英制textbook answer solutions

1、Chapter 12Given: dmax = 25 mm, bmin = 25.03 mm, l/d = 1/2, W = 1.2 kN, m = 55 mPa×s, and12-1N =1100 rev/min.= bmin - dmax= 25.03 - 25 = 0.015 mmcmin22r B 25/2 = 12.5 mmr/c = 12.5/0.015 = 833.3N = 1100/60 = 18.33 rev/sP = W/ (ld) = 1200/ 12.5(25) = 3.84 N/mm2 = 3.84 MPaé 55(10-3 )18.33ù

2、;æ r ö2 m NS = ç= 833.32 êêëú = 0.182úûEq. (12-7):÷3.84 (106 )è c øPh0 /c = 0.3Þf r/c = 5.4ÞFig. 12-16:h0 = 0.3(0.015) = 0.0045 mmAns.Fig. 12-18:f = 5.4/833.3 = 0.006 48T =f Wr = 0.006 48(1200)12.5(10-3) = 0.0972 N×mHloss = 2

3、p TN = 2p (0.0972)18.33 = 11.2 WAns.Q = 5.1(12.5)0.015(18.33)12.5 = 219 mm3/sQ/(rcNl) = 5.1ÞFig. 12-19:Qs /Q = 0.81ÞQs = 0.81(219) = 177 mm3/sFig. 12-20:Ans.Given: dmax = 32 mm, bmin = 32.05 mm, l = 64 mm, W = 1.75 kN, m = 55 mPa×s, and12-2N =900 rev/min.= bmin - dmax= 32.05 - 32 = 0.

4、025 mmcmin22r B 32/2 = 16 mmr/c = 16/0.025 = 640N = 900/60 = 15 rev/sChapter 12, Page 1/26P = W/ (ld) = 1750/ 32(64) = 0.854 MPal/d = 64/32 = 2é 55(10-3 )15 ùæ r ö2 m NEq. (12-7):S = ç= 6402 êêëú = 0.797÷è c øP0.854úûEq. (12-16),

5、Figs. 12-16, 12-19, and 12-21h0 = 0.92 c = 0.92(0.025) = 0.023 mmAns.pmax = P / 0.065 = 0.854/0.65 = 1.31 MPaAns.Q = 3.20 rcNl = 3.20(16)0.025(15)64 = 1.23 (103) mm3/sAns.12-3Given: dmax = 3.000 in, bmin = 3.005 in, l = 1.5 in, W = 800 lbf, N = 600 rev/min, and SAE 10 and SAE 40 at 150°F.bmin -

6、 dmax3.005 - 3.000c= 0.0025 inmin22r B 3.000 / 2 = 1.500 inl / d = 1.5 / 3 = 0.5r / c = 1.5 / 0.0025 = 600N = 600 / 60 = 10 rev/sW800P = 177.78 psild1.5(3)Fig. 12-12: SAE 10 at 150°F, µ¢ = 1.75 µreynæ r ö2 m Né1.75(10-6)(10) ùS = ç= 6002 ê= 0.0354

7、47;úè c øP177.78ëûFigs. 12-16 and 12-21: h0/c = 0.11 and P/pmax = 0.21h0 = 0.11(0.0025) = 0.000 275 inAns.Ans.= 177.78 / 0.21 = 847 psipmaxFig. 12-12: SAE 40 at 150°F,µ¢= 4.5 µreynChapter 12, Page 2/26l/dy¥y1y1/2y1/4yl/dh0/c20.980.830.610.360.92P/pma

8、x20.840.540.450.310.65Q/rcNl23.13.454.25.083.20S = 0.0354 æ4.5 ö =0.0910ç 1.75 ÷èøh0 / c = 0.19,P /= 0.275pmaxh0 = 0.19(0.0025) = 0.000 475 inAns.= 177.78 / 0.275 = 646 psipmaxAns.12-4Given: dmax = 3.250 in, bmin = 3.256 in, l = 3.25 in, W = 800 lbf, and N = 1000 rev/mi

9、n.bmin - dmax3.256 - 3.250c= 0.003min22r B 3.250 / 2 = 1.625 inl / d = 3 / 3.250 = 0.923r / c = 1.625 / 0.003 = 542N = 1000 / 60 = 16.67 rev/sW800P = 82.05 psild3(3.25)Fig. 12-14: SAE 20W at 150°F, m¢ = 2.85 m reynæ r ö2 m Né 2.85(10-6)(16.67) ùS = ç= 5422 ê=

