barcode generator vb.net source code CLUTCHES AND BRAKES 8.20 in Software

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CLUTCHES AND BRAKES 8.20
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MACHINE ELEMENTS THAT ABSORB AND STORE ENERGY
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Finally, the energy to be dissipated is E= W 987 2 (V o V f2) = [(80.7)2 02] 2g 2(32.17) = 99 900 lb ft (135.5 kN m) = 128.4 Btu [135.5 kilojoules (kJ)] Industrial Brakes. The approach is the same as for vehicular brakes. The heat energy the brake must dissipate equals the change in kinetic energy of the rotating machine: E= where, with n in rev/min, = 2 n 60 (8.15) I ( o2 f2) 2 (8.14)
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In many industrial applications, the brakes are applied frequently. The average rate of heat dissipation is, for S stops per hour, Hav = ES 3600 (8.16)
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Tensioning Applications. In tensioning applications, a continuous application of the brake is required, for example, in unwinding a roll of aluminum foil. The maximum torque occurs at the maximum roll diameter Dmax. It is Tmax = Dmax Ft 2 (8.17)
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The tension Ft is, for material width b and thickness t, Ft = wtb (8.18)
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Typical values of tension per unit width and per unit thickness for a few materials are given in Table 8.6.
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TABLE 8.6 Tension Data for Typical Materials
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CLUTCHES AND BRAKES 8.21
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CLUTCHES AND BRAKES
The rate at which heat is generated by the brake friction is Hav = FtVw (8.19)
Example 4. A printing press is to print on Mylar 0.002 in [0.051 millimeter (mm)] thick. The web velocity is 4000 ft/min (20.3 m/s). The maximum roll diameter is 55 in (1.4 m). The web is 54 in wide. Find the necessary braking torque and the rate at which heat is generated by braking. Solution. For Mylar the unit tension is 0.60 lb/mil per inch (379.2 kN/mm per meter). So the web tension is F = wtb = 0.60(2)(54) = 64.8 lb (288 N) The maximum brake torque is Tmax = Dmax F t 55(64.8) = = 1782 in lb (201 N m) 2 2
The rate at which the brake must dissipate heat is, by Eq. (8.19), Hav = FtVw The web velocity is Vw = So Hav is Hav = 64.8(66.67) = 4320 ft lb/s [5855 watts (W)] = 5.55 Btu/s 4000 = 66.67 ft/s (20.3 m/s) 60
8.3 TEMPERATURE CONSIDERATIONS
8.3.1 Intermittent Operation: Clutches and Brakes The temperature rise can be estimated as T = E Cm (8.20)
where m [pounds mass (lbm) or kilograms (kg)] = mass of the parts adjacent to the friction surfaces. The specific heat C for steel or cast iron is about 0.12 Btu/(lbm F) [500 J/(kg C)].
8.3.2 Frequent Operation: Caliper Disk Brakes The average rate at which heat must be dissipated can be calculated by Eq. (8.16). The disk is capable of dissipating heat by a combination of convection and radiation.
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CLUTCHES AND BRAKES 8.22
MACHINE ELEMENTS THAT ABSORB AND STORE ENERGY
And the convection-heat transfer is sensitive to the velocity of air moving over the disk. The rate at which the disk can dissipate heat is Hdiss = hA(Td Ta) The overall heat transfer coefficient h is h = h r + fv hc (8.22) (8.21)
The heat transfer coefficients for radiation hr and convection hc are plotted in Fig. 8.13 against the temperature rise of the disk above the surrounding air. The ventilation factor fv is plotted against the velocity of the moving air in Fig. 8.14. Example 5. An industrial caliper brake is used 19 times per hour on average to stop a machine with a rotating inertia of I = 328 lb in s2 (37.06 kg m2) from a speed of 315 rev/min. The mean air velocity over the disk will be 30 ft/s (9.14 m/s). What minimum exposed area on the disk is needed to limit the disk s temperature rise to 200 F (111 C) Solution. From Figs. 8.13 and 8.14, hr = 3.1 10 6 Btu/(in2 s F), hc = 2.0 10 6 Btu/(in2 s F), and fv = 5.25. The overall heat transfer coefficient is h = 3.1 10 6 + 5.25(2.0 10 6) = 13.6 10 6 Btu/(in2 s F) The energy the brake must dissipate per stop is, by Eq. (8.14), E= I ( o2 f2) 2
FIGURE 8.13 Heat transfer coefficients in still air. (Tol-o-matic.)
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