Question 13.2: Maximum Tension of Flat-Belt Drive A 12 hp, 2200 rpm electri...
Maximum Tension of Flat-Belt Drive
A 12 hp, 2200 rpm electric motor drives a machine through the flat belt (Figure 13.7). The size of belt is 5 in. wide and 0.3 in thick and weighs 0.04 lb/in.³ The center distance is equal to 6.5 ft. The pulley on motor shaft has a r_{1} = 2.5 in. radius and the driven pulley is r_{2} = 7.5 in. in radius.
Find: The belt tensions
Assumption: The coefficient of friction will be f = 0.2.
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.