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 iscosit! of the lubricating oil is 0.015 Kg.m"s. #ind$ %1& The co'e(cient of friction %2& The heat generated at the bearing due to friction. )ien data$
= 100mm
d
l = 175mm w = 28 KN = 28 ×10 3 N N = 250rpm C = 0.001 Z = 0.015 kg .m s *olution$ Let Bearing Pressure P
=
W
×
l d
=
25000
×
175 100
= 1.6 N
mm
2
Rubbing Velocity v
v
=
π
× d × N 60
=
3.14
× 100 × 250 60
= 1308.9 mm s
= 1.308 m s l d
= 1.73
As
is in between 0.75 to 2.8 so K=0.002. (A !"e #o$e%cient o& &riction' &riction'
2 N d × + K 108 p c 33
µ
=
=
0.015 × 250 × 1 + 0.002 108 1.6 0.001 33
= 7.73437 ×10− + 0.002 4
= 0.0007437 + 0.002 µ
= 0.0027
..Ans
(B A)ount o& *eat generate+'
Q g
= µ × w × v
= 0.00274 × 28000 ×1.309 Q g = 101.5 J s ..Ans
A natural oil feed bearing operates at 12 r/s in a 20
℃
environment. atio /d of
2 t!e bearing is 1.0d."onstant is e#ual to 15$/ m ℃ .t!e %ournal radius is 30mm are
t!e radial "learan"e is 0.024mm radial load $&60' spe"ifi" $eig!t of lubri"ant −6
&8.46 × 10
3
'/ mm
spe"ifi" !eat
c p=179.8
J ℃ lodaral area of bearing N
2 !ousing A o =36,000 m for lubri"ating oil&6.5 c p .
(etermine t!e follo$ing. a) Average film temperature. b) *n"rease in temperature.
c Frictiontorque. Solution:
+iven data, T a=20 ℃
ni=12
r s
adius & r &30mm l & 2r &2 × 30 =60 mm 2
A o =36,000 mm
Area of bearing !ousing -pe"ifi" !eat adial load
c p=179.8
J N ℃
$ &60' 2
V o=15 w / m ℃ 15
&
10
3
2
w / mm ℃
" & 0.024 Average film temperature 2
3
π n 2r T f =T a + 16 π × V o A o C 2
3
2
2
20 + 16 π × 60 × 30 × 12 15
&
10
&
×
× 36,000 × 0.024 3 ℃
20 + 57.95
T f −T a 2
=
57 −99
−6
2 3
ρ= 8.46 × 10 N / mm c p=179.8 J / w ℃
b) *n"rease in temperature. &l c p < ≠ μ n j c p
=28.98 ℃
6.5 10
9
3 −6 &60 × 0.024 × 8.46 × 10 × 3.145 × 30 × 12 × 179.8 × 10
&2476.02 V o A o
( T f −T a ) 2
=
15 10
× 36,000 × 28.98
3
&15649.2 *n"rease in temperature ∆ T =
15649.2 2476.02
=6.32 ℃
......Ans.
) ri"tion tor#ue 2
&
2
4 π r 2 πn
C 2
2
4 × 3.14 × 30 × 60 × 12
&
0.024
×
6.5 9
10
2 &2.076 × 10
&207.64 '.mm
.....Ans.
,L--/ B1AR-/
n a "y+rostatic conical t"rust bearing3 t"e a4ial t"rust is 00 K/3 an+ t"e s"a&t rotates at 6000 r). -& t"e s"a&t +ia)eter is 50 ))3 t"e recess +ia)eter is 950 ))3 se)i$cone angle is 45
°
3 :l) t"ic;ness is 0.62 ))3
60 sus. ,eci:c gra
uire)ent.
,L?!-/' i
n=6000 r)
o =225 )) i =675 )) ! = 45°
"o =0.62 )) t=60
# =0.8 pi = =
,?PPLC PR1,,?R1'$ ¿ p i
2 ×$ ×% n × o / i
=
2
2
π ( o − i ) 5
=
2 × 6 × 10 × %n × 225¿ 175 3.14 ( 225
2
96043.73
=
20000
=.8022
DL@ R1?-R1E1/!' V-,#,-!C Kine)atic
& ' = (0.22t$
180
t
= (0.22 × 60 $
c,t
180 160
c,t
−1752)
=9.07c,t
F=
# × & ' × 9.07
=0.8
=2G.90 #P
μ =
& 10
29.30
9
=
EPa
9
10
o / i ¿ =
¿
6 × μ × %n ¿
π × (i ×" o
3
¿
¿ 175
225¿ ¿
=
6×
29.30 10
9
× %n ¿
3.14 × 4.8022 × 0.12
3
¿
0.026 × 10
9
= 6 × 29.30 × 0.5313 6
3
= 0.5885 × 10 mm / sec = 0.5885
×
0
= 95.96 litresH)in
litresH)in
*+IDIN) ,-/IN) esign a suitable Iournal bearing &or a centri&ugal u) &ro) t"e &ollowing a
te)erature = 90 Jc. #alculate t"e loa+ing re>uire)ent i& any.
-V1/ A!A'
@ = 69.5 K/ =69.5
3
× 10
= 80)) 2 2 P = 0.7/H mm ! 6. /H mm
t a
= 90 Jc (at). te)erature
t o
= 75 Jc (oil :l) te)erature
•
F = 0.02 KgH).s & N
2 mm = 28 /H (
C )
= 0.0069
* )
= 6 ! 2
* •
As
is in between 0.75 to 2.8
)
= 0.002 •
Pressure intensity & N (
P=
= 28
0.025 × 1440 28
2 = 6.2G /H mm
•
@ = p× %× +
Loa+ ( %× +
( %
=
%× +
= 605.62 10465.12
+
= %
= 60.
10465.12 80
= 690.82))
3
× 10
•
Velocity
π×+×N
<=
60000
=
π × 1440 × 80 60000
<=.092 )Hsec •
Let co$e%cient o& &riction μ
=
33 10
8
+ ¿( ( p c ) K
= μ
•
10
8
( 28
(
1 0.0013
) 0.002
= 0.00G66
=
μ × × w
•
33
Let "eat generate+ ,-
,-
& N
=0.00G6
= 76.8 Hsec
*eat +issiate+ ,+
= #A (
T /−T a
×
69.5
3
× 10 ×
.092
Let
T /−T a
(
1
=
2
(75$90
= 22.5 •
%
Let
= 0.690 ) = 0.080 )
Area A = (0.060 An+
= #A (
T /−T a
2
= 6900
,+
2
m ℃
# = 6900 @H
,+
•
m
×
0.060
×
22.5
= 90.2 @ ,t
,o
= *eat generate+ M *eat +issiate+ = 76.8 M 90.2 = 97. @
•
Let )ass o& lubricating oil re>uire+ ()= , t
=
m×s×t
97. =
m× 1900 × 10
) = 0.02909 KgHsec
) = 6.98 KgHsec