10、0.1701÷úè c øP82.05ëûFrom Eq. (12-16), and Figs. 12-16 and 12-21:ho = 0.46c = 0.46(0.003) = 0.001 38 inAns.P82.05p= 191 psiAns.max0.430.43Fig. 12-14: SAE 20W-40 at 150°F, m¢ = 4.4 m reyn2 4.4(10-6)(16.67)S = 542= 0.26382.05From Eq. (12-16), and Figs. 12-16 and

11、 12-21:Chapter 12, Page 3/26l/dy¥y1y1/2y1/4yl/dho/c0.9230.910.60.380.20.58P/pmax0.9230.830.480.350.240.46l/dy¥y1y1/2y1/4yl/dho/c0.9230.850.480.280.150.46P/pmax0.9230.830.450.320.220.43h0 = 0.58c = 0.58(0.003) = 0.001 74 inAns.820582.05p= 178 psiAns.max0.460.4612-5Given: dmax = 2.000 in, bm

12、in = 2.0024 in, l = 1 in, W = 600 lbf, N = 800 rev/min, and SAE 20 at 130°F.bmin - dmax2.0024 - 2c= 0.0012 inmin22d2= 1 in,l / d = 1 / 2 = 0.50r B22r / c = 1 / 0.0012 = 833N = 800 / 60 = 13.33 rev/sW600P = 300 psild2(1)Fig. 12-12: SAE 20 at 130°F, µ¢ = 3.75 µreynæ r 

13、46;2 m Né 3.75(10-6)(13.3) ùS = ç= 8332 ê= 0.115÷úè c øP300ëûFrom Figs. 12-16, 12-18 and 12-19:h0 / c = 0.23,r f / c = 3.8,Q / (rcNl) = 5.3h0 = 0.23(0.0012) = 0.000 276 inAns.3.8833f = 0.004 56The power loss due to friction is2p f WrN778(12)2p (0.004

14、 56)(600)(1)(13.33)778(12)Ans.H = 0.0245 Btu/sQ = 5.3rcNl= 5.3(1)(0.0012)(13.33)(1)= 0.0848 in3 / sAns.Given: dmax = 25 mm, bmin = 25.04 mm, l/d = 1, W = 1.25 kN, m = 50 mPa×s, and N = 1200 rev/min.12-6Chapter 12, Page 4/26bmin - dmax25.04 - 25c= 0.02 mml / d = 1min22r B d / 2 = 25 / 2 = 12.5 m

15、m,r / c = 12.5 / 0.02 = 625N = 1200 / 60 = 20 rev/sW1250P = 2 MPald252æ r ö2 m Né 50(10-3)(20) ùS = ç= 6252 ê= 0.195For µ = 50 MPa · s,÷ú2(106)è c øPëûFrom Figs. 12-16, 12-18 and 12-20:h0 / c = 0.52,f r / c = 4.5,Qs / Q = 0.57h0 =

16、 0.52(0.02) = 0.0104 mmAns.4.5f = 0.0072625T = f Wr = 0.0072(1.25)(12.5) = 0.1125 N · mThe power loss due to friction isH = 2T N = 2 (0.1125)(20) = 14.14 WAns.Qs = 0.57Q The side flow is 57% of QAns.Given: dmax = 1.25 in, bmin = 1.252 in, l = 2 in, W = 620 lbf, m¢ = 8.5 m reyn, and12-7N =1

17、120 rev/min.bmin - dmax1.252 - 1.25c= 0.001 inmin22r = d / 2 = 1.25 / 2 = 0.625 inr / c = 0.625 / 0.001 = 625N = 1120 / 60 = 18.67 rev/sW620P = 248 psild1.25(2)æ r ö2 m Né8.5(10-6)(18.67) ùS = ç= 6252 ê= 0.250÷úè c øP248ëûl / d = 2 / 1.25 =

