budynas_SM_ch12.pdf

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Chapter 12
12-1 Given d max =
1
.
000 in and b min =
1
.
0015 in , the minimum radial clearance is
c min =
b min −
d max
=
1
.
0015
−
1
.
000
=
0
.
000 75 in
2
2
Also
l
/
d
=
1
r
=
1
.
000
/
2
=
0
.
500
r
/
c
=
0
.
500
/
0
.
000 75
=
667
N
=
1100
/
60
=
18
.
33 rev/s
P
=
(667 2 ) 8(10 − 6 )(18
/
( ld )
=
250
/
[(1)(1)]
=
250 psi
Eq. (12-7):
S
=
.
33)
=
0
.
261
250
Fig. 12-16:
h 0 /
c
=
0
.
595
Fig. 12-19:
Q
/
( rcNl )
=
3
.
98
Fig. 12-18:
fr
/
c
=
5
.
8
Fig. 12-20:
Q s
/
Q
=
0
.
5
h 0 =
0
.
595(0
.
000 75)
=
0
.
000 466 in Ans
.
f
=
5
.
c =
8
667 =
5
.
0
.
0087
r
/
The power loss in Btu/s is
H
=
2
fWrN
778(12)
π
=
2
π
(0
.
0087)(250)(0
.
5)(18
.
33)
778(12)
=
0
.
0134 Btu/s Ans
.
Q
=
3
.
98 rcNl
=
3
.
98(0
.
5)(0
.
000 75)(18
.
33)(1)
=
0
.
0274 in 3 /s
Q s
=
0
.
5(0
.
0274)
=
0
.
0137 in 3 /s Ans
.
12-2
b min −
d max
1
.
252
−
1
.
250
c min =
=
=
0
.
001 in
2
2
=
r
1
.
25
/
2
=
0
.
625 in
r
/
c
=
0
.
625
/
0
.
001
=
625
N
=
1150
/
60
=
19
.
167 rev/s
P
=
400
=
128 psi
1
.
25(2
.
5)
l
/
d
=
2
.
5
/
1
.
25
=
2
S
=
(625 2 )(10)(10 − 6 )(19
.
167)
=
0
.
585
128
W
8
672655887.003.png
Chapter 12
305
The interpolation formula of Eq. (12-16) will have to be used. From Figs. 12-16, 12-21,
and 12-19
For
/
d
=∞
, h o
/
c
=
0
.
96, P
/
p max =
0
.
84 ,
Q
rcNl =
3
.
09
l
/
d
=
1, h o
/
c
=
0
.
77, P
/
p max =
0
.
52,
Q
rcNl =
3
.
6
l
/
d
=
1
2 , h o
/
c
=
0
.
54, P
/
p max =
0
.
42,
Q
rcNl =
4
.
4
l
/
d
=
1
4 , h o
/
c
=
0
.
31, P
/
p max =
0
.
28,
Q
rcNl =
5
.
25
Equation (12-16) is easily programmed by code or by using a spreadsheet. The results are:
l
/
dy ∞
y 1
y 1 / 2
y 1 / 4
y l / d
h o /
c
2
0.96 0.77 0.54 0.31 0.88
P
/
p max
2
0.84 0.52 0.42 0.28 0.64
Q
/
rcNl 2
3.09 3.60 4.40 5.25 3.28
∴ h o =
0
.
88(0
.
001)
=
0
.
000 88 in Ans
.
p max =
128
0
.
64 =
200 psi Ans
.
Q
=
3
.
28(0
.
625)(0
.
001)(19
.
167)(2
.
5)
=
0
.
098 in 3 /s Ans
.
12-3
b min −
d max
3
.
005
−
3
.
000
c min =
=
=
0
.
0025 in
2
2
=
r
3
.
000
/
2
=
1
.
500 in
l
/
d
=
1
.
5
/
3
=
0
.
5
r
/
c
=
1
.
5
/
0
.
0025
=
600
N
=
600
/
60
=
10 rev/s
P
=
800
5(3) =
177
.
78 psi
1
.
Fig. 12-12: SAE 10,
µ =
1
.
75
µ
reyn
S
=
(600 2 ) 1
.
75(10 − 6 )(10)
177
=
0
.
0354
.
78
Figs. 12-16 and 12-21: h o /
c
=
0
.
11, P
/
p max =
0
.
21
h o =
0
.
11(0
.
0025)
=
0
.
000 275 in Ans
.
p max =
177
.
78
/
0
.
21
=
847 psi Ans
.
l
672655887.004.png
306
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Fig. 12-12: SAE 40,
µ =
4
.
5
µ
reyn
0354 4
S
=
0
.
.
5
=
0
.
0910
1
.
75
h o
/
c
=
0
.
19, P
/
p max =
0
.
275
h o
=
0
.
19(0
.
0025)
=
0
.
000 475 in Ans
.
p max =
177
.
78
/
0
.
275
=
646 psi Ans
.
12-4
b min −
d max
3
.
006
−
3
.
000
c min =
=
=
0
.
003
2
2
=
r
3
.
000
/
2
=
1
.
5in
l
/
d
=
1
r
/
c
=
1
.
5
/
0
.
003
=
500
N
=
750
/
60
=
12
.
