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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+
+clc
+//given
+ratio=1.25
+u=.675
+P=12//hp
+//W=P*%pi*(r1^2-r2^2); Total axal thrust.
+//M=u*W*(r1+r2); Total friction moemnt
+//reducing the two equations and using ratio=1.25(r1=1.25*r2) we get, M=u*21.2*r2^3
+ReqM=65//lb ft
+RM=ReqM*12//lb in
+r2=(RM/(u*P*%pi*(1.25^2-1)))^(1/3)
+r1=1.25*r2
+d1=r1*2
+d2=r2*2
+printf("The dimensions of the friction surfaces are:\nOuter Diameter= %.1f in\nInner Diameter= %.1f in\n",d1,d2)