mimo lqg and lqg ltr control design via 2 optimization
Handbook of Reliability, Availability, Maintainability and Safety in Engineering Design - Part 2 docx
Ngày tải lên :
02/07/2014, 10:20
... 433 4 .2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4. 12 4.13 4.14 4.15 4.16 xvi 4.17 4.18 4.19 4 .20 4 .21 4 .22 4 .23 4 .24 4 .25 4 .26 4 .27 4 .28 4 .29 4.30 4.31 4. 32 4.33 4.34 4.35 4.36 ... 1 12 3 .22 3 .23 3 .24 3 .25 3 .26 xiii xiv 3 .27 3 .28 3 .29 3.30 3.31 3. 32 3.33 3.34 3.35 3.36 3.37 3.38 3.39 3.40 3.41 3. 42 3.43 3.44 3.45 3.46 3.47 3.48 3.49 3.50 3.51 3. 52 3.53 3.54 3.55 ... 6 02 Fault tree of dormant failure of a high-integrity protection system (HIPS; Andrews 1994) 620 xviii 5 .23 5 .24 5 .25 5 .26 5 .27 5 .28 5 .29 5.30 5.31 5.32...
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Friction and Lubrication in Mechanical Design Episode 2 Part 1 potx
Ngày tải lên :
05/08/2014, 09:20
... 37.8"C[N/(sec-m2)] 0.0 I55 0.03I8 0.1059 0 .24 07 p at 93.3" [N/(sec-m2)] s 0.00376 0.0057 0.00168 0.0 1975 -0.0 1 42 -0.0 I 72 -0. 022 -0. 025 Design of Fluid Film Bearings 23 5 25 to 72 C The results ... no 10 11 12 13 S 0. 025 0.0361 0. 025 1 0.036 0. 025 1 0.036 0 .2 198 0 .2 198 0 .2 198 0.031 I 0.031 0 .26 49 0 .26 49 P , x (106) SP, x (106) 3. 72 4 .25 7.44 8.50 14.88 17.00 3.54 7.09 14.17 2. 853 5.705 ... 0.0036 0.0104 0.0007 0.0 020 0.644 0.700 0.93 0.994 1 .27 4 1.414 0 .23 8 0.336 0.483 0.455 0.686 0.196 0. 322 0.63 0. 72 0.966 0.959 1 .25 3 1.365 0 .25 2 0.357 0.470 0. 42 0.679 0 .24 5 0.343 The test conditions...
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Friction and Lubrication in Mechanical Design Episode 2 Part 1 docx
Ngày tải lên :
05/08/2014, 09:20
... 37.8"C[N/(sec-m2)] 0.0 I55 0.03I8 0.1059 0 .24 07 p at 93.3" [N/(sec-m2)] s 0.00376 0.0057 0.00168 0.0 1975 -0.0 1 42 -0.0 I 72 -0. 022 -0. 025 Design of Fluid Film Bearings 23 5 25 to 72 C The results ... no 10 11 12 13 S 0. 025 0.0361 0. 025 1 0.036 0. 025 1 0.036 0 .2 198 0 .2 198 0 .2 198 0.031 I 0.031 0 .26 49 0 .26 49 P , x (106) SP, x (106) 3. 72 4 .25 7.44 8.50 14.88 17.00 3.54 7.09 14.17 2. 853 5.705 ... 0.0036 0.0104 0.0007 0.0 020 0.644 0.700 0.93 0.994 1 .27 4 1.414 0 .23 8 0.336 0.483 0.455 0.686 0.196 0. 322 0.63 0. 72 0.966 0.959 1 .25 3 1.365 0 .25 2 0.357 0.470 0. 42 0.679 0 .24 5 0.343 The test conditions...
