University:
Duke University
Course:
PHYSICS 151L | Introductory Mechanics
Academic year:

2023
Views:

355

Pages:

2
Author:

Pemuda H

Can neglect v2 because it is much smaller than u^2 [pu(2)] + [pv (2)]dxdy+f(n) This looks like: pux+pv=-x+\u ax 2 Now to prove it +pv (2) ay]dxdy [pu (24)+pv +f(n)→ u^2 2x (pu) u^2ay (pv) = 0 +u dy Mass conservation [pu^2 + pura]dxdy+f(n) →(pun) + 3y (pvh) = (K) + (xu)-xtup -ydx + upay^2 ат 227 puCpx + pvcpay + hax (pu) + hay (pv) = Kayz 2 0= u ax+vay = pcp ayz (2) Setting PCP Thermal diffusivity. Thermal Dissipation r is called momentum diff. And divide (1) by (2) 2x 27 + = where *= 1 ay at ue E [0, 1] [0,1] TE [TW, Tos] TW-T Prandti's Number P=K PCP = P=0.72 close enough to 1 "Viscous Dissipation - Kinetic velocity If P= 10,000 then » Sr >> OIL If Pr <<1 then S « S 2() May y=0 de ulle K(TwT) TW-Too ay ple (TwT) y=0 hf (Tw-T) hex K Governing equation for u and I are identical for Pr1 so they Cancel out puê K(TW-T) plex Nusselt Number Nusselt Number = Nux = x K ↳ how much heat is being transferred through convection and conduction Nux 2 Cf Rex this is not the definition of the Nusselt number Relationship between Nux and Rex Nux = 2 Cf Rex 0.332 Rex Rex => Nux-0.332 Rex for Pr=1 For Tw=const. For qu-const. Nux = 0.332 Rex Pr This is all for laminar flows Nux 0.458 Rex Pr Turbulent Thermal Boundary Layers (4.3.2) pcp(ax+vay)... V thermal inertia (Q= mCPAT) For Laminar: qw-Kdy = -pCpxdy ατ For Turbulent: q = -p Cp (x + Endy; Eπα Eh-energy transfer diffusivity For flow in a pipe: == 9

Mass Conservation Equation

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