fundamentals of hydrogen safety engineering 2

... stoichiometric hydrogen- air mixtures (Babkin, 20 03); and Su0 is the laminar burning velocity at 29 8 K 4.0 3.8 3.6 3.4 3 .2 3.0 2. 8 2. 6 2. 4 2. 2 2. 0 1.8 1.6 1.4 1 .2 1.0 0.8 0.6 0.4 0 .2 0.0 8.0 7.6 7 .2 6.8 ... of hydrogen safety Part I 1.4 Hazards, risk, safety Part I 1.5 Hydrogen safety communication Part I 1.6 he subject and scope of hydrogen safety en...

Ngày tải lên: 09/03/2016, 10:20

... Hazards, risk, safety 13 1. 5 Hydrogen safety communication 15 1. 6 he subject and scope of hydrogen safety engineering 16 1. 7 he emerging profession of hydrogen safety engineering 17 1. 8 Knowledge ... bookboon.com Fundamentals of Hydrogen Safety Engineering I Contents Contents Introduction 1. 1 Why hydrogen? 1. 2 Public perception of hydrogen...

Ngày tải lên: 09/03/2016, 10:20

... 0. 625 0.375 0.5 0.5 Mab Mba Mbc Mcb Mcd 30 +15 +15 +7.50 −4.69 2. 81 2. 35 1. 41 +0. 71 +0.70 +0.36 −0 .22 −0 .14 −0 .11 −0.07 +0.04 +0.03 +0. 02 −0. 01 −0. 01 0.00 0.00 2. 46 −4. 92 +4. 92 +14 .27 +15 .73 ... FEM +2 2 DM +0.8 +1 .2 1 .2 COM +0.4 SUM +0.4 1 .2 +1 .2 +1 .2 (3) Post Moment-Distribution Operations The moment and deflection diagrams are shown below 0.4 −3.8...

Ngày tải lên: 05/08/2014, 09:20

... 0. 625 0.375 0.5 0.5 Mab Mba Mbc Mcb Mcd 30 +15 +15 +7.50 −4.69 2. 81 2. 35 1. 41 +0. 71 +0.70 +0.36 −0 .22 −0 .14 −0 .11 −0.07 +0.04 +0.03 +0. 02 −0. 01 −0. 01 0.00 0.00 2. 46 −4. 92 +4. 92 +14 .27 +15 .73 ... FEM +2 2 DM +0.8 +1 .2 1 .2 COM +0.4 SUM +0.4 1 .2 +1 .2 +1 .2 (3) Post Moment-Distribution Operations The moment and deflection diagrams are shown below 0.4 −3.8...

Ngày tải lên: 05/08/2014, 09:20

... a wL2 12 −(6−8a+3a2) aL (4-3a) aL w wL2 − 12 a wL2 12 wL2 12 L w − wL 20 wL2 12 L w − 5wL 96 5wL2 96 L M − b(2a-b)M a(2b-a)M aL bL Note: Positive moment acts clockwise 20 8 Beam and Frame Analysis: ... (1) (2) kN b a b a c 2m c 2EI EI 3m 2EI 2m kN-m kN kN/m EI 3m 2m 2m (3) 50 kN 50 kN a c b 2EI 2m d 2EI EI 2m 2m (4) 2m 4m (5) 50 kN 50 kN a c b 2EI 4m 2EI 4m EI c b 2EI 2EI d 4m a d 4m...

Ngày tải lên: 05/08/2014, 09:20

... a wL2 12 −(6−8a+3a2) aL (4-3a) aL w wL2 − 12 a wL2 12 wL2 12 L w − wL 20 wL2 12 L w − 5wL 96 5wL2 96 L M − b(2a-b)M a(2b-a)M aL bL Note: Positive moment acts clockwise 20 8 Beam and Frame Analysis: ... (1) (2) kN b a b a c 2m c 2EI EI 3m 2EI 2m kN-m kN kN/m EI 3m 2m 2m (3) 50 kN 50 kN a c b 2EI 2m d 2EI EI 2m 2m (4) 2m 4m (5) 50 kN 50 kN a c b 2EI 4m 2EI 4m EI c b 2EI 2EI d 4m a d 4m...

Ngày tải lên: 05/08/2014, 09:20

... (2) kN b a b a c 2EI EI 3m 2EI 2m 2m kN-m kN kN/m (3) EI 3m 2m 2m c (4) 50 kN 50 kN a a c b 2EI 2EI 2EI 2EI 4m EI 4m EI d d 4m c b 4m 4m 8m (5) 4m 8m (6) b c b c EI EI 2m 2m kN kN EI EI 2m 2m ... 1 . 23 kN c c 1.64 kN-m kN d 0.69 kN 0.69 kN 1 . 23 kN 1 . 23 kN 0. 82 kN-m c b a 2. 77 kN 1 . 23 kN 1 . 23 kN 1 . 23 kN 0.69 kN 1 . 23 kN 3. 36 kN-m 0.69 kN FBDs of each member and nodes b and...

