(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 9 pot

... hour) (inches) (inches) 12 × 6 0.40 60 31. 5 4.0 15 × 6 0. 49 73 41. 5 5.0 18 × 6 0.56 84 5.0 6.0 24 × 8 1. 16 17 4 10 .0 30 × 10 1. 60 240 14 .0 36 × 12 2.40 360 16 .0 Control Valves 19 3 Air introduced through this ... valve. When Type 2, 3, and 6 (see Figure 9 .15 ) are used, the upstream side of the valve must 19 2 Control Valves #1 #1 #1 #2 #3A #2 #3B #2 #3C Figure...

Ngày tải lên: 05/08/2014, 11:20

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Ngày tải lên: 05/08/2014, 11:20

... 11 9. 976 12 0.000 −0.002 11 9. 934 12 0.000 −0.044 11 9. 8 69 12 0.000 −0 .10 9 Min 11 9. 975 11 9. 978 −0.025 11 9. 955 11 9. 978 −0.045 11 9. 9 41 1 19 .97 8 −0.0 59 11 9. 899 11 9. 978 −0 .10 1 11 9. 834 11 9. 978 −0 .16 6 16 0 Max 16 0. 012 ... −0.036 99 .8 89 10 0.000 −0.0 89 Min 99 .97 5 99 .97 8 −0.025 99 .95 5 99 .97 8 −0.045 99 .9 41 99...

Ngày tải lên: 05/08/2014, 12:20

... 3,600 1, 750 1, 000 200 10 8,700 12 ,000 25,000 44,000 220,000 20 5,500 8,000 17 ,000 30,000 15 0,000 30 4,000 6,000 13 ,000 24,000 12 7,000 40 2,800 4,500 11 ,000 20,000 11 1,000 50 3,500 9, 300 18 ,000 97 ,000 60 ... 2,600 8,000 16 ,000 88,000 70 6,700 14 ,000 81, 000 80 5,700 12 ,000 75,000 90 4,800 11 ,000 70,000 10 0 4,000 10 ,000 66,000 Source: Marks’ Standard Handbo...

Ngày tải lên: 05/08/2014, 12:20

... (17 -12 ) and (17 -13 ) is F cosðf ÀpÞ¼kðeÞðe Àe tr ÞD À0:5J 12 e j U j 17 -15 Þ F sinðf ÀpÞ¼0:5J 11 e U 17 -16 Þ The integrals J 11 and J 12 are defined in Eqs. (7 -13 ). Equations (17 -15 ) and (17 - 16 ) ... force in Eqs. (17 -18 ) and (17 - 19 ) yields FðtÞcosðf À pÞ¼kðe Àe tr ÞD À 0:5eJ 12 jUjþJ 12 e _ ff þ J 22 _ ee þ m € ee Àm _ ff 2 17 -23Þ FðtÞsinðf À pÞ¼0:5eJ 11 U À J...

Ngày tải lên: 21/07/2014, 17:20

... 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 ... bearings. J 11 ¼ ð p 0 sin 2 y 1 þ e cos yÞ 3 dy ¼ p 2 1 Àe 2 Þ 3=2 ð7 -13 aÞ J 12 ¼ ð p 0 sin ycos y 1 þ e cos yÞ 3 dy ¼ À2e 1 Àe 2 Þ 2 ð7 -13 bÞ J...

Ngày tải lên: 21/07/2014, 17:20

... plane-slider is as follows: W ¼ 6mULB 2 h 2 2 1 b  1  2 ln b  2ðb  1 b þ 1 ! ð4 - 19 Þ where b is the ratio of the maximum and minimum film thickness, h 2 =h 1 .A similar derivation can be followed ... pressure wave along the x direction (between x ¼ h 1 =a and x ¼ h 2 =a): p ¼ 6mU a 3 ðh 1  axÞðax h 2 Þ ðh 1 þ h 2 Þx 2 ð4 -17 Þ At the boundaries h ¼ h 1 and h ¼ h 2 ,...

Ngày tải lên: 21/07/2014, 17:20

... (k G ) 5 = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −− −− −− −− 8 .12 6 .92 56 .9 6 .92 .76 .92 .7 256 .98 .12 6 .9 6 .92 .76 .92 .7 Member 6(for the second truss only): (k G ) 6 = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −− −− −− −− 8 .12 6 .98 .12 6 .9 6 .92 .76 .92 .7 8 .12 6 .98 .12 6 .9 6 .92 .76 .92 .7 (4) ... ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ −−− −−− −−− −−− −−− −−− −−− −−− 80.3760 .90 0.25080 .12 60 .9...

Ngày tải lên: 05/08/2014, 09:20

... solutions. P 1 2 3 θ 2 ’ M 1 M 1 θ 11 θ 1 ’ θ 3 ’ M 1 θ 21 M 1 θ 31 M 1 M 2 θ 21 M 2 θ 22 M 2 θ 32 M 3 M 3 θ 31 M 3 θ 23 M 3 θ 33 θ 2 =0 θ 1 =0 θ 3 =0 P Beam and Frame Analysis: Force Method, Part ... ∆ 2 ’ + R 1 δ 21 + R 2 δ 22 = 0 These two equations can be put in the following matrix form. L L w L ∆ 1 ’ ∆ 2 ’ L L w L R 1 δ 11 R 1 δ 21 R...

Ngày tải lên: 05/08/2014, 09:20

... Force Method, Part I by S. T. Mau 11 9 (11 ) (12 ) (13 ) (14 ) (15 ) (16 ) Problem 2. Frame problems. 5 m 5 m 10 kN 5 m 5 m 10 kN-m 5 m 5 m 10 kN 5 m 5 m 10 kN-m 5 m 5 m 10 kN 5 m 5 m 10 kN-m ... kN -60 kN 75 kN-m 9 kN - 39 kN 15 kN M -75 kN-m 75 kN-m 48 kN/5m =9. 6 kN/m 36 kN/5m=7.2 kN/m 0 .94 m 5m (9) /( 39+ 9) =0 .94 m 4.22 kN-m M (x=0 .94 )= (9) (0 .94 )- (9. 6)(0 .94 )...

Ngày tải lên: 05/08/2014, 11:20