Transcript Chapter 7:
Chapter 7: Belt Drives and Chain Drives CVT Overview – why used? 1.) Transfer power (torque) from one location to another. From driver: motor,peddles, engine,windmill,turbine to driven: conveyor belt, back wheels/bike,generator rock crusher,dryer. 2.) Used to span large distances or need flexible xmission elements. Gear drives have a higher torque capability but not flexible or cheap. 3.) Often used as torque increaser (speed reducer), max speed ratio: 3.5:1. Gear drives?? Virtually unlimited! Applications? Show rust abrader, glove factory, draw sample drive of rust abrader, show slides from mechanism book. Sometimes desirable to have both chain and belt drive (Fig 7.1) Belt: high speed/low torque Chain: Low speed/high torque Belts vs. Chains Belts Use When: Speed: Dis: Advs: High Speed, Low T 2500 < Vt < 7000 ft./min. Chains High T, Low Speed V < 1500 ft./min. Must design with standard lengths, wear, creep, corrosive environment, slip, temp., when must have tension need idler Must be lubricated, wear, noise, weight, vibration Quiet, flexible, cost Strength, length flexibility Types of Belts: a)V-belt most common for machine design, several types (Fig. 7.5 – 7.8) •Timing belt (c & d) have mating pulleys to minimize slippage •c) Pos retention due to mating pulleys •d) Pos retention due to increased contact area •Flat belt (rubber/leather) not shown, run on tapered pulleys Add notes Types of V-Belts V-belt Drive Design Process Need rated power of the driving motor/prime mover. BASE sizing on this. Service factor based on type of driver and driven load. Center distance (adjustment for center distance must be provided or use idler pulley) nominal range D2 < C < 3(D2 + D1) Power rating for one belt as a function of size and speed of the smaller sheave Belt length (then choose standard size) Sizing of sheaves (use standard size). Most commercially available sheaves should be limited to 6500 ft/min belt speed. Belt length correction factor Angle of wrap correction factor. Angle of wrap on smaller sheave should be greater than 120 deg. Number of belts Initial tension in belts Key Equations Belt speed (no slipping) = Speed ratio = D11 D22 b R11 R2 2 2 2 1 D2 Pitch dia’s of sheaves Pitch dia 2 D1 in inches Belt speed ft/min b rpm D1n1 12 Key Equations Belt length: ( D2 D1 ) 2 L 2C ( D2 D1 ) 4C Center Distance: 2 B B 2 32( D2 D1 ) 2 C 16 Where, B 4L 2 ( D2 D1 ) Recommended D2 < C < 3(D2+D1) Note: usually belt length standard (use standard belt length table 7-2), then calculate C based on fixed L Key Equations cont… Angle of contact of belt on each sheave D2 D1 1 180 2 sin 2C 1 D2 D1 2 180 2 sin 2C 1 Note: Select D’s and C’s so maximum contact (Ѳ1 + Ѳ2 = 180º). If less then smaller sheave could slip and will need reduction factor (Table 7-14). V-Belt Design Example Given: 4 cylinder Diesel runs @ 80hp, 1800 rpm to drive a water pump (1200 rpm) for less than 6 hr./day Find: Design V-belt drive V-belt Design Example Cont… 1.) Calculate design power: Use table 7-1(<6h/day, pump, 4 cyl. Engine) Design Power = input power x service factor = 80 hp x 1.1 = 88 hp V-belt Design Example Cont… 2.) Select belt type, Use table 7-9 Choose 5V Speed = 1800 rpm Design Power = 88 hp V-belt Design Example Cont… 3.) Calculate speed ratio SR = w1/w2 = 1800 rpm/1200 rpm = 1.5 V-belt Design Example Cont… 4.) Determine sheave sizes Choose belt speed of 4000 