Ngày tải lên: 05/08/2014, 09:20

... (2) kN b a b a c 2EI EI 3m 2EI 2m 2m kN-m kN kN/m (3) EI 3m 2m 2m c (4) 50 kN 50 kN a a c b 2EI 2EI 2EI 2EI 4m EI 4m EI d d 4m c b 4m 4m 8m (5) 4m 8m (6) b c b c EI EI 2m 2m kN kN EI EI 2m 2m ... 1 . 23 kN c c 1.64 kN-m kN d 0.69 kN 0.69 kN 1 . 23 kN 1 . 23 kN 0. 82 kN-m c b a 2. 77 kN 1 . 23 kN 1 . 23 kN 1 . 23 kN 0.69 kN 1 . 23 kN 3. 36 kN-m 0.69 kN FBDs of each member and nodes b and...

Ngày tải lên: 05/08/2014, 09:20

... K11 K 12 ⎢K K ⎢ 21 22 ⎢ K 31 K 32 ⎢ ⎢ K 41 K 42 ⎢ K 51 K 52 ⎢ ⎢ K 61 K 62 ⎢ 0 ⎢ ⎢ 0 ⎢ 0 ⎣ K13 K 23 K 33 K 43 K 14 K 24 K 34 K 44 K15 K 25 K 35 K 45 K16 K 26 K 36 K 46 0 K 47 0 K 48 K 53 K 54 K 55 ... 26 Member 1: 24 4 Beam and Frame Analysis: Displacement Method, Part II by S T Mau 22 ,500 ⎡ 11 ,25 0 ⎢ 1x10 ⎢ ⎢ 22 ,500 60,000 (kG)1= ⎢ - 22 ,500 ⎢- 11 ,25 0 ⎢ - 1...

Ngày tải lên: 05/08/2014, 09:20

... K11 K 12 ⎢K K ⎢ 21 22 ⎢ K 31 K 32 ⎢ ⎢ K 41 K 42 ⎢ K 51 K 52 ⎢ ⎢ K 61 K 62 ⎢ 0 ⎢ ⎢ 0 ⎢ 0 ⎣ K13 K 23 K 33 K 43 K 14 K 24 K 34 K 44 K15 K 25 K 35 K 45 K16 K 26 K 36 K 46 0 K 47 0 K 48 K 53 K 54 K 55 ... 26 Member 1: 24 4 Beam and Frame Analysis: Displacement Method, Part II by S T Mau 22 ,500 ⎡ 11 ,25 0 ⎢ 1x10 ⎢ ⎢ 22 ,500 60,000 (kG)1= ⎢ - 22 ,500 ⎢- 11 ,25 0 ⎢ - 1...

Ngày tải lên: 05/08/2014, 09:20

... location of the group load 2m kN 1kN 1m 1kN 1 . 25 0.41 FCJ 1 . 25 0. 6 25 7m 9m Placing the group load to maximize FCJ 26 7 Influence Lines by S T Mau (FCJ)max= − [2( 0. 6 25 )+1(0. 6 25 )(7/9)+1(0. 6 25 )(6/9)]= 2. 15 ... compression in member CJ (FCJ)max= −10[0 .5( 1. 82) (0. 6 25 )+0 .5( 9)(0. 6 25 )]= −33.8 kN (4) Distributed load of finite length 10 kN/m 6m 1 . 25 0.41 FCJ 1 . 25...

Ngày tải lên: 05/08/2014, 09:20

... location of the group load 2m kN 1kN 1m 1kN 1 . 25 0.41 FCJ 1 . 25 0. 6 25 7m 9m Placing the group load to maximize FCJ 26 7 Influence Lines by S T Mau (FCJ)max= − [2( 0. 6 25 )+1(0. 6 25 )(7/9)+1(0. 6 25 )(6/9)]= 2. 15 ... compression in member CJ (FCJ)max= −10[0 .5( 1. 82) (0. 6 25 )+0 .5( 9)(0. 6 25 )]= −33.8 kN (4) Distributed load of finite length 10 kN/m 6m 1 . 25 0.41 FCJ 1 . 25...

Ngày tải lên: 05/08/2014, 09:20

... 0.2L 0.6L Cab Cba Sab Sba 0 .69 1 0 .69 1 9.08EK 9.08EK 0.2L L /2 w P L /2 MFab L MFba -0.159PL MFab MFba -0.102wL2 0.159PL 0.102wL2 b a h h 0.8L Cab 0 .69 4 Cba Sab 0.475 0.2L Sba 4.49EK L /2 MFab 6. 57EK ... 0.4 (28 .8)= 11. 52 kN-m Mbc = 0 .6 (28 .8)= 17 .28 kN-m The carryover moments are: Mab = 0.5(11. 52) = 5. 76 kN-m Mcb = 0.5 (17 .28 )= 8 .64 kN-m The superposition of two solu...

Ngày tải lên: 05/08/2014, 09:20