ME 150 – Heat and Mass Transfer Chap. 14.5: Emperical Correlations – Internal Flow
Forced Convection in Internal Geometries No free flow condition (u∞ and T∞) exists in this case: Flow is determined by two boundary layers
Thermal entrance length (laminar): Prof. Nico Hotz
⎛ X fd ,t ⎜⎜ ⎝ D
⎞ ⎟⎟ ≈ 0.05 ⋅ Re D ⋅ Pr ⎠lam 1
ME 150 – Heat and Mass Transfer
Chap. 14.6: Emperical Correlations – Pipe Flow
For pipe flow: r0
Mean velocity:
m um = = ρ⋅A
2π ⋅ ρ ⋅ ∫ u (x, r ) ⋅ r ⋅ dr 0
ρ ⋅ π ⋅ r02
2 r0 = 2 ⋅ ∫ u ( x, r ) ⋅ r ⋅ dr r0 0
r
Mean temperature:
2 0 Tm ( x) = u (r ) ⋅ T ( x, r ) ⋅ r ⋅dr 2 ∫ um ⋅ r0 0
Note: For pipe flow, um und Tm are used instead of u∞ and T∞
Prof. Nico Hotz
2
ME 150 – Heat and Mass Transfer
Chap. 14.6: Emperical Correlations – Pipe Flow
Valid for laminar pipe flow in fully developed region.
Constant surface heat flux: Nu D =
h⋅ D = 4.36 k
Constant wall temperature: NuD = 3.66
Prof. Nico Hotz
3
ME 150 – Heat and Mass Transfer
Chap. 14.6: Emperical Correlations – Pipe Flow
Heat transfer within the entrance region Empirical solution for the case of identical hydrodynamic and thermal entrance length (Pr ≈ 1): 1.33 ⎡ D ⎞ ⎤ ⎛ 0.14 0.067 ⋅ ⎜ Re D ⋅ Pr⋅ ⎟ ⎥ ⎢ ⎛ ⎞ µ L ⎠ ⎥ ⎝ Nu D = ⎢3.66 + ⋅ ⎜⎜ ⎟⎟ 0.83 ⎢ ⎥ D ⎞ ⎛ ⎝ µ w ⎠ 1 + 0 . 1 Pr ⋅ Re ⋅ ⎜ D ⎟ ⎥ ⎢ L ⎠ ⎦ ⎝ ⎣
Valid for laminar pipe flow: ReD < 2300 !
Prof. Nico Hotz
4
ME 150 – Heat and Mass Transfer
Chap. 14.7: Correlations – Turbulent Pipe Flow
Turbulent pipe flow: Only experimental / emperical solutions possible: 23 ⎡ ⎤ ⎛ µ ⎞ D ⎛ ⎞ 0.8 0.3 Nu D = 0.0235 ⋅ (Re D − 230) ⋅ (1.8 ⋅ Pr − 0.8) ⋅ ⎢1 + ⎜ ⎟ ⎥ ⋅ ⎜⎜ ⎟⎟ ⎢⎣ ⎝ L ⎠ ⎥⎦ ⎝ µ w ⎠
0.14
Valid for thermal entrance length = 10-40 times diameter D
Simple solution for limited parameter range: 3000 0.6
< <
ReD Pr L/D
< < >
⎛ µ ⎞ Nu D = 0.027 ⋅ Re 0D.8 ⋅ Pr1 3 ⋅ ⎜⎜ ⎟⎟ ⎝ µ w ⎠ Prof. Nico Hotz
105 500 10 0.14
5
ME 150 – Heat and Mass Transfer
Chap. 14.8: Correlations – Non-Circular Channels
Laminar Flow in NonCircular Channels Nusselt number and friction coefficient comparable to laminar flow in circular pipes But: ReD calculated with hydraulic diameter Dh Dh = 4 ⋅
Geometry a a a a
b b b b
a a
b b
c f Re D
b a
q ʹ′wʹ′ = const
T w = const
-
4.36
3.66
64
1.0
3.61
2.98
57
1.43
3.73
3.08
50
2.0
4.12
3.39
62
3.0
4.79
3.96
69
4.0
5.33
4.44
73
8.0
6.49
5.60
82
∞ -
8.23 3.11
7.54 2.47
96 53
cross − sectional area wetted perimeter
Turbulent case: same equation as in laminar case, with Dh instead of D Prof. Nico Hotz
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ME 150 â&#x20AC;&#x201C; Heat and Mass Transfer
Prof. Nico Hotz
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