Handbook of Mechanical Engineering Calculations ar Episode 2 Part 4 docx

24.1 SECTION 24 MECHANICAL AND ELECTRICAL BRAKES Brake Selection for a Known Load 24.1 Mechanical Brake Surface Area and Cooling Time 24.3 Band Brake Heat Generation, Temperature Rise, and Required Area 24.6 Designing a Brake and Its Associated Mechanisms 24.8 Internal Shoe Brake Forces and Torque Capacity 24.15 Analyzing Failsafe Brakes for Machinery 24.17 BRAKE SELECTION FOR A KNOWN LOAD Choose a suitable brake to stop a 50-hp (37.3-kW) motor automatically when power is cut off. The motor must be brought to rest within 40 s after power is shut off. The load inertia, including the brake rotating member, will be about 200 lb ⅐ ft 2 (82.7 N ⅐ m 2 ); the shaft being braked turns at 1800 r/min. How many revolutions will the shaft turn before stopping? How much heat must the brake dissipate? The brake operates once per minute. Calculation Procedure: 1. Choose the type of brake to use Table 1 shows that a shoe-type electric brake is probably the best choice for stop- ping a load when the braking force must be applied automatically. The only other possible choice—the eddy-current brake—is generally used for larger loads than this brake will handle. 2. Compute the average brake torque required to stop the load Use the relation T a ϭ Wk 2 n/(308t), where T a ϭ average torque required to stop the load, lb ⅐ ft; Wk 2 ϭ load inertia, including brake rotating member, lb ⅐ ft 2 , n ϭ shaft speed prior to braking, r / min; t ϭ required or desired stopping time, s. For this brake, T a ϭ (200)(1800)/[308(40)] ϭ 29.2 lb ⅐ ft, or 351 lb ⅐ in (39.7 N ⅐ m). 3. Apply a service factor to the average torque A service factor varying from 1.0 to 4.0 is usually applied to the average torque to ensure that the brake is of sufficient size for the load. Applying a service factor of 1.5 for this brake yields the required capacity ϭ 1.5(351) ϭ 526 in ⅐ lb (59.4 N ⅐ m). Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 24.2 DESIGN ENGINEERING TABLE 1 Mechanical and Electrical Brake Characteristics 4. Choose the brake size Use an engineering data sheet from the selected manufacturer to choose the brake size. Thus, one manufacturer’s data show that a 16-in (40.6-cm) diameter brake will adequately handle the load. 5. Compute the revolutions prior to stopping Use the relation R s ϭ tn/120, where R ϭ number of revolutions prior to stopping; other symbols as before. Thus, R s ϭ (40)(1800)/120 ϭ 600 r. 6. Compute the heat the brake must dissipate Use the relation H ϭ 1.7FWk 2 (n/100) 2 , where H ϭ heat generated at friction sur- faces, ft ⅐ lb/min; F ϭ number of duty cycles per minute; other symbols as before. Thus, H ϭ 1.7(1)(200)(1800/100) 2 ϭ 110,200 ft ⅐ lb/ min (2490.2 N ⅐ m/ s). 7. Determine whether the brake temperature will rise From the manufacturer’s data sheet, find the heat dissipation capacity of the brake while operating and while at rest. For a 16-in (40.6-cm) shoe-type brake, one man- ufacturer gives an operating heat dissipation H o ϭ 150,000 ft ⅐ lb/ min (3389 5 N ⅐ m/s) and an at-rest heat dissipation of H v ϭ 35,000 ft ⅐ lb/ min (790.9 N ⅐ m/ s). Apply the cycle time for the event; i.e., the brake operates for 400 s, or 40/60 of the time, and is at rest for 20 s, or 20 / 60 of the time. Hence, the heat dissipation of the brake is (150,000)(40/60) ϩ (35,000)(20 / 60) ϭ 111,680 ft ⅐ lb/min (2523.6 N ⅐ m/s). Since the heat dissipation, 111,680 ft ⅐ lb/min (2523.6 N ⅐ m/s), exceeds the heat generated. 