Design and Optimization of Thermal Systems Episode 2 Part 8 pdf
... 7 n 10 n 15 1 33.3 20 .0 14.3 10.0 5.0 2 44.5 32. 0 24 .5 18. 0 9.5 3 14 .8 19 .2 17.5 14.4 8. 6 4 7.4 11.5 12. 5 11.5 7.7 5 11.5 8. 9 9 .2 6.9 6 5 .8 8.9 7.4 6 .2 7 8. 9 6.6 5.9 8 4.5 6.6 5.9 9 6.5 5.9 10 ... (6 .20 b). Then, i 0.1, n 2, and m 12, and we obtain S (136 ,83 3. 62) (./)( ./) (./) 0 1 12 1 0 1 12 101 12 1 24 24 Đ â ă ả á Ã $6,314. 18 Therefore,...
Ngày tải lên: 06/08/2014, 15:20
... equations 7c 1 c 2 2 c 3 c 4 3.7 c 1 2 8c 2 3c 3 c 4 4.9 2c 1 2c 2 5c 3 c 4 2 8. 8 c 1 c 2 c 3 2 14c 4 18 .2 Numerical Modeling and Simulation 28 7 (c) Set up this system of equations ... T a , as d d HF Q T QQ QQ 1 12 1 1 2 ()(,) (4.44a) d d HH G Q T QQQQQ 2 22 3 2 1 1 2 ()(,) (4.44b) where H hA CV H hA CV H hA CV 1 11 111 2 22 222...
Ngày tải lên: 06/08/2014, 15:20
... = 114.65 78 109.7 9 28 105. 380 3 101.3 981 97. 82 6 3 94.6467 91 .84 34 89 .4 022 87 .3109 85 .5 588 84 .1371 83 .0 388 82 . 2 581 81 .7914 81 .6360 81 .7914 82 . 2 581 83 .0 388 84 .1371 85 .5 588 87 .3109 89 .4 022 91 .84 34 94.6467 97. 82 6 3 101.3 981 105. 380 3 109.7 9 28 114.65 78 T(1) ... 0.00001 114.65 72 109.7915 105.3 784 101.3957 97. 82 3 4 94.6433 91 .83...
Ngày tải lên: 06/08/2014, 15:20
Design and Optimization of Thermal Systems Episode 2 Part 2 ppt
... 1 68. 69990 1 68. 69990 1 68. 69990 1 68. 69990 1 68. 69990 1 68. 69990 1 68. 69 980 1 68. 69970 1 68. 69950 1 68. 69900 1 68. 6 983 0 1 68. 69 680 1 68. 69410 1 68. 688 90 1 68. 67910 1 68. 66060 1 68. 625 10 1 68. 55670 1 68. 421 90 1 68. 14510 167. 527 70 165 .87 840 1 58. 95630 105.65 780 NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3...
Ngày tải lên: 06/08/2014, 15:20
Design and Optimization of Thermal Systems Episode 2 Part 4 doc
... solidication. 3 12 Design and Optimization of Thermal Systems 900.0 720 .0540.0360.0 180 .00.0 0.00 0.03 0.06 Mold wall thickness d Mold material Iron T pour = 185 0 K h = 80 W/m 2 K 3 cm 2. 5 cm 2 cm Solid ... (s) 1.0 k m /k s 0.4 0 .2 300. 020 0.0100.00.0 0.00 0.03 0.06 0.09 Solid thickness (m) 0. 12 0.15 FIGURE 5 .8 Effect of the thermal conductivity of the ma...
Ngày tải lên: 06/08/2014, 15:20
Design and Optimization of Thermal Systems Episode 2 Part 5 docx
... components. 1.00 .80 .6 Y (b) 0.40 .20 .0 0.0 50.0 100.0 150.0 T – T a (°C) 20 0.0 25 0.0 Temperature limit Q = 10 × 20 0 W Q = 8 × 20 0 W Q = 5 × 20 0 W Q = 3 × 20 0 W 1.00 .80 .60.40 .2 0.0 0.0 40.0 80 .0 120 .0 T ... components. 0.0 0.0 50.0 100.0 150.0 20 0.0 25 0.0 Width of board (m) w = 0.15 w = 0 .20 w = 0 .25 w = 0.30 Temperature limit T – T a (°C) 0 .2 0.4...
Ngày tải lên: 06/08/2014, 15:20
Design and Optimization of Thermal Systems Episode 2 Part 6 doc
... losses and may be written as p V gz p V gz fL D V K V h 1 1 2 12 2 2 2 22 22 2 R R R R RR 33 22 (5.30) where the subscripts 1 and 2 refer to two locations in the ow, g is the magnitude of ... Therefore, Nu D 0. 023 (Re D ) 0 .8 (Pr) 0.4 0. 023 (29 ,80 0) 0 .8 (5 .83 ) 0.4 176 .8 since the Prandtl number Pr MC p /k 5 .83 for water. This gives h i as h...
Ngày tải lên: 06/08/2014, 15:20
Design and Optimization of Thermal Systems Episode 2 Part 7 ppsx
... 110. 52 110. 52 2 120 .0 121 .0 122 .04 122 .14 122 .14 5 150.0 161.05 164.53 164 .86 164 .87 10 20 0.0 25 9.37 27 0.70 27 1.79 27 1 .83 20 300.0 6 72. 75 7 32. 81 7 38. 70 7 38. 91 30 400.0 1744.94 1 983 .74 20 07.73 20 08. 55 ... 12 % 15 % 5 1 28 .36 1 48. 98 164.53 181 .67 21 0. 72 10 164.71 22 1.96 27 0.70 330.04 444. 02 15 21 1.37 330.69 445.39 599. 58 935.63 20 2...
Ngày tải lên: 06/08/2014, 15:20
Design and Optimization of Thermal Systems Episode 2 Part 9 pps
... formulation of the problem are 1. Determination of the design variables, x i where i 1, 2, 3, . , n 2. Selection and denition of the objective function, U 424 Design and Optimization of Thermal Systems year ... rates of 4, 6, and 10%. Compare the results obtained with $150,000 Down payment $30,000 Salvage 10 987 654 321 Years FIGURE P6 .24 440 Design and...
Ngày tải lên: 06/08/2014, 15:21
Design and Optimization of Thermal Systems Episode 2 Part 10 ppsx
... measured as a function of time T over a day. For T of 2, 3, 6, 8, 10, 15, 18, 22 , and 24 hours, T is obtained as 86 .5, 97.7, 1 02. 0, 101.7, 92. 5, 62. 3, 55.0, 67.5, and 80 .0nC, respectively. ... d/D and x 2 V 2 /V 1 . Then, the objective function may be written as UxxAFxxFxxA(,)=+ (,)+ (,)= 12 1 12 2 12 +++ 21 Bx x Cx x nm 12 (7.13) where F 1 and F 2...
Ngày tải lên: 06/08/2014, 15:21