External Forced Convection :
The Flat Plate in Parallel Flow
Chapter 7
Section 7.1 through 7.3
7.1 Empirical Method
Nu L =
m
C Re L
Pr
n
7.2 The Flat Plate in Parallel Flow
• To determine the conditions, compute
Re L =
ρ u ∞ L u∞ L
=
µ
ν
and compare with the critical Reynolds number for transition to turbulence, Re x , c .
Re L < Re x , c → laminar flow throughout
Re L > Re x , c → transition to turbulent flow at xc / L ≡ Re x , c / Re L
• Value of Re x , c depends on free stream turbulence and surface roughness.
Nominally,
Re ≈ 5 × 105.
x,c
• If boundary layer is tripped at the leading edge Re x , c = 0 and the flow is
turbulent throughout.
• Surface thermal conditions are commonly idealized as being of uniform
temperature T or uniform heat flux q′′ . Is it possible for a surface to be
s
s
concurrently characterized by uniform temperature and uniform heat flux?
• Thermal boundary layer development may be delayed by an unheated starting
length.
Equivalent surface and free stream temperatures for x < ξ and uniform Ts
(or q′′s ) for x > ξ .
Similarity Solution for Laminar,
Constant-Property Flow over an Isothermal Plate
• Based on premise that the dimensionless x-velocity component, u / u∞ ,
and temperature, T * ≡ ⎡⎣(T − Ts ) / (T∞ − Ts ) ⎤⎦ , can be represented exclusively in
terms of a dimensionless similarity parameter
η ≡ y ( u∞ /ν x )
1/ 2
• Similarity permits transformation of the partial differential equations associated
with the transfer of x-momentum and thermal energy to ordinary differential
equations of the form
d3 f
d2 f
2 3+ f
=0
2
dη
dη
where ( u / u∞ ) ≡ df / dη ,
d 2T * Pr dT *
+
f
=0
2
2 dη
dη
• Subject to prescribed boundary conditions, numerical solutions to the momentum
and energy equations yield the following results for important local boundary layer
parameters:
- with u / u∞ = 0.99 at η = 5.0,
∂u
- with τ s = µ
∂y
and d f / dη
2
= µ u∞
y =0
2
η =0
δ=
d2 f
u∞ / vx
dη 2
= 0.332,
and dT * / dη
η =0
( u∞ / vx )
1/ 2
=
τ s, x
= 0.332 Pr1/ 3
hx x
Nu x =
= 0.332 Re1/x 2 Pr1/ 3
k
( Re x )
1/ 2
ρ u∞2 / 2
= 0.664 Re −x1/ 2
= k ( u∞ / vx )
1/ 2
y =0
5x
η =0
C f ,x ≡
- with hx = q′′s / (Ts − T∞ ) = k ∂T * / ∂y
5.0
dT * / dη
for Pr > 0.6,
and
δ
= Pr1/ 3
δt
η =0
• How would you characterize relative laminar velocity and thermal boundary layer
growth for a gas?
An oil? A liquid metal?
• How do the local shear stress and convection coefficient vary with distance from
the leading edge?
• Average Boundary Layer Parameters:
τ s, x
1 x
≡ ∫0 τ s dx
x
hx =
1 x
∫0 hx dx
x
C f , x = 1.328 Re −x1/ 2
Nu x = 0.664 Re1/x 2 Pr1/ 3
• The effect of variable properties may be considered by evaluating all properties
at the film temperature.
Ts + T∞
Tf =
2
Turbulent Flow
• Local Parameters:
C f , x = 0.0592 Re −x 1/ 5
Empirical
Correlations
Nu x = 0.0296 Re 4x / 5 Pr1/ 3
How do variations of the local shear stress and convection coefficient with
distance from the leading edge for turbulent flow differ from those for laminar flow?