18、 1.6From Eq. (12-16), and Figs. 12-16, 12-18, and 12-19Chapter 12, Page 5/26h0 = 0.69 c = 0.69(0.001) =0.000 69 inAns.f = 4.92/(r/c) = 4.92/625 = 0.007 87Ans.Q = 1.6 rcNl = 1.6(0.625) 0.001(18.57) 2 = 0.0833 in3/sAns.12-8Given: dmax = 75.00 mm, bmin = 75.10 mm, l = 36 mm, W = 2 kN, N = 720 rev/min,

19、and SAE 20 and SAE 40 at 60°C.bmin - dmax75.10 - 75c= 0.05 mmenough)min22l / d = 36 / 75 = 0.48 B 0.5r = d / 2 = 75 / 2 = 37.5 mmr / c = 37.5 / 0.05 = 750N = 720 / 60 = 12 rev/s(W2000P = 0.741 MPald75(36)Fig. 12-13: SAE 20 at 60°C, µ = 18.5 MPa · sæ r ö2 m Né18.5(1

20、0-3)(12) ùS = ç= 7502 ê= 0.169÷ú0.741(106)è c øPëûFrom Figures 12-16, 12-18 and 12-21:h0 / c = 0.29,f r / c = 5.1,P /= 0.315pmaxh0 = 0.29(0.05) = 0.0145 mmAns. f = 5.1 / 750 = 0.0068T = f Wr = 0.0068(2)(37.5) = 0.51 N · mThe heat loss rate equals the

21、 rate of work on the filmHloss = 2T N = 2(0.51)(12) = 38.5 Wpmax = 0.741/0.315 = 2.35 MPaAns.Ans.Fig. 12-13: SAE 40 at 60°C, µ = 37 MPa · sChapter 12, Page 6/26l/dy¥y1y1/2y1/4yl/dh0/c1.60.90.580.360.1850.69fr/c1.64.55.36.584.92Q/rcNl1.633.984.975.63.59S = 0.169(37)/18.5 = 0.338Fr

22、om Figures 12-16, 12-18 and 12-21:h0 / c = 0.42,f r / c = 8.5,= 0.38P /Ans.pmaxh0 = 0.42(0.05) = 0.021 mmf = 8.5 / 750 = 0.0113T = f Wr = 0.0113(2)(37.5) = 0.85 N · m= 2pTN = 2p (0.85)(12) = 64 WAns.Hlosspmax= 0.741 / 0.38 = 1.95 MPaAns.12-9Given: dmax = 56.00 mm, bmin = 56.05 mm, l = 28 mm, W

23、= 2.4 kN, N = 900 rev/min, and SAE 40 at 65°C.bmin - dmax56.05 - 56c= 0.025 mmmin22r = d / 2 = 56 / 2 = 28 mmr / c = 28 / 0.025 = 1120l / d = 28 / 56 = 0.5,N = 900 / 60 = 15 rev/s2400P = 1.53 MPa28(56)Fig. 12-13: SAE 40 at 65°C, µ = 30 MPa · sæ r ö2 m Né 30(10-3)(1

24、5) ùS = ç= 11202 ê= 0.369÷úè c øP1.53(106)ëûFrom Figures 12-16, 12-18, 12-19 and 12-20:h0 / c = 0.44,f r / c = 8.5,Qs / Q = 0.71,Ans.Q / (rcNl) = 4.85h0 = 0.44(0.025) = 0.011 mmf = 8.5 / 1000 = 0.007 59T = f Wr = 0.007 59(2.4)(28)= 0.51 N · mH = 2pTN

25、 = 2p (0.51)(15) = 48.1 WAns.Q = 4.85 rcNl = 4.85(28)(0.025)(15)(28) = 1426 mm3/sQ = 0.71(1426) = 1012 mm3/sAns.s12-10Consider the bearings as specified byd +0 ,+tminimum f :um W:bb-td-0d +0 ,+tbb-td-0Chapter 12, Page 7/26d ¢ .and differing only in d andPreliminaries:l / d = 1P = W/ (ld ) = 700

26、 / (1.252) = 448 psiN = 3600 / 60 = 60 rev/sFig. 12-16:minimum f :um W:S B 0.08S B 0.20µ = 1.38(10-6) reynµN/P = 1.38(10-6)(60/448) = 0.185(10-6)Fig. 12-12:Eq. (12-7):rcS=µN / PFor minimum f :rc c0.080.185(10-6)= 658= 0.625 / 658 = 0.000 950 B 0.001 inIf this is cmin,b - d = 2(0.001)

