cema belt book fifth edition chapter 6 belt tension, power, and ..., Schemes and Mind Maps of Engineering

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CEMA BELT BOOK FIFTH EDITION CHAPTER 6 BELT TENSION, POWER, AND DRIVE ENGINEERING AS REFERENCED OCCASIONALLY IN CEMA BELT BOOK SIXTH EDITION 85 CHAPTER 6 Belt Tension, Power, and Drive Engineering Basic power requirements Belt tension calculations CEMA horsepower formula Drive pulley relationships Drive arrangements Maximum and minimum belt tensions Tension relationships and belt sag between idlers Acceleration and deceleration forces Analysis of acceleration and deceleration forces Design considerations Conveyor horsepower determination — graphical method Examples of belt tension and horsepower calculations — six problems Belt conveyor drive equipment Backstops Brakes Brakes and backstops in combination Devices for acceleration, deceleration, and torque control Brake requirement determination (deceleration calculations) Belt Tension, Power, and Drive Engineering 88 T bc = tension resulting from belt pull required for belt-cleaning devices such as belt scrapers, lbs T e = effective belt tension at drive, lbs T m = tension resulting from the force needed to lift or lower the conveyed material, lbs: T p = tension resulting from resistance of belt to flexure around pulleys and the resistance of pulleys to rotation on their bearings, total for all pulleys, lbs T pl = tension resulting from the frictional resistance of plows, lbs T sb = tension resulting from the force to overcome skirtboard friction, lbs T tr = tension resulting from the additional frictional resistance of the pulleys and the flexure of the belt over units such as trippers, lbs T x = tension resulting from the frictional resistance of the carrying and return idlers, lbs: T yb = total of the tensions resulting from the resistance of the belt to flexure as it rides over both the carrying and return idlers, lbs: T yc = tension resulting from the resistance of the belt to flexure as it rides over the carrying idlers, lbs: T ym = tension resulting from the resistance of the material to flexure as it rides with the belt over the carrying idlers, lbs: T yr = tension resulting from the resistance of the belt to flexure as it rides over the return idlers, lbs: V = design belt speed, fpm T m H W m×±= T x L K x× Kt×= T yb T yc T yr+= T yc L K y× W b× Kt×= T ym L K y× W m×= T yr L 0.015× W b Kt××= 89 Belt Tension Calculations Wb = weight of belt in pounds per foot of belt length. When the exact weight of the belt is not known, use average estimated belt weight (see Table 6-1) Wm = weight of material, lbs per foot of belt length: Three multiplying factors, Kt , Kx , and Ky , are used in calculations of three of the components of the effective belt tension, Te . Kt — Ambient Temperature Correction Factor Idler rotational resistance and the flexing resistance of the belt increase in cold weather operation. In extremely cold weather the proper lubricant for idlers must be used to prevent excessive resistance to idler rotation. Figure 6.1 Variation of temperature correction factor, Kt , with temperature. Kt is a multiplying factor that will increase the calculated value of belt tensions to allow for the increased resistances that can be expected due to low temperatures. Fig- ure 6.1 provides values for factor Kt . W m Q 2 000,× 60 V× ------------------------ 33.33 Q× V -----------------------= = Operation at temperatures below –15ºF involves problems in addition to horsepower considerations. Consult conveyor manufacturer for advice on special belting, greasing, and cleaning specifications and necessary design modification. Ambient temperature ºF conveyor operation Belt Tension, Power, and Drive Engineering 90 Kx — Idler Friction Factor The frictional resistance of idler rolls to rotation and sliding resistance between the belt and the idler rolls can be calculated by using the multiplying factor Kx . Kx is a force in lbs/ft of conveyor length to rotate the idler rolls, carrying and return, and to cover the sliding resistance of the belt on the idler rolls. The Kx value required to rotate the idlers is calculated using equation (3). The resistance of the idlers to rotation is primarily a function of bearing, grease, and seal resistance. A typical idler roll equipped with antifriction bearings and sup- porting a load of 1,000 lbs will require a turning force at the idler roll periphery of from 0.5 to 0.7 lbs to overcome the bearing friction. The milling or churning of the grease in the bearings and the bearing seals will require additional force. This force, however, is generally independent of the load on the idler roll. Under normal conditions, the grease and seal friction in a well-lubricated idler will vary from 0.1 to 2.3 lbs/idler, depending upon the type of idler, the seals, and the condition of the grease. Sliding resistance between the belt and idler rolls is generated when the idler rolls are not exactly at 90 degrees to the belt movement. After initial installation, deliberate idler misalignment is often an aid in training the belt. Even the best installations have a small requirement of this type. However, excessive idler misalignment results in an extreme increase in frictional resistance and should be avoided. Some troughing idlers are designed to operate with a small degree of tilt in the direction of belt travel, to aid in belt training. This tilt results in a slight increase in sliding friction that must be considered in the horsepower formula. Table 6-1. Estimated average belt weight, multiple- and reduced-ply belts, lbs/ft. Belt Width Material Carried, lbs/ft3 inches (b) 30-74 75-129 130-200 18 3.5 4.0 4.5 24 4.5 5.5 6.0 30 6.0 7.0 8.0 36 9.0 10.0 12.0 42 11.0 12.0 14.0 48 14.0 15.0 17.0 54 16.0 17.0 19.0 60 18.0 20.0 22.0 72 21.0 24.0 26.0 84 25.0 30.0 33.0 96 30.0 35.0 38.0 1. Steel-cable belts — increase above value by 50 percent. 2. Actual belt weights vary with different constructions, manufacturers, cover gauges, etc. Use the above values for estimating. Obtain actual values from the belt manufacturer whenever possible. 93 Belt Tension Calculations Ky values in Tables 6-2 and 6-3 are applicable for conveyors up to 3,000 ft long with a single slope and a 3% maximum sag of the belt between the troughing and between the return idlers. The return idler spacing is 10 ft nominal and loading of the belt is uniform and continuous. 50 0.031 0.028 0.026 0.024 0.023 0.019 0.016 75 0.030 0.027 0.024 0.022 0.019 0.016 0.016 100 0.030 0.026 0.022 0.019 0.017 0.016 0.016 1000 150 0.033 0.024 0.019 0.016 0.016 0.016 0.016 200 0.032 0.023 0.017 0.016 0.016 0.016 0.016 250 0.033 0.022 0.017 0.016 0.016 0.016 0.016 300 0.033 0.021 0.018 0.018 0.018 0.018 0.018 50 0.029 0.026 0.024 0.022 0.021 0.016 0.016 75 0.028 0.024 0.021 0.019 0.016 0.016 0.016 100 0.028 0.023 0.019 0.016 0.016 0.016 0.016 1400 150 0.029 0.020 0.016 0.016 0.016 0.016 0.016 200 0.030 0.021 0.016 0.016 0.016 0.016 0.016 250 0.030 0.020 0.017 0.016 0.016 0.016 0.016 300 0.030 0.019 0.018 0.018 0.018 0.018 0.018 50 0.027 0.024 0.022 0.020 0.018 0.016 0.016 75 0.026 0.021 0.019 0.016 0.016 0.016 0.016 100 0.025 0.020 0.016 0.016 0.016 0.016 0.016 2000 150 0.026 0.017 0.016 0.016 0.016 0.016 0.016 200 0.024 0.016 0.016 0.016 0.016 0.016 0.016 250 0.023 0.016 0.016 0.016 0.016 0.016 0.016 300 0.022 0.018 0.018 0.018 0.018 0.018 0.018 50 0.026 0.023 0.021 0.018 0.017 0.016 0.016 75 0.025 0.021 0.017 0.016 0.016 0.016 0.016 100 0.024 0.019 0.016 0.016 0.016 0.016 0.016 2400 150 0.024 0.016 0.016 0.016 0.016 0.016 0.016 200 0.021 0.016 0.016 0.016 0.016 0.016 0.016 250 0.021 0.016 0.016 0.016 0.016 0.016 0.016 300 0.020 0.018 0.018 0.018 0.018 0.018 0.018 50 0.024 0.022 0.019 0.017 0.016 0.016 0.016 75 0.023 0.019 0.016 0.016 0.016 0.016 0.016 100 0.022 0.017 0.016 0.016 0.016 0.016 0.016 3000 150 0.022 0.016 0.016 0.016 0.016 0.016 0.016 200 0.019 0.016 0.016 0.016 0.016 0.016 0.016 250 0.018 0.016 0.016 0.016 0.016 0.016 0.016 300 0.018 0.018 0.018 0.018 0.018 0.018 0.018 Table 6-2. Factor Ky values. Conveyor Length (ft) Wb + Wm (lbs/ft) Percent Slope 0 3 6 9 12 24 33 Approximate Degrees 0 2 3.5 5 7 14 18 Idler spacing: The above values of Ky are based on the following idler spacing (for other spacing, see Table 6-3). (Wb+Wm), lbs per ft Less than 50 50 to 99 Si , ft (Wb+Wm), lbs per ft 100 to 149 150 and above Si , ft 4.5 3.5 4.0 3.0 Belt Tension, Power, and Drive Engineering 94 Equation (4) provides Ky values for the carrying idlers of belt conveyors whose length, number of slopes, and/or average belt tensions exceed the limitations speci- fied above for the conveyors covered by Tables 6-2 and 6-3. This equation is applica- ble for conveyors in which the average belt tension is 16,000 lbs or less. To determine the Ky factor for use in calculating conveyors of this class, it is necessary, first, to assume a tentative value for the average belt tension. The graphical method for deter- mining conveyor horsepower (pages 141 through 145) may be of assistance in esti- mating this initial tentative value of average belt tension. After estimating the average belt tension and selecting an idler spacing, refer to Table 6-4 to obtain values for A and B for use in the following equation: By using equation (4), an initial value for Ky can be determined and an initial average belt tension can be subsequently calculated. The comparison of this calcu- lated average belt tension with the original tentative value will determine the need to select another assumed belt tension. Recalculate Ky and calculate a second value for the average belt tension. The process should be repeated until there is reasonable agreement between the estimated and final calculated average belt tensions. There are no tabulated Ky values or mathematical equations to determine a Ky for conveyors having an average belt tension exceeding 16,000 lbs. A reasonably accurate value that can be used for calculations is Ky equals 0.016. It is suggested that this value for Ky be considered a minimum, subject to consultation with a CEMA member company on any specific applications. The force that results from the resistance of the belt to flexure as it moves over the idlers for the return run is calculated in the same manner as the resistance to flexure for the carrying run, except a constant value of 0.015 is used in place of Ky . The resis- tance of the belt flexure over idler rolls is a function of the belt construction, cover thickness and indentation by the idler rolls, type of rubber compound, idler roll diameter, temperature, and other factors. The belt flexing resistance increases at lower temperatures. Figure 6.2 Effect of belt tension on resistance of material to flexure over idler rolls. K y W m W b+( ) A 10 4– B 10 2–×+××= (4) 95 Belt Tension Calculations The resistance of the material load to flexure over idler rolls is a function of belt tension, type of material, shape of the load cross section, and idler spacing. Measure- ments indicate that the most important factor is belt tension, because this controls the amount of load flexure. Figure 6.2 shows this relationship for a typical idler spac- ing. For a given weight per foot of belt and load, the running resistance, in pounds per ft of load, decreases with increases in belt tension. For a given belt tension, running resistance, in pounds per ft of load, increases with increases in the amount of load. However, the running resistance is not proportional to the weight of the load. Table 6-3. Corrected factor Ky values when other than tabular carrying idler spacings are used. Wb + Wm (lbs/ft) Si, (ft) Reference Values of Ky for Interpolation 0.016 0.018 0.020 0.022 0.024 0.026 0.028 0.030 0.032 0.034 3.0 0.0160 0.0160 0.0160 0.0168 0.0183 0.0197 0.0212 0.0227 0.0242 0.0257 Less than 50 3.5 0.0160 0.0160 0.0169 0.0189 0.0207 0.0224 0.0241 0.0257 0.0274 0.0291 4.0 0.0160 0.0165 0.0182 0.0204 0.0223 0.0241 0.0259 0.0278 0.0297 0.0316 4.5 0.016 0.018 0.020 0.022 0.024 0.026 0.028 0.030 0.032 0.034 5.0 0.0174 0.0195 0.0213 0.0236 0.0254 0.0273 0.0291 0.0031 0.0329 0.0348 3.0 0.0160 0.0162 0.0173 0.0186 0.0205 0.0221 0.0239 0.026 0.0274 0.029 3.5 0.0160 0.0165 0.0185 0.0205 0.0222 0.024 0.0262 0.0281 0.030 0.0321 50 to 99 4.0 0.016 0.018 0.020 0.022 0.024 0.026 0.028 0.030 0.032 0.034 4.5 0.0175 0.0193 0.0214 0.0235 0.0253 0.0272 0.0297 0.0316 0.0335 0.035 5.0 0.0184 0.021 0.023 0.0253 0.027 0.029 0.0315 0.0335 0.035 0.035 3.0 0.0160 0.0164 0.0186 0.0205 0.0228 0.0246 0.0267 0.0285 0.0307 0.0329 100 to 149 3.5 0.016 0.018 0.020 0.022 0.024 0.026 0.028 0.030 0.032 0.034 4.0 0.0175 0.0197 0.0213 0.0234 0.0253 0.0277 0.0295 0.0312 