BASIC COASTAL ENGINEERING Part 2 pptx
... Inserting the pressure and the particle velocity from Eq. (2. 21) leads to (Longuet-Higgins and Stewart, 1964) S xx ¼ rgH 2 8 1 2 þ 2kd sinh 2kd ¼ EE 2n À 1 2 (2: 47) For a wave traveling in ... wave transformation. 26 / Basic Coastal Engineering 22 . A wave having a height of 2. 4 m and a period of 8 s in deep water is propagating toward the shore without refracting....
Ngày tải lên: 09/08/2014, 11:20
BASIC COASTAL ENGINEERING Part 4 pptx
... 0.13 0 .20 0. 52 1.13 1.06 0.98 1.00 u ¼ 135 1 /2 0 .24 0 .24 0 .25 0 .26 0 .28 0. 32 0.36 0. 42 0. 52 0.63 0.76 0.90 1.00 1 0.18 0.17 0.18 0.19 0 .21 0 .23 0 .28 0.34 0.44 0.59 0.78 0.95 1.00 2 0. 12 0. 12 0.13 ... 0.13 0 .22 0.54 1.10 1.00 u ¼ 165 1 /2 0 .23 0 .23 0 .23 0 .24 0 .26 0 .28 0.31 0.35 0.41 0.50 0.63 0.79 1.00 1 0.16 0.16 0.17 0.17 0.19 0 .20 0 .23 0 .26 0. 32 0.4...
Ngày tải lên: 09/08/2014, 11:20
BASIC COASTAL ENGINEERING Part 9 pptx
... (mm) Weight Retained (grams) 2. 000 0 1.414 0 1.000 0.3 0.707 1.7 0.500 6 .2 0.353 27 .8 0 .25 0 24 .1 0.177 17.7 0. 125 15.3 0.088 5.0 0.0 62 1.9 Coastal Zone Processes / 28 5 9.1 Field Investigations For ... Long- Term Rise in Water Level,’’ Coastal Engineering Technical Aid 79 2, U.S. Army Coastal Engineering Research Center, Ft. Belvoir, VA. Wentworth, C.K. (1 922 ), ‘‘A S...
Ngày tải lên: 09/08/2014, 11:20
... mạch kênh. Switching Engineering Page 2 Nội dung ! Giới thiệu. ! Chuyển mạch thời gian T. ! Chuyển mạch không gian. ! Ghép các cấp chuyển mạch. Switching Engineering Page 26 Chuyển mạch khung ! ... chuyển đổi nội dung giữa các khe thời gian ngõ vào và ngõ ra. Hình 2- 1 Chuyển mạch T và chuyển mạch S Switching Engineering Page 23 Chuyển mạch tin ! Thời gian trễ: T d =t nhận +t xử...
Ngày tải lên: 27/07/2014, 12:20
BASIC COASTAL ENGINEERING Part 1 doc
... (kd) ¼ 1 kd (2: 28) sinh k(d þ z) sinh kd ¼ 1 þ z d (2: 29) Deep Shallow SWL L /2 Figure 2. 2. Deep and shallow water surface proWles and particle orbits. 20 / Basic Coastal Engineering BASIC COASTAL ENGINEERING Third ... the particle acceleration. This yields a x ¼ 2p 2 H T 2 cosh k(d þ z) sinh kd ! sin (kx Àst) (2: 23) for the horizontal component and a z ¼À 2p 2 H T...
Ngày tải lên: 09/08/2014, 11:20
BASIC COASTAL ENGINEERING Part 3 doc
... Waves / 63 z ¼À H 2 cosh k(d þz) sinh kd sin (kx À st) þ pH 2 8L sinh 2 (kd) 1 À 3 cosh 2k(d þz) 2 sinh 2 (kd) sin 2( kx À st) þ pH 2 4L cosh 2k (d þz) sinh 2 (kd) st (3:11) « ¼ H 2 sinh k(d þz) sinh ... (kx À st) þ 3p 3 H 2 T 2 L cosh 2k(d þ z) sinh 4 kd sin 2 (kx À s t) (3:9) a z ¼À 2p 2 H T 2 sinh k(d þz) sinh kd cos (kx À s t) À 3p 3 H 2 T 2 L sinh 2k(d þ...
Ngày tải lên: 09/08/2014, 11:20
BASIC COASTAL ENGINEERING Part 5 pdf
... 34400 0.44 22 .69 24 .79 18340 0 .21 25 . 32 26.51 6 820 0.07 26 .97 32. 00 13 820 0. 12 32. 34 45. 72 229 30 0.14 45. 92 64.00 36690 0.17 64.03 82. 29 9170 0.03 82. 29 190 320 2. 22 Thus, the total wind-induced ... Depth (m) Distance (m) 7.31 27 50 10.05 4590 13. 72 229 30 14. 62 45860 20 . 12 48150 23 .97 825 50 25 .60 100890 27 .43 107710 36.57 121 530 54.86 144460 73.1...
Ngày tải lên: 09/08/2014, 11:20
BASIC COASTAL ENGINEERING Part 6 doc
... meters): 1.05 1. 92 3. 72 2.04 2. 54 1.47 3.50 0.96 2. 54 1.36 1.15 3.05 1.80 1.70 2. 05 2. 07 2. 27 1.39 2. 39 2. 51 0. 82 1.86 2. 07 3.07 1.66 3.10 2. 94 3.00 1.39 3.11 1.39 1.53 2. 37 1.16 2. 56 2. 84 4.53 3.31 ... (6.40) to (6. 42) yield H mo ¼ 0:0016 9:81 (20 ,000) (46:6) 2 0:5 (46:6) 2 9:81 ¼ 3:4m T p ¼ 0 :28 6 9:81 (20 ,000) (46:6) 2 0:33 46:6 9:81 ¼ 6:1s...
Ngày tải lên: 09/08/2014, 11:20
BASIC COASTAL ENGINEERING Part 7 ppsx
... ¼ Z h Àd C d 2 rDu 2 þ C m r pD 2 4 @u @t (d þ z)dz Integration, with the same assumptions yields M ¼ C d 8 rgDH 2 n cos 2 ( À st)d 1 2 þ 1 2n þ 1 2 þ 1 Àcosh 2kd 2kd sinh 2kd ! þ C m 8 rgpD 2 Hd ... Nonbreaking Quarrystone—smooth rounded 1 .2 2.4 1.1 1.9 Quarrystone—rough angular 2. 0 4.0 1.6 2. 8 Riprap 2. 2 2. 5 — — Dolos 15.8 31.8 8.0 16.0 Tetrapod 7.0 8.0 4.5 5...
Ngày tải lên: 09/08/2014, 11:20
BASIC COASTAL ENGINEERING Part 8 docx
... b)(a 1 þ a 2 cos 2 b)gH max p 2 ¼ a 3 p 1 a 1 ¼ 0:6 þ0:5 2 kd à sinh 2kd à ! a 2 ¼ d b À d 3d b H max d 2 "# or 2d H max ! a 3 ¼ 1 À d 0 d à 1 À 1 cosh kd à ! where, in the term a 2 the ... 40,040 N=m 2 Thus, the total force per unit length is 40; 040 2 (3:5 þ0:95 þ0:15)(1) ¼ 92, 0 92 N=m This force acts at a distance of 0:5 þ (3:5 þ0:95 þ:015) 3 ¼ 2: 03 m 23 0...
Ngày tải lên: 09/08/2014, 11:20