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External Flow:External Flow:
Flow over Bluff ObjectsFlow over Bluff Objects
(Cylinders, Sphere, Packed Beds)(Cylinders, Sphere, Packed Beds)
andand
Impinging JetsImpinging Jets
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The Cylinder in Cross Flow
• Conditions depend on special features of boundary layer development, including
onset at a stagnation point and separation, as well as transition to turbulence.
– Stagnation point: Location of zero velocity and maximum pressure.( )0u∞ =
– Followed by boundary layer development under a favorable pressure gradient
and hence acceleration of the free stream flow .
( ) / 0dp dx < ( ) / 0du dx∞ >
– As the rear of the cylinder is approached, the pressure must begin to increase.Hence, there is a minimum in the pressure distribution, p(x), after which boundary
layer development occurs under the influence of an adverse pressure gradient
( ) / 0, / 0 .dp dx du dx∞> <
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– Separation occurs when the velocity gradient reduces to zero.0 / ydu dy =
and is accompanied by flow reversal and a downstream wake.
– Location of separation depends on boundary layer transition.
Re DVD VD ρ
ν ≡ =
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– What features differentiate boundary development for the flat plate in
parallel flow from that for flow over a cylinder?
• Force imposed by the flow is due to the combination of friction and form drag.
The dimensionless form of the drag force is
( )2Figure 7.8
/ 2
D D
f
F C
A V ρ = →
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• Heat Transfer Considerations
– The Local Nusselt Number:
– How does the local Nusselt number vary with for ?θ 5Re 2 10 D < x
What conditions are associated with maxima and minima in the variation?
– How does the local Nusselt vary with 5for Re 2 10 ?θ >D
x What conditions
are associated with maxima and minima in the variation?
– The Average Nusselt Number ( ) / : D Nu hD k ≡
– Churchill and Bernstein Correlation:
( )
4 / 55 / 81/ 2 1/ 3
1/ 42 / 3
0.62Re Pr Re0.3 1
282,0001 0.4 / Pr
D D D Nu
⎡ ⎤⎛ ⎞= + +⎢ ⎥⎜ ⎟
⎝ ⎠⎡ ⎤ ⎢ ⎥+ ⎣ ⎦⎣ ⎦
– Cylinders of Noncircular Cross Section:
1/ 3Re Prm D D Nu C =
, Table 7.3C m →
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Flow Across Tube Banks
• A common geometry for
two-fluid heat exchangers.
max
ST V V S DT
=−
• Aligned and Staggered Arrays:
Aligned:
Staggered: ( ) ( )if 2maxST V V S D S D D T S DT
= − ≥ −−
( )( ) ( )if 2max
2
ST V V S D S D D T S D D
= − ≤ −−
or,
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• Flow Conditions:
How do convection coefficients vary from row-to-row in an array?
How do flow conditions differ between the two configurations?
Why should an aligned array not be use for ST /SL < 0.7?
• Average Nusselt Number for an Isothermal Array:
( )1/ 40.36
2 ,maxRe Pr Pr/ Prm D D s Nu C C ⎡ ⎤= ⎣ ⎦
2
, Table 7.7
Table 7.8
C m
C
→
→
All properties are evaluated at except for Pr s.( ) / 2i oT T +
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• Fluid Outlet Temperature (T o) :
s o
s i T T p
T T DNhexp
T T VN S c
⎛ ⎞− π= −⎜ ⎟
⎜ ⎟− ρ⎝ ⎠
T L N N N = x
What may be said about T o as ? N → ∞
• Total Heat Rate:
s mq hA T = Δl
( )s
N DLπ =
( ) ( )s i s o
m
s i
s o
T T T T T
T T n
T T
− − −Δ =
⎛ ⎞−⎜ ⎟−⎝ ⎠
l
l
• Pressure Drop:2
max
2 L
V p N f
ρ χ
⎛ ⎞Δ = ⎜ ⎟
⎝ ⎠, Figures 7.13 and 7.14 f χ →
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The Sphere and Packed Beds
• Flow over a sphere
– Boundary layer development is similar to that for flow over a cylinder,involving transition and separation.
