Post on 26-Oct-2014
transcript
BKF2422
Chapter 4
Principles of Steady-state Heat Transfer in Radiation
Topic Outcomes
It is expected that students will be able to: Utilize the basic equation of radiation for black and gray
bodies Solve problems related to combined radiation and
convection heat transfer mechanism Apply view factors to determine radiation heat transfer
rate for various geometries Analyze and solve problems fro radiation heat transfer
involves absorbing gases medium
electromagnetic radiation emitted by a body due to large
temperature difference.
transmitted through space & vacuum
medium not heated (medium heated by convection)
1. Thermal energy of a hot surface at T1 is converted into
energy of electromagnetic radiation waves
2. Waves travel through intervening space in a straight lines
& strike a cool object at T2
3. Electromagnetic waves are absorbed by the body and
converted to thermal energy or heat
ransmittedfraction t ivity transmiss
absorbedfraction ty absorptivi
reflectedn irradiatio offraction ty reflectivi
1
reflection
absorption
reflection
transmission
Opaque bodies, τ = 0 α + ρ = 1
bodyblack ofpower emissive total
surface a ofpower emissive total ,Emissivity
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ing transmitt,reflecting without hs wavelengtallat directions all from
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1 body,Gray
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Kirchhoff’s law:A body at the same temperature, T1
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:Simplified
EXAMPLE 4.10-1 Page 303 Radiation to a Metal Tube
A small oxidized horizontal metal tube with an OD of 0.0254 M and being 0.61 m long with a surface temperature at 588 K is in a very large furnace enclosure with fire-brick walls and the surrounding air at 1088 K. The emissivity of the metal tube is 0.60 at 1088 K and 0.46 at 588K. Calculate the heat transfer to the tube by radiation.
K 5881 T m 0.0254
K 1088 air, 2 T
m 0.61
12qK 10882 T
6.0@
46.0@
21
11
T
T
2m )61.0)(0254.0( DLAi
W2130
1088588)10676.5)(6.0)(0254.0(
448
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:bodyblack perfect a fromradiation by fer Heat trans
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(K)body black theof re temperatu
./10676.5 =constant sBoltzmann' -Stefan =
body theof area surface
where
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,When 2TT
radconv qqq :ferheat trans Total 211
211
where
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rrad
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,1 when ,1 with unitsEnglish in 2-4.10 Figure
EXAMPLE 4.10-2 Page 305 Combined Convection Plus Radiation from a Tube
Recalculate Example 4.10-1 for combined radiation plus natural convection to the horizontal 0.0254-m tube.
.K W/m3.87
5881088100
588
100
1088
676.56.0
100100676.5
2
44
21
4
2
4
1
TT
TT
hr
.K W/m64.15
0254.0
588108832.1
32.1
2
4/1
4/1
D
Thc
W2507
)1088588)(0487.0)(3.8764.15(
211
TTAhhq rc
View Factors
There is net radiant exchange between surfaces.
A net flow of energy will occur from hotter surface to the colder surface
The shape, size, and orientation of two radiating surfaces determine the heat flow.
42
4112112
:2 plane to1 plane fromradiation Net
TTFAq
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4121121
:1 plane to2 plane fromradiation Net
TTFAq
1 plane of area surface where 1 A
1 planes, parallel infiniteFor 2112 FF
1 body,black For 21
1
111
21
42
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TTAq
:2 plane to 1 plane from radiationNet
1 plane of emissivity Where 1
12112
1
FF
EXAMPLE 4.11-1 Radiation Between Parallel Planes
Two parallel gray planes which are very large he emmissivities of ε1 = 0.8 and ε2 = 0.7 and surface 1 is at 1100oF and surface 2 at 600oF. Use English units for the following: What is the net radiation from 1 to 2? If the surface are both black, what is the net radiation?
6.0@7.0
46.0@8.0
21
11
T
T
BodyGray (a)
2
448
21
42
41
12
21
42
41112
W/m15010
17.0
1
8.0
11
8.5885.86610676.5
111
1
111
1
TTA
q
TTAq
BodyBlack (b)
.K W/m25110
8.5885.86610676.5 2
448
42
41
1
12
42
41112
TTA
q
TTAq
The example shows the influence of surfaces with emissivities less than 1 on radiation.
