# L an 075 0125 7855 1000 12 270 w 60 60 1227 kw ip

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Unformatted text preview: e, n = N / 2 = 2000 / 2 = 1000 and cross-sectional area of the cylinder, π D2 π (100) 2 = = 7855 m m 2 4 4 We know that indicated power, A= p m . L. A.n 0.75 × 0.125 × 7855 × 1000 = = 12 270 W 60 60 = 12.27 kW IP = ∴ Brake power, BP = IP × ηm = 12.27 × 0.8 = 9.8 kW ...(∴ ηm = BP / IP) We know that the heat flowing through the piston head, H = C × HCV × m × BP = 0.05 × 42 × 103 × 41.7 × 10–6 × 9.8 = 0.86 kW = 860 W ....(Taking C = 0.05) ∴Thickness of the piston head on the basis of heat dissipation, 860 H tH = 12.56 k (T − T ) = 12.56 × 46.6 × 220 = 0.0067 m = 6.7 mm C E ...(∵ For cast iron , k = 46.6 W/m/°C, and TC – TE = 220°C) Taking the larger of the two values, we shall adopt tH = 16 mm Ans. Since the ratio of L / D is 1.25, therefore a cup in the top of the piston head with a radius equal to 0.7 D (i.e. 70 mm) is provided. 2. Radial ribs The radial ribs may be four in number. The thickness of the ribs varies from tH / 3 to tH / 2. ∴ Thickness of the ribs, tR...
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## This note was uploaded on 09/30/2012 for the course MECHANICAL 403 taught by Professor A.thirumoorthy during the Spring '11 term at Anna University Chennai - Regional Office, Coimbatore.

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