Question

11- A standard V belt is used to connect two sheaves with pitch diameters of 8,4 in. and 27.7 in. with a center distance of no more than 60.0 in.

a-Specify an acceptable belt.

b-If the small sheave is rotating at 1160 rpm, compute the linear speed of the belt.

c-Determine the contact angle for each sheave if this is an open belt drive.

d-Find the rated horsepower.

e-What is the actual center distance?

Answer 1

Given that
The pitch diameter of larger sheave
=d(L) =27.7 in
The pitch diameter of smaller sheave
=d(S) =8.4 in

The maximum distance between centres
=60.0 in

a)Length of the belt drive for an open configuration
L=(pi)*(d(S)+d(L))/2+2C+[(d(L)+d(S))^2]/4C
L=(pi)*(8.4+27.7)/2+120+[(27.7+8.4)^2]/240
L=182.135789 is approximately ~ 182.2 in

b)Linear velocity of the belt
V=(pi)*d(S)*N(S)/60=(pi)*(8.4)*1160/60
=510.194647 in/sec
V is approximately ~ 510.2 in/sec

c)beta=sin^-1 ((d(L)-d(S))/2C)
=sin^-1 ((27.7-8.4)/120)=0.16153493 rad
=9.25526917 deg approximately ~ 9.3 deg

contact angle for small sheave
alpha(S)=180-2*(beta) = 161.4 deg

contact angle for large sheave
alpha(L)=180+2*(beta) = 198.6 deg

d)Rated horse power= P(max)
We know that power transmitted by a blet drive P=(T1-T2)*V

Where T1 and T2 are tensions on either sides of the belt

Also initial tension Ti=(T1+T2)/2
and for max. Power transmission T2=0 N
Ti=T1/2

The relation between T1 and T2 is given by the relation

(T1-mv^2)/(T2-mv^2)=e^(mu)*(alpha)

And for design,the sheave with the smaller (mu)*(alpha) value is considered
where (mu) is the coefficient of friction between a sheave and belt.

In (mV^2) m is the linear mass density of the belt=(breadth*thickness*volume density) in lb/in and V is the linear Velocity of the belt.

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