Design and Optimization of Thermal Systems Episode 2 Part 1 doc

Design and Optimization of Thermal Systems Episode 2 Part 1 doc

Design and Optimization of Thermal Systems Episode 2 Part 1 doc

... = 11 4.6578 10 9.7 928 10 5.3803 10 1.39 81 97. 826 3 94.6467 91. 8434 89.4 022 87. 310 9 85.5588 84 .13 71 83.0388 82. 25 81 81. 7 914 81. 6360 81. 7 914 82. 25 81 83.0388 84 .13 71 85.5588 87. 310 9 89.4 022 91. 8434 94.6467 97. 826 3 10 1.39 81 105.3803 10 9.7 928 11 4.6578 T (1) ... = T (11 ) = T( 12 ) = T (13 ) = T (14 )= T (15 ) = T (16 ) = T (17 ) = T (18 ) = T (...

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Design and Optimization of Thermal Systems Episode 2 Part 4 doc

Design and Optimization of Thermal Systems Episode 2 Part 4 doc

... as Y |x 1 | |x 2 | |x 3 |orY |(x 1 d 1 )| |(x 2 d 2 )| |(x 3 d 3 )| (5.9) or as Y x 1 2 x 2 2 x 3 2 or Y (x 1 d 1 ) 2 (x 2 d 2 ) 2 (x 3 d 3 ) 2 (5 .10 ) A square root of the expressions ... s Evaporator rottling valve Condenser 30°C 25 °C Temperature, T 1 4 T(°C) P(kPa) Saturated vapor 25 25 18 8.7 30 15 1.5 15 1.5 11 67 .1 116 7 .1 Saturated...

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Design and Optimization of Thermal Systems Episode 2 Part 5 docx

Design and Optimization of Thermal Systems Episode 2 Part 5 docx

...   11 1 1 2 22 2 () ( ) ,, , , (5 .16 a) QUAT UA TT T T TT m io oi i    $ ()() ln[( ,, , , , 12 1 2 1 22 12 , , , )/( )] oo i TT (5 .16 b) where U is the overall heat transfer coefcient, and ... distance L 1 , so that the plastic cords are heated up to this distance, and then cooled to 15 010 0500 25 50 75 10 0 Temperature T(°C) 12 5 1. 0 1. 0 q s = 0.5...

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Design and Optimization of Thermal Systems Episode 2 Part 6 doc

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 ...  m and P to obtain the following two equations for  m 1 and  m 2 : Fm m H Cm m P Am(, ) ( ) () .     12 1 2 2 11 27 5    (d) Gm m H Cm m...

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Design and Optimization of Thermal Systems Episode 2 Part 2 ppt

Design and Optimization of Thermal Systems Episode 2 Part 2 ppt

... 16 8.69990 16 8.69990 16 8.69990 16 8.69990 16 8.69990 16 8.69990 16 8.69980 16 8.69970 16 8.69950 16 8.69900 16 8.69830 16 8.69680 16 8.69 410 16 8.68890 16 8.67 910 16 8.66060 16 8. 625 10 16 8.55670 16 8. 4 21 90 16 8 .14 510 16 7. 527 70 16 5.87840 15 8.95630 10 5.65780NH3: NH3: NH3: NH3: 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

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 ... 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 14 c 4 18 .2 Numerical Modeling and...

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Design and Optimization of Thermal Systems Episode 2 Part 7 ppsx

Design and Optimization of Thermal Systems Episode 2 Part 7 ppsx

... Comp. Continuous Comp. 1 110 .0 11 0.0 11 0.47 11 0. 52 11 0. 52 2 12 0 .0 12 1 .0 12 2 .04 12 2 .14 12 2 .14 5 15 0.0 16 1.05 16 4.53 16 4.86 16 4.87 10 20 0.0 25 9.37 27 0.70 27 1. 79 27 1. 83 20 300.0 6 72. 75 7 32. 81 738.70 738. 91 30 400.0 17 44.94 ... 10 7.6 3.8 19 96 15 6.9 3.3 19 86 10 9.6 1. 1 19 97 16 0.5 1. 7 19 87 11 3.6 4.4 19 98 16 3.0 1....

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Design and Optimization of Thermal Systems Episode 2 Part 8 pdf

Design and Optimization of Thermal Systems Episode 2 Part 8 pdf

... (6 .20 b). Then, i 0 .1, n 2, and m 12 , and we obtain S (13 6,833. 62) (./)( ./) (./) 0 1 12 1 0 1 12 10 1 12 1 24 24 Đ â ă ả á Ã $6, 314 .18 Therefore, a monthly payment of $6, 314 .18 will pay off ...  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...

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Design and Optimization of Thermal Systems Episode 2 Part 9 pps

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 denition of the objective function, U 424 Design and Optimization of Thermal Systems year ... payment $30,000 Salvage 10 987654 3 21 Years FIGURE P6 .24 440 Design and Optimization of Thermal Systems and H i (x 1 , x 2 , x 3...

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Design and Optimization of Thermal Systems Episode 2 Part 10 ppsx

Design and Optimization of Thermal Systems Episode 2 Part 10 ppsx

... 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 represent the costs for the die and ... 8, 10 , 15 , 18 , 22 , and 24 hours, T is obtained as 86.5, 97.7, 1 02. 0, 10 1.7, 92. 5, 62. 3, 55.0, 67.5, and 80.0nC, respectively. Obtain a best...

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