sec11-022410

# sec11-022410 - ChE 120B Estimating h Boundary Layer...

This preview shows pages 1–5. Sign up to view the full content.

ChE 120B 11 - 1 Estimating h — Boundary Layer Equations Before, we just assumed a heat transfer coefficient, but can we estimate them from first principles? Look at steady laminar flow past a flat plate, again: Clearly, the fluid motion is coupled to the heat transfer processes. Momentum Equations x comp: 22 x xx x xy x vv v v p g x yx x y        y comp: yy y y y v v p g x x y Continuity: 0 y x v v Thermal: px y x TT T T Cv v kv x y (assuming negligible viscous dissipation) Boundary conditions , x vu 0 y v x  , x y  [email protected]  0 y 0 x p p x p p y 0 x x 0 0 y y 0 x

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

View Full Document
ChE 120B 11 - 2 ,, upT  are conditions far from plate. To get anywhere — need some idea of which terms are important. Hence we non- dimensionalize: / xx Uv u / yy u  22 // Pp pp u u   xH H X vu L  p u u  Re / Nu L Pr p NC k v  00 / TT T T This choice of (logical) variables gives the dimensionless B.L. Eqn: Momentum Re Re 1 1 x x xy y y UU U U P XY X N X Y U U P Y N X Y           Continuity x U X + 0 y U Y Energy Re Pr 1 X YN NX Y  The real coupling between these equations is the temperature dependence of the viscosity. If we are not concerned about this, then the momentum equations are decoupled. Now we can say something about magnitudes to identify the important terms: 0(1) means order of 1. That is, the expected value of this quantity is about 1.   / 01, Uu u X  and we can estimate the important derivatives so 01 x U X     10 1 x u and from continuity
ChE 120B 11 - 3  00 1 yy x UU U XY Y    01 y U Y but changes in U y occur only over the boundary layer so y H U yH U  2 22 2 1 1 x x U U X X  2 1 1 x x H H U U Y Y  We can now put these estimates into the momentum B.L. Eqn. we also know for the T.E. Eqn. 2 11 0 , 0 TT X Y      In general , , TH L   so 1 T  so ,, a n d xx YX      2 1 1 / H H H Re 1 x xy U dP d x N Y 1 HH H  2 2 Re 1 y U dP x yd y N y    The second equation is always 0 H , and the first is 0(1). Hence, the pressure gradient PP x y  . Hence, let 0 P y and hence P only changes in the x direction.

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

View Full Document
ChE 120B 11 - 4 P X function only of X.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 11

sec11-022410 - ChE 120B Estimating h Boundary Layer...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online