Question 13.2: Maximum Tension of Flat-Belt Drive A 12 hp, 2200 rpm electri...

The cross-sectional area of the belt is 5(0.3) = 1.5 in.² and its unit weight equals w = 0.04(1.5) = 0.0.06 lb/in. = 0.72 lb/ft. The belt velocity, using Equation 13.4,

V=\frac{\pi d n}{12}             (13.4)

V=\frac{\pi d_1 n_1}{12}=\frac{\pi(5)(2200)}{12}=2880  fpm

By Equation 13.3,

hp =\frac{\left(F_1-F_2\right) V }{33,000}=\frac{T n}{63,000}       (13.3)

F_1-F_2=\frac{33,000 hp }{V}=\frac{33,000(12)}{2880}=137.5  lb      (a)

Centrifugal force acting on the belt, applying Equation 13.13,

F_c=\frac{w}{g} V^2=\frac{0.72}{32.2}\left(\frac{2880}{60}\right)^2=51.52  lb

Through the use of Equation 13.6, we find

\sin \alpha=\frac{r_2-r_1}{C}      (13.6)

\alpha=\sin ^{-1}\left[\frac{r_2-r_2}{c}\right]=\sin ^{-1}\left[\frac{7.5-2.5}{6.5(12)}\right]=3.675^{\circ}

Then the angle of wrap, \phi=\pi-2 \alpha , equals

\phi=180^{\circ}-2\left(3.675^{\circ}\right)=172.65^{\circ}

We have e^{f \phi}=e^{(0.2)(172.65)(\pi / 180)}=1.827 . Substituting these into Equation 13.16 leads to

\frac{F_1-F_c}{F_2-F_c}=e^{f \phi}        (13.16)

\frac{F_1-51.52}{F_2-51.52}=1.827

or

F_1=1.827 F_2-42.6        (b)

Solving Equations (a) and (b) results in

F_1=355.3  lb \quad \text { and } \quad F_2=217.8  lb

Comment: Equation 13.2 estimates the initial tension as (355.3 + 217.8)/2 = 286.6 lb and so the largest belt tension in the belt drive is F_{1} = 355.3 lb.