0.033 0.035 4.5 0.0188 0.0213 0.0232 0.0253 0.0273 0.0295 0.0314 0.033 0.0346 0.035 5.0 0.0201 0.0228 0.0250 0.0271 0.0296 0.0316 0.0334 0.035 0.035 0.035 3.0 0.016 0.018 0.020 0.022 0.024 0.026 0.028 0.030 0.032 0.034 150 to 199 3.5 0.0172 0.0195 0.0215 0.0235 0.0255 0.0271 0.0289 0.031 0.0333 0.0345 4.0 0.0187 0.0213 0.0235 0.0252 0.0267 0.0283 0.0303 0.0325 0.0347 0.035 4.5 0.0209 0.023 0.0253 0.0274 0.0289 0.0305 0.0323 0.0345 0.035 0.035 5.0 0.0225 0.0248 0.0272 0.0293 0.0311 0.0328 0.0348 0.035 0.035 0.035 3.0 0.016 0.018 0.020 0.022 0.024 0.026 0.028 0.030 0.032 0.034 200 to 249 3.5 0.0177 0.0199 0.0216 0.0235 0.0256 0.0278 0.0295 0.031 0.0327 0.0349 4.0 0.0192 0.0216 0.0236 0.0256 0.0274 0.0291 0.0305 0.0322 0.0339 0.035 4.5 0.021 0.0234 0.0253 0.0276 0.0298 0.0317 0.0331 0.0347 0.035 0.035 5.0 0.0227 0.0252 0.0274 0.0298 0.0319 0.0338 0.035 0.035 0.035 0.035 To use this table to correct the value of Ky for idler spacing other than shown in bold type, apply the procedure shown in the two examples on page 91. Belt Tension, Power, and Drive Engineering 98 6. Tam — from force to accelerate the material continuously as it is fed onto the belt When material is discharged from chutes or feeders to a belt conveyor, it cannot be assumed that the material is moving in the direction of belt travel, at belt speed, although this may be the case in some instances. Normally, the material loaded onto the belt is traveling at a speed considerably lower than belt speed. The direction of material flow may not be fully in the direction of belt travel. Therefore, the material must be accelerated to the speed of the belt in the direction of belt travel, and this acceleration requires additional effective tension. The belt tension Tam can be derived from the basic equation where: M = mass of material accelerated per second, slugs W = weight of material accelerated Q = tph g = 32.2 ft/sec2 Vc = velocity change, fps V = design belt speed, fpm Vo = initial velocity of material as it is fed onto belt, fpm Table 6-5. Belt tension to rotate pulleys. Location of Pulleys Degrees Wrap of Belt Pounds of Tension at Belt Line Tight side 150o to 240o 200 lbs/pulley Slack side 150o to 240o 150 lbs/pulley All other pulleys less than 150o 100 lbs/pulley Note: Double the above values for pulley shafts that are not operating in antifriction bearings. F MV c= T am F MV c== Q 2000× 3600 --------------------- , lbs/sec= M W g ---- Q 2000× 3600 32.2× ----------------------------= = V V o– 60 ---------------= T am Q 2000× 3600 32.2× ---------------------------- V V 0– 60 ---------------×= 2.8755 10 4– Q V V 0–( )×××= 99 Belt Tension Calculations The graph in Figure 6.3 provides a convenient means of estimating the belt tension, Tam , for accelerating the material as it is fed onto the belt. 7. Tac — from the resistance generated by conveyor accessories Conveyor accessories such as trippers, stackers, plows, belt cleaning equipment, and skirtboards usually add to the effective tension, Te . The additional belt tension requirements may come from frictional losses caused by the accessory. If the acces- sory lifts the conveyed material a force will be added to belt tension. Ttr — from trippers and stackers Figure 6.3. Effective tension required to accelerate material as it is fed onto a belt conveyor. The additional belt pull to flex the belt over the pulleys and rotate the pulleys in their bearings can be calculated from Table 6-5 or Tables C-l and C-2. The force needed to lift the material over the unit can be calculated from the for- mula, Tm = H x Wm lbs. Frictional resistance of the idlers, belt, and material should be included with that of the rest of the conveyor. To use this chart: • Enter chart at belt velocity and read Tam per 1,000 tph. • Again enter chart at material velocity in direction of belt travel and read Tam per 1,000 tph. This may be positive, zero, or negative. • Subtract the second Tam reading from the first Tam reading and convert the difference from 1,000 tph to the value for the actual tonnage. This will be the Tam desired, lbs. Belt Tension, Power, and Drive Engineering 100 Tpl — from frictional resistance of plows The use of a plow on a conveyor will require additional belt tension to overcome both the plowing and frictional resistances developed. While a flat belt conveyor may be fitted with a number of plows to discharge material at desired locations, seldom is more than one plow in use at one time on one run of the belt conveyor. However, when proportioning plows are used — with each plow taking a fraction of the load from the belt — two or even three separate plows may be simultaneously in contact with the carrying run of the belt. To approximate the amount of additional belt pull that normally will be required by well-adjusted, rubber-shod plows, the values given in Table 6-6 can be used. Tbc — from belt-cleaning devices Belt scraper cleaning devices add directly to the belt pull. The additional belt pull required for belt cleaning devices can vary from 2 to 14 lbs/in. width of scraper blade contact. This wide variance is due to the different types of cleaning blades and single cleaner system vs. multiple cleaner systems that are available. In lieu of data on spe- cific cleaning system being used, use 5 lbs/in. width of scraper blade contact for each blade or scraper device in contact with the belt. Rotary brushes and similar rotating cleaning devices do not impose appreciable belt pull, if independently driven and properly adjusted. If such devices are driven from the conveyor drive shaft, suitable additional power should be incorporated in the drive to operate them. Consult a CEMA member company for horsepower required. Tsb — From skirtboard friction The force required to overcome skirtboard friction is normally larger per foot of skirtboarded conveyor than the force to move the loaded belt over the idlers. In some cases, this force can be significant. When the total conveyor length is many times that portion of the length provided with the skirtboards, the additional power require- ments for the skirtboards is relatively small, and in some cases negligible. However, if a large portion of the conveyor is equipped with skirtboards, the additional belt pull Table 6-6. Discharge plow allowance. Type of Plow Additional Belt Pull per Plow, at Belt Line (lbs/in belt width) Full V or single slant plow, removing all material from belt 5.0 Partial V or single slant plow, removing half mate- rial from belt 3.0 103 CEMA Horsepower Formula + Tp , pulley resistance + Tam , accelerated material + Tac , accessories Subtotal (C) CEMA Horsepower Formula Equation 1, page 86, provides the means for calculating the horsepower (hp) required by a belt conveyor having an effective tension, Te , at the drive pulley and a design velocity, V, of the belt, as follows: Combining equations (1) and (2) on page s 86-87, the hp load can be expressed as follows: The motor that will drive a fully loaded belt conveyor without becoming over- heated may not be able to accelerate the loaded conveyor from rest to the design speed. To insure adequate starting capabilities, the following conditions must exist. First, the locked rotor torque of the motor should exceed the sum of the torque required to lift the material, plus approximately twice the torque required to over- come total conveyor friction, despite any possible voltage deficiencies that may exist during the acceleration period. This may not be true for long, horizontal conveyors or for declined conveyors. Second, the motor speed-torque curve should not drop below a line drawn from the locked rotor torque requirement to the torque of the running horsepower require- ment at full speed. This is further explained in Chapter 13, “Motors and Controls.” For examples illustrating the use of the equations in determining the effective belt tension, Te , at the drive pulley and the horsepower to operate the belt conveyor, refer to the two problems on pages 145 through 152. It is also possible to arrive at a close approximation of the horsepower required to operate a belt conveyor by means of a graphical solution. This method, used under proper circumstances, is quick and relatively simple. Generally, a graphical solution will provide a somewhat conservative value of required horsepower. However, it must be recognized that it is impractical to incorporate all elements of belt conveyor design into a simple graphical solution. Therefore, the graphs should be used based on a T tr T pl T bc T sb+ + +( ) T p T am T ac+ +( )= T e Σ Subtotals (A), (B), and (C)= LKt K x K yW b 0.015W b+ +( ) W m LK y H±( ) T p T am T ac+ + + += hp T e V× 33 000, -----------------= hp V 33 000, ----------------- LKt K x K yW b 0.015W b+ +( ) W m LK y H±( ) T p T am T ac+ + + +[ ]= Belt Tension, Power, and Drive Engineering 104 complete understanding of all aspects of the analytical method of calculating belt conveyor tension and horsepower, in order to allow for adjustment of the results to account for unusual situations. It is recommended that final design be based on cal- culations made by the analytical method. The graphical method of designing belt conveyors is described on pages 141-145. Drive Pulley Relationships The force required to drive a belt conveyor must be transmitted from the drive pulley to the belt by means of friction between their two surfaces. The force required to restrain a downhill regenerative conveyor is transmitted in exactly the same man- ner. In order to transmit power, there must be a difference in the tension in the belt as it approaches and leaves the drive pulley. This difference in tensions is supplied by the driving power source. Figures 6.4 and 6.5 illustrate typical arrangements of single pulley drives. It should be noted that if power is transmitted from the pulley to the belt, the approaching portion of the belt will have the larger tension, T1 , and the departing portion will have the smaller tension, T2 . If power is transmitted from the belt to the pulley, as with a regenerative declined conveyor, the reverse is true. Wrap is used here to refer to the angle or arc of contact the belt makes with the pulley’s circumference. Wrap Factor, Cw The wrap factor, Cw , is a mathematical value used in the determination of the effective belt tension, Te , that can be dependably developed by the drive pulley. The Te that can be developed is governed by the coefficient of friction existing between the pulley and the belt, wrap, and the values of T1 and T2 . The following symbols and formulas are used to evaluate the drive pulley rela- tionships: Te = T1 - T2 = effective belt tension, lbs T1 = tight-side tension at pulley, lbs T2 = slack-side tension at pulley, lbs e = base of naperian logarithms = 2.718 f = coefficient of friction between pulley surface and belt surface (0.25 rub- ber surfaced belt driving bare steel or cast iron pulley; 0.35 rubber sur- Figure 6.4 Inclined or horizontal conveyor, pulley driving belt. Figure 6.5 Declined conveyor. Lowering load with regeneration, belt driving pulley. 