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An open belt connects two flat pulleys. The pulley diameters are 300 mm and 450 mm and the corresponding angles of lap are 160° and 210°. The smaller pulley runs at 200 rpm. The coefficient of friction between the beit and pulley is 0.35. It is found that the beit is on the point of slipping when 3 kW is transmitted. To increase the power transmitted, two alternattves are suggested, namely (1) (i) increasing the coefficient of friction by 15% by the application of a suitable dressing increasing the initial tension by 15%, and to the belt. Which of these two methods would be more effective? Find the percentage increase in power possible in each case.

An open belt connects two flat pulleys. The pulley diameters are 300 mm and 450 mm and the corresponding angles of lap are 160° and 210°. The smaller pulley runs at 200 rpm. The coefficient of friction between the beit and pulley is 0.35. It is found that the beit is on the point of slipping when 3 kW is transmitted. To increase the power transmitted, two alternattves are suggested, namely (1) (i) increasing the coefficient of friction by 15% by the application of a suitable dressing increasing the initial tension by 15%, and to the belt. Which of these two methods would be more effective? Find the percentage increase in power possible in each case.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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An open belt connects two flat pulleys. The pulley diameters are 300 mm and 450 mm and the corresponding angles of lap are 160° and 210°. The smaller pulley runs at 200 rpm. The coefficient of friction between the beit and pulley is 0.35. It is found that the beit is on the point of slipping when 3 kW is transmitted. To increase the power transmitted, two alternatives are suggested, namely increasing the initial tension by 15%, and (i) increasing the coefficient offriction by 15% by the application of a suitable dressing to the belt. Which of these two methods would be more effective? Find the percentage increase in power possible in each case. Consider a driven pulley rotating in the clockwise direction as shown in Figure TQ3.3 80 80 - Ry RN Driven 80 pulley T+ 8T 80 2 T+8T (a) (b) Figure TQ3.3 Ratio of driving tension for flat belt T, = Tension in the belt on the tight side, T, = Tension in the belt on the slack side, and e = Angle of contact in radians (i.e. angle subtended by the are AB, along which the belt touches the pulley, at the centre). Let The above can be simply expressed as 图) log. = µ.0 or ..(v) The equation (v) can be expressed in terms of corresponding logarithm to the base 10, ie. Logo (T/T) - μ.0 Loge)-μ.0 (0.4343) (2.3026) (Log. (Ti/T:)) = µ.0 2.3 log = 4.0
Transcribed Image Text:An open belt connects two flat pulleys. The pulley diameters are 300 mm and 450 mm and the corresponding angles of lap are 160° and 210°. The smaller pulley runs at 200 rpm. The coefficient of friction between the beit and pulley is 0.35. It is found that the beit is on the point of slipping when 3 kW is transmitted. To increase the power transmitted, two alternatives are suggested, namely increasing the initial tension by 15%, and (i) increasing the coefficient offriction by 15% by the application of a suitable dressing to the belt. Which of these two methods would be more effective? Find the percentage increase in power possible in each case. Consider a driven pulley rotating in the clockwise direction as shown in Figure TQ3.3 80 80 - Ry RN Driven 80 pulley T+ 8T 80 2 T+8T (a) (b) Figure TQ3.3 Ratio of driving tension for flat belt T, = Tension in the belt on the tight side, T, = Tension in the belt on the slack side, and e = Angle of contact in radians (i.e. angle subtended by the are AB, along which the belt touches the pulley, at the centre). Let The above can be simply expressed as 图) log. = µ.0 or ..(v) The equation (v) can be expressed in terms of corresponding logarithm to the base 10, ie. Logo (T/T) - μ.0 Loge)-μ.0 (0.4343) (2.3026) (Log. (Ti/T:)) = µ.0 2.3 log = 4.0
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