110,200 ft ⅐ lb/min (2490.2 N ⅐ m/s), the temperature of the Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MECHANICAL AND ELECTRICAL BRAKES MECHANICAL AND ELECTRICAL BRAKES 24.3 brake will remain constant. If the heat generated exceeded the heat dissipated, the brake temperature would rise constantly during the operation. Brake temperatures high than 250 ЊF (121.1ЊC) can reduce brake life. In the 250 to 300 ЊF (121.1 to 148.9ЊC) range, periodic replacement of the brake friction sur- faces may be necessary. Above 300 ЊF (148.9ЊC), forced-air cooling of the brake is usually necessary. Related Calculations. Because electric brakes are finding wider industrial use, Tables 2 and 3, summarizing their performance characteristics and ratings, are pre- sented here for easy reference. The coefficient of friction for brakes must be carefully chosen; otherwise, the brake may ‘‘grab,’’ i.e., attempt to stop the load instantly instead of slowly. Usual values for the coefficient of friction range between 0.08 and 0.50. The methods given above can be used to analyze brakes applied to hoists, ele- vators, vehicles, etc. Where Wk 2 is not given, estimate it, using the moving parts of the brake and load as a guide to the relative magnitude of load inertia. The method presented is the work of Joseph F. Peck, reported in Product Engineering. MECHANICAL BRAKE SURFACE AREA AND COOLING TIME How much radiating surface must a brake drum have if it absorbs 20 hp (14.9 kW), operates for half the use cycle, and cannot have a temperature rise greater than 300 ЊF (166.7ЊC)? How long will it take this brake to cool to a room temperature of 75 ЊF (23.9ЊC) if the brake drum is made of cast iron and weighs 100 lb (45.4 kg)? Calculation Procedure: 1. Compute the required radiating area of the brake Use the relation A ϭ 42.4hpF/K, where A ϭ required brake radiating area, in 2 ; hp ϭ power absorbed by the brakes; F ϭ brake load factor ϭ operating portion of use cycle; K ϭ constant ϭ Ct r , where C ϭ radiating factor from Table 4, t r ϭ brake temperature rise, ЊF. For this brake, assuming a full 300ЊF (166.7ЊC) temperature rise and using data from Table 4, we get A ϭ 42.4(20)(0.5)/[(0.00083)(300)] ϭ 1702 in 2 (10,980.6 cm 2 ). 2. Compute the brake cooling time Use the relation t ϭ (cW ln t r )/(K c A), where t ϭ brake cooling time, min; c ϭ specific heat of brake-drum material, Btu/(lb ⅐ ЊF); W ϭ weight of brake drum, lb; t r ϭ drum temperature rise, ЊF; ln ϭ log to base e ϭ 2.71828; K c ϭ a constant varying from 0.4 to 0.8; other symbols as before. Using K c ϭ 0.4, c ϭ 0.13, t ϭ (0.13 ϫ 100 ln 300)/[(0.4)(1702)] ϭ 0.1088 min. Related Calculations. Use this procedure for friction brakes used to stop loads that are lifted or lowered, as in cranes, moving vehicles, rotating cylinders, and similar loads. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MECHANICAL AND ELECTRICAL BRAKES 24.4 TABLE 2 Performance Characteristics of Electric Brakes Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MECHANICAL AND ELECTRICAL BRAKES 24.5 TABLE 3 Representative Range of Ratings and Dimensions for Electric Brakes Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MECHANICAL AND ELECTRICAL BRAKES 24.6 DESIGN ENGINEERING TABLE 4 Brake Radiating Factors Temperature rise of brake ЊF ЊC Radiating factor C 100 200 300 400 55.6 111.1 166.7 222.2 0.00060 0.00075 0.00083 0.00090 BAND BRAKE HEAT GENERATION, TEMPERATURE RISE, AND REQUIRED AREA A construction-industry hoisting engine is to be designed to lower a maximum load of 6000 lb (2724 kg) using a band brake, Fig. 1, having a 48-in (121.9-cm) drum diameter and a drum width of 8-in (20.3-cm). The brake band width is 6-in (15.2- cm) with an arc of contact of 300 Њ. Maximum distance for lowering a load is 200 ft (61 m). The cycle of the engine is 1.5 min hoisting and 0.75 min lowering, with 0.3 min for loading and unloading. If a temperature rise of 300 ЊF (166.5ЊC) is permitted, determine the heat generated, radiating surface required, and the actual temperature rise of the drum if the brake is made of cast iron and weighs 600 lb (272.4 kg). Calculation Procedure: 1. Determine the heat generated, equivalent to the power developed Use the relation, H g ϭ (wd / T L )(33,000), where H g ϭ heat generated during low- ering of the load of w, lb (kg), hp (kW); d ϭ distance the load is lowered, ft (m); T L ϭ time for lowering, minutes. Substituting for this hoisting engine, we have, H g ϭ (6000)(200)/(0.75)(33,000) ϭ 48.48 hp (36.2 kW). 