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authorpriyanka2015-06-24 15:03:17 +0530
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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)