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Friction and Lubrication in Mechanical Design Episode 2 Part 2 docx
Ngày tải lên :
05/08/2014, 09:20
... of 0.09Pa-s at 26 °C; the loads were 94,703, 189,406 ,28 4,109, and 378,8 12 N/m; the slide/roll ratios were 0,0.08,0.154,0 .22 2,0.345; rolling speeds varied the from 0.3 to 2. 76 m/s, and the sliding ... Steel Copper Chromium Tin 10W30 0. 42 0.38 0.17 0 .20 Material Properties 43 40 94 67 0.145 7800 8930 7135 728 0 888 473 386 450 22 2 1880 20 3.4 103 25 0 46 - Chapter 26 6 The experimental results cover ... 0.05 c 26 0.04 c 0.03 0.5 oq0.03 1.0 1.5 2. 0 2. 5 3.0 3.5 ~ LL+ +- 0.00 O.O1 0.5 40 w (WmXlO', (8) 1.0 2. 0 1.5 2. 5 3.0 3.5 w (Nlm x1oq @) 0.03 - -W cW +W - 0. 02 = 94703 Nlm = 198408 Nln = 28 4109...
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Friction and Lubrication in Mechanical Design Episode 2 Part 2 doc
Ngày tải lên :
05/08/2014, 09:20
... of 0.09Pa-s at 26 °C; the loads were 94,703, 189,406 ,28 4,109, and 378,8 12 N/m; the slide/roll ratios were 0,0.08,0.154,0 .22 2,0.345; rolling speeds varied the from 0.3 to 2. 76 m/s, and the sliding ... Steel Copper Chromium Tin 10W30 0. 42 0.38 0.17 0 .20 Material Properties 43 40 94 67 0.145 7800 8930 7135 728 0 888 473 386 450 22 2 1880 20 3.4 103 25 0 46 - Chapter 26 6 The experimental results cover ... 0.05 c 26 0.04 c 0.03 0.5 oq0.03 1.0 1.5 2. 0 2. 5 3.0 3.5 ~ LL+ +- 0.00 O.O1 0.5 40 w (WmXlO', (8) 1.0 2. 0 1.5 2. 5 3.0 3.5 w (Nlm x1oq @) 0.03 - -W cW +W - 0. 02 = 94703 Nlm = 198408 Nln = 28 4109...
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Friction and Lubrication in Mechanical Design Episode 2 Part 3 docx
Ngày tải lên :
05/08/2014, 09:20
... Chapter 19 20 21 22 23 7.9 stokes), substitute ml, T l and m2, T2 into Eq (7. 32) , respectively, and solve these two linear equations simultaneously to get a and (If lubricant is SAE 10, SAE 20 , SAE ... Calculate D2 by using Eq (7 .29 ) where K, c, p, U are substituted by K , c2,p2, E2 Use Eq (7.30) to calculate p2 where h, = hr2, D = 02 17 Calculate AT, by using Eq (7.31) where /? = (PI P2) /2 18 Use ... (7. 32) to calculate a and for theparticular lubricant as follows Suppose that the viscosity is rnl at T1 and m2 at T2 (where T1 and T2 are absolute temperature, say, K = 27 3.16 "C, ml and m2 are...
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Friction and Lubrication in Mechanical Design Episode 2 Part 4 ppsx
Ngày tải lên :
05/08/2014, 09:20
... ~ ) 27 0 29 6 22 4 25 2 27 0 27 0 22 4 29 6 58 63 40 50 58 58 40 63 I99 468 27 0 397 359 160 I80 384 27 6 24 2 26 0 22 0 21 6 746 22 0 33 106 58 90 80 27 32 82. 5 65 47 55 40 40 150 40 22 6 40 26 2 340 27 0 29 6 ... 0 .20 32 Dry A B 0 .20 0.54 0.54 Dry A B 0 .20 0.54 0 .20 440 c 521 00 vs steel 1045 Dry A B 0 .20 0.45 0 .20 1060 Dry A B 0 .20 0 .20 0 .20 4140 LL Dry A B 0 .20 0 .20 0 .20 521 00 Dry A B 0 .20 0 .20 0 .20 ... Dry A B 3 02 0.54 A Dry 0.54 Dry A B 440 c 0 .20 0 .20 B 32 0 .20 0 .20 0.54 Material G p Dry 0 .20 0 .20 0 .20 0 .20 0 .20 0.66 A 0 .20 0. 12 0.73 0 .22 B Dry Brass vs steel 1045 Oila 0 .20 0 .20 0 .24 0.80 A...