105 Drive Pulley Relationships faced belt driving rubber lagged pulley surface). Values apply to normal running calculations. = wrap of belt around the pulley, radians (one degree = 0.0174 radians) Cw = wrap factor (see Table 6-8) It should be noted that the wrap factors do not determine T2 but only establish its safe minimum value for a dry belt. A wet belt and pulley will substantially reduce the power that can be transmitted from the one to the other because of the lower coeffi- cient of friction of the wet surfaces. Various expedients, such as grooving the lagging on the pulley, lessen this problem. However, the best solution is to keep the driving side of the belt dry. If this is impractical, increasing the wrap is helpful, or providing some means of increasing the slack side tension, T2 . This can be done, for example, by increasing the counterweight in a gravity takeup. Wrap Factor with Screw Takeup When a screw takeup is used, Table 6-8 indicates an increased wrap factor. This increased wrap factor is necessary to provide sufficient slack side tension, T2 , to drive the conveyor in spite of the amount of stretch in the conveyor belt, for which the screw takeup makes no automatic provision. θ T 2 T e ----- 1 e fθ 1– ---------------== Table 6-8. Wrap factor, Cw (Rubber-surfaced belt). Type of Pulley Drive θ Wrap Automatic Takeup Manual Takeup Bare Pulley Lagged Pulley Bare Pulley Lagged Pulley Single, no snub 180o 0.84 0.50 1.2 0.8 Single with snub 200o 0.72 0.42 1.0 0.7 210o 0.66 0.38 1.0 0.7 220o 0.62 0.35 0.9 0.6 240o 0.54 0.30 0.8 0.6 Dual* 380o 0.23 0.11 0.5 0.3 420o 0.18 0.08 — — *Dual values based on ideal distribution between primary and secondary drive. For wet belts and smooth lagging, use bare pulley factor. For wet belts and grooved lagging, used lagged pulley factor. If wrap is unknown, assume the following: Type of Drive Assumed Wrap Single–no snub 180o Single –with snub 210o Dual 380o Belt Tension, Power, and Drive Engineering 108 by definition by definition Substituting: solving for Cw , and noting that TesCws = T2 , For example, if the angles of wrap of the primary and secondary drive pulleys are 180 degrees and 220 degrees, respectively, the factors are as follows for lagged pulleys (see Table 6-8): Cws = 0.35 for 220° angle of wrap Cwp = 0.50 for 180° angle of wrap Cw = 0.095 for 400° total angle of wrap by interpolating between 380° and 420°, or: T es T 2 Cws --------= T ep T 3 Cwp --------= T 3 T 2 T es+ , whence= T ep T 2 T es+ Cwp -------------------= T 2 T 2 Cws -------- T 2 T es+( ) Cwp ------------------------+ Cw= T 2 Cws -------- T 2 Cwp -------- T es Cwp --------+ +    Cw= T 2 T 2Cwp T 2Cws T esCws+ + CwsCwp -------------------------------------------------------------    Cw= Cw T 2CwsCwp T 2Cwp T 2Cws T 2+ + ------------------------------------------------- , and because T esCws T 2,= = Cw CwsCwp Cwp Cws 1+ + ---------------------------------= 109 Drive Arrangements The tensions exterior to a dual-pulley drive are the same as those for a single-pul- ley drive. A part of the effective tension, Te is taken on the primary drive pulley and a part on the secondary drive pulley. Using two motors, the ratio of Tep to Tes is the ratio of the horsepower ratings of the two motors. For example, if the total calculated horsepower is 250, this could be supplied, allowing for drive losses, by using a 200-horsepower primary drive and a 75-horse- power secondary drive, with a drive efficiency of 90 percent. The primary pulley would take The secondary pulley would take If the belt velocity, V, is 400 fpm, then and (see Problem 1, page 145.) Drive Arrangements The final selection and design of a conveyor drive arrangement is influenced by many factors, including the performance requirements, the preferred physical loca- tion, and relative costs of components and installation. Figures 6.6A through 6.7F illustrate some of the drive combinations that have been furnished. Other arrangements may be better suited to a particular conveyor in a particular location. CEMA member companies can assist in final recommendations. Note that the illustrated arrangements that are on a substantially downhill run are usually regenerative and are so indicated in the title. Cw CwpCws 1 Cwp Cws+ + --------------------------------- 0.5( ) 0.35( ) 1 0.35 0.50+ + ------------------------------------ 0.0945== = 200 275⁄( ) 250( ) 182hp= 75 275⁄( ) 250( ) 68hp= T ep 182( ) 33 000,( ) 400 ------------------------------------ 15,000 lbs= = T es 68( ) 33 000,( ) 400 --------------------------------- 5,625 lbs= = T ep T es ------- 15 000, 5 625, ----------------- 2.67= = Belt Tension, Power, and Drive Engineering 110 Figure 6.6 Single-pulley/drive arrangements. Figure 6.6A Single-pulley drive at head end of conveyor without snub pulley. Figure 6.6B Single-pulley drive at head end of conveyor with snub pulley. Figure 6.6C Single-pulley drive at tail end without snub pulley. Used when head end drive cannot be applied. Figure 6.6D Single-pulley drive at tail end of conveyor without snub pulley; regenerative. Figure 6.6E Single-pulley drive at tail end of conveyor with snub pulley; regenerative. 113 Maximum and Minimum Belt Tensions Maximum and Minimum Belt Tensions For the illustrated common conveyor profiles and drive arrangements, minimum and maximum tensions will be discussed and procedures given for calculating the belt tension at any point in the conveyor. The applicable formulas are indicated with the various profile and drive arrangements where single-pulley drives are involved. The tensions involved in multiple-pulley drives are treated separately. Maximum Belt Tension Operating Maximum Belt Tension. The operating maximum belt tension is defined as the maximum belt tension occurring when the belt is conveying the design load from the loading point continuously to the point of discharge. Operating maxi- mum tension usually occurs at the discharge point on horizontal or inclined convey- ors and at the loading point on regenerative declined conveyors. On compound conveyors, the operating maximum belt tension frequently occurs elsewhere. Because the operating maximum belt tension must be known to select a belt, its location and magnitude must be determined. For details on belt tensions, refer to Figures 6.8 through 6.16. Conveyors having horizontal and lowering, or horizontal and elevating, sections can have maximum tensions at points other than a terminal pulley. In this case, belt tensions can be calculated by considering the horizontal and sloping sections as sepa- rate conveyors. Temporary Operating Maximum Belt Tension. A temporary operating maximum belt tension is that maximum tension which occurs only for short periods. For exam- ple, a conveyor with a profile that contains an incline, a decline, and then another incline, may generate a higher operating tension when only the inclines are loaded and the decline is empty. These temporary operating maximum belt tensions should be considered in the selection of the belt and the conveyor machinery. Starting and Stopping Maximum Tension The starting torque of an electric motor may be more than 2½ times the motor full-load rating. Such a torque transmitted to a conveyor belt could result in starting tensions many times more than the chosen operating tension. To prevent progressive weakening of splices and subsequent failure, such starting maximum tensions should be avoided. Refer to Chapter 13. Likewise, if the belt is brought to rest very rapidly, especially on decline conveyors, the inertia of the loaded belt may produce high ten- sions. The generally recommended maximum for starting belt tension is 150 percent of the allowable belt working tension. On conveyors with tensions under 75 lbs/ply in or the equivalent, the maximum can be increased to as high as 180 percent. For final design allowances, conveyor equipment or rubber belt manufacturers should be con- sulted. Minimum Belt Tension, Tmin For conveyors that do not overhaul the drive, the minimum belt tension on the carrying run will usually occur at the tail (feed) end. For conveyors that do overhaul their drive, the minimum belt tension will usually occur at the head (discharge) end. The locations and magnitude of minimum belt tensions are given in connection with the conveyor profiles and drives shown in Figures 6.8 through 6.16. It will be seen that the minimum tension is influenced by the T2 tension required to drive, without slippage of the belt on the pulley, and by the T0 tension required to Belt Tension, Power, and Drive Engineering 114 limit the belt sag at the point of minimum tension. The minimum tension is calcu- lated both ways and the larger value used. If T0 to limit belt sag is larger than the Tmin produced by the T2 tension necessary to drive the belt without slippage, a new T2 ten- sion is calculated, using T0 and considering the slope tension, Tb , and the return belt friction, Tyr . Formulas for calculating T2 , having T0 , Tb , and Tyr are given for each of the conveyor profiles and drive arrangements. Tension Relationships and Belt Sag Between Idlers Chapter 5, “Belt Conveyor Idlers,” presents the basic facts on the subject of idler spacing. A major requirement, noted in Chapter 5, is that the sag of the belt between idlers must be limited to avoid spillage of conveyed material over the edges of the belt. The sag between idlers is closely related to the weight of the belt and material, the idler spacing, and the tension in the belt. Graduated Spacing of Troughing Idlers For belt conveyors with long centers, it is practical to vary the idler spacing so as to equalize the catenary sag of the belt as the belt tension increases. The basic equation for the sag in a catenary can be written: where : W = weight, (Wb + Wm), lbs/ft of belt and material Si = idler spacing, ft T = tension in belt, lbs The basic sag formula can also be expressed as a relation of belt tension, T , idler spacing, Si , and the weight per foot of belt and load, (Wb + Wm), in the form: y = vertical drop (sag) between idlers, ft Experience has shown that when a conveyor belt sags more than 3 percent of the span between idlers, load spillage often results. For 3 percent sag the equation becomes: Sag, ft W Si 2 8T ------------= y Si 2 W b W m+( ) 8T ----------------------------------= Si 2 W b W m+( ) 8T ---------------------------------- 3Si 100 --------= 115 Tension Relationships and Belt Sag Between Idlers While pure catenary equations are used, the allowable percent sag takes into account such factors as stiffness of the belt carcass, strength of the belt span due to the “channel” shape of a troughed belt, etc. Simplifying for minimum tension to produce various percentages of belt sag yields the following formulas: For 3 percent sag, For 2 percent sag, For 1½ percent sag, See Table 6-10 for recommended belt sag percentages for various full load condi- tions. The graduated spacing should be calculated to observe the following limitations: (1) A maximum of 3 percent sag should be maintained when belt is operating with a normal load. (2) A maximum of 4.5 percent sag should be maintained when the loaded belt is standing still. (3) The idler spacing should not exceed twice the sug- gested normal spacing of the troughing idlers listed in Table 5-2. (4) The load on any idler should never exceed the idler load ratings given in Chapter 5. Moreover, the number of spacing variations must be based on practical consider- ations, such as the number of different stringer sections in the conveyor support structure, so that the fabrication cost of the support structure does not become exces- sive. Usually, the spacing of troughing idlers is varied in 6-inch increments. Limiting the calculated belt sag to 3 percent of the idler spacing, at any point on the conveyor, will generally prevent spillage of the material from conveyor belts oper- ating over 20 degrees troughing idlers. When handling lumpy material on belts operating on 35 degrees (or deeper) troughing idlers, belt tension should be increased to reduce the percent of sag. Deep- troughed conveyor belts normally carry a relatively large cross-sectional loading and corresponding heavy weight of material per foot of length. Therefore, the material exerts a greater pressure against the side of the trough, tending to cause greater trans- verse belt flexure. The purpose of increasing the minimum belt tension in belts oper- ating on idlers of greater than 20 degrees troughing angle is to keep this transverse belt flexure to an acceptable minimum and thus prevent spillage. Similarly, when frequent surge loads are encountered or a substantial percentage of large lumps is expected, the material weight per foot of conveyor will be increased. Consideration of increased minimum belt tension at, or closely adjacent to, the load- ing points is recommended. T 0 4.2Si W b W m+( )= T 0 6.25Si W b W m+( )= T 0 8.4Si W b W m+( )= Belt Tension, Power, and Drive Engineering 118 Figure 6.8 Head pulley drive — horizontal or elevating.* Use the larger value of T2 Use the larger value of Tt *With decline conveyors the Te tension required for an empty conveyor may some- times be greater than the Te for the loaded conveyor. T e T 1 T 2–= T 2 Cw T e or T 2× T t T b T yr–+= = T t T 0 = or T t T 2 T b– T yr+= T t T min T 1 T max= = T cx T t T wcx T fcx+ += T rx T t T wrx T frx–+= Figure 6.8A Inclined conveyor with head pulley drive. Figure 6.8B Horizontal belt conveyor with concave vertical curve, and head pulley drive. Figure 6.8C Horizontal belt conveyor with convex vertical curve, and head pulley drive. NOTE: Two takeups are shown only to illustrate alternative. Two automatic takeups cannot function properly on the same conveyor. 119 Tension Relationships and Belt Sag Between Idlers Figure 6.9 Head pulley drive — lowering without regenerative load.