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Unformatted text preview: I N (1 . ‘1:
Known: A petroleum 011 is at 135 0F. Find: Determine the weight per cubic inch. Schematic and Given Data: Pctmlcum DiE
at 185 0F Assumplinn: Equation 13.64:: is valid. Analysis:
1. From Eq. [3.151; for T = 185 nF. Hausa, p = 0.89 — {1.1300350 85  '50) = H.846
gfcm3 2. Wﬁting pin lbmﬁn},
p = H.846 ga’un‘l3 = [1.846 gfcmg' [10‘3 kgfgﬂlbmfﬁﬂﬂ kg}(cm3f1ﬂ'ﬁ 1'113}(l].3l]t131r
m3fft3)(f13H121'in.3]= [mama lbmr’in.3 I _ E. _ 3.22 _ JR.
3. v _ g: p _ 312 (0.0306) _ M306 ing SOL { 3'}
Known: SAE 3D oil is at 160%. Find: Datcrmine the weight par cubic inch. Schematic and Given Data: SAE 3D Oil
at 133°]: Assumption: Equation 13.6h is valid. Analysis:
l. From 13.61:: fur T = 1613 “F. Hence, p = 0.39 — UﬂﬂﬂSSUﬁﬂ d 60} = 0.355
gx‘cm3 2. Writing [[1 in ibWili.3.
p = 0.855 @9ch = 0.855 ghemFl ( Iii—3 kgﬁgmbmJUASii kgﬂcnﬂ! iii"6 m3][O.3ﬂ433
mﬁffl3){f[3r'i‘23 inﬁ) = [103139 :bmﬁnﬁ l 3 512.2 = 1b
3‘ 1,ng ammmnm 0.03mi“; SOLUTION (13.11}
Known: Oil is at a known temperature of 30 “C. Find: Determine the kinematic viscosity at 30 DC. Schematic and Given Data: Oil data: Specific gravityn
= 0.8? at 15.6 C p. =5.~5mPasatSU “c 1ir’iscmneter Assumption: The absolute viscosity can be determined from Fig. 13.6 by
interpolation. Analysis:
1. From Eq. 13.6a but with [1.39 replaced by 0.81“,
pm of”. = 0.37  Oﬂﬂﬂﬁﬁﬂﬂ — 15.6) = 0.329 gram. 2. From Eq. 13.3, _ iter _ M
“30 DC — Emma _ CI. 829 (gr'cm3) 5.6 x lﬂ'3(Nim2) s
o. 829 (m3 kgrloﬁ 1113) 5.6 x lﬂ'3lfkg W931 eeezmz] 3
0.829 x liﬁkgfmi) = 6.836 X 113‘” I11sz or {3.836 x IIZI'I'ﬁ St. SOLUTION (13.12) Known: A11 oil has a known specific gravity at 60 0F. Also we know its vigitzositj.r at
two other known temperatures.
Find: Estimato tho viscosity of this oil in nroyn at 1311} “F. Schematic and Given Data: Gil (iota: Specific gravityO
= 0.337 at 60 F
Oil viscosity: 65 sus at 210 “F 550 sos at too “F Saybolt
Viscomotor Amumptions:_ Equation 13.5 is valid. 2. Tho absolute viscosity can be determined from Fig. 13.6 by interpolation.
3. Equation 13.6]: is valid. but [139 ix replaced by [1.881 Analysis:. From Eq. l3.6:r,
pump: (133'?  [3.00035 (210  60) = {1.3345 gicrtli‘s
Pinon} = 0.38?  0.00035 {100  (SCI) = 0.8?3 gﬁcmil
2. From Eq. 13.5,
palm.ch = X  X = Illfey“ WWF = 0.145 {0.22 x 550  130550) x 0.373 =15.23 mayo
3. With tht: interpolation method, we obtain from Fig. 13.6, than n}: = 2. 3 LLI'Eju'Ii I W'— Known: A lightly loaded 360“ journal hearing has a known diameter, length and radial
clearance. Thejounial has a given rotational speed and is lubricated with a SAE 10 oil
at a known average temperature. Find: Determine the power loss and the frictional torque. Schematic and Given Data: SAE 10 Oil _ 'D
Tavg—r 150 F Assumptions: , . . .
1. The effect of eccentricity between thejournal hearing and the Journal is negligible.
2. There is no lubricant ﬂow in the axial direction. Analysis: 1. From Fig. 13.6.1.1. =1.6 x {04” urcyn
2. From Petrost equation, _ 43r3unLR7" _ 41131.6 1: 164’ reynmj reviaiio in.](2 in. Tl _ C '" {0.1102 in.)
= 22.75 lhil't. = 1.90 lb‘fl. I
3 Power loss W = “Tl = {9m rEV'J_min}(l'9ﬂ [ME
1 ‘ 5252 5252 [Ilb'ft'l'ev]
.tninhp
= I}. 326 hpihearing  Comments: 1. In an actual bearing, we would need to verify that while dissipating llZI.32IS
hpihearing the average oil temperature in the hearing would he comment with the
value of viscosity used in the calcuiatiort. 2. If we double the radius then the frictional torque is increased by ii. Hui increasing
the hearing radum rcqmrea more clearance and eventually more oil ITltl$l be
supplied to prevent hearing failure. SOLUTION (13.17)
Known: A. shaft with known diameter, rotational speed. and radial load is supported by
an oil lubricated bearing of specified length and diametrical clearance. 'l'hc oil viscosin is known.
Find: Estimate the bearing coefﬁ cient of friction and the power loss. Schematic and Given Data:
12m} lb lI——L= 3.0 in —l Assumptions: I _
1. The load creates a negligible eccentricity between journal and bearing. 2. There is negligible lubricant ﬂow in the axial directron.
3. The Petroft‘s approach is suitable. Analysis:
1 . The load intensity, P=w— “00"? =l33.3psi LD ‘ t3 in.)(3 in. t
2. From Eq. (13.7): r = WE 11 =2a2__—({5 x “3'6 mum“ %')'._..l_.5 in_. J: acts? I
P C Bill—bk 0.002 in. m.1
3. Tr: Mr: (Dﬂlﬁ’rl'HlZﬂD Ib)(1.5 in.) = 29.98in.1b = 2.5 ft lb
Tﬂft lb) mire—y) (2.5 ft lb}[ 1300 E
mm _ \_ = _M—__ 1111“ = (1856 h
4. Power loss Slﬁﬂﬂ 1b Ev} SEﬂmJ P I
 —mm hp min hp Comment: Petrost equation can be checked by using Fig. 13.14 to calculate f. For%= %=1 and s = (%)1%"
Hence. f = D.{lIT33 which is 4% larger than Uﬂlﬁﬁ'i'. = 13.632, we have %f = 13. ...
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This note was uploaded on 10/26/2008 for the course MEM 431 taught by Professor Zhou during the Spring '08 term at Drexel.
 Spring '08
 ZHOU

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