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Friction and Lubrication in Mechanical Design Episode 2 Part 4 ppsx
Ngày tải lên :
05/08/2014, 09:20
... ~ ) 27 0 29 6 22 4 25 2 27 0 27 0 22 4 29 6 58 63 40 50 58 58 40 63 I99 468 27 0 397 359 160 I80 384 27 6 24 2 26 0 22 0 21 6 746 22 0 33 106 58 90 80 27 32 82. 5 65 47 55 40 40 150 40 22 6 40 26 2 340 27 0 29 6 ... 0 .20 32 Dry A B 0 .20 0.54 0.54 Dry A B 0 .20 0.54 0 .20 440 c 521 00 vs steel 1045 Dry A B 0 .20 0.45 0 .20 1060 Dry A B 0 .20 0 .20 0 .20 4140 LL Dry A B 0 .20 0 .20 0 .20 521 00 Dry A B 0 .20 0 .20 0 .20 ... Dry A B 3 02 0.54 A Dry 0.54 Dry A B 440 c 0 .20 0 .20 B 32 0 .20 0 .20 0.54 Material G p Dry 0 .20 0 .20 0 .20 0 .20 0 .20 0.66 A 0 .20 0. 12 0.73 0 .22 B Dry Brass vs steel 1045 Oila 0 .20 0 .20 0 .24 0.80 A...
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Friction and Lubrication in Mechanical Design Episode 2 Part 5 doc
Ngày tải lên :
05/08/2014, 09:20
... v) 20 0 4oo 120 0 1400- 1600- 1800- 20 00- w (3 22 00 24 00 - 26 00 - 28 00 - 3000 Figure 9 .2 9.1.5 Design chart for gear lubrication Lubrication Factor for Surface Durability Most gear rating and design ... Technology,” NASA, SP-3 18, 1973 27 Almen, J O., “Lubricants and False Brinelling of Ball and Roller Bearings,” Mech Eng., 1937, Vol 59, pp 415 422 28 Temlinson, G A., Thorpe, P L., and Gough, J H., “An Investigation ... Protection and Passivity, E Arnold, London, England, 1946 22 Avery, H S., Surface Protection Against Wear and Corrosion, American Society for Metals, 1954, Chapter 23 Larsen, R G., and Perry,...
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Friction and Lubrication in Mechanical Design Episode 2 Part 6 pot
Ngày tải lên :
05/08/2014, 09:20
... (2) (2) I Uniform thickness and Uniform thickness and Uniform thickness and sharing Uniform thickness and sharing (2) = 0 .25 external loading internal loading equal load = 0.50 73,456.4 27 2 ,28 3.7 ... Uniform thickness and equal load sharing Uniform thickness and optimal load sharing )( : = 0 .25 (2) = 0.50 ):( = 0.75 448.9 1345.8 465 .2 787.9 0.75 469.8 724 .3 395 .2 543.8 338.9 29 5.9 3 02. 8 36 Case ... material Outer and inner race material Surface finish for rollers and races 4.3305 in 1.96 82 in 1.06 in 3.719 in 2. 5658 in 2. 5637 in 2. 5 620 in 0.5766 in 0.659 in 12 0.1719 in 0.3 0.0 021 in 0.0038...
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Friction and Lubrication in Mechanical Design Episode 2 Part 6 doc
Ngày tải lên :
05/08/2014, 09:20
... (2) (2) I Uniform thickness and Uniform thickness and Uniform thickness and sharing Uniform thickness and sharing (2) = 0 .25 external loading internal loading equal load = 0.50 73,456.4 27 2 ,28 3.7 ... Uniform thickness and equal load sharing Uniform thickness and optimal load sharing )( : = 0 .25 (2) = 0.50 ):( = 0.75 448.9 1345.8 465 .2 787.9 0.75 469.8 724 .3 395 .2 543.8 338.9 29 5.9 3 02. 8 36 Case ... material Outer and inner race material Surface finish for rollers and races 4.3305 in 1.96 82 in 1.06 in 3.719 in 2. 5658 in 2. 5637 in 2. 5 620 in 0.5766 in 0.659 in 12 0.1719 in 0.3 0.0 021 in 0.0038...
