77-7 Sliding Bearing [PDF]

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77-7 Sliding Bearing [PDF]

The main bearing of a steam engine is 100mm in Diameter and 175mm long .The main bearing support a load of 28KN at 250rp

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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