2. Find the load factor for the brake The load factor, q, for a brake ϭ T L /(T H ϩ T L ϩ T LU ), where T H ϭ hoisting time, minutes; T LU ϭ time to load and unload, minutes; other symbols as before. Sub- stituting, q ϭ (0.75)/(1.5 ϩ 0.75 ϩ 0.3) ϭ 0.294. 3. Compute the required radiating area for the brake Use the relation, A R ϭ 42.4(q)(H g )/Ct r ), where A R ϭ required radiating area, in 2 (cm 2 ); Ct r ϭ brake radiating factor from Table 4. Assuming a temperature rise of 300 ЊF (166.7ЊC), and substituting, A R ϭ (42.4)(0.294)(48.5) / 0.25 ϭ 2418.3 in 2 (15,601.9 cm 2 ). It is necessary to assume a temperature rise when analyzing a brake because the rise is a factor in the computation. Without such an assumption the required radi- ating area cannot be determined. Using the given dimensions for this brake, we can find the area of the drum from A d ϭ 2( ␲ )(48 ϫ 8) Ϫ ( ␲ )(48 ϫ 6)(300Њ/360Њ) ϭ 1657.9 in 2 (10,696.1 cm 2 ). Since the required radiating area is 2418 in 2 ., as computed above, the excess area required will be 2418 Ϫ 1658 ϭ 760 in 2 (4903.2 cm 2 ). This area can be provided Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MECHANICAL AND ELECTRICAL BRAKES MECHANICAL AND ELECTRICAL BRAKES 24.7 FIGURE 1 (a) Self-locking band brake. (b) Pressure var- iation along the surface of a band brake. by the brake flanges and web. To be certain that sufficient area is available for heat radiation, check the physical dimensions of the brake flanges and webs to see if the needed surface is present. 4. Find the temperature rise during brake operation The actual temperature of the brake drum will vary slightly above and below the assumed 300 ЊF (166.7ЊC) temperature rise during operation. The reason for this is because heat is radiated during the whole cycle of operation but is generated only during the lowering cycle, which is 29.4 percent of the total cycle. The temperature change of the drum during the braking operation will be T r ϭ [1/(778)(W r )(c)][Wh Ϫ Ct r A r m(778)], where c ϭ specific heat of the drum material ϭ 0.13 for cast iron; m ϭ lowering time in minutes; other symbols as given earlier. Substituting, T r ϭ [1/(778)(600)(0.13)][(6000 ϫ 200) Ϫ (0.25)(2418)(0.75)(778)] ϭ 14ЊF (7.78ЊC). Thus, the drum temperature can range Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MECHANICAL AND ELECTRICAL BRAKES 24.8 DESIGN ENGINEERING about 14Њ, or about 7Њ above and 7Њ below the average operating temperature of 300 ЊF (148.9ЊC). Related Calculations. The actual temperature attained by a brake drum, and the time required for it to cool, cannot be accurately calculated. But the method given here is suitable for preliminary calculations. In the final design of a new brake, heating should be checked by a proportional comparison with a brake already known to give good performance in actual service. An approximation of the time required for a brake to cool can be made using the relation given in step 2 of the previous procedure. Note that the value of K selected in that relation will directly influence brake cooling time. Thus, a lower value chosen for K will increase the estimated cooling time while a higher value will decrease the time. For safety reasons, engineers will often select lower K values so the brake will be given more time to cool, or will be provided with a larger capacity cooling mechanism. This procedure is the work of Alex Vallance, Chief Designer, Reed Roller Bit Co. and Venton L. Doughtie, Professor of Mechanical Engineering, University of Texas. DESIGNING A BRAKE AND ITS ASSOCIATED MECHANISMS Design a hoist for the building industry to lift a cubic yard (0.76 m 