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Friction and Lubrication in Mechanical Design Episode 2 Part 7 pdf
Ngày tải lên :
05/08/2014, 09:20
... specific heat 1. 128 Qfi dz (9.53) 396 Chapter Torqi~ed2.6'( -cos(!?pi'f"t))M I N-m I I -2W'(l-~s (2' pi~f't)) /2 lb-in 5- (v E I Frea=lSOHz 0 .25 'T g3 - 0 .2 ! E c Is I" I 0.3 4- - I c 0.15 2- I 0.1 1- ... method due to convection to the lubricant I 70.4 - I r I I I I I I Torque =22 .6*(1-cos (2' pi'Pt)) /2 N-m =20 0'( 1-cos (2' pi'f9)) /2 Ib-in f=100 HZ Figure 9.37 The heat flux at four key points under the ... fretting wear depth H , = Vickers hardness I I I I I I Torque =22 .6‘(1-cos (2* pi’ft)) /2 N-m =20 0’(1 -ws(P*pi’ft))l2 Ib-in 0 Figure 9.41 0.1 0 .2 0.3 0.4 0.5 0.6 Time (sec) Fretting wear prediction using...
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Friction and Lubrication in Mechanical Design Episode 2 Part 7 ppsx
Ngày tải lên :
05/08/2014, 09:20
... specific heat 1. 128 Qfi dz (9.53) 396 Chapter Torqi~ed2.6'( -cos(!?pi'f"t))M I N-m I I -2W'(l-~s (2' pi~f't)) /2 lb-in 5- (v E I Frea=lSOHz 0 .25 'T g3 - 0 .2 ! E c Is I" I 0.3 4- - I c 0.15 2- I 0.1 1- ... method due to convection to the lubricant I 70.4 - I r I I I I I I Torque =22 .6*(1-cos (2' pi'Pt)) /2 N-m =20 0'( 1-cos (2' pi'f9)) /2 Ib-in f=100 HZ Figure 9.37 The heat flux at four key points under the ... fretting wear depth H , = Vickers hardness I I I I I I Torque =22 .6‘(1-cos (2* pi’ft)) /2 N-m =20 0’(1 -ws(P*pi’ft))l2 Ib-in 0 Figure 9.41 0.1 0 .2 0.3 0.4 0.5 0.6 Time (sec) Fretting wear prediction using...
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Friction and Lubrication in Mechanical Design Episode 2 Part 8 docx
Ngày tải lên :
05/08/2014, 09:20
... Vol 2, pp 22 2 -23 5 110 Jaeger, J C., “Moving Sources of Heat and the Temperature at Sliding Contacts,” Proc Roy Soc (N.S.W.), 19 42, Vol 56, p 20 3 1 Meng, H C., and Ludema, K C., “Wear Models and ... Eng Res., 1960, Vol 5 (2) , pp 22 6 -23 2 78 Verma, B., “Oscillating Soil Tools - A Review,” Trans ASAE, 1971, pp 1107-1 115 79 Choa, S., and Chanceller, W., “Optimum Design and Operation Parameters ... f0.05 Increase to 0.45f0.05 0 .20 f0. 02 Decrease to Increase to 0.15f0. 02 0.75f0.05 Decrease to Increase to 0 .20 f0. 02 0.55f0.04 R-N2' R-Ozc - 0.6 - - 1.0 - R-( 02/ N2)d 1.6 1.3 - 0.8 3.8 - 0.5 -...
