PS8_sol_S08 - Problem Set 8 (2.006, Spring 08) Problem 1...

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Problem Set 8 (2.006, Spring 08) Problem 1 Make a 100×101 array for hot stream and 101×100 array for cold stream to designate inlet and outlet temperatures of each cell. For convenience, the entire device’s inlet boundary conditions for hot and cold streams are placed at j=1 column and i=1 row, respectively. If we assume that flow properties are identical to every cell, we can see that ,, min, , max, , min, min, min, ** max, max, max, min, min, 0.01 / , 0.011 / 100 100 /100 0.9091 /10000 h tot c tot cell h cell cell c cell cell tot tot cell tot cell tot tot cell tot cell cell tot CC W K W K C C UA UA NTU === == = = = = 1 ,1 100 tot tot NTU NTU = Apply the first law to the element shown by assuming that the bulk temperature of each stream is approximately an average of both inlet and outlet temperatures. Then () (, ) (, 1 ) ( 1 , ) 1 ) 22 hh cc cell h cell h h c cell c c T i jT i j T i i j UA C Tij Tij C Ti j Tij ++ + + ⎛⎞ ⋅−= + = + ⎜⎟ ⎝⎠ * ( ,) ( , 1 ) ( 100 1 100 1 ) c c tot tot tot T i i j T i T i i j T i jT i j NTU C NTU + + ⇒−= + = + * * 1 100 1 1 100 1 1 ) 2 2 1 1 100 1 1 100 (1 , ) ( , ) 2 2 h h tot tot c c tot tot tot tot Tij NTU NTU j CN T U T U +− + + ⇒= + −+ ⋅⋅ 1 1 100 1 1 100 1 1 ) 2 2 11 1 0 0 1 1 1 0 0 , ) 2 2 h h tot tot c c tot tot tot tot NTU NTU j T U T U + + = + T h (1,1) =100 o C T h (2,1) =100 o C T h (100,1) =100 o C T c (1,1) =1 o C T c (1,2) =1 o C T c (1,100) =1 o C T h (1,101) T h (2,101) T h (100,101) T c (101,1) T c T c (101,100) T c (i,j) T c (i+1,j) T h (i,j) T h (i,j+1)
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Once you know the inlet temperature of hot and cold stream into a cell, you can calculate both of the outlet temperatures explicitly by using the underlined matrix which is determined by a given set of NTU and C*. And you start calculations from the top-left corner (i=j=1) since you know both inlet temperatures there. There is no iteration necessary, and the temperatures of all inlets and outlets of the cells are sequentially calculated. Use Matlab or MS-Excel to perform the calculations. Bulk temperature for the total hot outlet flow and cold outlet flow is determined by averaging all of the outlet temperatures. 100 100 ,, 11 ( ,101) , (101, ) 100 100 h out h c out c ij TT i j == ∑∑ Effectiveness is: , hto t C ε = () min, hin hou t tot C ( ) 99 t o cin C = And effectiveness-NTU formula for single-pass cross-flow (unmixed) found in the class note is: 0.22 *0 . 7 8 * 1e x p x p NTU CN T U C ⎧⎫ ⎡⎤ =− ⋅ − − ⋅ ⎨⎬ ⎣⎦ ⎩⎭ (a)-(d) Bulk temperature , 51.66 o h out TC = , , 44.95 o c out = , * 0.9091, 1, 0.4883 tot tot T U = 0 20 40 60 80 100 Hot flow temperature distribution T h,out T h,in =100 o C Cold flow outlet side Cold flow inlet side Temperature ( o C) 0 20 40 60 80 100 Cold flow temperature distribution T c,out T c,in =1 o C Hot flow outlet side Hot flow inlet side (e)-(g) 0 5 10 15 20 0.0 0.2 0.4 0.6 0.8 1.0 (effectiveness) NTU C*=0.1, calculation C*=1, calculation C*=0.1, formula C*=1, formula Note: In part (a)-(d), heat capacity of hot flow is C mim and all of the formulations have been done accordingly.
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This note was uploaded on 10/06/2011 for the course MECHANICAL 2.006 taught by Professor Blah during the Spring '08 term at MIT.

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PS8_sol_S08 - Problem Set 8 (2.006, Spring 08) Problem 1...

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