Post on 30-May-2018
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BAJAJ PULSAR 180 DTS-i
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DTS-i ENGINE
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Determination of important engine parameters
Mean piston speedSp = 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
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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 (/4) x B2 x L
where n = no. of cylinders = 1
B = bore = 63.5 mm
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L = stroke = 56.4 mm
Vd = (/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
= /4 x D2x pmax
/4 x (Do2 D)2
where D = cylinder inside dia.
Do= cylinder outside dia.
and pmax= max. gas pressure
sc= circumferential stress = pmaxx D
2t
Net sl = sl sc/m
and sc = sc - sl/m
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where1/m = Poissons ratio
To calculate wall thickness:
t = pmax x 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 = [(80x 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 =
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.
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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
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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 ./4.D2.p = lDpt
= 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 /4 x 6.352 x 80 = I x 6.35 x 2.5
l = 8 cm.\