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The main bearing of a steam engine is 100mm in Diameter and 175mm long .The main bearing support a load of 28KN at 250rpm. If the Diameter clearance to the diameter is 0.001 and the absolute viscosity of the lubricating oil is 0.015 Kg.m/s. Find: (1) The co-efficient of friction (2) The heat generated at the bearing due to friction. Given data: d 100mm
l 175mm w 28 KN 28 103 N N 250rpm C 0.001 Z 0.015 kg.m s Solution: Let Bearing Pressure
P
W 25000 1 .6 N mm2 l d 175 100
Rubbing Velocity
v
d N 3.14 100 250 1308.9 mm s 60 60
v 1.308 m l As
d
s
1.73 is in between 0.75 to 2.8 so K=0.002.
(A) The Co-efficient of friction:
33 2 N d K 108 p c
33 0.015 250 1 0.002 108 1 .6 0.001
7.73437 10 4 0.002
0.0007437 0.002
0.0027
……………………………………..Ans
(B) Amount of Heat generated:
Qg w v 0.00274 28000 1.309
Qg 101.5 J s
…………………………………..Ans
A natural oil feed bearing operates at 12 r/s in a 20 ℃ environment. Ratio R/d of 2
the bearing is 1.0d.constant is equal to 15w/ m ℃ .the journal radius is 30mm are the radial clearance is 0.024mm radial load w=60N specific weight of lubricant J −6 3 =8.46 ×10 N/ mm specific heat c p=179.8 N ℃ lodaral area of bearing 2 housing A o =36,000 m for lubricating oil=6.5 c p .
Determine the following. a) Average film temperature. b) Increase in temperature. c) Friction torque . Solution: Given data: T a=20 ℃
ni=12
r s
Radius = r =30mm l = 2r =2 ×30=60 mm 2 Area of bearing housing A o =36,000 mm
J Specific heat c p=179.8 N ℃
Radial load
w =60N 2
V o=15 w/ m ℃ 15 w /mm 2 ℃ 3 10
=
c = 0.024 Average film temperature T f =T a +16 π 2 ×
=
π n2 2 r 3 V o Ao C
20+ 16 π 3 ×60 × 302 ×122 6.5 × 9 15 10 ×36,000 ×0.024 3 10
= 20+57.95
℃
T f −T a 57−99 = =28.98 ℃ 2 2 ρ=8.46× 10−6 N / mm3 c p=179.8 J /w ℃
(b) Increase in temperature. =l c p