* Use the larger value of T2 *With decline conveyors the Te tension required for an empty conveyor may some- times be greater than the Te for the loaded conveyor. *T e T 1 T 2–= T 2 Cw T e or T 2× T o T e or T 2– T o T b– T yr–= = = T 1 T e T 2+= T t T 2 T b T yr or T t+ + T o T b T yr T e or T t–+ + T o= = = T max T tor T 1= T min T t T 1+= T cx T t T wcx– T fcx+= T rx T t T wrx– T frx–= Figure 6.9A Declined belt conveyor with head pulley drive. Lowering without regenerative load. Figure 6.9B Conveyor with convex vertical curve, head pulley drive. Lowering without regenerative load. Figure 6.9C Conveyor with concave vertical curve, head pulley drive. Lowering without regenerative load. NOTE: Two takeups are shown only to illustrate alternative. Two auto- matic takeups cannot function prop- erly on the same conveyor. Belt Tension, Power, and Drive Engineering 120 Figure 6.10 Head pulley drive — lowering with regenerative load. * Use the larger value of T2 . *See page 119. Takeup on return run not recommended to avoid driving through the takeup. T e T 1 T 2–= T 2 Cw T e or T 2× T o= = T t T max T 1 T b T yr or T t+ + T e T o T b T yr+ + += = = T min T 2= T cx T t T wcx– T frx+= T rx T t Tywrx– T frx–= Figure 6.10A Declined belt conveyor with head pulley drive. Lowering with regenerative load. Figure 6.10B Conveyor with convex vertical curve, head pulley drive. Lowering with regenerative load. Figure 6.10C Conveyor with concave vertical curve, head pulley drive. Lowering with regenerative load. 123 Tension Relationships and Belt Sag Between Idlers Figure 6.13 Tail pulley drive — lowering with regenerative load.* Use the larger value of T2 *Calculate belt tension required at takeup during acceleration and make takeup ade- quate to prevent lift-up. See page 119. *T e T 1 T 2–= T 2 Cw T e or T 2× T o T b T yr+ += = T hp T 2 T b– T yr or T hp– T o= = T hp T min= T 1 T max T e T 2+= = T cx T 1 T wcx– T fcx+= T rx T 2 T wrx– T frx–= Figure 6.13A Declined belt conveyor with tail pulley drive. Lowering with regenerative load. Figure 6.13B Conveyor with concave vertical curve, tail pulley drive. Lowering with regenerative load. Figure 6.13C Conveyor with convex vertical curve, tail pulley drive. Lowering with regenerative load. NOTE: Two takeups are shown only to illustrate alternatives. Two automatic takeups cannot function properly on the same conveyor. Belt Tension, Power, and Drive Engineering 124 Figure 6.14 Drive on return run — horizontal or elevating.* here Hd = lift to the drive pulley Use the larger value of T2 Use the larger value of Tt Takeups on return run or at tail pulley. *See page 119 T e T 1 T 2–= T 2 Cw T e or ×= T 2 T o 0.015W bLs– W bHd+= T t T min and T t T o= = T t T 2 0.015W bLs W bHd–+= T hp T e T 2 L Ls– L -------------- T b T yr–( ) or T hp+ + T e T o T b T yr–+ += = T hp T max= T cx T t T fcx T wcx+ += T rx T t T frx– T wrx+= Figure 6.14A Inclined conveyor with drive on return run. Figure 6.14B Horizontal belt conveyor with concave vertical curve, and drive on return run. Figure 6.14C Horizontal belt conveyor with convex vertical curve, and drive on return run. NOTE: Two takeups are shown only to illustrate alternatives. Two automatic take- ups cannot function properly on the same conveyor. 125 Tension Relationships and Belt Sag Between Idlers Figure 6.15 Drive on return run — lowering without regenerative load.* Use the larger value of T2 here Use the larger value of Tt here Takeup on return run or at tail pulley. *See page 119. *T e T 1 T 2–= T 2 Cw T e or T 2× T o T e or T 2– T o 0.015W b ts( )– W bHd–= = = T 1 T e T 2+= T t T 2 0.015W bLs W bHd+ += Hd lift to the drive pulley, or T t T o= = T max T t or T 1= T min T t or T 1= T hp T t T fcx T wcx–+= Lx L= T cx T t T fcx T wcx–+= T rx T t T frx– T wrx–= Figure 6.15A Declined conveyor, with drive on return run. Lowering without regenerative load. Figure 6.15B Conveyor with convex vertical curve, drive on return run. Lowering without regenerative load. Figure 6.15C Conveyor with concave vertical curve, drive on return run. Lowering without regenerative load. NOTE: Two takeups are shown only to illustrate alternatives. Two automatic take- ups cannot function properly on the same conveyor. Belt Tension, Power, and Drive Engineering 128 Example 2 Conveyor length = 1,200 ft Belt speed = 400 fpm Te at the drive pulleys = 20,625 lbs Required hp at the drive pulleys = 250 hp Total motor horsepower = 275 hp Primary drive motor = 200 hp Secondary drive motor = 75 hp Belt weight, Wb = 20 lbs/ft Cw = 0.11, according to Table 6-8 (380° wrap, lagged pulleys) Step 1: Calculate Tep and Tes : Step 2: Calculate T2, which is the minimum value that avoids slippage of the belt on the secondary pulley: T2 = TeCw = (20,625)(0.11) = 2,269 lbs Step 3: Calculate T3 : T3 = T2 + Tes = 2,269 + 5,625 = 7,894 lbs Step 4: Calculate T1 : T1 = T3 + Tep = 7,894 + 15,000 = 22,894 lbs Step 5: Calculate Cwp and Cws : Step 6: Check T2, using formula in Figure 6.14A. Assume conveyor is 1,200 ft long, the lift is 60 ft, Wb = 20, Wm = 80, idler spacing 3½ ft, drive at head of the conveyor. T0 = Tmin = 6.25(Wb + Wm)Si = 6.25(20 + 80)3.5 = 2,188 lbs (See “Minimum Belt Tension,” page 113.) Tyr = 0.015LWbKt = (.015)(1,200)(20)(1) = 360 lbs (See “Return Belt Friction Tension,” pages 116-117; assume temperature above 32°, Kt = 1.0) Tb = HWb = (60)(20) = 1,200 lbs (See “Tension, Tb” page 116) T ep 200 275 --------    250( ) 33 000, 400 -----------------    15 000 lbs,== T es 75 275 --------    250( ) 33 000, 400 -----------------    5 625 lbs,== Cwp T 3 T ep ------- 7 894, 15 000, ----------------- 0.53, requiring 180°wrap angle= = = Cws T 3 T es ------- 1 7 894, 5 625, -------------- 1–=– 0.40, requiring 205°wrap angle= = 129 Tension Relationships and Belt Sag Between Idlers Then T2 = Tmin + Tb -Tyr = 2,188 + 1,200 - 360 = 3,028 lbs Therefore, as T2 based on Tmin is larger than T2 minimum to prevent slip- page (3,028 is greater than 2,269), use T2 = 3,028 lbs. Step 7: Calculate corrected values of T3 , T1 , Cws , and Cwp : Based on wrap factors from Step 5, which provide the minimum T2 tension to drive without slippage, the secondary drive pulley would require a 205-degree angle of wrap, and the primary drive pulley would require a 180 degree angle of wrap. The revised value of T2 and the correspondingly revised values of Cwp and Cws , per Step 7, indicate that both drives could have 180 degree angles of wrap. In order to have equal resistance to slip, both drives should have approximately the same wrap angle. Belt Tension Calculations Five illustrative examples are offered to make clear the use of the formulas in determining the belt tensions at point X on the belt conveyor. Example 1 The basis for this example is the conveyor profile in Figure 6.8A from page 118, repeated below. 48-in. belt conveyor Wb = weight of belt = 15 lbs/ft Wm = weight of material = 106.6 lbs/ft T 3 T 2 T es+ 3 028, 5 625,+ 8 653 lbs,= = = T max T 1 T 3 T ep+ 8 653, 15 000,+ 23 653 lbs,= = = = Cws T 3 T es ------- 1 8 653, 5 625, -------------- 1–=– 0.54= = Cwp T 3 T ep ------- 8 653, 15 000, -----------------= 0.58= = Figure 6.8A. Belt Tension, Power, and Drive Engineering 130 Troughing idlers, 20-degree angle, Class E6, 6-inch diameter, spaced at 3½ ft, factor Ai = 2.8 Return idlers, Class C6, 6-inch diameter, spaced at 10 ft Kt = temperature correction factor = 1.0 Tt = T0 = 1,788 lbs, as T0 = Tmin here Lx = 1,000 ft Hx = 31.3 ft To find the belt tension at point X on the carrying run: Tcx = tension at point X on carrying run Twcx = tension resulting from weight of belt and material at point X Tfcx = tension resulting from friction on carrying run at point X Tcx = Tt + Twcx + Tfcx Twcx = Hx(Wb + Wm) = (31.3) (121.6) = 3,806 lbs Tfcx = LxKt(Kx + KyWb) + LxKyWm when Kt = 1.0, then Tfcx = Lx[Kx + Ky(Wb + Wm)] = 0.883 (for value of Ai , see tabulation on page 91) Ky = 0.025 when conveyor length is 1,000 ft (see Table 6-2). At 3.13% slope, Wb + Wm = 121.6 (use 125 in tables); and 3½ ft standard idler spacing. Tfcx = 1,000 [0.883 + 0.025(121.6)] = 3,923 lbs Tcx = 1,788 + 3,806 + 3,923 = 9,517 lbs To find the belt tension at point X on the return run: Trx = tension at point X on the return run Twrx = tension at point X on the return run resulting from the weight of the belt Tfrx = tension at point X on the return run resulting from return run friction Trx = Tt + Twrx - Tfrx Twrx = HxWb = (31.3)(15) = 470 lbs Tfrx = Lx.015 WbKt = (1,000)(0.015)(15)(1) = 225 lbs Trx = 1,788 + 470 - 225 = 2,033 lbs K x 0.00068 W b W m+( ) Ai Si ----+= 0.00068 15 106.6+( ) 2.8 3.5 ------+= 133 Tension Relationships and Belt Sag Between Idlers 1. Calculate Te , T1 , T2 , and Tt , the same as for a conveyor driven at a terminal pul- ley, using the horsepower formula on page 86, and appropriate tension formulas indicated with the conveyor profiles, Figure 6.14A. The Te , T1 , and T2 tensions so calculated apply at the drive pulley regardless of its location along the return run. 2. Calculate the tension, Thp , at the head pulley using the appropriate formula for Tcx as indicated for conveyor profile with drive on the return run, Figure 6.14. Cal- culate Twcx and Tfcx from the formula for determining belt tension at any point. See page 117. Conveyor data: Wm = 120 lbs/ft Wb = 15 lbs/ft Kt = 1.0 Kx = 0.35 Ky = 0.0243 Cw = 0.35 Si = 3.5 ft, idler spacing 36-in. belt conveyor, 600-ft centers, drive located midway of return run, lift 54 ft, slope 9 percent Calculate the head pulley tension, Thp , and the tail pulley tension, Tt . Te = LKt(Kx + KyWb + 0.015 Wb) + Wm(LKy +H) = 600 [0.35 + (0.0243)(15) + (0.015)(15)] + 120[(600)(0.0243) + 54] = 8,794 lbs For 3 percent belt sag, T0 = 4.2 Si(Wb + Wm) = (4.2)(3.5)(135) = 1,985 lbs T2 minimum to drive = TeCw = (8.794)(0.35) = 3,078 lbs Corresponding Tt = T2 + (L/2)(0.015 Wb) - (H/2) Wb = 3,078 + (300)(0.015)(15) - (27)(15) = 2,741 lbs This total tail pulley tension, 2,741 lbs, is greater than 1,985 lbs Therefore: Tt = 2,741 lbs, and T2 = 3,078 lbs T1 = Te + T2 = 8,794 + 3,078 = 11,872 lbs The tension at any point on the carrying run is: Tcx = Tt + Twcx + Tfcx Belt Tension, Power, and Drive Engineering 134 Now let Lx = L Kt = 1.0 Then, tension at the head pulley, Thp = Tt + Twcx + Tfcx Twcx= Hx(Wb + Wm) = (54)(135) = 7,290 lbs Tfcx = Lx[KtKx + KyWb] + LxKyWm, and since L = Lx = 600 [0.35 + (0.0243)(15)] + (600)(0.0243)(120) = 2,178 lbs Therefore, Thp = 2,741 + 7,290 + 2,178 = 12,209 lbs This is the maximum belt tension. The T1 tension at the drive pulley may be checked as follows: T1 = Thp - 27Wb + L/2(0.015Wb) = 12,209 - (27)(15) + (300)(0.015)(15) = 11,872 lbs This checks with the 11,872 lbs calculated for T1 from the formula, T1 = Te + T2 . Example 5 This example calculates the belt tension at any point in a declined regenerative conveyor. The calculation is substantially the same as that for a non-regenerative con- veyor except that ⅔Ky is used in place of Ky , and the factor Ai is eliminated in the for- mula for Kx . The value of Ky is for the length Lx . Figure 6.10A from page 120 is repeated below. Conveyor data: 36-in. belt conveyor, 1,000-ft centers, head pulley drive, drop 90 ft, slope 9 per- cent Wb = 15 lbs/ft Figure 6.10A. 135 Tension Relationships and Belt Sag Between Idlers Wm = 120 lbs/ft Si = 3.5 ft idler spacing Kx = 0.00068 (Wb + Wm) = 0.00068(135) = 0.0918 Ky = (0.0169)(0.666) = 0.01126 for 1,000 ft, 135 for (Wb + Wm) and 9 percent slope Kt = 1.0 Cw = 0.35 Tcx = Tt - Twcx + Tfcx Trx = Tt - Twrx - Tfrx Te = LKt(Kx + KyWb + 0.015Wb) + Wm(LKy + H) T2 = TeCw, or T2 = T0. If T0 is the greater T1 = Te + T2 Tt = T1 + 0.015WbL +WbH Te = 1,000 [0.0918 + (0.01126)(15) + (0.015)(15)] + (1,000)(0.01126)(120) - (90)(120) = 485.7 + 1,351.2 - 51.2 - 10,800 = -8,963 lbs The minus sign merely means that the belt drives the pulley (the conveyor is regenerative). T2 = (8,963)(0.35) = 3,137 lbs T0 for 3 percent sag = (4.2)(3.5)(135) = 1,985 lbs Therefore, T2 can be taken at 3,137 lbs T1 = Te + T2 = 8,963 + 3,137 = 12,100 lbs Tt = 12,100 + (1,000)(0.015)(15) + (90)(15) = 13,675 lbs To calculate Tcx at a point 500 ft from the tail shaft Tcx = Tt - Twcx + Tfcx Twcx= Hx(Wb + Wm) = (45)(135) = 6,075 lbs Tfcx = Lx(Kx + KyWb) + LxKyWm when Kt = 1.0 = 500 [0.0918 + (0.0182)(15)] + (500)(0.0182)(120)= 182 + 1,092 = 1,274 lbs when Ky = ⅔Ky for a 500-ft conveyor at 9 percent slope Then, Tcx = 13,675 - 6,075 + 1,274 = 8,874 lbs To calculate Trx at a point 500 ft from the tail shaft Trx = Tt - Twrx - Tfrx Twrx= HxWb = (45)(15) = 675 lbs Tfrx = Lx(0.015) Wb = (500)(0.015)(15) = 113 lbs when Kt = 1.0 Then Trx = 13,675 - 675 - 113 = 12,887 lbs For conveyor profiles per Figures 6.10B and 6.10C, the portion of the conveyor on a given slope is calculated separately, as in Example 2, page 131, and Problems 5 and 6, pages 172-176. Belt Tension, Power, and Drive Engineering 138 Analysis of Acceleration and Deceleration Forces The accelerating and decelerating forces that act on a belt conveyor during the starting and stopping intervals are the same in either case. However, their magnitude and the algebraic signs governing these forces change, as do the