27、= 0.002 inThe median clearance istd + tbtd + tbc = c+= 0.001 +min22and the clearance range for this bearing istd + tbDc =2which is a function only of the tolerances.Forum W:rc0.20.185(10-6)= 1040c = 0.625 / 1040 = 0.000 600 B 0.0005 inIf this is cminChapter 12, Page 8/26b - d¢ = 2(0.0005) = 0.0

28、01 in2cmintd + tbtd + tbc = c+= 0.0005 +min22td + tbDc =2The difference (mean) in clearance between the two clearance ranges, crange, istd + tb- æ 0.0005 + tb ötdc= 0.001 +ç÷range2è2ø= 0.0005 inFor the minimum f bearingb - d = 0.002 inord = b - 0.002 inFor theum W beari

29、ngd¢ = b - 0.001 inFor the same b, tb and td, we need to change the journal diameter by 0.001 in.d¢ - d = b - 0.001 - (b - 0.002)= 0.001 ind ¢ of theIncreasing d of the minimum friction bearing by 0.001 in, definesumload bearing. Thus, the clearance range provides for bearing dimensio

30、ns which are attainable in manufacturing.Ans.12-11 Given: SAE 40, N = 10 rev/s, Ts = 140°F, l/d = 1, d = 3.000 in, b = 3.003 in, W = 675lbf.bmin - dmax3.003 - 3c= 0.0015 inmin22r = d / 2 = 3 / 2= 1.5 inr / cP =1.5 / 0.00Wldpsi3(3)Trial #1: From Figure 12-12 for T = 160°F, µ = 3.5 

31、1; reyn,DT = 2(160 - 140) = 40°Fæ r ö2 m Né 3.5(10- 6 )(10) ùS = ç= 10002 ê= 0.4667÷úè c øP75ëûFrom Fig. 12-24,Chapter 12, Page 9/269.70DT= 0.349 109 + 6.009 40(0.4667) + 0.047 467(0.4667)2 = 3.16PP75DT = 3.16= 3.16= 24.4°F9.709.7

32、0Discrepancy = 40 - 24.4 = 15.6°FTrial #2: T = 150°F, µ = 4.5 µ reyn,DT = 2(150 - 140) = 20°Fé 4.5 (10- 6 )10 ùS = 10002 êêëFrom Fig. 12-24,ú = 0.6úû759.70DT= 0.349 109 + 6.009 40(0.6) + 0.047 467(0.6)2= 3.97PP75DT = 3.97= 3.97= 30.7&#

33、176;F9.709.70Discrepancy = 20 - 30.7 = - 10.7°FTrial #3: T = 154°F, µ = 4 µ reyn,DT = 2(154 - 140) = 28°Fé 4 (10-6 )10 ùS = 10002 êêëFrom Fig. 12-24,ú = 0.533úû759.70DT= 0.349 109 + 6.009 40(0.533) + 0.047 467(0.533)2= 3.57PP75DT = 3.5

34、7= 3.57= 27.6°F9.709.70Discrepancy = 28 - 27.6 = 0.4°FO.K.Tav = 140 +28/2 = 154°FAns.T1 = Tav - DT / 2 = 154 - (28 / 2) = 140°F T2 = Tav + DT / 2 = 154 + (28 / 2) = 168°F S = 0.4From Figures 12-16, 12-18, to 12-20:Chapter 12, Page 10/26h0 cf rcQQs Q= 0.75,= 11,= 3.6,Ans.= 0.

35、33rcN lh0 = 0.75(0.0015)= 0.001 13 in11f = 0.0111000T = f Wr = 0.0075(3)(40) = 0.9 N · m2p f WrN2p (0.011)675(1.5)10H= 0.075 Btu/sAns.778(12)778(12)lossQ = 3.6rcN l = 3.6(1.5)0.0015(10)3 = 0.243 in3/sAns.Q = 0.33(0.243) = 0.0802 in3/sAns.s12-12Given: d = 2.5 in, b = 2.504 in, cmin = 0.002 in, W

36、 = 1200 lbf, SAE = 20, Ts = 110°F,N = 1120 rev/min, and l = 2.5 in.P = W/(ld) = 1200/(2.5)2 = 192 psi,N = 1120/60 = 18.67 rev/sFor a trial film temperature, let Tf = 150°FTable 12-1:m¢ = 0.0136 exp1271.6/(150 + 95) = 2.441 m reynæ 2.5 / 2 ö2 2.441(10-6 )18.67æ r ö2

