fundamentals of hydrogen safety engineering 1

... 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

... 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

... from each of the three nodes NODE FBD Py2 (Fx2 )1+ (Fx2)2 Px2 (Fy2 )1+ (Fy2)2 (Fy2 )1 (Fy2)2 2 (Fy1 )1 (Fy3)2 (Fx1 )1 (Fx3)2 3 Py1 (Fx1 )1+ (Fx1)3 (Fx2)2 (Fx2 )1 Px1 (Fy1 )1+ (Fy1)3 (Fy1)3 (Fx1)3 Py3 (Fx3)2+ ... 93 94 10 1 10 2 10 9 11 8 Beam and Frame Analysis: Force Method, Part II Deflection of Beam and Frames Problem 12 1 12 1 13 2 iii Problem Problem 14 3 15 2 Beam an...

Ngày tải lên: 05/08/2014, 09:20

... from each of the three nodes NODE FBD Py2 (Fx2 )1+ (Fx2)2 Px2 (Fy2 )1+ (Fy2)2 (Fy2 )1 (Fy2)2 2 (Fy1 )1 (Fy3)2 (Fx1 )1 (Fx3)2 3 Py1 (Fx1 )1+ (Fx1)3 (Fx2)2 (Fx2 )1 Px1 (Fy1 )1+ (Fy1)3 (Fy1)3 (Fx1)3 Py3 (Fx3)2+ ... 93 94 10 1 10 2 10 9 11 8 Beam and Frame Analysis: Force Method, Part II Deflection of Beam and Frames Problem 12 1 12 1 13 2 iii Problem Problem 14 3 15 2 Beam an...

Ngày tải lên: 05/08/2014, 09:20

... 31 ⎪ ⎪ ⎢ ⎪ Fy ⎪ ⎢k 41 ⎩ ⎭ ⎣ 0 0 0 k 11 k 21 k 31 k 41 k 12 k 22 k 32 k 42 k13 k 23 k 33 k 43 0 0 0 k13 k 23 0 k 12 k 22 0 k 32 k 42 0 0 0 k 33 k 43 ⎤ ⎧u1 ⎫ ⎥ ⎪ v1 ⎪ ⎥ ⎪ ⎪ ⎪ ⎪ k14 ⎥ ⎪u ⎪ ⎥ ⎨ ⎬ k 24 ... equation: ⎡ K 11 ⎢K ⎢ 21 ⎢ K 31 ⎢ ⎢ K 41 ⎢ K 51 ⎢ ⎢ K 61 ⎣ K 12 K 22 K 32 K 42 K 52 K 62 K 13 K 23 K 33 K 43 K 53 K 63 K 14 K 24 K 34 K 44 K 54 K 64 K 15 K 25...

Ngày tải lên: 05/08/2014, 09:20

... (2-c) kN 1. 5m kN kN 3m 3m 3m 2m 4m 4m (3- a) (3- b) 1. 2 m kN 4kN 5kN kN 1. 6 m (4-a) 0.7 m 0.9 m 0.9 m 0.7 m 0.9 m 0.9 m 0.9 m 2m (3- c) 1. 2 m 1. 2 m 3m 3m (4-b) (4-c) kN kN 2m 2m kN 1m 1m 1m 1m 2m 2m ... Fy = 0, F5 (3/ 5) – F4 (3/ 5) –F1= 0, F1= –6 kN 38 Truss Analysis: Force Method, Part I by S T Mau Joint Σ Fx = 0, 4 F6(4/5) + F3(4/5) =0, F3 = kN Σ Fy = 0, F6 F6 (3/ 5) – F3 (3/...

Ngày tải lên: 05/08/2014, 09:20

... (2-c) kN 1. 5m kN kN 3m 3m 3m 2m 4m 4m (3- a) (3- b) 1. 2 m kN 4kN 5kN kN 1. 6 m (4-a) 0.7 m 0.9 m 0.9 m 0.7 m 0.9 m 0.9 m 0.9 m 2m (3- c) 1. 2 m 1. 2 m 3m 3m (4-b) (4-c) kN kN 2m 2m kN 1m 1m 1m 1m 2m 2m ... Fy = 0, F5 (3/ 5) – F4 (3/ 5) –F1= 0, F1= –6 kN 38 Truss Analysis: Force Method, Part I by S T Mau Joint Σ Fx = 0, 4 F6(4/5) + F3(4/5) =0, F3 = kN Σ Fy = 0, F6 F6 (3/ 5) – F3 (3/...