means for dealing with them. Acceleration The acceleration of a belt conveyor is accomplished by some form of prime mover, usually by an electric motor. The resulting forces in a horizontal conveyor are determined by inertia plus friction; in an inclined conveyor, by inertia plus friction plus elevating of the load; in a declined conveyor, by inertia plus friction minus low- ering of the load. Deceleration The deceleration of a belt conveyor is accomplished by some form of brake. The resulting forces in a horizontal conveyor are determined by inertia minus friction; in an inclined conveyor, by inertia minus friction minus elevating of the load; in a declined conveyor, by inertia minus friction plus lowering of the load. If the conveyor contains several portions with different (positive or negative) slopes, a combination of these conditions may result. Calculation of Acceleration and Deceleration Forces The belt conveyor designer is then confronted with the necessity to compute for the conveyor in question the inertia of all its moving parts, the inertia of the load on the belt, total frictional forces, and forces caused by elevating or lowering the load and belt. To be useful, the first two quantities have to be converted to a pound force at the belt line. Inasmuch as acceleration is defined as the second derivative of displacement with respect to time, and deceleration is simply negative acceleration, time is the basic variable in computing the force. To compute the time, Newton’s second law is used. The basic approach is as follows: Fa = Ma where : Fa = accelerating or decelerating force, lbs M = mass, in slugs = We = equivalent weight of moving parts of the conveyor and load, lbs g = acceleration by gravity = 32.2 ft/sec2 a = acceleration, ft per sec per sec (ft/sec2) The force necessary to achieve the acceleration or deceleration is always directly proportional to the mass (or the weight) of the parts and material in motion. For purposes of calculation, it can be assumed that the belt and the load on it move in a straight line. Other important parts of the system, however, rotate. This is true for all pulleys (including those on takeups and belt trippers), all idlers, and all the rotating parts of the drive. W e g ------ 139 Design Considerations It appears convenient to use the equation for linear motion as the basis for calcu- lating the acceleration and deceleration forces. This makes it necessary to convert the physical properties of the rotating components of the system to a form in which they can be used in the basic linear relationship: In other words, one must find the “Equivalent Weight” of the rotating parts. For rotating bodies, the mass actually distributed around the center of rotation is equivalent in its effect to the whole mass concentrated at a distance, K, (the polar radius of gyration, in ft) from that center. The WK2 is the weight of the body multiplied by the square of the radius of gyra- tion. If WK2 is known for the rotating conveyor components, the Equivalent Weight of these components, at the belt line, can be found by solving the equation where: V = belt velocity, fpm Values of WK2 (expressed in lb-ft2), which are difficult to compute, except for very simple shapes, must be obtained for each component from the manufacturers of the conveyor components, motors, transmission elements, etc. So far considered have been the forces in the system caused by inertia of the mov- ing parts of the conveyor, the moving parts of the drive, and the moving load. Two other forces also are involved, as mentioned on page 138. These are: (1) Forces result- ing from friction. (2) Forces resulting from the elevating or lowering of the load and belt. These simply represent the components of the weight of the material and belt, in the direction of motion of the belt, in the various portions of the conveyor. Design Considerations The belt conveyor designer is confronted with two problems: (1) The necessity to provide a prime mover powerful enough to start the conveyor, sometimes under adverse conditions. (2) To make sure, for the reasons outlined under “Acceleration and Deceleration Forces,” page 138, that the maximum force exerted on the conveyor is within safe limits. In long, level, high-speed conveyors, a motor large enough for continuous full- load operation may be unable to start the fully loaded conveyor, particularly in cold weather. On the other hand, a motor capable of continuous full-load operation of an inclined conveyor may overstress the belt during starting, unless preventive measures are taken. The maximum permissible accelerating forces are determined by the factors listed in page 138 of this chapter. Minimum accelerating forces may be dictated by the time during which the prime mover, which usually is an electric motor, can exert its start- ing torque without being damaged. This limitation is also affected by the frequency of starting the conveyor system. F W e g ------    a= Equivalent Weight lbs W K2 2π rpm V -----------------    2 = Belt Tension, Power, and Drive Engineering 140 In the case of deceleration, maximums are governed by the same factors. A mini- mum deceleration may be dictated by safety or may be necessary because of the mate- rial flow at transfer points. In all deceleration calculations involving brakes, the energy-dissipating capacity of the brake will be an important factor to consider. Necessary Assumptions As in all engineering investigations of this type, the first question is, “To what degree of accuracy will the computations have to be carried out?” The answer is not simple. Important factors are the overall size, the importance of the installation, and the type and sensitivity of the equipment adjacent to it. In any case, numerous simplifying assumptions will have to be made to keep the engineering work within reasonable limits. For examples of simplifying assumptions, refer to the problems connected with belt stretch (elastic elongation from accelerating or decelerating forces) and takeup reactions. During both the acceleration and deceleration cycles, the transient forces imposed result in extra stretch not encountered during steady state operation. This may result in early splice failure, excessive takeup travel, and other difficulties. Because of the vast differences in carcass construction, from the standpoint of both materials used and methods of manufacture, no single numerical value can express belt stretch as a function of the applied force. Most manufacturers have listed values of Bm (elastic constant) for their line of belts. These vary from 1.3 x 106 lbs per in. of belt width for steel-cable belt to 2.3 x 103 lbs per ply in. of belt width for cotton fabric belts. Other rubber manufacturers may list different values, but they also would vary over the same wide range. For this reason, as well as many others, the calculations for acceleration and deceleration treat the system as a rigid body. This is a common practice in the solu- tion of problems in dynamics. And while the results usually are quite satisfactory, there is more cause for concern over the accuracy of results in the case of belt convey- ors. No further attempt will be made to justify simplifying the assumption, because this usually is not of major significance. However, the belt conveyor designer should be aware that, for conveyor systems with very long center belts, stretch considerations should not be overlooked. Calculations While the calculations are relatively simple for a conveyor with only one slope, they become increasingly complex for belt conveyors which change slope several times, or which are loaded and unloaded at different points, or which have belt trip- pers operating on them. All this results in a great number of possible combinations of load distribution, tripper position, etc. Although theoretically it suffices to investigate only the worst combination of conditions, without analysis, it is usually impossible for even the experienced designer to tell which combination of factors will lead to this extreme case. In the more complicated cases, it will be necessary to divide the conveyor into portions or sections—within which neither the slope nor the conditions of loading change more than is permitted by the required accuracy of the calculations—and to determine the physical properties for each such portion discussed under “Analysis of Acceleration and Deceleration Forces,” page 138. Any really large rotating member of the belt conveyor, because of its very magnitude, may have to be considered a portion or section by itself. 143 Conveyor Horsepower Determination — Graphical Method Figure 6.18 Horsepower required to elevate material. Conveyor specifications: Length, L = 2,000 ft Lift, H = 75 ft Capacity, Q = 1,600 tph Belt speed, V = 500 fpm Material density, dm = 100 lbs/cu ft Belt width, b = 48 in. Graphical analysis. Referring to Figure 6.17, the weight per foot of belt and revolv- ing idler parts for a 48-in. wide conveyor and 100 pounds per cubic foot of material is given as 51 pounds per foot. Using this value, the horsepower required to drive the empty conveyor at a speed of 100 fpm is 6.5. Belt Tension, Power, and Drive Engineering 144 Therefore, the horsepower to drive the empty conveyor at 500 fpm is equal to: Figure 6.19 Horsepower required to convey material horizontally. From using Figure 6.18, the horsepower required to elevate the material can be determined. The horsepower per foot of lift for the 1,600 tph capacity is 1.62. There- fore, the horsepower required to elevate the material 75 ft is given as: 1.62 x 75 = 121.5 hp The horsepower needed to convey the material horizontally is determined through the use of Figure 6.19. Using the given conveyor specification of 2,000 ft of conveyor length, the horsepower required to convey 100 tph of material is equal to 5.5 hp. Therefore, the horsepower required for the 1,600 tph capacity is equal to: The total required horsepower at the belt line is the sum of the above, and equals 32.5 + 121.5 + 88 = 242 hp. Assuming a standard 5 percent loss of power through the drive components due to their inefficiencies, the required motor horsepower is: A comparison of the horsepower derived by the analytical method, shown on pages 145 to 148 and the above illustrated graphical method shows that the results of the two methods are comparatively very close. This is coincidental inasmuch as the 6.5 500× 100 ---------------------- 32.5 horsepower= 5.5 1,600× 100 -------------------------- 88 horsepower= 242 .95 -------- 254.7horsepower= 145 Examples of Belt Tension and Horsepower Calculations — Six Problems degree of accuracy of determinations made with the graphical solution is dependable only for estimating purposes. Final design should be made by the analytical method for greatest accuracy. Examples of Belt Tension and Horsepower Calculations — Six Problems Application of the CEMA horsepower (hp) formula and analysis of belt tensions and power requirements will be illustrated by the following six problems: Problem 1 — inclined conveyor; Problem 2 — declined conveyor with regenerative characteristics; Problem 3 — horizontal conveyor; Problem 4 — conveyor with a horizontal section, an inclined section, and vertical curves; Problems 5, 6 — comparison of tension and hp values on two similar conveyors. Problems 3 and 4 also include calculation of acceleration and deceleration forces. Problem 1 Inclined Belt Conveyor Figure 6.20 Inclined belt conveyor. Problem: Determine effective tension, Te ; slack-side tension, T2 ; maximum tension, T1 ; tail tension, Tt ; belt and motor horsepower requirements; and type and location of drive. In this problem, only two accessories are considered, pulley friction from non- driving pulleys and skirtboard friction. The belt speed is too low to involve any appre- ciable material acceleration force. The discharge is made freely over the head pulley and no cleaning devices are used. Belt Tension, Power, and Drive Engineering 148 T2 , by choice = 2,463 Return run friction (2,000)(0.015)(15) = + 450 2,913 Less weight of return belt (75)(15) = -1,125 lbs Tail tension, Tt = 1,788 lbs Final Tensions: Te = 15,816 lbs T2 = 2,463 lbs Tt = Te + T2 = 15,816 + 2,463 = 18,279 lbs Tt = 1,788 lbs Horsepower at Motor Shafts: = 239.64 = 6.06 Add 5 percent for speed reduction loss = 0.05 (239.64 + 6.06) = 12.29 Total hp at motor shafts = 257.99 hp Problem 2 Declined Belt Conveyor Figure 6.21 Declined belt conveyor. Belt hp T eV 33 000, ----------------- 15 816,( ) 500( ) 33 000, ------------------------------------= = Drive pulley friction loss hp 2( ) 200( ) 500( ) 33 000, -----------------------------------= Belt