15 d mm 15 d 1000 m assume we know that the indicated

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Unformatted text preview: in m. = 1.5 D mm = 1.5 D / 1000 m ....(Assume) We know that the indicated power, I.P = B.P. / ηm = 5000 / 0.8 = 6250 W We also know that the indicated power (I.P.), pm . l . A . n 0.35 × 1.5D × π D 2 × 600 = = 4.12 × 10–3 D3 60 60 × 1000 × 4 ...( ∵ For four stroke engine, n = N/2) ∴ D3 = 6250 / 4.12 × 10–3 = 1517 × 103 or D = 115 mm Ans. and l = 1.5 D = 1.5 × 115 = 172.5 mm Taking a clearance on both sides of the cylinder equal to 15% of the stroke, therefore length of the cylinder, L = 1.15 l = 1.15 × 172.5 = 198 say 200 mm Ans. 2. Thickness of the cylinder head Since the maximum pressure ( p) in the engine cylinder is taken as 9 to 10 times the mean effective pressure ( pm), therefore let us take p = 9 pm = 9 × 0.35 = 3.15 N/mm2 We know that thickness of the cyclinder head, 0.1 × 3.15 C.p th = D σ = 115 = 9.96 say 10 mm Ans. 42 t 6250 = ...(Taking C = 0.1 and σt = 42 MPa = 42 N/mm2) 3. Size of studs for the cylinder head Let d = Nominal diameter of the stud in mm, dc = Core diamete...
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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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