PS6_sol_S08

# PS6_sol_S08 - Problem Set 6(2.006 Spring 08 Problem 1 a A...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ) ( ) ( = − × ⋅ Ω × × = ⇒ − = − ⋅ = − = ∴ = = − − − − ∞ ∞ ∞ ∞ π π ρ π π ρ ρ ρ 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 μ We need to find Reynolds number that gives the Nusselt number = 3.674. Since we don’t 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 μ ρ − − × ⋅ ∴ = = = × 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 ρ & 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 ρ π π ρ ρ τ The frequency of fluctuation possible to measure is up to ~ τ / 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 ρ μ − = = = × > = × × ⋅ Boundary layer is mixed (laminar and turbulent) (From 1/7 0.16Re x x δ − = , ~ 0.352 0.1 x L m D m δ = >> = . 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 π π = = = & , ( )( ) ( ) 5 1.84 10 1005 / Pr 0.693 0.0267 / p Pa s J kgK c k W m K μ − × ⋅ = = = 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 μ ρ − × ⋅ × = = = 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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online