tension 18,279 48 --------------- 381lbs per inch of belt width== 149 Examples of Belt Tension and Horsepower Calculations — Six Problems Before attempting the solution of a declined conveyor, certain peculiar conditions must be considered. A declined conveyor, which delivers material below the elevation at which it is received, will generate power if the net change in elevation is more than 2½ percent of the conveyor length. It may generate power at a lower slope, depending on conditions. An electric motor, acting as a generator, is used to retard the conveyor. A brake is used to stop the conveyor. The motor size is determined by the maximum horsepower, either positive or negative, that it will be called on to produce, and it usually is set by the horsepower generated. The drive is usually located at the tail (feed) end of the conveyor, involving special design problems. One of these is that the motor must start the conveyor by driving through the gravity takeup without lifting the takeup pulley. Care must be taken to check the horsepower and belt tensions for an empty and partially loaded belt. The brake must be large enough to absorb the torque generated and to decelerate the load. However, the retarding torque must be limited so that it does not overstress the belt. Frequently, on large conveyors, the limiting factor in brake selection will be its holding power within its ability to absorb and dissipate heat. Refer to “Break Heat Absorption Capacity,” pages 191-192. When a conveyor runs downhill, friction forces increase the belt tension in the direction of motion, while gravity forces decrease the belt tension, by the weight per foot of belt and load, for every foot that the belt and load are lowered. Reduced friction. The belt, load, and idler friction absorb some of the power that the motor or brake would be compelled to absorb if these quantities did not exist. Therefore, it is important not to overestimate the friction forces or else the selected size of motor or brake might be too small. In order to avoid overestimating the friction forces, the effective tension, Te is calculated as follows: Te = LKt(Kx + C1KyWb + C10.015Wb) + C1WmLKy - HWm + C1Tac in which factor C will vary from 0.5 to 0.7 and, for average conditions, will be 0.66. For declined conveyors only, determine Kx by the formula Kx = 0.00068 (Wb + Wm). The additive term Ai/Si is omitted because the allowance for grease and seal fric- tion, represented by the factor Ai , is no longer on the safe side. It may under some conditions approach zero, so the safe course in declined conveyors is to make Ai = O. Problem: Determine effective tension (Te), slack-side tension (T2), maximum tension (T1), tail tension (Tt), and belt and motor horsepower requirements. In this problem, only two accessories are considered, pulley friction of nondriving pulleys and skirtboard friction. The belt speed is too low to involve any appreciable material acceleration and no cleaning devices are employed. Belt Tension, Power, and Drive Engineering 150 Conveyor Specifications: Wb = 10 lbs/ft from Table 6-1 L = length = 1,200 ft V = speed = 450 fpm H = drop = 200 ft Q = capacity = 1,000 tph Si = spacing = 4 ft Ambient temperature = 32°F, minimum Belt width = 36 in. Material = limestone at 85 lbs/cu ft, 4-in. maximum lumps Drive = lagged grooved tail pulley, wrap = 220 degrees Troughing idlers = Class C6, 6-in. diameter, 20-degree angle, Ai = 1.5 Return idlers = Class C6, 6-in. diameter, 10 ft spacing Analysis: From Table 6-8, drive factor, Cw = 0.35 From Figure 6.1, for 32°F, Kt = 1.0 Formula (not considering C1 factor): Te = LKt(Kx + KyWb + 0.015Wb) + Wm(LKy - H) + Tac To find Kx and Ky it is necessary to find Wb + Wm = 10 + 74 = 84 lbs/ft Kx must be calculated for two cases. In the first calculation, Kx is taken at its nor- mal value, so that the tension for the full friction can be determined. In the second calculation, Kx is taken at its reduced value, so that the tension for the reduced fric- tion can be calculated. Reduced Kx = 0.00068(84) = 0.05712 Ky also must be determined for two cases. In the first instance, Ky will have its normal value as selected from Tables 6-2 and 6-3. This value is then employed for the full friction calculation of the tension. In the second instance, Ky is modified by the reduced friction factor C. The slope is (200/1,200)(100%) = 16.6 percent . From Table 6-2, it can be seen that 4 ft is not a tabular spacing. For the 16.6 percent slope, L = 1,200 and Wb + W m 33.3Q V -------------- 33.3( ) 1 000,( ) 450 ---------------------------------- 74 lbs per foot= = = Normal K x 0.00068 W b W m+( ) Ai Si ----+= K x 0.00068( ) 84( ) 1.5 4 ------+ 0.05712 0.375+ 0.4321= = = 153 Examples of Belt Tension and Horsepower Calculations — Six Problems Problem 3 Horizontal Belt Conveyor Figure 6.22 Horizontal belt conveyor. Problem: Determine effective tension (Te), slack-side tension (T2), maximum tension (T1), tail tension (Tt), belt and motor horsepower requirements. In this problem, only two accessories are considered, pulley friction from non- driving pulleys and skirtboard friction. Material acceleration force has been omitted in this example. The discharge is made freely over the head pulley. No belt-cleaning devices are employed. Conveyor Specifications: Wb = 17 lbs/ft, from Table 6-1 L = length = 2,400 ft V = speed = 500 fpm H = lift = 0 Q = capacity = 3,400 tph Si = spacing = 3 ft Ambient temperature = 60°F Belt width = 48 in. Material = iron ore at 150 lbs/cu ft 10-in. maximum lumps from a gyratory crusher Drive = lagged and grooved head pulley, 220-degree wrap Troughing idlers = Class E6, 6-in. diameter, 20-degree angle Return idlers = rubber-disc type, Class C6, 6-in. diameter, 10 ft spacing Analysis: From Table 6-8, drive factor, Cw = 0.35 From Figure 6.1 for 60°F, Kt = 1.0 Formula: Te = LKt(Kx + KyWb + 0.015Wb) + Wm(LKy + H) + Tac W m 33.3Q V -------------- 33.3( ) 3 400,( ) 500 ---------------------------------- 226.4 lbs per ft= = = Belt Tension, Power, and Drive Engineering 154 To find Kx and Ky it is necessary to find Wb + Wm = 17 + 226.4 = 243.4 lbs/ft thus: Kx = 1.099, for 3.0 spacing Ai = 2.8 and Wb + Wm = 243.4 lbs from equation (3) Ky = 0.021, for L = 2,400, slope 0° and Wb + Wm = 243.4 lbs. Refer to Table 6-2. Minimum tension, T0 , for 3 percent sag = 4.2 Si (Wb + Wm) = (4.2)(3)(243.4) = 3,067 lbs Determine Accessories: In this case the only accessories are pulley friction and a loading chute plus skirtboards. Assume that skirtboards are 10 ft long and spaced apart, two- thirds the width of the belt. The pull on the belt to overcome skirtboard fric- tion is T = CsLbhs 2 . From the calculation of skirtboard friction, from page 101, 90, hs = (0.1)(48) = 4.8 inches. Cs is (safely).276, from Table 6-7, for iron ore @ 150 pounds per cubic foot. Therefore, T = (0.276)(10)(4.8)2 = 64 lbs. For the additional 20 ft of rubber edging on the skirtboards, additional resistance is (3)(20) = 60 lbs. Total skirtboard resistance, Tsb = 64 + 60 = 124 lbs. LKtKx = (2,400)(1)(1.099) = 2,638 LKtKyWb = (2,400)(1)(0.021)(17) = 857 LKt 0.015 Wb = (2,400)(1)(0.015)(17) = 612 KyLWm = (0.021)(2,400)(226.4) = 11,411 HWm = (0)(226.4) = 0 Nondriving pulley friction = (4)(100) + (2)(150) = 700 Skirtboard resistance, Tsb = 124 Effective tension, Te = 16,342 lbs T2 = CwTe = (0.35)(16,342) = + 5,720 Maximum tension, T1 = Te + T2 = 22,062 lbs Tail tension, Tt = T2 +.015LKtWb + pulley friction = 5,720 + 612 + 700 = 7,032 lbs Final tensions: Te = 16,342 lbs T2 = 5,720 lbs T1 = 22,062 lbs Tt = 7,032 lbs 155 Examples of Belt Tension and Horsepower Calculations — Six Problems Horsepower at Motor Shaft: Add 5% for speed reduction loss = 0.05(247.61 + 3.03) = 12.53 Horsepower at motor shaft = 263.17 hp Horsepower at motor shaft = 263 (select 300 hp, 1,750 rpm, motor) Acceleration Calculations: The following calculations are used to determine the acceleration forces and times: WK2 of drive (all values are taken at motor speed and should be obtained from the equipment manufacturer): WK2 of motor = 101 lb-ft2 Equivalent WK2 of reducer = 20 lb-ft2 (common practice is to take ⁄ of WK2 of motor) WK2 of coupling = 4 lb-ft2 Equivalent WK2 of drive pulley = 5 lb-ft2 Total WK2 of drive = 130 lb-ft2, at motor speed Converting this WK2 value by using the equation for equivalent weight, page 139, 62,870 lbs is calculated as follows: For purposes of calculating the equivalent weights, the pulley diameters must first be estimated. The diameters of the head and tail pulleys are assumed to be 42 in.; the rest of the pulleys are assumed to be 30 inches. The actual required pulley diameters are a function of the characteristics of the belt to be used and the belt tension at the pulley. The assumed pulley diameters should be reviewed after this information is known. Belt hp T eV 33,000 --------------- 16 342,( ) 500( ) 33,000 ------------------------------------ 247.61= = = Drive pulley hp 200( ) 500( ) 33,000 --------------------------- 3.03= = Belt tension T 1 width ------------- 22,062 48 --------------- 460 lbs per inch of belt width= = = Drive equivalent weight is 130 lb-ft 2( ) 1,750 rpm 500fpm ------------------------    2 2π( )2 62,870 lbs= Belt Tension, Power, and Drive Engineering 158 The total equivalent mass = 768,926/32.2 = 23,880 slugs From the equation, Fa = Ma The time needed is: Therefore: The time required by the motor to accelerate the loaded conveyor, 11.46 seconds, is greater than the minimum acceleration time to stay within the maximum allowable belt tension, 7.43 seconds. Therefore, the conveyor is safe to start, fully loaded, with the equipment selected. Had the starting belt stress been limited to 120 percent of the normal belt rating instead of 180 percent, the allowable extra belt tension would have been (1.2)(25,920) – (22,062) = 9,042 lbs and the acceleration time, t = (21,927)(500 – 0)/(9,042)(60) = 20.21 sec minimum. This is more than the time calculated for the motor to accelerate the loaded system, 11.46 seconds. So, if such limitation had been placed on the start- ing belt stress, the system would not have been safe to start with the equipment selected. In fact, the belt stress during acceleration must be: The foregoing assumes that the mass between the slack side of the drive pulley and the takeup is negligible. If the takeup is far removed from the drive, this should be taken into account in the calculations. In Chapter 13 it is indicated that the acceleration time for NEMA Type C motors, in general, be considered as 10 seconds or less. It, therefore, would be prudent to Fa 300( ) 1.8( ) 33 000,( ) 500 ------------------------------------------------- 263.17( ) 33 000,( ) 500 -------------------------------------------– 0.95= 0.95 300( ) 1.8( ) 263.17–[ ]33 000, 500 -----------------–= 263( ) 33 000,( ) 500 ------------------------------------ 17 358 lbs,== the acceleration, a Fa M ----- 17,358 23,880 ---------------- 0.727 ft per sec2= = = t V t V o– 60a -----------------= t 500 0– 60( ) 0.727( ) ---------------------------- 11.46 seconds.