3 ) of concrete at the rate of 200 ft/min (61 m / min). A cubic yard of concrete weighs approxi- mately 400 lb (1816 kg) and the bucket weighs 1250 lb (568 kg). Since the hoist may be used for other construction purposes, the design capacity will be 6000 lb (2725 kg). There will be no counterweights in this hoist and the cable drum will be connected to a 1750-r/min electric motor through a reduction gear train. Low- ering will be controlled by manual operation of a brake. This brake must automat- ically hold the load at any position when the motor is not driving the hoist. Figure 2 shows the proposed arrangement of parts and the limiting dimensions of this hoist control. It has been proposed that a spring-loaded band brake be used that will be manually released by the operator during lowering of the load. An overrunning clutch is to be provided to disengage the brake automatically when the torque from the motor through the gear train is sufficient to raise the load. Determine (a) the dimensions q and n, Fig. 2, for maximum self-energization without self-locking if the coefficient of friction, ␮ , varies between 0.20 and 0.50—the wide range is selected to cover possible changes in service due to unavoidable entrance of small amounts of water, oil, or dirt into the brake; (b) the spring force required to ensure that the brake will not slip when ␮ is between 0.20 and 0.50; (c) the force exerted by the operator when the rated load first starts to lower under normal conditions; i.e., when ␮ ϭ 0.35; (d) the minimum width of brake lining; and (e) the maximum lowering speed for a reasonable wear life. Calculation Procedure: 1. Determine the actuating arm dimensions The force F 1 and F 2 must act as shown, Fig. 2, for the braking torque to oppose the load torque. This brake will be self-energizing when the F 1 moment tends to Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MECHANICAL AND ELECTRICAL BRAKES MECHANICAL AND ELECTRICAL BRAKES 24.9 SI Values 12 in. 30.5 cm 48 in. 121.9 cm 24 in. 60.9 cm 6000 lb 2724 kg FIGURE 2 Band brake used in hoisting application. apply the brake, that is q is less than n, and becomes self-locking when F 1 q Ն F 2 n. The appropriate equation is F 1 ␮␪ ϭ e F 2 and the critical design condition will be when F 1 /F 2 is a maximum, ␮ ϭ 0.50. Thus, Fn 1 ␮␪ 0.50 ϫ (3 / 2) ␲ ϭϭe ϭ e ϭ 10.55 Fq 2 This unit will be most compact when q is as small as possible. The strengths of the pin and the lever will be the major factors in determining the minimum dimen- sion for q. However, if q is estimated to be 1.5 in (3.8 cm), it can be seen that there is not enough space in which to place the lever, as shown, when n ϭ 10.55q. Therefore, under the specified conditions, it is not practical to use self-energization, Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MECHANICAL AND ELECTRICAL BRAKES 24.10 DESIGN ENGINEERING SI Values 48 in. 121.9 cm 12 in. 30.5 cm 2 in. 5.1 cm FIGURE 3 Lever for band brake actuation. and the proposed design will be modified to make q ϭ 0, as shown in Fig. 3. Choose the dimension n as 2.0 in (5.08 cm). 2. Find the required spring force Maximum spring force will be required when the coefficient of friction is a mini- mum, that is when ␮ ϭ 0.20. For this case, F 1 ␮␪ 0.20 ϫ (3 / 2) ␲ ϭ e ϭ e ϭ 2.57 F 2 and F ϭ 2.57F 12 D T ϭ (F Ϫ F ) 12 2 and Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. MECHANICAL AND ELECTRICAL BRAKES [...]... 