• Gas Flow through a Packed Bed
– Flow is characterized by tortuous paths through a bed of fixed particles.
– Large surface area per unit volume renders configuration desirable for
the transfer and storage of thermal energy.
( ) ( )1/ 41/ 2 2 / 3 0.42 0.4Re 0.06Re Pr / D D D s Nu μ μ = + + –
Figure 7.8 DC → –
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– For a packed bed of spheres:
0.5752.06 Re
void fraction (0.3 < < 0.5)
H D jε
ε ε
−=
→
, p t mq hA T = Δl
, total surface area of particles p t A →
–
,
,
expp t s o
s i c b p
hAT T
T T VA c ρ
⎛ ⎞−= −⎜ ⎟⎜ ⎟− ⎝ ⎠
, cross-sectional area of bedc b →
–
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Problem: 7.78 Measurement of combustion gas temperature with a spherical
thermocouple junction.
KNOWN: Velocity and temperature of combustion gases. Diameter and emissivity of thermocouple
unction. Combustor temperature.
FIND: (a) Time to achieve 98% of maximum thermocouple temperature rise for negligible radiation, (b)
Steady-state thermocouple temperature, (c) Effect of gas velocity and thermocouple emissivity on
measurement error.
SCHEMATIC:
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ASSUMPTIONS: (1) Validity of lumped capacitance analysis, (2) Constant properties, (3) Negligible
conduction through lead wires, (4) Radiation exchange between small surface and a large enclosure (parts b
and c).
PROPERTIES: Thermocouple: 0.1 ≤ ε ≤ 1.0, k = 100 W/m⋅K, c = 385 J/kg⋅K, ρ = 8920 kg/m3; Gases:
k = 0.05 W/m⋅K, ν = 50 × 10-6
m2 /s, Pr = 0.69.
ANALYSIS: (a) If the lumped capacitance analysis may be used, it follows from Equation 5.5 that
( )i
s
T TVc D ct ln ln 50
T ThA 6h
ρ ρ ∞
∞
−= =
−.
Neglecting the viscosity ratio correlation for variable property effects, use of V = 5 m/s with the Whitaker
correlation yields
( ) ( )1/ 2 2 / 3 0.4D D DNu hD k 2 0.4 Re 0.06 Re Pr= = + +
( ) ( )( )( )1/ 2 2 / 3 0.4 20.05 W m Kh 2 0.4 100 0.06 100 0.69 328W m K
0.001m
⋅ ⎡ ⎤= + + = ⋅⎢ ⎥⎣ ⎦
Since Bi = ( )oh r 3 k = 5.5 × 10-4, the lumped capacitance method may be used.
( )( )
3
2
0.001m 8920kg m 385J kg Kt ln 50 6.83s
6 328W m K
⋅= =
× ⋅
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(b) Performing an energy balance on the junction, qconv = qrad.
Hence, evaluating radiation exchange from Equation 1.7 and with ε = 0.5,
( ) ( )4 4s s chA T T A T Tε σ ∞ − = −
( ) ( )8 2 4
44 4
2
0.5 5.67 10 W m K1000 T K T 400 K
328W m K
−× × ⋅ ⎡ ⎤− = −
⎢ ⎥⎣ ⎦⋅T = 936 K
Parametric calculations to determine the effects of V and ε yield the following results:
0 5 10 15 20 25
Velocity, V(m/s)
900
950
1000
T e m p e r
a t u r e ,
T ( K )
Emissivity, epsilon = 0.5
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Emissivity
890
910
930
950
970
990
T e m p e r a t u r e ,
T ( K )
Velocity, V = 5 m/s
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Since the temperature recorded by the thermocouple junction increases with increasing V and decreasing ε,
the measurement error, T∞ - T, decreases with increasing V and decreasing ε. The error is due to net
radiative transfer from the junction (which depresses T) and hence should decrease with decreasing ε.
For a prescribed heat loss, the temperature difference (T∞ - T) decreases with decreasing convection
resistance, and hence with increasing h(V).
COMMENTS: To infer the actual gas temperature (1000 K) from the measured result (936 K),correction would have to be made for radiation exchange with the cold surroundings.
What measures may be taken to reduce the error associated with radiation effects?