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P1-14 An 1800-rpm electric motor drives a belt conveyor through a worm gear speed reducer and flexible coupling.

The belt speed is 3 m/s, and the net driving force is 1200 N. If the combined efficiency of the speed reducer and cou- pling is 0.78, specify the standard-size motor required. The conveyor serves a limestone quarry (Fig. 1-10). Duration of service: 8 hr/day, 5 days/wk. Bin Coupling 1200 N Motor 1800 rpm 3 m/s Worm and gear

NATURE AND COMPOSITION OF MACHINES 15 anufac- experi- oads on utches In each is mul- an part is must catalog ulated Power source of service 2 (1-12) Engines 2.5 r is ob- One cylinder Two cylinders Four cylinders 2.0 greater are de- ble ca- or 1.5 Steam angine Reciprocating hydraulic or oneumatic motor Water turbine Steam turbine Electric motor Rotary hydraulic or pneumatic motor roxi on 0.6 the en- condi Figure 1-6 Service factor diagram. Service factors help to ensure optimal service life of machine components. (Developed by W. Richter and H. Ohlendorf.) ates and 16 head- indus- As might be expected, the service factor n creases when 1. 4. The duration of services increases from 1 to power nce a type re ac- gen- from g point the 24 h/day. Smoothly running power sources based on rotary motion are replaced by reciprocating tive slope, thus effectively indicating relative engines The starting load becomes heavier and the number of starts increases. The frequency of full-load conditions and shock phenomena increases Only the curves for sensitivities have a nega- smoothness of operation for the seven most com- mon means of power transmission. Spur gears for instance, require a service factor nearly twice that of worm gearing and 50 to 60% higher than 2- ers 3. those of belt and chain drives

NATURE AND COMPOSITION OF MACHINES 11 rti ng, TABLE 1-3 Standard Capacities and Speeds for Three-Phase, 60-Hz, Alternating-Current Induction or dis- Motors engine hp pm rpm power e tor- result 0.25 1800 1200 3600 1800 1200 100 1800 1200 900 1800 1200 900 600 450 ad of rations0.50 1800 1200 1800 1200 3600 1800 1200 900 30 125 1800 1200 900 720 to the utches well as 600 450 1800 1200 0.75 7.5 1800 1200 3600 800 1200 40 1800 1200 900 150 rs are 450 3600 1800 1200 10 50 200 1800 1200 1800 1200 1200 rs are lifting Is in 600 450 1.5 3600 1800 1200 3600 1800 1200 60 1800 1200 250 1800 1200 n types 600 can be arting ven at ume oad is 450 360 1800 1200 3600 1800 3600 1800 1200 20 300 1200 600 up this stored energy at an equally high rate s said to vibrate when it describes an oscillating motion about a reference point. Such oscillation occurs because of the dynamic effects of manu facturing tolerances, clearances, rolling and rub- bing contact between machine parts, and out-of Mechanical vibration is motion that is usually balanc forces in rotating and reciprocating nintentional and unwanted. A machine member members. Apparently insignificant vibrations func- while slowing down. en 1-9 EFFECTS OF VIBRATION give

8 PRIMARY CONSIDERATIONS through three machine members with effi- ez, e. By passing through the frst TABLE 1-2 Mechanical Efficier Typical Values member, E loses a small amount of energy and thereby reduces the output to (e)(E). The output of the first member becomes the input of the second member. Again, energy is lost, and the output of the second member (e)(e2)(E) be- comes the input of the third member. Total out- put E, is therefore Efficiency, e Machine Member High Low 1. Ball bearings 2. Silent chains 0.999 0.99 0.99 0.97 0.985 _ 3. Spur and helical gears (including bearings) 4. Roller bearings 0.98 For n machine members with efficiencies e5 e,6. el gears Synchronous belts ez . . . e, total output is E(e)(e2). .. 6. Bevel gears (e)Ei, and total efficiency becomes the product, not the sum, of the individual efficiencies. 7. V-belts 8. Roller chains 9. Worm gearing 0.97 0.95 0.95 e = (e)(ez) . . . (en) 10. Bal bearing screws 0.90 Power screws (multithreads 0.84 0.38 -single threads) Table 1-2 presents typical values of the effi-1 ciency for common machine elements inge 13. Screw fasteners 0 EXAMPLE 1-1 An electric motor drives a machine by means of a gear train with three sets of spur gears and a roller chain drive. Find the efficiency of the drive system 1-7 POWER Power is the rate of doing work or transmitting energy. The SI power unit is the watt (W). Be- cause the watt is such a small unit, the kilowatt (1000 W) is preferred in most technical calcu- lations. With proper units, the power equations SOLUTION 7 Using values from Table 1-2 in Eq. 1-5, we| obtain: ku14 P 1000 Tn 9550 e (0.9850.96) 0.92 kW Efficiency can be measured directly or calcu lated, but it can also be evaluated by means of noise or heat generated. Although not always recognized, noise represents an excellent me dium for evaluating machine performance. En ergy expended in unwanted machine motion or vibration reduces the useful output and results in noise. Generally, the condition of least noise corresponds to minimum loss in efficiency and 1000 33,000 (1-10) 63,000 1-8 where P- power; kW, hp F - force; N, lb minimum wear.

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