37、 m NEq. (12-7):S = ç c ÷= ç÷P0.002= 0.927èøèø192Fig. 12-24:1929.70ëé0.349 109 + 6.009 40 (0.0927) + 0.047 467 (0.09272 )ùûDT = 17.9°FDT= 110 + 17.9 = 119.0°FT= T +avs22Tf - Tav = 150 - 119.0 = 31.0°Fwhich is not 0.1 or less,

38、therefore try averaging for the new trial film temperature, let150 + 119.0(T )= 134.5°Ff new2Proceed with additional trials using a spsheet (table also shows the first trial)Chapter 12, Page 11/26TrialTfNewTfm'DTTf -TavSTav150.0134.5127.9125.3124.3124.0123.82.4413.4664.0844.3694.4854.5214.5

39、450.09270.13170.15510.16590.17040.17170.172617.922.625.426.727.227.427.5119.0121.3122.7123.3123.6123.7123.731.013.25.22.00.70.30.1134.5127.9125.3124.3124.0123.8123.8Note that the convergence begins rapidly. There are ways to speed this, but at this point they would only add complexity.µ¢ =

40、 4.545(10-6),S = 0.1726(a)h0 c= 0.482,h = 0.482(0.002) = 0.000 964 inFrom Fig. 12-16:0From Fig. 12-17: f = 56°Ans.(b) e = c - h0 = 0.002 - 0.000 964 = 0.001 04 inAns.f rc= 4.10,f = 4.10(0.002 /1.25) = 0.006 56(c) From Fig. 12-18:Ans.(d) T = f Wr = 0.006 56(1200)(1.25) = 9.84 lbf · in 2p T

41、N = 2p (9.84)(1120 / 60)H = 0.124 Btu/sAns.778(12)778(12)Q= 4.16(e) From Fig. 12-19:rcNlQ = 4.16(1.25)(0.002) æ 1120 ö (2.5) = 0.485 in3/sAns.ç÷è60øQs Q= 0.6,Q = 0.6(0.485) = 0.291 in3/sFrom Fig. 12-20:Ans.s/ (ld )W1200 / 2.52P= 0.45,= 427 psi(f) From Fig. 12-21:pmaxAns

42、.p0.450.45maxFrom Fig. 12-22: f p= 16° Ans.maxChapter 12, Page 12/26(g) From Fig. 12-22: f p = 82°0Ans.(h) From the trial table, Tf = 123.8°FAns.(i) With DT = 27.5°F from the trial table, Ts + DT = 110 + 27.5 = 137.5°FAns.12-13Given: d = 1.250 in, td = 0.001 in, b = 1.252 in

43、, tb = 0.003 in, l = 1.25 in, W = 250 lbf,N = 1750 rev/min, SAE 10 lubricant, sump temperature Ts = 120°F.P = W/(ld) = 250/1.252 = 160 psi,N = 1750/60 = 29.17 rev/sFor the clearance, c = 0.002 ± 0.001 in. Thus, cmin = 0.001 in, cmedian = 0.002 in, andcmax = 0.003 in.For cmin = 0.001 in, st

44、art with a trial film temperature of Tf = 135°Fm¢ = 0.0158 exp1157.5/(135 + 95) = 2.423 m reynTable 12-1:æ 1.25 / 2 ö2 2.423(10-6 )29.17æ r ö2 m NS = ç c ÷= ç= 0.1725Eq. (12-7):÷èøPè0.001ø160Fig. 12-24:1609.70éë0.349 109

45、 + 6.009 40 (0.1725) + 0.047 467 (0.17252 )ùûDT = 22.9°FDT22.9T= T += 120 += 131.4°Favs22Tf - Tav = 135 - 131.4 = 3.6°Fwhich is not 0.1 or less, therefore try averaging for the new trial film temperature, let135 + 131.4(T )= 133.2°Ff new2Proceed with additional trials u