• Average Parameters:
(
1 xc
L
hL =
h
dx
+
∫0 1am
∫xc hturb dx
L
Substituting expressions for the local coefficients and assuming
C f ,L
0.074 1742
= 1/ 5 −
Re L
Re L
(
)
Re x ,c = 5 × 10 ,
5
)
Nu L = 0.037 Re 4L / 5 − 871 Pr1/ 3
For Re x ,c = 0 or L xc
( Re
L
Re x ,c ) ,
C f , L = 0.074 Re −L1/ 5
Nu L = 0.037 Re 4L/ 5 Pr1/ 3
Special Cases: Unheated Starting Length (USL)
and/or Uniform Heat Flux
For both uniform surface temperature (UST) and uniform surface heat flux (USF),
the effect of the USL on the local Nusselt number may be represented as follows:
Laminar
Nu x =
Nu x ξ = 0
a b
⎡1 − (ξ / x ) ⎤
⎣
⎦
Nu x ξ = 0 = C Re mx Pr1/ 3
Turbulent
UST
USF
UST
USF
a
3/4
3/4
9/10
9/10
b
1/3
1/3
1/9
1/9
C
0.332
0.453
m
1/2
1/2
0.0296 0.0308
4/5
4/5
Sketch the variation of hx versus ( x − ξ ) for two conditions: ξ > 0 and ξ = 0.
What effect does an USL have on the local convection coefficient?
• UST:
q′′s = hx (Ts − T∞ )
q = hL As (Ts − T∞ )
L ⎡
( p +1) / ( p + 2 ) ⎤ p / ( p +1)
Nu L = Nu L
1 − (ξ / L )
ξ =0 ( L − ξ ) ⎣
⎦
p = 2 for laminar flow throughout
p = 8 for turbulent flow throughout
hL → numerical integration for laminar/turbulent flow
1 ⎡ xc
L
hL =
h
dx
+
∫ξ 1am
∫xc hturb dx ⎤⎦
⎣
L
• USF:
q′′s
Ts = T∞ +
hx
q = q′′s As
• Treatment of Non-Constant Property Effects:
Ts + T∞
Evaluate properties at the film temperature. T f =
2
7.3 Methodology for a Convection Calculation
• Become immediately cognizant (認定) of the flow geometry.
• Specify the appropriate reference temperature and evaluate the
pertinent fluid properties at that temperature.
• Calculate the Reynolds number.
• Decide whether a local or surface average coefficient is
required.
• Select the appropriate correlation.
Problem: Orientation of Heated Surface
Problem 7.21: Preferred orientation (corresponding to lower heat loss) and the
corresponding heat rate for a surface with adjoining smooth and
roughened sections.
SCHEMATIC:
ASSUMPTIONS: (1) Surface B is sufficiently rough to trip the boundary layer when in the upstream position
(Configuration 2); (2) Re x, c ≈ 5 × 105 for flow over A in Configuration 1.
Orientation of Heated Surface (cont.)
-6
PROPERTIES: Table A-4, Air (Tf = 333K, 1 atm): ν = 19.2 × 10
W/m⋅K, Pr = 0.7.
ANALYSIS: With
u L
Re L = ∞ =
ν
20 m/s × 1m
19.2 ×10-6 m 2 / s
2
-3
m /s, k = 28.7 × 10
= 1.04 × 106.
transition will occur just before the rough surface (xc = 0.48m) for Configuration 1. Hence,
)
(
4/5
⎡
⎤
Nu L,1 = ⎢0.037 1.04 × 106
− 871⎥ 0.71/3 = 1366
⎣
⎦
)
(
4/5
Since Nu L,2 = 0.037 1.04 ×106
( 0.7 )
transfer is associated with Configuration 1.
For Configuration 1:
Hence
(
h L,1L
k
1/ 3
= 2139 > Nu L,1 , it follows that the lowest heat
= Nu L,1 = 1366.
)
h L,1 = 1366 28.7 × 10−3 W/m ⋅ K /1m = 39.2 W/m 2 ⋅ K
q1 = h L,1A ( Ts − T∞ ) = 39.2 W/m 2 ⋅ K ( 0.5m × 1m )(100 − 20 ) K = 1568 W
<
Comment: For a very short plate, a lower heat loss may be associated with Configuration 2. In
fact, parametric calculations reveal that for L< 30 mm, this configuration provides the preferred
orientation.
Problem: Conveyor Belt
Problem 7.24: Convection cooling of steel plates on a conveyor by
air in parallel flow.