= = extra belt tension 21,927( ) 500 0–( ) 11.46( ) 60( ) -------------------------------------------- 15,752 lbs= = % of normal belt rating 15,752( ) 22,062+ 25,920 -------------------------------------------- 100%( ) 146%= 159 Examples of Belt Tension and Horsepower Calculations — Six Problems check with the motor manufacturer to make sure that the calculated acceleration time of 11.46 seconds would not cause the motor to overheat during starting. In the case of this particular problem, the motor manufacturer was asked what maximum safe accel- eration time would be for his 300 hp NEMA Type C motor. The manufacturer stated that any time up to 20 seconds would be permissible. Therefore, the conveyor in this problem could be safely started with the equip- ment selected, provided the allowable belt tension during starting was 146 percent of normal belt rating, or greater. The use of the 300 hp NEMA Type C motor is justified, provided that operating conditions of this particular conveyor are such that abnormal starting conditions (which would require forces considerably in excess of those calcu- lated) are unlikely to occur. If abnormal starting conditions are likely to occur, even infrequently, consideration should be given to the use of a different means of starting that would satisfy all the requirements described earlier in the section, “Acceleration and deceleration forces.” Deceleration Calculations: In the preceding example on acceleration, it was found that the total equivalent mass of the conveyor system under normal conditions of operation is equal to 23,880 slugs (see page 158). As these calculations are based on the belt speed of 500 fpm or 8.33 fps, the kinetic energy of the system is: Earlier in this problem (3), it was found that 263.17 hp is required to operate this conveyor at its rated speed of 500 fpm. Because the conveyor is horizontal, this repre- sents the product of the friction forces and the distance traveled in unit time. This means that the frictional retarding force is: The average velocity of the conveyor during the deceleration period would be Because the total work performed has to be equal to the kinetic energy of the total mass, (t)(250 fpm)(l7,369) = 828,503 ft lbs where: t = time in minutes MV 2 2 ----------- 23,880( ) 8.33( )2 2 --------------------------------------- 828,503 ft lbs= = 263.17( ) 33,000( ) 500 ------------------------------------------ 17,369 lbs= 500 0+ 2 ----------------- 250 fpm= Belt Tension, Power, and Drive Engineering 160 Therefore: and the belt will have moved (0.191)(250 fpm) = 47.75 ft in this time. As the belt is fully loaded (by assumption), the belt will discharge the following amount of material: If 5.41 tons of material discharged is objectionable, the use of a brake has to be considered. Such a step, however, can be justified only if the reduced deceleration time is still greater than, or at least equal to, the deceleration cycle of whatever piece of equipment delivers to the conveyor in this example. Also, another difficulty arises. Suppose it is desirable or necessary to reduce the deceleration time from 11.45 seconds to 7 seconds. Since the total retarding force is inversely proportional to the deceleration time, the additional braking force required must be: If the brake is connected to the drive pulley shaft, the drive pulley is required to transmit to the belt a braking force equal to The difference between the 11,042 lbs and the 10,139 lbs is the braking force required to decelerate the drive and drive pulley and is not transmitted to the belt. However, under coasting conditions, the belt tension is principally governed by the gravity takeup which, if located adjacent to the head pulley, would provide a max- imum tension equal to T2, or 5,720 lbs. Obviously, it is impossible to secure a braking force of 10,139 lbs on the head pulley. Even a much smaller force than this would result in looseness of the belt around the head pulley. The solution is to provide the braking action on the tail pulley, where it would increase rather than decrease the contact pressure between the belt and pulley. How- ever, a further check on the tail pulley indicates that with 11,042 lbs braking tension, a plain bare tail pulley with 180-degree wrap angle could not produce a sufficient ratio of tight-side to slack-side tension. Therefore, it would be necessary to do one or a combination of the following: increase the takeup tension weight, lag the tail pulley, or snub the tail pulley for a greater wrap angle. If the increasing takeup weight should result in a heavier and more costly belt carcass, the second and third remedies are preferable and more eco- nomical. t 828,503 250( ) 17,369( ) ----------------------------------- 0.191 minutes, or 11.45 seconds= = 3 400 tph, 60 -----------------------    47.75 500 ------------    5.41 tons= 17 369 11.45 7– 7 -------------------   ×, 11 042 lbs,= 11,042 768,926 62 870,( )– 768 926, ---------------------------------------------    10,139 lbs= 163 Examples of Belt Tension and Horsepower Calculations — Six Problems Initial horizontal section, 3000 ft long, 3 percent sag in belt, where: Kt = 1.0 Kx = 0.427 Wb = 10 lbs/ft Wm = 66.6 lbs/ft (Wb + hm)= 76.6 lbs/ft Ky from Table 6-2 would be 0.023. The average tension is: Here, Tt is at least equal to T0 . And T0, for 3 percent sag is 4.2 Si (Wb + Wm) = (4.2)(4)(76.6) = 1,287 lbs Thus, the average tension is: or, Equation (4) indicates Ky = 0.0255, for 4,570 lbs average tension, and (Wb+ Wm) = 76.6 lbs. Re-estimate using Ky = 0.0255. Average tension is: or, Equation (4) checks Ky = 0.0255, for an average tension of 4,858 lbs, and (Wb + Wm) = 76.6 lbs. The formula for the actual tension because of friction in the initial horizontal portion (see page 117) is: Tfcx = Lx[Kt(Kx + KyWb)] + LxKyWm T t Kt K xL K yLW b+[ ] K yLW m T t+ + + 2 ---------------------------------------------------------------------------------------------- 1,287 0.427( ) 3,000( ) 0.023( ) 3,000( ) 76.6( ) 1,287+ ++ 2 --------------------------------------------------------------------------------------------------------------------------------------- 1,287 1,281 5,860 + 1,287++ 2 ------------------------------------------------------------------------ 9,140 2 ------------ 4,570 lbs= = 1,287 1,281 0.0255( ) 3,000( ) 76.6( ) 1,287+ ++ 2 ------------------------------------------------------------------------------------------------------------------ 1,287 1,281 + 5,860 + 1,287 + 2 -------------------------------------------------------------------------- 9,715 2 ------------ 4,858 lbs= = Belt Tension, Power, and Drive Engineering 164 where: Lx = 3,000 ft Kx = 0.427 Kt = 1.0 Ky = 0.0255 Wb = 10 lbs/ft Wm = 66.6 lbs/ft Therefore: Tfcx = LX[Kx + Ky(Wb + Wm)], since Kt = 1.0 Tfcx = 3,000[0.427 + 0.0255(76.6)] = 3,000(2.38) = 7,141 lbs The tension at the beginning of the vertical concave curve is calculated using the formula for belt tension at any point on the conveyor length, for point X on the carrying run (page 117), at the intersection of the initial horizontal run and the inclined run: Tcx = Tt + Twcx + Tfcx Twcx = Hx(Wb + Wm) = (0)(76.6) = 0 so Tcx = 1,287 + 7,141 = 8,428 lbs The tension at the bottom of the incline, therefore, is 8,428 lbs. The estimated Ky is 0.024, for the first approximation of the calculation for the upper end of the incline from Table 6-2 for a value of (Wb + Wm) = 76.6, and a slope of (70/ 800)(100%) = 8.8%. Average tension is: in which Tt is the tension at the bottom of the incline, or 8,428 lbs, so, or, As stated on page 92, the minimum Ky value = .016 for 12,016 lbs tension and Wb + Wm = 76.6. Re-estimate using Ky = .016. Average tension then is: T t Kt K xL K yLW b+( ) K yLW m H W b W m+( ) T t+ + + + 2 ------------------------------------------------------------------------------------------------------------------------------------- 8,428 0.427( ) 800( ) 0.024( ) 800( ) 76.6( ) 70( ) 76.6( ) 8,428+ + + + 2 --------------------------------------------------------------------------------------------------------------------------------------------------------------- 8,428 + 342 + 1,471 + 5,362 + 8,428 2 ---------------------------------------------------------------------------------------- 24 031, 2 ----------------- 12 016 lbs,= = 165 Examples of Belt Tension and Horsepower Calculations — Six Problems or, This checks Ky = 0.016 minimum value, for average tension of 11,770 lbs, and Wb + Wm = 76.6 lbs/ft. From page 117, Tcx = Tt + Twcx + Tfcx . Here, Tt is 8,428 lbs, the tension at the bottom of an incline. Twcx = Hx(Wb + Wm) = (70)(76.6) = 5,362 lbs Tfcx = LxKt[Kx + Ky(Wb + Wm)] where Lx = 800 ft and Kt = 1.0 = (800)(1.0)[0.427 + 0.016(76.6)] = 800(1.653) = 1,322 lbs Tcx = 8,428 + 5,362 + 1,322 = 15,112 lbs The tension at the top of the incline, then, is 15,112 lbs. The final horizontal portion is 200 ft long: Kt = 1.0 Kx = 0.427 Wb = 10 lbs/ft Wm= 66.6 lbs/ft Wb + Wm = 76.6 lbs/ft Ky will be at a minimum value because of the high tension that is obvious in this portion of the belt. From page 91, the minimum Ky of 0.016 is applicable at the indicated average tension (obviously more than 15,112 lbs) and with Wb + Wm = 76.6 lbs/ft. From page 117, Tcx = T1 + Twcx + Ffcx . Here, Tt is the tension at the beginning of this horizontal section, or 15,112 lbs and Twcx = 0, since Hx = 0. Tfcx = LxKt(Kx + KyWb) + LxKyWm where Lx = 200 ft and Kt = 1.0 = (200)(1)(0.427 + (0.016)(10)) + (200)(0.016)(66.6) = 117 + 213 = 330 lbs Tcx = 15,112 + 330 = 15,442 lbs In this case, Tcx = T1 = 15,442 lbs 8,428 + 342 0.016( ) 800( ) 76.6( ) 5,362 + 8,428+ + 2 --------------------------------------------------------------------------------------------------------------------------- 8,428 + 342 + 980 + 5,362 + 8,428 2 ----------------------------------------------------------------------------------- 23,540 2 --------------- 11,770 lbs= = Belt Tension, Power, and Drive Engineering 168 From page 156, ⅔(2,540 lbs + 2,860 lbs) = 3,600 lbs Belt, carrying run, from Table 6-1,10 lbs/ft x 4,000 ft = 40,000 lbs Subtotal = 43,600 lbs Belt, return run, 10 lbs/ft x (4,000 ft + 50 ft) = 40,500 lbs Idlers, troughing, from Table 5-11, for 36-in. belt width and class C6, the weight is 43.6 lbs. = 43,600 lbs Idlers, return, from Table 5-12, for 36-in. belt width and Class C6, the weight is 37.6 lbs. = 15,040 lbs Total conveyor equivalent weight = 142,740 lbs Material load (66.6 lbs/ft)(4,000 ft) = 266,400 lbs Total equivalent weight of system = 55,615 lbs + 142,740 lbs + 266,400 lbs = 464,755 lbs Percent of total within the conveyor: (142,740 + 266,400)/464,755 x 100% = 88% Having selected a belt for T1 = 15,602 lbs, as explained in Chapter 7, at an allow- able rating of 70 lbs/in./ply, 7 plies are required and the rated tension is 17,640 lbs. If the starting tension is limited to 180 percent of the rated tension (see page 113), then the allowable extra tension is: (1.80)(17,640) - (15,602) = 31,752 - 15,602 = 16,150 lbs The time for acceleration is found from the equation: Where: Fa = the allowable extra accelerating tension = 16,150 lbs t = time, seconds V1 = final velocity = 400 fpm Vo = initial velocity = 0 fpm M = mass of conveyor system = = 12,706 slugs Solving for t: 43.6 4,000 ft 4-ft idler spacing ----------------------------------------    37.6 4,000 ft 10-ft idler spacing -------------------------------------------    Fat M V 1 V 0– 60 ------------------= 409,140 32.2 ------------------ t M Fa ----- V 1 V 0–( ) 60 ----------------------- 12,706 16,150 --------------- 400 0–( ) 60 ---------------------- 5.24 seconds= = = = 169 Examples of Belt Tension and Horsepower Calculations — Six Problems This means that in order not to exceed the maximum permissible belt tension at 31,752 lbs, the time used for acceleration should not be less than 5.24 seconds. In determining the starting tension in the belt, the first step is to find the total horsepower available, in the form of tension. From this value, subtract the total ten- sion to operate the loaded conveyor. The result will be the force available to accelerate the total system. The total horsepower, in the form of tension, available to accelerate the entire system comes from the 75 hp and 125 hp motors. The starting torque avail- able from these NEMA Type C motors is a variable that should be confirmed by the motor manufacturer. For this example, the value is assumed to be 200 percent of the motor rating. Then, the total tension available is: From this value we subtract the tension required to operate the loaded conveyor: The resulting tension available to accelerate the loaded conveyor is 18,048 lbs. The acceleration of the total system consists of the acceleration of the drive (12 per- cent of the total system) and of the conveyor (88 percent of the total system). How- ever, in the process of acceleration, some amount of the available force (tension) is absorbed by the frictional losses (heat) in the drive machinery. Since this is a small amount compared to the total, a conservative approach is to ignore these losses because it is our aim to determine the effect of acceleration on the belt and its capac- ity to withstand tensile forces. Therefore, 0.88 x 18,048 = 15,882 lbs, which is the acceleration force (expressed in lbs of belt tension). The operating T1 is then added to this value in order to obtain the actual starting tension in the belt, which is 15,602 + 15,882 = 31,484 lbs. This is not excessive when compared to the 31,752 lbs belt ten- sion allowable at 180 percent of belt rating. Another limiting factor may be the time that the motor needs to accelerate the system. The average torque available during acceleration of the chosen motor taken from its speed torque curve is 180 percent of full-load torque. For a drive efficiency of 94 percent , it was found in Problem 4, page 166, that the horsepower at the motor shaft to operate the loaded conveyor is 181.23 hp. Therefore: 2 75 125+( ) 33,000( ) 400 -------------------------------------------------- 33,000 lbs= 33,000 14,055 0.94 drive efficiency( ) ----------------------------------------------------– 18,048 lbs= Horsepower pull in lbs( ) belt speed, fpm( ) 33,000 ----------------------------------------------------------------------= pull in lbs( ) hp( ) 33,000( ) belt