300 ϫ 24 ⁄ 30 ϭ 24 0 ft / min (73 .2 m / min) Summarizing this design we have the following: Brake lever—compound lever, tight side of band to pivot—see Fig 4; Spring force ϭ 510 lb (22 68 N); release force ϭ 18.8 lb (80.1 N) with rated load of 6000 lb (27 24 kg) and ␮ ϭ 0.35; lining width ϭ 4 in (10 .2 cm); lowering velocity ϭ 24 0 ft / min (91 .4 m / min) with rated load of 6000 lb (27 24 kg) Related Calculations. .. be 510 lb (22 68 N) 3 Compute the operating force for rated load under normal conditions The operator must push the lever to the right to lower the load When ␮ ϭ 0.35, F1 ϭ e ␮␪ ϭ e0.35ϫ(3 / 2) ␲ ϭ 5 .20 F2 and F1 ϭ 5 .20 F2 Solving the equations for force simultaneously, we find F1 ϭ 5 940 lb (26 42 1 N) F2 ϭ 1 140 lb (5071 N) Then, ͚MO ϭ 0 ϪP ϫ 48 ϩ Fs ϫ 12 Ϫ F2 ϫ 2 ϭ 0 ϪP ϫ 48 ϩ 510 ϫ 12 Ϫ 1 140 ϫ 2 ϭ 0 Solving.. .MECHANICAL AND ELECTRICAL BRAKES 24 .11 MECHANICAL AND ELECTRICAL BRAKES T ϭ Fcable Dcable drum ϭ 6000 ϫ 24 / 2 ϭ 72, 000 lb ⅐ in 2 D ϭ 30 / 2 ϭ 15 in 2 (8136 N ⅐ m) (38.1 cm) Solving for F1, we find F1 ϭ F2 ϩ 48 00 lb Solving the two equations above simultaneously, we find F2 ϭ 3060 lb (13,611 N) F1 ϭ 7860 lb ( 34, 961 N) and ͚MO ϭ 0 Fs ϫ 12 Ϫ F2 ϫ 2 ϭ Fs ϫ 12 Ϫ 3060 ϫ 2 ϭ 0 Solving for... ENGINEERING SI Values 48 in 121 .9 cm 34 in 86 .4 cm 2 in 5.1 cm 12 in 30.5 cm FIGURE 4 Compound lever for band brake should be noted that the operator must now pull the lever to the left to release the brake Taking moments we have: ͚MO2 ϭ 0 ϪFB ϫ 14 ϩ Fs ϫ 12 ϭ ϪFB ϫ 14 ϩ 510 ϫ 12 ϭ 0 from which FB ϭ 43 7 lb ͚MO3 ϭ 0 Pb Ϫ FB a ϭ 30 ϫ ( 34 Ϫ a) Ϫ 43 7a ϭ 0 from which a ϭ 2. 18 in and b ϭ 34 Ϫ 2. 18 ϭ 31. 82 in Downloaded... This procedure is the work of Richard M Phelan, Associate Professor of Mechanical Engineering, Cornell University Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 20 06 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website MECHANICAL AND ELECTRICAL BRAKES 24 .15 MECHANICAL AND ELECTRICAL BRAKES... Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 20 06 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website MECHANICAL AND ELECTRICAL BRAKES MECHANICAL AND ELECTRICAL BRAKES 24 .17 This procedure is the work of Allen S Hall, Jr., Professor of Mechanical Engineering, Purdue University; Alfred R Holowenko, Professor... consider 2 in (5.08 cm) as the value for the distance from the point of spring attachment to the pin B on lever 2, Fig 4 It Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 20 06 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website MECHANICAL AND ELECTRICAL BRAKES 24 . 12 DESIGN ENGINEERING. .. Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 20 06 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website MECHANICAL AND ELECTRICAL BRAKES MECHANICAL AND ELECTRICAL BRAKES 24 .13 The operating force for rated load with the compound lever under normal conditions is: ͚MO2 ϭ 0 ϪFB ϫ 14 ϩ Fs ϫ 12 Ϫ F2 ϫ 2 ϭ ϪFB... Copyright © 20 06 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website MECHANICAL AND ELECTRICAL BRAKES 24 .16 DESIGN ENGINEERING (3.8 cm) For a coefficient of friction of 0.3 and a maximum permissible pressure of 150 lb / in2 (1033.5 kPa), with ␪1 ϭ 0, 2 ϭ 130Њ, ␪m ϭ 90Њ, a ϭ 5 in ( 12. 7 cm), and c ϭ 9 in (22 .9 cm), determine the value of the actuating... Calculation Procedure: 1 Find the moment of the frictional forces about the brake-shoe pivot The moment of the frictional forces, Mƒ , about the right-hand pivot of the brake is given by Mƒ ϭ ƒpmwr sin ␪m ͵ 2 0 (sin ␪)(r Ϫ a cos ␪) d␪ ϭ ϭ 340 0 in ⅐ lb ͫ ͬ ƒpmwr 1 r Ϫ r cos 2 Ϫ a sin2 2 sin ␪m 2 (3 84 .2 N ⅐ m)(0.03)(150)(1.5)(6)[6 ϩ 6(0. 643 ) Ϫ 2. 5(0.766 )2] where ƒ ϭ coefficient of friction; pm ϭ maximum permissible . normal con- ditions is: ͚M ϭ 0 O 2 ϪF ϫ 14 ϩ F ϫ 12 Ϫ F ϫ 2 ϭϪF ϫ 14 ϩ 510 ϫ 12 Ϫ 1, 140 ϫ 2 ϭ 0 Bs2 B F ϭ 27 4 lb B ͚M ϭ 0 O 3 P ϫ 31. 82 Ϫ F ϫ 2. 18 ϭ P ϫ 31. 82 Ϫ 27 4 ϫ 2. 18 ϭ 0 B from which Solving. / 2) ␲ ϭ e ϭ e ϭ 5 .20 F 2 and F ϭ 5 .20 F 12 Solving the equations for force simultaneously, we find F ϭ 5 940 lb (26 42 1 N) F ϭ 1 140 lb (5071 N) 12 Then, ͚M ϭ 0 O ϪP ϫ 48 ϩ F ϫ 12 Ϫ F ϫ 2 ϭ 0 s 2 ϪP. radiating factor from Table 4. Assuming a temperature rise of 300 ЊF (166.7ЊC), and substituting, A R ϭ ( 42 . 4) (0 .29 4) (48 .5) / 0 .25 ϭ 24 18.3 in 2 (15,601.9 cm 2 ). It is necessary to assume a temperature