KNOWN: Plate dimensions and initial temperature. Velocity and temperature of air in parallel flow
over plates.
FIND: Initial rate of heat transfer from plate. Rate of change of plate temperature.
Problem: Conveyor Belt (cont.)
SCHEMATIC:
ASSUMPTIONS: (1) Negligible radiation, (2) Negligible effect of conveyor velocity on boundary
layer development, (3) Plates are isothermal, (4) Negligible heat transfer from edges of plate, (5)
5
Re x,c = 5 × 10 , (6) Constant properties.
PROPERTIES: Table A-1, AISI 1010 steel (573K): kp = 49.2 W/m⋅K, c = 549 J/kg⋅K, ρ = 7832
3
-6
2
kg/m . Table A-4, Air (p = 1 atm, Tf = 433K): ν = 30.4 × 10 m /s, k = 0.0361 W/m⋅K, Pr = 0.688.
ANALYSIS: The initial rate of heat transfer from a plate is
q = 2 h As ( Ti − T∞ ) = 2 h L2 ( Ti − T∞ )
With Re L = u ∞ L / ν = 10 m / s × 1m / 30.4 × 10 −6 m 2 / s = 3.29 × 105 , flow is laminar over the entire
surface.
Hence,
(
2 Pr1/ 3 = 0.664 3.29 × 105
Nu L = 0.664 Re1/
L
)
1/ 2
( 0.688)1/ 3 = 336
h = ( k / L ) Nu L = ( 0.0361W / m ⋅ K /1m ) 336 = 12.1W / m 2 ⋅ K
q = 2 × 12.1W / m 2 ⋅ K (1m )
2
( 300 − 20 ) °C = 6780 W
Problem: Conveyor Belt (cont.)
Performing an energy balance at an instant of time for a control surface about the plate, − E out = E st ,
ρ δ L2 c
dT
= − h 2L2 ( Ti − T∞ )
dt i
(
)
2 12.1W / m 2 ⋅ K ( 300 − 20 ) °C
dT
=−
= −0.26°C / s
3
dt i
7832 kg / m × 0.006m × 549 J / kg ⋅ K
−4
COMMENTS: (1) With Bi = h ( δ / 2 ) / k p = 7.4 × 10 , use of the lumped capacitance method is
appropriate.
(2) Despite the large plate temperature and the small convection coefficient, if adjoining plates are in
close proximity, radiation exchange with the surroundings will be small and the assumption of negligible
radiation is justifiable.
External Forced Convection :
Flow over Bluff Objects
(Cylinders, Spheres, Packed Beds)
and
Impinging Jets
Chapter 7
Sections 7.4 through 7.8
7.4 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 ( u∞ = 0 )
and maximum pressure.
– Followed by boundary layer development under a favorable pressure gradient
( dp / dx < 0 ) and hence acceleration of the free stream flow ( du∞ / dx > 0 ) .
– 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
( dp / dx > 0, du∞ / dx < 0 ) .
– Separation occurs when the velocity gradient du / dy
y =0
reduces to zero
and is accompanied by flow reversal and a downstream wake.
– Location of separation depends on boundary layer transition.
Re D ≡
ρVD VD
=
µ
ν
– 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
CD =
(
FD
A f ρV / 2
2
)
→ Figure 7.8
• Heat Transfer Considerations
– The Local Nusselt Number:
– How does the local Nusselt number vary with
θ
5
for Re D < 2 x 10 ?
What conditions are associated with maxima and minima in the variation?
– How does the local Nusselt number vary with θ for ReD > 2 × 105 ?
What conditions are associated with maxima and minima in the variation?
