Choose a leather belt to transmit 50 hp (37.8 kW) from a 1750-r/min squirrel-cage compensator-starting motor through a 12-in (30.5-cm) diameter pulley in an oily atmosphere. What belt width is needed with a 50-hp (37.3-kW) internal-combustion engine fitted with a 1750-r/min 12-in (30.5-cm) diameter pulley operating in an oily atmosphere?
Calculation Procedure:
The speed of a belt S is found from S = p RD, where R = rpm of driving or driven pulley; D = diameter, ft, of driving or driven pulley. Thus, for this belt, S = p(1750)(12/12) = 5500 ft/min (27.9 m/s).
Use the National Industrial Leather Association recommendations. Enter Table 1 at the bottom at a belt speed of 5500 ft/min (27.9 m/s), i.e., between 4000 and 6000 ft/min (20.3 and 30.5 m/s); and project horizontally to the next smaller pulley diameter than that actually used. Thus, by entering at the line marked 4000-6000 ft/min (20.3-30.5 m/s) and projecting to the 10-in (25.4-cm) minimum diameter pulley, since a 12-in (30.5-cm) pulley is used, we see that a 23/64-in (0.91-cm) thick double-ply heavy belt should be used. Read the belt thickness and type at the top of the column in which the next smaller pulley diameter appears.
Enter the body of Table 1 at a belt speed of 5500 ft/min (27.9 m/s), i.e., between 4000 and 6000 ft/min (20.3 and 30.5 m/s); then project to the double-ply heavy column. Interpolating by eye gives a belt capacity factor of Kc = 14.8.
Table 2 lists motor, pulley diameter, and operating correction factors, respectively. Thus, from Table 2, the motor correction factor M = 1.5 for a squirrel-cage compensatorostarting motor.
TABLE 1 Leather-Belt Capacity Factors
Belt Speed |
Double Ply |
||||
20/64 in (7.9 mm) |
23/64 in (9.1 mm) |
||||
ft/min |
m/s |
Medium |
Heavy |
||
4000 5000 6000 |
20.3 25.4 30.5 |
10.9 12.5 13.2 |
12.6 14.3 15.2 |
||
Minimum pulley diameters |
|||||
Up to 2500 2500-4000 4000-6000 |
Up to 12.7 12.7-20.3 20.3-30.5 |
5 in° 6° 7° |
1.27 cm° 15.2° 17.8° |
8 in° 9° 10° |
20.3 cm° 22.9° 25.4° |
°For belts 8 in (20.3 cm) and over, add 2 in (5.1 cm) to pulley diameter.
Table 2 Leather-Belt Correction Factors
Correction Factor |
|
Characteristics or condition of motor and starter: Squirrel-cage, compensator-starting motor Squirrel-cage, line-starting Slip-ring, high starting torque Diameter of small pulley, in (cm): 4 and under (10.2 and under 4.5 to 8 (11.4 to 20.3) 9 to 12 (22.9 to 30.5) 13 to 16 (33.0 to 40.6) 17 to 30 (43.2 to 76.2) Over 30 (over 76.2) Operating conditions: Oily, wet, or dusty atmosphere Vertical drives Jerky loads Shock and reversing loads |
M = 1.5 M = 2.0 M = 2.5 P = 0.5 P = 0.6 P = 0.7 P= 0.8 P = 0.9 P = 1.0 F = 1.35 F = 1.2 F = 1.2 F = 1.4 |
Also from Table 2, the smaller pulley diameter correction factor P = 0.7; and F = 1.35 for an oily atmosphere.
The required belt width, in, is W = hpMF/(KcP), where hp = horsepower transmitted by the belt; the other factors are as given above. For this belt, then, W = (50)(1.5)(1.35)/[14.8(0.7)] = 9.7 in (24.6 cm). Thus, a 10-in (25.4-cm) wide belt would be used because belts are commercially available in 1-in (2.5-cm) increments.
For a double-ply belt driven by a driver other than an electric motor, W = 2750 hp/dR, where d = driving pulley diameter, in; R = driving pulley, r/min. Thus, W = 2750(50)/[(12)(1750)] = 6.54 in (16.6 cm). Hence, a 7-in (17.8-cm) wide belt would be used.
For a single-ply belt the above equation becomes W = 1925 hp/dR.
Related Calculations: Note that the relations in steps 1, 5, and 6 can be solved for any unknown variable when the other factors in the equations are known. Where the hp rating of a belt material is available from the manufacturer's catalog or other published data, find the required width from W = hpbF/KcP, where hpb = hp rating of the belt material, as stated by the manufacturer; other symbols as before. To find the tension Tb lb in a belt, solve Tb = 33,000 hp/S where S = belt speed, ft/min. The tension per inch of belt width is Tbi = Tb/W. Where the belt speed exceeds 6000 ft/min (30.5 m/s), consult the manufacturer.