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Friction and Lubrication in Mechanical Design Episode 2 Part 8 ppsx
Ngày tải lên :
05/08/2014, 09:20
... Vol 2, pp 22 2 -23 5 110 Jaeger, J C., “Moving Sources of Heat and the Temperature at Sliding Contacts,” Proc Roy Soc (N.S.W.), 19 42, Vol 56, p 20 3 1 Meng, H C., and Ludema, K C., “Wear Models and ... Eng Res., 1960, Vol 5 (2) , pp 22 6 -23 2 78 Verma, B., “Oscillating Soil Tools - A Review,” Trans ASAE, 1971, pp 1107-1 115 79 Choa, S., and Chanceller, W., “Optimum Design and Operation Parameters ... f0.05 Increase to 0.45f0.05 0 .20 f0. 02 Decrease to Increase to 0.15f0. 02 0.75f0.05 Decrease to Increase to 0 .20 f0. 02 0.55f0.04 R-N2' R-Ozc - 0.6 - - 1.0 - R-( 02/ N2)d 1.6 1.3 - 0.8 3.8 - 0.5 -...
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Friction and Lubrication in Mechanical Design Episode 2 Part 9 pps
Ngày tải lên :
05/08/2014, 09:20
... 1986, Vol 52( 481), pp 24 63 -24 71 Lyon, R., “Noise Reduction and Machine Diagnostic and Educational Challenge,” Noise Vibr Control Wldwide, Sept 1985, Vol 16(8), pp 22 1 -22 4 Fielding, B., and Skorecki, ... and different gear ratios was determined Figures 1.7 and 1.8 present the effect of change of load for surface contact stresses 68.9, 689, 1378, and 1 722 .5MPa ( 10,000, 100,000, 20 0,000, and 25 0,OOOpsi) ... in ( 127 mm) center distance was calculated in both dry and lubricated regimes for different design and operating parameters The teeth are standard with a normal pressure angle equal to 25 " and...
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Friction and Lubrication in Mechanical Design Episode 2 Part 9 pptx
Ngày tải lên :
05/08/2014, 09:20
... 1986, Vol 52( 481), pp 24 63 -24 71 Lyon, R., “Noise Reduction and Machine Diagnostic and Educational Challenge,” Noise Vibr Control Wldwide, Sept 1985, Vol 16(8), pp 22 1 -22 4 Fielding, B., and Skorecki, ... and different gear ratios was determined Figures 1.7 and 1.8 present the effect of change of load for surface contact stresses 68.9, 689, 1378, and 1 722 .5MPa ( 10,000, 100,000, 20 0,000, and 25 0,OOOpsi) ... in ( 127 mm) center distance was calculated in both dry and lubricated regimes for different design and operating parameters The teeth are standard with a normal pressure angle equal to 25 " and...
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Friction and Lubrication in Mechanical Design Episode 2 Part 10 pps
Ngày tải lên :
05/08/2014, 09:20
... parameter Specific heat Sources: Refs 3-5 Value 1.0 104 > 1 .2 > 110 0.03 1.8 104 3. 52 1 .22 0 .2 1.1 x 10-6 20 3.0 x 108 0.853 Units kg/mm2 GPa GPa Dimensionless mls glcm3 GPa Dimensionless K-‘ W/(cm-K) ... a silicon nitride interface layer (U = 12. 7m/sec, I = 0 .25 mm.) Figure 2. 7 For i = to n: (1 2. 3) eL A= S S ( 2. 4) % For i = to n: ( I 2. 5) T = -4 Kd ( 2. 6) ... coating rate and low substrate temperature Parameters such as coating thickness, coating composition and substrate temperature are easily controlled 12. 2.3 Comparison between the CVD and PVD Processes...
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Friction and Lubrication in Mechanical Design Episode 2 Part 10 ppsx
Ngày tải lên :
05/08/2014, 09:20
... parameter Specific heat Sources: Refs 3-5 Value 1.0 104 > 1 .2 > 110 0.03 1.8 104 3. 52 1 .22 0 .2 1.1 x 10-6 20 3.0 x 108 0.853 Units kg/mm2 GPa GPa Dimensionless mls glcm3 GPa Dimensionless K-‘ W/(cm-K) ... a silicon nitride interface layer (U = 12. 7m/sec, I = 0 .25 mm.) Figure 2. 7 For i = to n: (1 2. 3) eL A= S S ( 2. 4) % For i = to n: ( I 2. 5) T = -4 Kd ( 2. 6) ... coating rate and low substrate temperature Parameters such as coating thickness, coating composition and substrate temperature are easily controlled 12. 2.3 Comparison between the CVD and PVD Processes...
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