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