Ch13 - 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

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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 gfcm-g' [10‘3 kgfgfllbmffiflfl kg}(cm3f1fl'fi 1'113}(l].3l]t-13-1r m3fft3)(f13H12-1'in.3]= [mama lbmr’in.3 I _ E. _ 3.2-2 _ 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 kgfigmbmJUAS-ii 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 (m-3 kgrlo-fi 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 = Ill-fey“ 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 lh-il'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] .tnin-hp = 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 l-I——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.

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Ch13 - 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

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