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## Q. 7.2

Design a V-belt drive that has the input sheave on the shaft of an electric motor (normal torque) rated at 50.0 hp at 1160-rpm, full-load speed. The drive is to a bucket elevator in a potash plant that is to be used 12 hours (h) daily at approximately 675 rpm.

## Verified Solution

Objective: Design the V-belt drive.
Given: Power transmitted = 50 hp to bucket elevator
Speed of motor = 1160 rpm; output speed = 675 rpm

Analysis: Use the design data presented in this section. The solution procedure is developed within the Results section of the problem solution.

Results:

Step 1. Compute the design power. From Table 7–1, for a normal torque electric motor running 12 h daily driving a bucket elevator, the service factor is 1.30. Then the design power is 1.30(50.0 hp) = 65.0 hp.
Step 2. Select the belt section. From Figure 7–13, a 5V belt is recommended for 70.0 hp at 1160-rpm input speed.
Step 3. Compute the nominal speed ratio:
Ratio = 1160/675 = 1.72
Step 4. Compute the driving sheave size that would produce a belt speed of $v_b$ = 4000 ft/min, as a guide to selecting a standard sheave, adapted from Equation (7–2) ($v \frac{\text { belt }}{\text { chain }}=\frac{P D_{\text {driving }}}{2} \cdot \omega_{\text {driving }}=\frac{P D_{\text {driven }}}{2} \cdot \omega_{\text {driven }}$):

$\text { Belt speed }=v_{b}=\frac{D_{1} n_{1}}{2}$
Then the required diameter to give $v_{b}=4000 \mathrm{ft} / \mathrm{min}$ is
$D_{1}=\frac{2 v_{b}}{n_{1}}=\frac{2(4000 \mathrm{ft} / \mathrm{min})}{1160 \mathrm{rev} / \mathrm{min}} \times(12 \mathrm{in} / \mathrm{ft})(1 \mathrm{rev} /(2 \pi \mathrm{rad})=13.17 \mathrm{in}$
Step 5. Select trial sizes for the input sheave, and compute the desired size of the output sheave. Select a standard size for the output sheave, and compute the actual ratio and output speed.

For this problem, the trials are given in Table 7–3 (diameters are in inches).
The two trials in boldface in Table 7–3 give only about 1% variation from the desired output speed of 675 rpm, and the speed of a bucket elevator is not critical. Because no space limitations were given, let’s choose the larger size; $D_1$ = 12.4 in; $D_2$ = 21.1 in.
We can now compute the actual belt speed:

$v_{b}=D_{1} n_{1} / 2=(12.4 \mathrm{in} / 2)(1160 \mathrm{rev} / \mathrm{min})(2 \pi \mathrm{rad} / \mathrm{rev})(1 \mathrm{ft} / 12 \mathrm{in})=3766 \mathrm{ft} / \mathrm{min}$

Step 6. Determine the rated power from Figures 7–14 to 7–16.
For the 5V belt that we have selected, Figure 7–15 is appropriate. For a 12.4-in sheave at 1160 rpm, the basic rated power is 26.4 hp. Multiple belts will be required. The ratio is relatively high, indicating that some added power rating can be used. This value can be estimated from Figure 7–15 or taken directly from Figure 7–17 for the 5V belt. Power added is 1.15 hp. Then the actual rated power is 26.4 + 1.15 = 27.55 hp.

