Design and Optimization of Thermal Systems Episode 2 Part 6 doc
Design and Optimization of Thermal Systems Episode 2 Part 6 doc
... chapters. 70 060 0500400300 5.1 5 .2 5.3 5.4 5.5 5 .6 5.7 5.8 5.9 6. 0 Total flow rate m(kg/s) Total flow rate m(kg/s) 6. 1 6 .2 6. 3 6. 4 6. 5 5.5 5 .6 5.7 5.8 5.9 6. 0 6. 1 6 .2 6. 3 6. 4 6. 5 6. 6 6. 7 6. 8 Flow rate ... 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 an...
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Design and Optimization of Thermal Systems Episode 2 Part 1 doc
... root. T 581.53 02 f (T ) 460 6.784180 T 63 1.7 920 f (T ) 1 066 .578 125 T 64 6.9347 f (T ) –77.774414 T 64 5.90 56 f (T ) 1 .22 265 6 T 64 5. 921 5 f (T ) 0.005859 T 64 5. 921 6 f (T ) 0.004883 Therefore, ... = T( 12) = T(13) = T(14)= T(15) = T( 16) = T(17) = T(18) = T(19) = T (20 ) = T (21 ) = T (22 ) = T (23 ) = T (24 )= T (25 ) = T ( 26 ) = T (27 ) = T (2...
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Design and Optimization of Thermal Systems Episode 2 Part 4 doc
... Modeling of the system 5. Simulation of the system 6. Evaluation of the design 7. Selection of an acceptable design Optimization of the design follows the determination of a domain of accept- able designs ... states. 304 Design and Optimization of Thermal Systems Library of Previous Designs Any industry involved with the design of systems and equip...
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Design and Optimization of Thermal Systems Episode 2 Part 5 docx
... 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 0 .6 Y 0.8 1.0 Acceptable Design of a Thermal System 323 applications and vary ... W/(m · K) and that for aluminum as 22 6 W/(m · K). Because of the small thick- ness of the board, the Biot number Bi is found to be...
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Design and Optimization of Thermal Systems Episode 2 Part 2 ppt
... 168 .69 990 168 .69 990 168 .69 990 168 .69 990 168 .69 990 168 .69 990 168 .69 980 168 .69 970 168 .69 950 168 .69 900 168 .69 830 168 .69 680 168 .69 410 168 .68 890 168 .67 910 168 .66 060 168 . 62 5 10 168 .5 567 0 168 . 421 90 168 .14510 167 . 527 70 165 .87840 158.9 563 0 105 .65 780NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: NH3: N...
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Design and Optimization of Thermal Systems Episode 2 Part 3 pdf
... 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 3 11 22 2 RRR , Then the analytical ... Example 4.8. 27 6 Design and Optimization of Thermal Systems where KFyz KGyz KF iii iii1 (, , ) (, , ) ` $$ $ TT TT T 1 2 TT T TT T iii i y K z K KG y...
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Design and Optimization of Thermal Systems Episode 2 Part 7 ppsx
... % 12 % 15 % 5 128 . 36 148.98 164 .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 27 1 . 26 4 92. 68 7 32. 81 1089 . 26 1971.55 25 348.13 734. 02 120 5 .69 1978.85 ... 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 30...
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Design and Optimization of Thermal Systems Episode 2 Part 8 pdf
... 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. 5 5.9 11 ... (6 .20 b). Then, i 0.1, n 2, and m 12, and we obtain S (1 36, 833. 62 ) (./)( ./) (./) 0 1 12 1 0 1 12 101 12 1 24 24 Đ â ă ả á Ã $6, 314.18 Therefor...
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Design and Optimization of Thermal Systems Episode 2 Part 9 pps
... rates of 4, 6, and 10%. Compare the results obtained with $150,000 Down payment $30,000 Salvage 1098 765 4 321 Years FIGURE P6 .24 440 Design and Optimization of Thermal Systems and H i (x 1 , x 2 , ... present worth of buying $300,000 123 4 567 8910 Years FIGURE P6 .22 $20 0,000 123 4 56 Years Final payment FIGURE P6 .23 Problem Formulation for Optimizatio...
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Design and Optimization of Thermal Systems Episode 2 Part 10 ppsx
... 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. Obtain a best t assuming a variation of the form A sin (2 T /24 ) B cos (2 T /24 ) ... used to develop a trade-off curve in optimization. 470 Design and Optimization of Thermal Systems 7. 12. For the thermal systems considered in Example 5.5 a...
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