5rev/s
P
=
600
3(3)
=
66
.
7 psi
Fig. 12-14: SAE 10W,
µ =
2
.
1
µ
reyn
(500 2 ) 2
S
=
.
1(10 − 6 )(12
.
5)
=
0
.
0984
66
.
7
From Figs. 12-16 and 12-21:
h o
/
c
=
0
.
34, P
/
p max =
0
.
395
h o
=
0
.
34(0
.
003)
=
0
.
001 020 in Ans
.
p max =
66
395 =
.
7
169 psi Ans
.
0
.
Fig. 12-14: SAE 20W-40,
µ =
5
.
05
µ
reyn
S
=
(500 2 ) 5
.
05(10 − 6 )(12
.
5)
=
0
.
237
66
.
7
From Figs. 12-16 and 12-21:
h o
/
c
=
0
.
57, P
/
p max =
0
.
47
h o
=
0
.
57(0
.
003)
=
0
.
001 71 in Ans
.
p max =
66
47 =
.
7
142 psi Ans
.
0
.
672655887.005.png
Chapter 12
307
12-5
b min −
d max
2
.
0024
−
2
c min =
=
=
0
.
0012 in
2
2
r
=
d
2
=
2
2 =
1in, l
/
d
=
1
/
2
=
0
.
50
r
/
c
=
1
/
0
.
0012
=
833
N
=
800
/
60
=
13
.
33 rev/s
P
=
600
2(1)
=
300 psi
Fig. 12-12: SAE 20,
µ =
3
.
75
(833 2 ) 3
µ
reyn
S
=
.
75(10 − 6 )(13
.
3)
=
0
.
115
300
From Figs. 12-16, 12-18 and 12-19:
h o /
c
=
0
.
23, rf
/
c
=
3
.
8, Q
/
( rcNl )
=
5
.
3
h o =
0
.
23(0
.
0012)
=
0
.
000 276 in Ans
.
f
=
8
833 =
3
.
0
.
004 56
The power loss due to friction is
H
=
2
fWrN
778(12)
π
=
2
π
(0
.
004 56)(600)(1)(13
.
33)
778(12)
=
0
.
0245 Btu/s Ans
.
Q
=
5
.
3 rcNl
=
5
.
3(1)(0
.
0012)(13
.
33)(1)
=
0
.
0848 in 3 /s Ans.
12-6
c min =
b min −
d max
=
25
.
04
−
25
=
0
.
02 mm
2
2
r
=
d
/
2
=
25
/
2
=
12
.
5mm, l
/
d
=
1
r
/
c
=
12
.
5
/
0
.
02
=
625
N
=
1200
/
60
=
20 rev/s
P
=
1250
25 2
=
2MPa
(625 2 ) 50(10 − 3 )(20)
2(10 6 )
For
µ =
50 mPa
·
s,
S
=
=
0
.
195
From Figs. 12-16, 12-18 and 12-20:
h o /
c
=
0
.
52, fr
/
c
=
4
.
5, Q s /
Q
=
0
.
57
h o =
0
.
52(0
.
02)
=
0
.
0104 mm Ans
.
f
=
5
625 =
4
.
0
.
0072
T
=
fWr
=
0
.
0072(1
.
25)(12
.
5)
=
0
.
1125 N
·
m
672655887.006.png
308
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
The power loss due to friction is
H
=
2
π
TN
=
2
π
(0
.
1125)(20)
=
14
.
14 W Ans
.
Q s
=
0
.
57 Q The side flow is 57% of Q Ans
.
12-7
b min −
d max
30
.
05
−
30
.
00
c min =
=
=
0
.
025 mm
2
2
r
=
d
2
=
30
2
=
15 mm
r
c =
15
025 =
600
0
.
N
=
1120
60
=
18
.
67 rev/s
P
=
2750
30(50)
=
1
.
833 MPa
(600 2 ) 60(10 − 3 )(18
S
=
.
67)
=
0
.
22
1
.
833(10 6 )
l
d
=
50
30 =
1
.
67
d requires use of the interpolation of Raimondi and Boyd, Eq. (12-16).
From Fig. 12-16, the h o
/
/
c values are:
y 1 / 4 =
0
.
18, y 1 / 2 =
0
.
34, y 1 =
0
.
54, y ∞ =
0
.
89
Substituting into Eq. (12-16),
h o
c
=
0
.
659
From Fig. 12-18, the fr
/
c values are:
y 1 / 4 =
7
.
4, y 1 / 2 =
6
.
0, y 1 =
5
.
0, y ∞ =
4
.
0
Substituting into Eq. (12-16),
fr
c
=
4
.
59
From Fig. 12-19, the Q
/
( rcNl ) values are:
y 1 / 4 =
5
.
65, y 1 / 2 =
5
.
05, y 1 =
4
.
05, y ∞ =
2
.
95
Substituting into Eq. (12-16),
Q
rcN l =
3
.
605
h o
=
0
.
659(0
.
025)
=
0
.
0165 mm Ans
.
f
=
4
.
59
/
600
=
0
.
007 65 Ans
.
Q
=
3
.
605(15)(0
.
025)(18
.
67)(50)
=
1263 mm 3 /s Ans
.
This l
672655887.001.png 672655887.002.png

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