46、sing a spsheet (table also shows the first trial)TrialTfNewTfm'DTTf-TavTavS135.0133.2132.5132.2132.12.4232.5212.5602.5782.5830.17250.17950.18230.18360.184022.923.623.924.024.0131.4131.8131.9132.0132.03.61.40.60.20.1133.2132.5132.2132.1132.1Chapter 12, Page 13/26With Tf = 132.1°F, DT = 24.0&

47、#176;F, m¢ = 2.583 m reyn, S = 0.1840,Tmax = Ts + DT = 120 + 24.0 = 144.0°FFig. 12-16:h0/c = 0.50, h0 = 0.50(0.001) = 0.000 50 inò = 1 - h0/c = 1 - 0.50 = 0.05 inFig. 12-18:r f /c = 4.25, f = 4.25/(0.625/0.001) = 0.006 8Q/(rcNl) = 4.13, Q = 4.13(0.625)0.001(29.17)1.25 = 0.0941 in3/sQs

48、/Q = 0.58, Qs = 0.58(0.0941) = 0.0546 in3/sFig. 12-19:Fig. 12-20:The above can be repeated for cmedian = 0.002 in, and cmax = 0.003 in. The results are shown below.cmin0.001incmedian0.002 in125.63.0020.053411.1131.10.230.000690.771.80.00584.550.2070.820.170cmax0.003inTfm¢SDTTmax h0/c h0ò f

49、r/c fQ/(rcNl)QQs /QQs132.12.5830.18424.0144.00.50.000500.504.250.00684.130.09410.580.0546124.13.1120.02468.2128.20.1250.000380.881.220.00594.70.3210.900.28912-14Computer programs will vary.12-15Note to the Instructor: In the first printing of the 9th edition, the l/d ratio and the lubrication consta

50、nt a were omitted. The values to use are l/d = 1, and a = 1. This will be updated in the next printing. We apologize for any inconvenience this may have caused.Chapter 12, Page 14/26In a step-by-step fashion, we are building a skill for natural circulation bearings. Given the average film temperatur

51、e, establish the bearing properties. Given a sump temperature, find the average film temperature, then establish the bearing properties. Now we acknowledge the environmental temperatures role in establishing the sump temperature. Sec. 12-9 and Ex. 12-5 address this problem.Given: dmax = 2.500 in, bm

52、in = 2.504 in, l/d = 1, N = 1120 rev/min, SAE 20 lubricant, W = 300 lbf, A = 60 in2, T¥ = 70°F, and a = 1.600 lbf load with minimal clearance: We will start by using W = 600 lbf (nd = 2). The task is to iteratively find the average film temperature, Tf , which makes Hgen andHlossequal.bmin

53、 - dmax2.504 - 2.500c= 0.002 inmin22N = 1120/60 = 18.67 rev/sW600P = 96 psild2.52ö2 m¢(10-6 )18.67æ r ö2 m Næ1.25= 0.0760m¢= ç c ÷= ç÷P0.002Sèøèø96m¢ = 0.0136 exp1271.6/(Tf + 95)Table 12-1:2545 WNc ær ö =f2545 (600)

54、18.67 (0.002)f rcH=ç÷gen1050èøc1050f r c= 54.32.7 (60 / 144) (Th CR A (T- T ) =- 70)H=f¥lossa + 11 + 1f= 0.5625(Tf - 70)Start with trial values of Tf of 220 and 240°F.As a linear approximation, let Hgen = mTf + b. Substituting the two sets of values ofTf and Hgen we fin

55、d that Hgen = - 0.545 Tf +223.1. Setting this equal to Hloss andfor Tf gives Tf = 237°F.solvingChapter 12, Page 15/26Trial Tfm¢Sf r/cHgenHloss2200.7700.0591.9103.284.42400.6050.0461.792.395.6which is satisfactory.Table 12-16:h0/c = 0.21,h0 = 0.21 (0.002) = 000 42 inFig. 12-24:DT = 96 é

56、;ë0.349 109 + 6.009 4 (0.048) + 0.047 467 (0.0482 )ùû9.7= 6.31° FT1 = Ts = Tf - DT = 237 - 6.31/2 = 233.8°FTmax = T1 + DT = 233.8 + 6.31 = 240.1°FTrumplers design criteria:0.002 + 0.000 04d = 0.002 + 0.000 04(2.5) = 0.000 30 in < h0O.K.Tmax = 240.1°F < 250°F O.K.Wst300= 48 psi < 3