PS6_sol_S08 - Problem Set 6 (2.006, Spring 08) Problem 1 a....

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Unformatted text preview: Problem Set 6 (2.006, Spring 08) Problem 1 a. A steady state temperature is reached when the electrical energy dissipated in the wire is balanced by the heat transfer between the wire and the flowing fluid. Hence, ( ) ( ) K m W K K m m A h T T D I T T DL D L I T T A A L I h A L I R I T T A h elec elec s c elec c elec elec s 2 3 6 2 8 2 3 3 2 2 2 2 2 2 2 / 4831 300 313 10 20 ) 10 51 . 5 ( ) 10 150 ( 4 ) ( 4 ) ( 4 ) ( ) ( = = = = = = = We have ( ) ( ) ( ) 674 . 3 / 0263 . 10 20 / 4831 6 2 = = = K m W m K m W k D h Nu air D 7068 . / 0263 . ) / 1007 )( 10 6 . 184 ( Pr 7 = = = mK W K kg J s Pa k c air p We need to find Reynolds number that gives the Nusselt number = 3.674. Since we dont know the range of Re, try any of the known correlations first. From the class note, 1/ 2 1/3 4 1/ 4 2/3 0.62Re Pr 0.3 ; Re 10 , 0.5 Pr 1 (0.4/ Pr) D D D Nu = + < < + (flow across a cylinder) By solving it, get Re 48.4 D = <10 4 . Our choice of correlation is thus valid. ( ) ( ) ( ) 7 3 6 184.6 10 48.4 Re 38.5 / 1.1614 / 20 10 air air Pa s V m s D kg m m = = = b. 1 000278 . / 174 ) 2 / 10 20 )( / 4831 ( ~ 6 2 < < = = mK W m K m W k hL Bi tungsten char The wire can be treated as an isothermal object. By applying first law, ) ( ) ( 2 = = T T hA R I dt dT cV Q dt dU s elec tungsten & The time constant of the device is (for this configuration), ( ) sec 00271 . ) / 4831 ( 4 ) 10 20 )( / 132 )( / 19800 ( 4 2 / 2 6 3 2 = = = = = K m W m kgK J m kg h cD DL h L D c hA cV s The frequency of fluctuation possible to measure is up to ~ / 1 =369Hz. Problem 2 (a) Assume flat plate flow on the outside surface of the pipe. ( )( )( ) ( ) 3 6 5 5 1.177 / 4 / 20 Re 5.117 10 Re 5 10 1.84 10 L tr kg m m s m U L Pa s = = = > = Boundary layer is mixed (laminar and turbulent) (From 1/7 0.16Re x x = , ~ 0.352 0.1 x L m D m = >> = . We cannot neglect the effect of curvature in reality) (b) We are dealing with uniform heat flux problem on flat surface. From , '' w x x q T T h = , wall temperature will be the highest where heat transfer coefficient is the smallest. Since boundary layer thickness is growing, heat transfer coefficient is smaller at downstream. Calculate h for the end of laminar boundary layer and for the end of the pipe where boundary layer is turbulent. 2 300 '' 47.75 / (0.1 )(20 ) Q W q W m DL m m = = = & , ( )( ) ( ) 5 1.84 10 1005 / Pr 0.693 0.0267 / p Pa s J kgK c k W m K = = = And ( )( ) ( ) ( ) 5 5 3 1.84 10 5 10 Re 1.95 1.177 / 4 / tr tr Pa s x m U kg m m s = = = At x=x tr , laminar: ( ) ( ) 1/ 2 1/3 1/ 2 1/3 5 0.453Re Pr 0.453 5 10 0.693 tr tr x x Nu = = = 283.5 2 3.88 / tr tr x x x x tr k Nu h W m K x = = = = At x=L , turbulent: ( ) ( ) 4/5 1/3 4/5 1/3 6 0.0308Re0....
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PS6_sol_S08 - Problem Set 6 (2.006, Spring 08) Problem 1 a....

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