(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 6 pps

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Ngày tải lên: 05/08/2014, 12:20

... 16 .14 4 10 1 11 .584 11 .5 36 11 . 264 15 6 14 .11 5 14 .077 14 . 013 211 16 .10 0 16 .13 3 16 .14 4 10 2 11 .584 11 .5 36 11 . 264 15 7 14 .11 5 14 .077 14 . 317 212 16 .6 21 16 .13 3 16 .14 4 10 3 11 .584 11 .5 36 12 . 0 16 15 8 14 .11 5 ... 16 .2 76 10 6 11 .584 11 . 817 12 . 0 16 16 1 14 .11 5 14 .229 14 . 317 2 16 16 .6 21 16. 395 16 ....

Ngày tải lên: 05/08/2014, 12:20

... 0 .19 40 1. 91 0 .68 41 0 .65 33 2. 41 1.3742 1. 65 59 1. 42 0.2 811 0 .19 96 1. 92 0 .69 49 0 .66 71 2.42 1. 3 914 1. 68 36 1. 43 0.28 71 0.2053 1. 93 0.7058 0 .6 811 2.43 1. 4087 1. 711 6 1. 44 0.29 31 0. 211 1 1. 94 0. 7 16 8 0 .69 53 ... 594 810 10 .5 11 3 .65 0 5 96. 660 35 4209.24 7 36 61 . 8 59.5 2 068 0.0 61 5 230 11 13 0 .6 71 718 .68 8 35.5 4392.23 77...

Ngày tải lên: 05/08/2014, 12:20

... 1. 88382 1. 964 36 9 1. 390 76 1. 47437 1. 55753 1. 64 0 21 1.72240 1. 80405 1. 88 5 16 1. 965 70 10 1. 39 2 16 1. 475 76 1. 558 91 1 .6 415 9 1. 723 76 1. 805 41 1.8 86 51 1. 967 04 11 1. 393 56 1. 47 715 1. 560 29 1. 64 2 96 1. 72 513 1. 8 067 7 ... 1. 568 58 1. 6 512 0 1. 733 31 1. 814 90 1. 89594 1. 9 763 9 18 1. 40333 1. 4 868 7 1. 569 9...

Ngày tải lên: 05/08/2014, 12:20

... ¼ mn P R C  2 ¼ D L  2 1 Àe 2 Þ 2 pe½p 2 1 Àe 2 Þ 16 e 2  1= 2 From the right-hand side of the equation, S ¼ð2Þ 2 1 À 0:8 2 Þ 2 p0:8½p 2 1 À0:8 2 Þ 16  0:8 2  1= 2 ¼ 0: 36 2 p  0:8  3: 71 ¼ 0: 013 9 b. Minimum ... yield f ðeÞ¼ 4WC 2 mUL 3 ¼ 4W 10 À3  D=2Þ 2 mpnDðD=4Þ 3 ¼ 4  16  4900  10 6 0: 018 5p  63 :3ð0:075Þ 2 ¼ 15 :15 According to the curve of f ðeÞ v...

Ngày tải lên: 21/07/2014, 17:20

... Â7000  1 3 h ¼ 0: 266 mm b. The maximum pressure at the center is p max ¼ 3mV h 3 R 2 ¼ 3ð5:5 Â 10 À2 N-s=m 2 Þ 10 msÞ 15 Â 10 À3 mÞ 2 ð0: 266 10 À3 mÞ 3 ¼ 1: 97 Â 10 7 Pa ¼ 19 :7 MPa Problems 5 -1 Two ... ¼ UrC m ¼ 26 :18 Â 0:995 Â25 10 6 2:08 10 À5 ¼ 31: 3 ðlaminar flowÞ c. The Reynolds number for water as lubricant is calculated as follows: Re ¼ UrC m ¼ 26 :18 Â 978 Â...

Ngày tải lên: 21/07/2014, 17:20

... relative motion is given by @ @x h 3 m @p @x  þ @ @z h 3 m @p @z  6 U 1 À U 2 Þ @h @x þ 6 @ @x ðU 1 þ U 2 Þ 12 ðV 2 À V 1 Þ ð5 -11 Þ The velocity components of the two surfaces that form the film ... ¼ ð " xx 1 " xx 2 À " xx 2 0 1 þ " xx 2 Þ 3 d " xx ð4- 41 For a numerical integration of the pressure wave according to Eq. (4- 41) , the boundary "...

Ngày tải lên: 21/07/2014, 17:20

... the results of the derivations are: T f ¼ p 2 m R 4 h 0 1 À R 4 0 R 4  o ðP2-1aÞ _ EE f ¼ p 2 m R 4 h 0 1 À R 4 0 R 4  o 2 ðP2-1bÞ 2-1b A hydrostatic circular pad as shown in Fig. 2-5 operates ... is between 15 0 and 290, while the viscosity index of commercial mineral oils ranges from 90 to 14 0. In comparison, the viscosity index of commercial polyol esters ranges from 12 0 to 1...

Ngày tải lên: 21/07/2014, 17:20

... m 5kN 1. 2 m 1m 1m 1 kN 2 m2 m 1m 1m 2 kN 2 m2 m 1m 1m 2 kN 1 kN Truss Analysis: Force Method, Part I by S. T. Mau 49 This particular cut separates the truss into two parts. We shall use the left part ... –F 1 = 0, F 1 = 6 kN. 2 1 2 3 6 kN 4 5 6 R y1 R x1 1 3 4 5 R x5 R y5 F 5 3 4 5 6 kN 3 F 6 3 4 5 2 F 5 F 4 F 1 3 5 4 3 4 5 Truss Analysis: Force Method, Part I b...

Ngày tải lên: 05/08/2014, 09:20

... –F 1 = 0, F 1 = 6 kN. 2 1 2 3 6 kN 4 5 6 R y1 R x1 1 3 4 5 R x5 R y5 F 5 3 4 5 6 kN 3 F 6 3 4 5 2 F 5 F 4 F 1 3 5 4 3 4 5 Truss Analysis: Force Method, Part I by S. T. Mau 49 This particular cut ... chord member forces. ΣM E =0, (12 ) 8 – (3) 16 – (8) F GJ =0 F GJ = 6 kN. ΣM G =0, (12 ) 8 – (3) 16 + (8) F EH =0 F EH = 6 kN. B D 8 kN C A F EH E H G F GJ J 8m 6m...

Ngày tải lên: 05/08/2014, 09:20