Step 7. Specify a trial center distance, $C D$.
We can use Equation (7-17) to determine a nominal acceptable range for $C D$ :
$\begin{gathered}D_{2}
In the interest of conserving space, let’s try $C D=24.0$ in.
Step 8. Compute the required belt length for our preliminary choices for pitch diameters of the sheaves and this center distance from Equation (7-12):
\begin{aligned}&L=2 C D+1.57\left(D_{2}+D_{1}\right)+\frac{\left(D_{2}-D_{1}\right)^{2}}{4 C} \\&L=2(24.0)+1.57(21.1+12.4)+\frac{(21.1-12.4)^{2}}{4(24.0)}=101.4\mathrm{in}\end{aligned}
Step 9. Select a standard belt length from Table 7-2, and compute the resulting actual center distance from Equation (7-13) ($C D=\frac{B+\sqrt{B^{2}-32\left(D_{2}-D_{1}\right)^{2}}}{16}$).
In this problem, the nearest standard length is 100.0 in. Then, from Equation (7-13),
\begin{aligned}&B=4 L-6.28\left(D_{2}+D_{1}\right)=4(100)-6.28(21.1+12.4)=189.6 \\&CD=\frac{189.6+\sqrt{(189.6)^{2}-32(21.1-12.4)^{2}}}{16}=23.30 \mathrm{in}\end{aligned}
Step 10. Compute the angle of wrap of the belt on the small sheave from Equation (7-14):
$\theta_{1}=180^{\circ}-2 \sin ^{-1}\left[\frac{D_{2}-D_{1}}{2 C D}\right]=180^{\circ}-2\sin ^{-1}\left[\frac{21.1-12.4}{2(23.30)}\right]=158^{\circ}$

Step 11. Determine the correction factors from Figures 7-18 and 7-19 . For $\theta=158^{\circ}, C_{\theta}=0.94 .$ For $L=100 \mathrm{in}, C_{L}=0.96 .$

Step 12. Compute the corrected rated power per belt and the number of belts required to carry the design power:
Corrected power rating per belt $=C_{\theta} C_{L} P=(0.94)(0.96)(27.55 \mathrm{hp})=24.86 \mathrm{hp}$
Minimum number of belts = Design power/Corrected power rating
Minimum number of belts $=65.0 \mathrm{hp} / 24.86 \mathrm{hp}=2.61$ belts (Use 3 belts.)

Input: Electric motor, 50.0 hp at 1160 rpm
Service factor: 1.3
Design power: 65.0 hp
Belt: 5V cross section, 100-in length, 3 belts
Sheaves: Driver, 12.4-in pitch diameter, 3 grooves, 5V. Driven, 21.1-in pitch diameter, 3 grooves, 5V
Actual output speed: 682 rpm
Center distance: 23.30 in

 TABLE 7–1 V-Belt Service Factors Driver type AC motors: Normal torque DC motors: Shunt-wound Engines: Multiple-cylinder AC motors: High torque DC motors: Series-wound, or compound-wound Engines: 4-cylinder or less Driven machine type <6 h per day 6–15 h per day >15 h per day <6 h per day 6–15 h per day >15 h per day Smooth loading 1 1.1 1.2 1.1 1.2 1.3 Agitators, light conveyors, centrifugal pumps fans and blowers under 10 hp (7.5 kW) Light shock loading 1.1 1.2 1.3 1.2 1.3 1.4 Generators, machine tools mixers, fans and blowers over 10 hp (7.5 kW) gravel conveyors Moderate shock loading 1.2 1.3 1.4 1.4 1.5 1.6 Bucket elevators, piston pumps textile machinery, hammer mills heavy conveyors, pulverizers Heavy shock loading 1.3 1.4 1.5 1.5 1.6 1.8 Crushers, ball mills, hoists rubber mills, and extruders Machinery that can choke 2 2 2 2 2 2
 TABLE 7–2 Standard Belt Lengths for 3V, 5V, and 8V Belts (in) 3V only 3V and 5V 3V, 5V, and 8V 5V and 8V 8V only 25 50 100 150 375 26.5 53 106 160 400 28 56 112 170 425 30 60 118 180 450 31.5 63 125 190 475 33.5 67 132 200 500 35.5 71 140 212 37.5 75 224 40 80 236 42.5 85 250 45 90 265 47.5 95 280 300 165 315 335 355
 TABLE 7–3 Trial Sheave Sizes  for Example Problem 7–1 Standard driving sheave size, $D_1$ Approximate driven sheave size (1.72$D_1$) Nearest standard sheave, $D_2$ Actual output speed (rpm) 13.1 22.5 21.1 720 12.4 21.3 21.1 682 11.7 20.1 21.1 643 10.8 18.6 21.1 594 10.2 17.5 15.9 744 9.65 16.6 15.9 704 9.15 15.7 15.9 668 8.9 15.3 14.9 693       