm',N1)

+printf('\n SPEED OF THE SMALLER PULLEY =%f rpm',N2)