ft/min (Recall 2500ft./min. < vb < 7000 ft./min) D1n1 12vb 12* 4000 vb D1 8.488in 12 n1 *1800 So… D1 = 8.488in D2 = SR * D1 = 1.5 * 8.488 D2 = 12.732in V-belt Design Example Cont… 5.) Find sheave size (Figure 7-11) Must find acceptable standard sheave 1, then corresponding acceptable sheave 2 Engine (D1) 8.4 8.4 8.9 X 1.5 12.6 12.6 13.35 Standard D2 12.4 13.1 13.1 Actual n2 1219 1154 1223 n1 D2 Dn n2 1 1 n2 D1 D2 **All look OK, we will try the first one V-belt Design Example Cont… 6.) Find rated power (use figure 7-11 again) Rated Power = 21 hp V-belt Design Example Cont… Adjust for speed ratio to get total power/belt Total power = 21hp +1.55hp = 21.55hp V-belt Design Example Cont… 7.) Find estimated center distance D2 < C < 3(D2+D1) 12.4 < C < 3 (12.4 + 8.4) Notice – using standard sheave sizes found earlier, not calculated diameters 12.4 < C < 62.4 To provide service access will try towards long end, try C = 40” V-belt Design Example Cont… 8.) Find belt length ( D2 D1 ) 2 L 2C ( D2 D1 ) 4C (4in) 2 L 2(40in) 1.57(20.8in) 112.765in 4(40in) 2 V-belt Design Example Cont… 9.) Select standard belt length Lcalc = 112.765 Choose 112” V-belt Design Example Cont… 10.) Calculate actual center distance B 4L 2 ( D2 D1 ) B 4(112) 6.28(20.8) B 317.367" B B 2 32( D2 D1 ) 2 C 16 317.367 317.3672 32(4) 2 C 16 C 39.62" V-belt Design Example Cont… 11.) Find wrap angle, small sheave D2 D1 1 180 2 sin 2C 4 1 1 180 2 sin 2 ( 39 . 62 ) 1 1 174.2 V-belt Design Example Cont… 12.) Determine correction factors C .98 CL .98 V-belt Design Example Cont… 13.) Calculate corrected power Corrected Power C CL P (.98)(.98)(22.5hp) 21.61hp V-belt Design Example Cont… 14.) Belts needed Design P ower (hp) # belts CorrectedP ower/belt 88hp # belts 4.07 21.61hp / belt Use 4 belts! V-belt Design Example Cont… 15.) Summary D1=8.4” D2=12.4” Belt Length = 112” Center Distance = 39.62” 4 Belts Needed Chain Drives Chain Drives Types of Chains Chain Drives Roller Chain Construction (Most common Type) Chain Design Process 1.) # of sprocket teeth, N1 (smaller sprocket) > 17 (unless low speed < 100 rpm.) 2.) Speed ratio = n1/n2 7 3.) 30 x Pitch Length < Center Distance < 50 x Pitch Length 4.) Angle of contact of chain on smaller sprocket > 120° 5.) # sprocket teeth, N2 (longer sprocket) < 120 Chain Drives Chain Drives Design Example Given: Driver: Hydraulic Motor Driven: Rock Crusher ni = 625 rpm, 100 hp no = 225 rpm Find: Design belt drive Chain Drives Design Example 1.) Design Power DP = SF x HP DP = 1.4 ( Table 7-8) x 100 hp DP = 140 hp Chain Drives Design Example 2.) Calculate Velocity Ratio ni N o VR no N i n = speed N = teeth 625rpm VR 225rpm VR = 2.78 Heavy Requirement!! Chain Drives Design Example 3.) Choose: Size - (40, 60, 80) 80 (1in) # Strands – use 4 Required HP/chain = 140hp/3.3 = 42.42 hp/chain VR 2.78 No No Ni 25 No = 69.5 use 70 teeth Number of Roller Chain Strands Multiple Strand Factor 2 1.7 3 2.5 4 3.3 5 3.9 6 4.6 Chain Drives Design Example Conclusion: 4 Strands No. 80 Chain Ni = 25 Teeth No= 70 Teeth Chain Drive Design Example Guess center distance: 40 Pitches N 2 N1 ( N 2 N1 ) 2 L( pitches) 2C 2 4 2C 70 25 (70 25) 2 L( pitches) 2(40) 2 4 2 (40) L = 128.8 pitches use 130 pitches Chain Drives Design Example Actual Center Distance, C 1 N N1 C ( pitches) L 2 4 2 1 70 25 C ( pitches) 130 4 2 N 2 N1 8( N 2 N1 ) 2 L 2 4 2 2 75 25 8(70 25) 130 2 4 2 2 C = 40.6 use 40 Pitches 2