speed, fpm( ) -----------------------------------------= Belt Tension, Power, and Drive Engineering 170 The force available for acceleration of the total equivalent mass of the loaded con- veyor system, for a belt speed of 400 fpm, is = = 13,864 lbs The total equivalent mass = 464,755/32.2 = 14,433 slugs. From the equation, Fa = Ma The time needed is: Therefore: The time required by the motor to accelerate the loaded conveyor, 6.94 seconds, is greater than the minimum acceleration time to stay within the maximum allowable belt tension, 5.24 seconds. Therefore, the conveyor is safe to start, fully loaded, with the equipment selected. Had the starting belt stress been limited to 140 percent of the normal belt rating instead of 180 percent, the allowable extra belt tension would have been (1.4)(17,640) – (15,602) = 9,094 lbs and the acceleration time t =(12,706)(400 – 0)/(9,094)(60) = 9.31 sec minimum. This is more than the time calculated for the motor to accelerate the loaded system, 6.94 seconds. If such a limitation had been placed on the starting belt stress, the system would not have been safe to start with the equipment selected. In fact, the belt stress during acceleration must be: The foregoing assumes that the mass between the slack side of the drive pulley and the takeup is negligible. If the takeup is far removed from the drive, this should be taken into account in the calculations. In Chapter 13, “Accelerating time,” it is indicated, in general, that the acceleration time for NEMA Type C motors be considered as 10 seconds or less. It is always pru- dent to check with the motor manufacturer to make sure that the calculated accelera- tion time will not cause the motor to overheat during starting. Fa 200( ) 1.80( ) 33,000( ) 400 -------------------------------------------------- 181.23( ) 33,000( ) 400 ------------------------------------------– 0.94= 0.94 33,000× 400 -------------------------------- 200 1.8 181.23–×( ) the acceleration, a Fa M ----- 13,864 14,433 --------------- 0.96 ft per sec2= = = t V 1 V 0– 60a ------------------= t 400 0– 60( ) 0.96( ) ------------------------- 6.94 seconds= = extra belt tension 12,706( ) 400 0–( ) 6.94( ) 60( ) ------------------------------------------- 12,206 lbs= = % of normal belt rating = 12,206 + 15,602 17 640,( ) -------------------------------------- 100%( ) 158%= 173 Examples of Belt Tension and Horsepower Calculations — Six Problems In Problem 5, Figure 6.24, the Ky factor for the tail half, L1 , of the conveyor is selected for a 300-foot horizontal conveyor. The Ky factor for the inclined drive half, L2 , is selected for the total conveyor length of 600 ft with an average slope of lift/total length = 36/600 = 6 percent , because the belt tension is higher than it would be for a 300-foot inclined conveyor, due to the belt pull at the end of the horizontal portion. Figure 6.24 Belt conveyor with concave vertical curve. In Problem 6, Figure 6.25, the Ky factor for the tail half, L1, of the conveyor is selected for a 300-foot conveyor inclined at a slope of lift/300 ft = 36/300 = 12 percent. The Ky factor for the drive half, L2, of the conveyor is less than it would be for a 300-ft horizontal conveyor, because of the high belt tension existing at the top of the inclined portion. The criterion for determining the Ky value to use for the horizontal drive half, L2, of this conveyor is the Ky value of a 600-foot inclined conveyor at a 6 percent slope. The Ky value for the horizontal portion cannot be more, and probably is a little less than this Ky value. Figure 6.25 Belt conveyor with convex vertical curve. The difference in the calculated effective tensions in Problems 5 and 6 is small. But larger and longer conveyors would entail more significant differences. Wb = 10 lbs/ft from Table 6-1 H = 36 ft L = 600 ft L1 = 300 ft L2 = 300 ft V = 500 fpm Q = 1,000 tph Si = 4.5 ft Ambient temperature = 60°F Belt width = 36 in. Material = 100 lbs/cu ft Drive = lagged head pulley, wrap = 220° Belt Tension, Power, and Drive Engineering 174 Troughing idlers = Class E6, 6-in. diameter 20° angle, Ai = 2.8 Return idlers = Class C6, 6-in. diameter, 10-foot spacing To simplify the calculations, all accessories are omitted. Analysis (Problem 5, Figure 6.24): From Table 6-8, wrap factor, Cw = 0.35. From Figure 6.1, for 60°F, Kt = 1.0 Wb + Wm = 10 + 66.6 = 76.6 lbs/ft T0 , minimum tension for 3 percent sag = 4.2 Si (Wb + Wm) = (4.2)(4.5)(76.6) = 1,448 lbs T1 is taken as T0 = 1,448 lbs T2 = Tt - 0.015LWb + HWb = 1,448 – (0.015)(600)(10) + (36)(10) = 1,448 – 90 + 360 = 1,718 lbs The effective belt tension, Te , is figured individually for each half of the conveyor. Horizontal portion, 300 ft long, Ky from Table 6-2, for 0° slope, 300 ft and (Wb + Wm) = 76.6, is 0.0347. Corrected for 4.5-foot idler spacing, Table 6-3, gives Ky = 0.0349. From “Belt Tension at Any Point, X, on Conveyor Length,” page 117, tension is: Tfcx = Tt + Twcx + Tfcx but, Twcx = Hx(Wb + Wm) = 0, for a horizontal belt and Tfcx = Lx[Kt(Kx + KyWb)] + LxKyWm and Kt = 1.0 for 60°F Therefore: Tfcx = LxKx + LxKyWb + LxKyWm = LxKx + LxKy(Wb + Wm) Thus, Tcx = Tt + 0 + LxKx + LxKy(Wb + Wm) W m 33.3Q V -------------- 33.3( ) 1,000( ) 500 --------------------------------- 66.6 lbs per ft= = = K x 0.000568( ) W b W m+( ) Ai Si ----+ 0.00068( ) 76.6( ) 2.8 4.5 ------+ 0.6743= = = 175 Examples of Belt Tension and Horsepower Calculations — Six Problems Calling Lx equal to L1 for the first (horizontal) half of the conveyor: Tcx = Tt + L1Kx + L1Ky(Wb + Wm) = 1,448 + (300)(0.6743) + (300)(0.0349)(76.6) = 1,448 + 202 + 802 = 2,452 lbs For the inclined portion, drive half L2. Kx = 0.6743 Ky is 0.028, for a slope of (36/600)(100%) = 6%, and Wb + Wm = 76.6, and a length of 600 ft, from Table 6-2 for the tabular idler spacing. The corrected value is 0.0298, from Table 6-3, for a 4½-foot spacing. Tcx = Tt + Twcx + Tfcx . However, Tt is the tension existing at the bottom of the incline, so: Tcx = 2,452 + Twcx + Tfcx = 2,452 + Hx(Wb + Wm) + LxKx + LxKy (Wb + Wm) substituting L2 for Lx, and 36 for Hx Tcx = 2,452 + 36(Wb + Wm) + L2Kx + L2Ky(Wb + Wm) =2,452 + (36)(76.6) + (300)(.6743) + (300)(.0298)(76.6) =2,452 + 2,757.6 + 202.3 + 684.8 =6,097 lbs Adding to Tcx the nondriving pulley friction, (2)(150) + (4)(100) = 700 lbs, the tension in the belt at the head pulley = Tcx + 700 = T1 = 6,097 + 700 = 6,797 lbs Te = T1 – T2 = 6,797 – 1,718 = 5,079 lbs Analysis (Problem 6, Figure 6.25): T0 has been calculated in Problem 5 as 1,448 lbs. Take Tt = T0 = 1,448 lbs Tt = T2 + L(0.015Wb) – Hwb, T2 = Tt - L(0.015Wb) + HWb = 1,448 – (600)(0.015)(10) + (36)(10) +(36)(10) = 1,448 – 90 + 360 = 1,718 lbs Inclined portion, 300 ft long. The slope of the incline is (36/300)(100%) = 12%. For this slope, a length of 300 ft and Wb + Wm = 76.6, the value of Ky , from Table 6-2, Belt horsepower T eV 33,000 --------------- 5,079( ) 500( ) 33,000 ------------------------------- 77 hp= = = Belt Tension, Power, and Drive Engineering 178 • Figure 6.27 — Gear motor combined with chain drive to drive shaft — is one of the lowest cost flexible arrangements and is substantially reliable. • Figure 6.28 — Parallel-shaft speed reducer directly coupled to the motor and to drive shaft — is versatile, reliable, and generally heavier in construction and easy to maintain. • Figure 6.29 — Parallel-shaft speed reducer coupled to motor, and with chain drive, to drive shaft — provides flexibility of location and also is suitable for the higher horsepower requirements. Figure 6.27 Gearmotor combined with chain drive or synchronous belt drive to drive shaft — is one of the lowest cost flexible arrangements, provides additional reduction, and is substantially reliable. Figure 6.28 Parallel-shaft speed reducer directly coupled to the motor and to drive shaft — is particularly well suited to large conveyors, is versatile, reliable, and generally heavier in construction and easy to maintain. Figure 6.29 Parallel-shaft speed reducer coupled to motor, and with chain drive, to drive shaft — provides flexibility of location and also is suitable for low speed high torque requirements. 179 Belt Conveyor Drive Equipment • Figure 6.30 — Spiral-bevel helical speed reducer, or worm-gear speed reducer, directly coupled to motor and to drive shaft — is often desirable for space-saving rea- sons and simplicity of supports. The spiral-bevel speed reducer costs substantially more than the worm-gear speed reducer but is considerably more efficient. • Figure 6.31 — Spiral-bevel helical speed reducer, or worm-gear speed reducer, coupled to motor and, with chain drive, to drive shaft — is a desirable selection for high reduction ratios in the lower horsepower requirements. This drive is slightly less efficient, but has lower initial costs and is most flexible in terms of location. • Figure 6.32 — Drive-shaft-mounted speed reducer with V-belt reduction from motor — provides low initial cost, flexibility of location, and the possibility of some speed variation and space savings where large speed reduction ratios are not required and where horsepower requirements are not too large. Figure 6.30 Spiral-bevel helical speed reducer, helical- worm speed reducer, or worm-gear speed reducer, directly coupled to motor and to drive shaft — is often desirable for space-saving reasons and simplicity of supports. The spiral-bevel speed reducer costs somewhat more than the helical-worm speed reducer and considerably more than the worm-gear speed reducer but is more efficient than the helical-worm and considerably more efficient than the worm gear. Figure 6.31 Spiral-bevel helical speed reducer, helical-worm speed reducer, or worm-gear speed reducer, coupled to motor and, with chain drive to drive shaft — is a desirable selection for high reduction ratios in the lower horsepower requirements. This drive is slightly less efficient, but has lower initial costs and is most flexible in terms of location. Figure 6.32 Drive-shaft-mounted speed reducer with direct drive of V-belt reduction from reducer mounted motor — provides low initial cost, flexibility of location, and the possibility of some speed variation and space savings where large speed reduction ratios are not required and where horsepower requirements are not too large. Belt Tension, Power, and Drive Engineering 180 • Figure 6.33 — Dual-pulley drive, shown in Figure 6.33, is used where power requirements are very large, and use of heavy drive equipment may be economical by reducing belt tensions. Selection of the type of speed-reduction mechanism can be determined by pref- erence, cost, power limitations, limitations of the speed-reduction mechanism, limi- tations of available space, or desirability of drive location. The use of speed reducers in the drives for belt conveyors is almost universal today. However, space-saving con- siderations and low initial cost sometimes may dictate the use of countershaft drives with guarded gear or chain speed reductions. All of the drives shown can be assembled in either left- or right-hand arrange- ment. Drive Efficiencies To determine the minimum horsepower at the motor, it is necessary to divide the horsepower at the drive shaft by the overall efficiency of the speed reduction machin- ery. To determine the overall efficiency, the efficiencies of each unit of the drive train are multiplied together. The final product is the overall efficiency. The efficiencies of various speed-reduction mechanisms are listed in Table 6-11. These efficiencies represent conservative figures for the various types of drive equip- ment as they apply to belt conveyor usage. They do not necessarily represent the spe- cific efficiencies of the drive units by themselves. Rather, they take into account the possible unforeseen adverse field conditions involving misalignment, uncertain maintenance, and the effects of temperature changes. While there are some variations in efficiency among different manufacturers’ products, the data in Table 6-11 gener- ally cover the efficiencies of the various speed-reduction mechanisms. As an example of the application of the overall drive efficiency — the result of combining equipment unit efficiencies — consider a belt conveyor drive consisting of a double-helical-gear speed reducer and an open-guarded roller chain on cut sprock- ets. The approximate overall efficiency, according to Table 6-11, is (0.94)(0.93) = 0.874. If the calculated minimum horsepower at the drive shaft is 13.92 hp, then the required motor horsepower is 13.92/0.874 = 15.9 hp. Therefore, it is necessary to use at least a 20-hp motor. Figure 6.33 Two motors (dual-pulley drive) coupled to helical or herringbone gear speed reducers, directly coupled to drive shafts.