Ngày tải lên: 05/08/2014, 09:20

... Denoting the elongation of memebr as V3 We have (a) V3= α (∆T)L= 1. 2 (10 -5)/oC ( 14 oC) (6,000 mm)= 1mm (b) V3=1mm (c) V3= 16 kN (6 m)/[200 (10 6) kN/m2 (500) (10 -6)m2]=0.0 01 m= 1mm Next we need to find ... Example 11 only in the externally applied loads Use the force transfer matrix of Eq to find the solution 1. 0 kN 0.5 kN y x 4m 3 3m 3m Problem 4- 2 66 1. 0 kN Truss Analysi...

Ngày tải lên: 05/08/2014, 09:20

... Denoting the elongation of memebr as V3 We have (a) V3= α (∆T)L= 1. 2 (10 -5)/oC ( 14 oC) (6,000 mm)= 1mm (b) V3=1mm (c) V3= 16 kN (6 m)/[200 (10 6) kN/m2 (500) (10 -6)m2]=0.0 01 m= 1mm Next we need to find ... Example 11 only in the externally applied loads Use the force transfer matrix of Eq to find the solution 1. 0 kN 0.5 kN y x 4m 3 3m 3m Problem 4- 2 66 1. 0 kN Truss Analysi...

Ngày tải lên: 05/08/2014, 09:20

... -56 .56 40.00 -56 .56 -40.00 25, 000 25, 000 25, 000 17 ,680 25, 000 17 ,680 25, 000 17 ,680 25, 000 3.20 3.20 1. 60 -6.40 4.80 -3.20 1. 60 -3.20 -1. 60 0.00 -0. 71 0.00 0.00 -0. 71 1.00 -0. 71 0.00 -0. 71 -0.33 -0.33 ... -0.8 -0.032 0.026 0 .50 33,333 0. 0 15 -0.6 -0.009 -0. 018 0. 011 -0.83 20,000 -0.042 1. 0 -0.042 0. 050 0. 050 20,000 1. 0 0. 050 0. 050 (kN/m) Vι (mm) -0.33 25,...

Ngày tải lên: 05/08/2014, 09:20

... -56 .56 40.00 -56 .56 -40.00 25, 000 25, 000 25, 000 17 ,680 25, 000 17 ,680 25, 000 17 ,680 25, 000 3.20 3.20 1. 60 -6.40 4.80 -3.20 1. 60 -3.20 -1. 60 0.00 -0. 71 0.00 0.00 -0. 71 1.00 -0. 71 0.00 -0. 71 -0.33 -0.33 ... -0.8 -0.032 0.026 0 .50 33,333 0. 0 15 -0.6 -0.009 -0. 018 0. 011 -0.83 20,000 -0.042 1. 0 -0.042 0. 050 0. 050 20,000 1. 0 0. 050 0. 050 (kN/m) Vι (mm) -0.33 25,...

Ngày tải lên: 05/08/2014, 09:20

... M=2, N=3, C =1 R =1 R =1 Number of unknowns = 3M+ΣR =12 , Number of equations = 3N+ΣC= 11 Indeterminate to the 1st degree Number of unknowns = 3M+ΣR =10 , Number of equations = 3N+C= 10 Statically ... M=5, N =6, C=2 M=3, N=4, C =1 R=3 R=2 R=3 R=3 R=2 R =1 Number of unknowns = 3M+ΣR = 21, Number of equations = 3N+C = 20 Indeterminate to the 1st degree Number of unknowns 3M...

Ngày tải lên: 05/08/2014, 09:20

... M=2, N=3, C =1 R =1 R =1 Number of unknowns = 3M+ΣR =12 , Number of equations = 3N+ΣC= 11 Indeterminate to the 1st degree Number of unknowns = 3M+ΣR =10 , Number of equations = 3N+C= 10 Statically ... M=5, N =6, C=2 M=3, N=4, C =1 R=3 R=2 R=3 R=3 R=2 R =1 Number of unknowns = 3M+ΣR = 21, Number of equations = 3N+C = 20 Indeterminate to the 1st degree Number of unknowns 3M...

Ngày tải lên: 05/08/2014, 09:20