– The Average Nusselt Number
( Nu
D
)
≡ hD / k :
– Flow over a circular cylinder [Churchill and Bernstein Correlation]:
Nu D = 0.3 +
1/ 2
D
0.62 Re
1/ 3
Pr
⎡1 + ( 0.4 / Pr )
⎣
2 / 3 1/ 4
⎤
⎦
⎡ ⎛ Re
⎞
D
+
1
⎢ ⎜
⎟
282,000
⎝
⎠
⎢⎣
for ( Re D Pr) ≥ 0.2
5/8 4/5
⎤
⎥
⎥⎦
– Cylinders of Noncircular Cross Section:
Nu D
hD
=
= C Re mD Pr1/ 3
k
for Pr ≥ 0.7
– Circular Cylinders [Zukauskas]:
Nu D
Pr 1/ 4
= C Re Pr ( )
Prs
m
D
0.7 ≤ Pr ≤ 500
n
1 ≤ Re D ≤ 106
If Pr ≤ 10, n = 0.37
;
If Pr ≥ 10, n = 0.36.
7.5 The Sphere
•
See Eqs. (7.47) - (7.49)
Nusph
hD
=
k
14
⎛ µ∞ ⎞
= 2 + ⎣⎡0.4 Re + 0.06 Re ⎦⎤ Pr ⎜
⎟
µ
⎝ s⎠
12
23
0.4
(7.48)
7.6 Flow Across Tube Banks
• A common geometry for
two-fluid heat exchangers.
Aligned:
• Aligned and Staggered Arrays:
ST
Vmax =
V
ST − D
ST
Staggered: Vmax = S − D V
T
ST
V
=
V
or, max
2 SD − D
(
)
(
) (
)
(
) (
)
if 2 S D − D ≥ ST − D
if 2 S D − D ≤ ST − D
Nu D = C1 Re mD ,max
N L ≥ 10
2000 ≤ Re D ,max ≤ 40, 000
Pr = 0.7
where
Re D ,max ≡
ρVmax D
µ
more general form:
Nu D = 1.13 C1 Re mD ,max Pr1/ 3
N L ≥ 10
2000 ≤ Re D ,max ≤ 40, 000
Pr ≥ 0.7
If N L < 10,
Nu D
( N L < 10)
= C2 Nu D
( N L ≥ 10)
• 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 used for ST/SL < 0.7?
• Average Nusselt Number for an Isothermal Array [Zukauskas]:
Nu D = C2 ⎡C Re mD ,max Pr 0.36 ( Pr/ Prs )
⎣
1/ 4
C , m → Table 7.7
All properties are evaluated at
;
⎤
⎦
C2 → Table 7.8
(Ti + To ) / 2
except for Prs.
⎛
Ts − To
πDNh ⎞
= exp ⎜ −
⎟⎟
⎜
Ts − Ti
⎝ ρVNT ST c p ⎠
• Fluid Outlet Temperature (To) :
N = NT x N L
What may be said about To as N → ∞ ?
• Total Heat Rate:
∆TAm =
• Pressure Drop:
As = N (π DL )
q = hAs ∆TAm
(Ts − Ti ) − (Ts − To )
⎛ Ts − Ti ⎞
An ⎜
⎟
T
−
T
⎝ s o⎠
2
⎛ ρVmax
∆p = N L χ ⎜
⎝ 2
χ, f
⎞
⎟f
⎠
→ Figures 7.13 and 7.14
7.8 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.
–
(
)
3
0.4
NuD = 2 + 0.4Re1/D 2 + 0.06Re2/
Pr
( µ / µs )
D
1/ 4
– What are the limiting values of the Nusselt number and the convection
coefficient for slow flows over small spheres.
–
CD → Figure 7.8
• 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.
– For a packed bed of spheres:
ε j H = 2.06 Re−D0.575
ε → void fraction (0.3 < ε < 0.5)
–
q = hAp,t ∆TAm
Ap,t → total surface area of particles
–
⎛
hAp ,t
Ts − To
= exp ⎜ −
⎜ ρVAc ,b c p
Ts − Ti
⎝
Ac , b
⎞
⎟⎟
⎠
→ cross-sectional area of bed
7.7
Impinging Jet
• Characterized by large convection coefficients and used for cooling and heating
in numerous manufacturing, electronic and aeronautic applications.
• Flow and Heat Transfer for a Circular or Rectangular Jet:
Significant Features:
– Mixing and velocity profile development in the free jet.
– Stagnation point and zone.
– Velocity profile development in the wall jet.
