Estimating h Boundary Layer Equations

Transcription

1 Estimating h Boundar Laer Equations ChE 0B Before, we just assumed a heat transfer coefficient, but can we estimate them from first rinciles? Look at stead laminar flow ast a flat late, again: Clearl, the fluid motion is couled to the heat transfer rocesses. Momentum Equations com: v v v v v v g com: v v v v v v g Continuit: v v 0 hermal: Cv v kv (assuming negligible viscous dissiation) Boundar conditions v v v u, v 0 u, v

2 u,, are conditions far from late. ChE 0B o get anwhere need some idea of which terms are imortant. ence we nondimensionalize: v / u v / u / / P u u X v / u / L / / P u u N Re ul/ NPr C / k v/ / 0 0 his choice of (logical) variables gives the dimensionless B.L. Eqn: P X Y X NRe X Y Momentum P X Y Y N Re X Y Continuit + 0 X Y Energ X Y NReNPr X Y he real couling between these equations is the temerature deendence of the viscosit. If we are not concerned about this, then the momentum equations are decouled. Now we can sa something about magnitudes to identif the imortant terms: 0() means order of. hat is, the eected value of this quantit is about. u / u 0, X 0, and we can estimate the imortant derivatives so X 0 0 u 0 0 and from continuit -

3 0 0 X Y Y ChE 0B Y 0 but changes in occur onl over the boundar laer so X 0 X 0 Y Y 0 We can now ut these estimates into the momentum B.L. Eqn. we also know for the.e. Eqn , 0 0 X Y X Y In general,, L so so,, and Y X Y X dp X Y d NRe Y / dp Re d N he second equation is alwas 0, and the first is 0(). ence, the ressure gradient P P P. ence, let 0 and hence P onl changes in the direction. - 3

4 P X function onl of X. ChE 0B P Finall, we know that for. So, 0 outside boundar laer; however, P is dp a function onl of, so it is indeendent of, and 0 d for all (to order ). his gives the Prandtl B.L. eqns. () X Y NRe Y () 0 X Y (3) X Y N N Y Re Pr 0 0 X X 0 Note that, is indeendent of (3). If we retained the temerature deendence of, in the st eqn, the three equations would be couled and ver difficult to solve so we sa things are constant. Now to solve these equations. Eact Solution: Requires use of stream function Similarit transform NRe Solutions to f ff N f 0 mess, but ossible R As an alternative we can develo aroimate solutions, in dimensional form - 4

5 v v v v v v v v 0 v v Again, we assume, L ChE 0B We now solve for an aroimate velocit field for these conditions: v can be aroimated as v v 0 We also want the stress to be continuous at boundar laer v 0@ Also, v v 0 so, if we recognize that, because ever equation holds at the boundar, we find: v v v v v v v 0@ Let s aroimate v b a olnomial in / ( is function of ) (higher order requires additional assumtions) / / / 3 v ab c d and use our boundar conditions to solve for a,b,c,d. v 0@ 0a 0,and v v v 0@ 0@ 0means v u 3 3 to determine v we can integrate the continuit equation. v v v d u d v 4 d Now we can integrate the equations of motion: - 5

6 v v v v v d v d 0 0 he RS is eas: v v v v d v v 0 0 ChE 0B and substituting the aroimate v, v into the original integral and evaluate 39 d 3v, gives ordinar D.E. for 80 d u 0 gives 4.64 v / u which determines v, v aroimatel. We can now use our velocit field to determine the temerature field v v d 0 0 and from our icture = = 0 = δ from continuit = δ continuit of q from original eqn: v = v = 0 0@ 0 as a olnomial We can now substitute the temerature distribution and velocit distribution into integral d u 0 0 d d After reorganizing we get 3d 3 4d R R R R 5 35 d d u 0 R / assume indeendent of, then - 6

7 d d d R d 3 d R R 0 40 d u 3 R R 0 40 u recalling that R R R 4 u 4.64 v / u R R.93 N Pr v 0asPr ence in general, for small R R 5 << R 3 3 R.93 N R 3 P r /3 93 N P r and / v / NP u r and from this we can determine h! q h 0 0 local q k 3 k 0 / 0 0 k Nu.33 N N / /3 Re Pr / /3 u v Nu.33k v and the average over the late is L k hav h c d.646 Ne N L L 0 L Define Nu h L / k / /3 0 Pr L ChE 0B Nu.646N N / /3 av e L Pr L is a ver good aroimation for P r > 0.6 R e > 0 3-7

8 For P r 0 we can show that / / N..03N N Nu, av e L Pr L which works for liquid metals ChE 0B eat ransfer Coefficients in Pies eat transfer coefficients are alwas defined relative to some far awa temerature which for ie flow might be difficult to measure. What is eas to measure is b, the bulk or miing cu temerature. Generall, this is what is done to define h. b R 0 r0 C, r r rdr R R r0 b C r rdr R, r r rdr he local heat transfer coefficient can then be written as h D d s b w C db w = mass flow rate wc db h D d s b hd wc db k k d s b D b C 4 db k d s b D C b D db k 4 d s b D db NNo NR N e Pr 4 d s b Once we know (,r) we know N Nu Eamle: Rod-like flow of constant velocit. Assume rod-like flow V = b, constant, then the thermal energ equation is: - 8

9 r r r b ChE 0B v 0 v constant, stead flow r b Assume eventuall b a = constant far from the entrance of the ie db a d ub and wc a h D u b s b w u D b 4 akd hd ad h or r t t k 4t t s b s b We still don t know a or b o evaluate a: r a r r r r r r r ra r r r r a C r r C a r We know that 0atr 0so C 0 r r a r r r a C 4 D s a C 6-9

10 D C s a 6 r D r a 4 6 to find b we integrate s rubr dr r i r D b a 0 s ubr dr ru i b 4 6 r a i r D rdr s r 0 i 4 6 ad b s 3 ad s b 3 and hd ad ad 3 Nu 8 k 4 4 ad s b ChE 0B From our analsis of Laminar and urbulent Boundar-Laer Equations, we have seen Nu a Re b Nu c nder the assumtion that the late was ver long comared to the boundar laer, and that hsical roerties are constant. Dimensional analsis of flows involving forced convection shows that in addition to Re and Pr, we eect for ies the additional arameters L/D and b / 0 should be imortant start with Eamle: Laminar Flow; constant roerties r vz vz R v z z r r r 0 r v z r R z r r r initial temerature z 0 0 r R Constant wall temerature 0 r R z 0 r 0-0

11 ChE 0B 0 his equation is searable 0 and has been solved to show that Nu = L/D -