Bajaj Pulsar 180 Dts-i

BAJAJ PULSAR 180 DTS-i

DTS-i ENGINE

Determination of important engine parameters Mean piston speed Sp = 2LN Now L = stroke = 56.4mm = 0.0564m N = 8000rev/min = 133.33rev/s Sp = 2x.0564x133.33 m/s = 15m/s

Mean Effective Pressure

mep = P(kW)nrx103 Vd(dm3)xN(rev/s) Now P = Power = 11.7 kW N = 8000 rev/min = 133.33 rev/s Vd = Displacement = 178.6cc mep = 11.7 x 2 x 1000 .1786 x 133.33 = 982.6 kPa

Specific Power P/Ap = 1 x mep x Sp 2nr where mep = mean effective pressure Sp = mean piston speed P/Ap = 0.25 x 982.6 x 15 = 36.85 kW/dm2.

Volumetric Displacement Vd = n x (π/4) x B2 x L where n = no. of cylinders = 1 B = bore = 63.5 mm

L = stroke = 56.4 mm Vd = (π/4) x (0.0635)2 x 0.0564 = 0.18 dm3.

Detailed design of (a) Cylinder The cylinder wall is subjected to gas pressure and the piston side thrust. The gas pressure produces two types of stresses: longitudinal and circumferential which act at right angle to each other and so the net stress in each direction is reduced. sl = longitudinal stress = Force/Area = π/4 x D2 x pmax π/4 x (Do2 – D)2 where D = cylinder inside dia. Do= cylinder outside dia. and pmax= max. gas pressure

sc= circumferential stress = pmax x D 2t

Net

sl = sl – sc/m

and sc = sc - sl/m

where1/m = Poisson’s ratio

To calculate wall thickness: t = pmaxx D + k 2sc where t = wall thickness in cm. sc= max. hoop stress = 350 kg/cm2 k = reboring factor = 1.25 mm pmax = 80 kg/cm2 Hence t = [(80 x 6.35)/(2 x 350)] + 0.125 = 8.5 mm Now D = 63.5 mm and Do= 63.5 +2 x 8.5 = 80.5 mm

Apparent longitudinal stress sl = [ (6.35)2 x 80] / (8.052 – 6.352) = 131.8 kg/cm2. Apparent circumferential stress sc = (80 x 6.35) / 1.7 = 298.8 kg/cm2. Taking 1/m = ¼ Net

sl = 131.8 – 298.8/4 = 131.8 – 74.7 = 57.1 kg/cm2. sc = 298.8 – 131.8/4 = 298.8 – 32.95 = 265.8 kg/cm2.

Length of cylinder = 10 to 15% of stroke = 1.15 x 5.64 = 6.5 cm.

(b) Piston Thickness of piston head t = ( 3pD2/16st )1/2 where

t = piston head thickness D= cylinder bore = 6.35 cm p = max. gas pressure = 80 kg/cm2 st = allowable stress in bending = 400 kg/cm2

Hence

t = [(3x80x6.352) / (16 x 200)]1/2 = 1.72 cm

Piston Rings The radial width of the ring is selected as to limit the wall pressure to 0.25 to 0.42 kg/cm2 wr = D (3pw/ st)1/2 where wr = radial width of the ring pw = wall pressure = 0.25 kg/cm2 st = allowable stress in bending = 850 kg/cm2 Hence wr = 6.35 x (3x0.25/850)1/2 = 0.2 cm. Piston Skirt The portion of the piston barrel below the ring section upto the open end is known as skirt and it takes the side thrust of the connecting rod. Its length

should be such that the side thrust pressure does not exceed 2.5 kg/cm2 for low speed engines and 5.0 kg/cm2 for high speed engines. Side thrust µ.π/4.D2.p = lDpt µ = coefficient of friction (0.03 to 0.10) D = piston diameter = 6.35 cm p = gas pressure = 80 kg/cm2 l = skirt length pt = side thrust pressure = 2.5 kg/cm2 Hence 0.05 x π/4 x 6.352 x 80 = I x 6.35 x 2.5 l = 8 cm.\