– Local Nusselt number distribution:
– Average Nusselt number:
hD
Nu = h = f ( Re, Pr, Ar ,H / Dh )
k
VD
Re = e h , Ar from Fig. 7.17
ν
Correlations → Section 7.7.2
• Jet Arrays
Slot Jets
– What is the nature and effect of jet interactions and discharge conditions?
– Nusselt number correlations for arrays of circular and slot jets → Section 7.7.2.
Problem: Extrusion Process
Problem 7.63: Cooling of extruded copper wire by convection and radiation.
KNOWN: Velocity, diameter, initial temperature and properties of extruded wire. Temperature and
velocity of air. Temperature of surroundings.
FIND: (a) Differential equation for temperature distribution T(x) along the wire, (b) Exact solution for
negligible radiation and corresponding value of temperature at prescribed length ( x = L = 5m) of wire,
(c) Effect of radiation on temperature of wire at prescribed length. Effect of wire velocity and
emissivity on temperature distribution.
Problem: Extrusion Process (cont.)
SCHEMATIC:
ASSUMPTIONS: (1) Negligible variation of wire temperature in radial direction, (2) Negligible effect
of axial conduction along the wire, (3) Constant properties, (4) Radiation exchange between small
surface and large enclosure, (5) Motion of wire has a negligible effect on the convection coefficient (Ve
<< V).
3
PROPERTIES: Copper: ρ = 8900 kg / m , c p = 400 J / kg ⋅ K, ε = 0.55. Air: k = 0.037 W / m ⋅ K,
ν = 3 × 10
−5
2
m / s, Pr = 0.69.
ANALYSIS: (a) Applying conservation of energy to a stationary control surface, through which the
wire moves, steady-state conditions exist and E in − E out = 0.
Hence, with inflow due to advection and outflow due to advection, convection and radiation,
ρ Ve A c c p T − ρ Ve A c c p ( T + dT ) − dq conv − dq rad = 0
Problem: Extrusion Process (cont.)
)
(
)
(
4 ⎤
− ρ Ve π D 2 / 4 c p dT − π Ddx ⎡ h ( T − T∞ ) + εσ T 4 − Tsur
⎥⎦ = 0
⎣⎢
dT
4
=−
dx
ρ Ve D c p
)
(
⎡ h ( T − T ) + εσ T 4 − T 4 ⎤
sur ⎥
∞
⎢⎣
⎦
(1)
<
Alternatively, if the control surface is fixed to the wire, conditions are transient and the energy balance
is of the form, − E out = E st , or
⎛ π D 2 ⎞ dT
4
4
⎡
⎤
dx ⎟ c p
−π D dx h ( T − T∞ ) + εσ T − Tsur = ρ ⎜
⎢⎣
⎥⎦
⎜ 4
⎟
dt
⎝
⎠
dT
4 ⎡
4 ⎤
=−
h ( T − T∞ ) + εσ T 4 − Tsur
⎥⎦
dt
ρ D cp ⎢⎣
)
(
)
(
Dividing the left- and right-hand sides of the equation by dx/dt and Ve = dx / dt, respectively, Eq. (1) is
obtained.
(b) Neglecting radiation, separating variables and integrating, Eq. (1) becomes
T dT
x
4h
=−
dx
Ti T − T∞
ρ Ve D cp 0
∫
∫
⎛ T − T∞
ln ⎜
⎝ Ti − T∞
⎞
4h x
⎟=−
ρ Ve D cp
⎠
⎛
4hx
T = T∞ + ( Ti − T∞ ) exp ⎜ −
⎜ ρ Ve D cp
⎝
⎞
⎟
⎟
⎠
(2)
<
Problem: Extrusion Process (cont.)
−5 2
With Re D = VD / ν = 5 m / s × 0.005 m / 3 × 10 m / s = 833, the Churchill-Bernstein correlation yields
1/ 2
1/ 3
5/8 4/5
⎡ ⎛ 833 ⎞
⎢1 + ⎜
Nu D = 0.3 +
⎟
1/
4
⎢ ⎝ 282, 000 ⎠
⎡1 + ( 0.4 / 0.69 )2 / 3 ⎤
⎣
⎣⎢
⎦⎥
0.62 ( 833)
h=
( 0.69 )
⎤
⎥
⎥
⎦
= 14.4
0.037 W / m ⋅ K
k
Nu D =
14.4 = 107 W / m 2 ⋅ K
D
0.005m
Hence, applying Eq. (2) at x = L,
⎛
⎞
4 × 107 W / m 2 ⋅ K × 5m
⎟
To = 25°C + ( 575°C ) exp ⎜ −
⎜ 8900 kg / m3 × 0.2 m / s × 0.005m × 400 J / kg ⋅ K ⎟
⎝
⎠
To = 340°C
<
(c) Numerically integrating from x = 0 to x = L = 5.0m, we obtain
To = 309°C
Hence, radiation makes a discernable contribution to cooling of the wire.
<
Problem: Extrusion Process (cont.)
Parametric conditions reveal the following distributions.
600
Wire tem perature, T(C)
Wire tem perature, T(C)
600
500
400
300
200
100
550
500
450
400
350
300
0
1
2
3
4
Distance from extruder exit, x(m)
Ve=0.5 m/s
Ve=0.2 m/s
Ve=0.1 m/s
5
0
1
2
3
4
5
Distance from extruder exit, x(m)
eps=0.8
eps=0.55
eps=0
The speed with which the wire is drawn from the extruder has a significant influence on the temperature
distribution. The temperature decay decreases with increasing Ve due to the increasing effect of
advection on energy transfer in the x direction.
The effect of the surface emissivity is less pronounced, although, as expected, the temperature decay
becomes more pronounced with increasing ε.
COMMENTS: (1) A critical parameter in wire extrusion processes is the coiling temperature, that is,
the temperature at which the wire may be safely coiled for subsequent storage or shipment. The larger
the production rate (Ve), the longer the cooling distance needed to achieve a desired coiling temperature.
(2) Cooling may be enhanced by increasing the cross-flow velocity, and the specific effect of V may also
be explored.
Problem: Thermocouple Junction
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
junction. 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:
Problem: Thermocouple Junction (cont.)
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
t=
ρ Vc
T − T∞ Dρ c
ln i
ln ( 50 ) .
=
T − T∞
hAs
6h
Neglecting the viscosity ratio correlation for variable property effects, use of V = 5 m/s with the Whitaker
correlation yields
VD 5 m s ( 0.001m )
2/3
=
= 100
Re D =
Nu D = ( hD k ) = 2 + 0.4 Re1D/ 2 + 0.06 Re D
Pr 0.4
−6 2
ν
50 × 10 m s
)
(
h=
(
)
0.05 W m ⋅ K ⎡
1/ 2
2/3
0.4
2 + 0.4 (100 )
0.69 ) ⎤ = 328 W m 2 ⋅ K
+ 0.06 (100 )
(
⎥⎦
0.001m ⎢⎣
Since Bi = h ( ro 3 ) k = 5.5 × 10-4, the lumped capacitance method may be used.
t=
(
)
0.001m 8920 kg m3 385 J kg ⋅ K
6 × 328 W m 2 ⋅ K
ln ( 50 ) = 6.83s
Problem: Thermocouple Junction (cont.)
(b) Performing an energy balance on the junction, qconv = qrad.
Hence, evaluating radiation exchange from Equation 1.7 and with ε = 0.5,
(
hAs ( T∞ − T ) = ε Asσ T 4 − Tc4
(1000 − T ) K =
)
0.5 × 5.67 × 10−8 W m 2 ⋅ K 4 ⎡ 4
4
T − ( 400 ) ⎤ K 4
⎢
⎥⎦
⎣
328 W m 2 ⋅ K
T = 936 K
Parametric calculations to determine the effects of V and ε yield the following results:
990
970
Temperature, T(K)
Temperature, T(K)
1000
950
950
930
910
900
890
0
5
10
15
Velocity, V(m/s)
Emissivity, epsilon = 0.5
20
25
0.1
0.2
0.3
0.4
0.5
0.6
Emissivity
Velocity, V = 5 m/s
0.7
0.8
0.9
1
Problem: Thermocouple Junction (cont.)
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?