This fact is used to reduce radiation loss or gain from a surface using planes as radiation shield
For example, two parallel surfaces of emissivity at T1 and T2, the exchange is
12
)()( 42
41012
TT
A
q
12
)(
1
1)( 42
4112
TT
NA
q N
Suppose we insert one or more radiation planes between the original surfaces,
General equation for view factor between black bodies
For finite size, some of the radiation from surface 1 does not strike surface 2, and vice versa.
Hence the net radiation interchange is less, since some is loss to the surroundings.
The fraction of radiation leaving surface 1 in all directions which is intercepted by surface 2 is called F12 and must be determined for each geometry by taking differential surface element and intergrating over the entire surface.
Two important quantities..
Solid angle, ω
dω1= dA1cosθ2/r2
Intensity of radiation, IB
IB = dq/(dA cos dω)
For general case for fraction of the total radiant heat that leaves a surface (black body) and arrives at the second surface, dq1→2 = ImdA cosθ1dω1
dq2→1 = ImdA cosθ2dω2
Finally,
q12= A1F12σ(T14 – T2
4) = A2F21σ(T14 – T2
4)
Where
A1F12 = A2F21
dω1= dA1cosθ2/r2
2 1
22121
112
coscos1A A r
AdA
AF
The reciprocity relationship of A1F12 = A2F21 can be applied to any two surfaces.
AiFij = AjFji
And if surface A1 sees a number of surfaces A2, A3,….
F11 + F12 + F13 + … = 1.0
EXAMPLE 4.11-2 View Factor from a plane to a Hemisphere
Determine the view factors between a plane A1 covered by a hemisphere A2 as shown in Fig. 4.11-5
212121 FAFA
1.0. factor view the,only see surface since 1221 FAA
2
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2)0.1(
2
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2
1
2
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EXAMPLE 4.11-3 Radiation Between Parallel Disks
In figure 4.11-6, a small disk of area A1 is parallel to a large disk of area A2 and A1 is centered directly below A2. The distance between the centers of the disks is R and the radius of A2 is a. Determine the view factor for radiant heat transfer from A1 to A2
a
A
A A
A A
dxxr
dxxrA
A
dxxr
dA
A
r
dAdA
AF
0 21
2
21
2
1
1
2121
1
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112
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cos
2
cos
2
coscos1
coscos1
1
2 1
2 1
dxxdA 22
21
2/122 xR
R
1 cos
22
2
12 aR
aF
View factors when black bodies surfaces are connected by reradiating walls
Example: furnace
A larger fraction of the radiation from surface 1 is intercepted by 2. View factor =
For the case of two black bodies surfaces A1 & A2 connected with reradiating walls:
1 2
Refractory reradiating wall
12F
122121
21221
12121
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/1
2 FAAAA
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4111212 TTAFq
View factors when gray bodies surfaces are connected by reradiating walls
• For the case of two gray bodies surfaces A1 & A2 which cannot see themselves and connected with reradiating walls:
F12
q12 = F12 A1 σ (T14 – T2
4)
A1F12 = A2 F21
• For black or gray bodies, with no reradiating walls, again .
1
11
11
1
122
1
12 AA
F
1212 FF
View Factors F12 = fraction of the radiation leaving surface 1 that is
intercepted/reaching by surface 2.
2 1
22121
112
coscosA A r
dAdAAF
General equation:
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EXAMPLE 4.11-5 Complex View Factor for Perpendicular Rectangles
Find the view factor F12 for the configuration shown in Fig. 4.11-9 of the rectangle with area A2 displaced from the common edge of rectangle A1 and perpendicular to A1. The temperature of A1 is T1 and that of A2 and A3 is T2
11 1
2
3 3
1A
2A
3A
1312)23(1
131121)23(11
42
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FFF
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+
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2r
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Example
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2
122
2
2
12
2
21
22
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121
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1211
1
4
4
1
1
1 ,0
1
r
rF
r
r
r
r
r
r
A
AF
FAFA
FF
FF
FF
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sphereFor
Radiation in Absorbing Gas
• Most mono atomic & diatomic gases are transparent to thermal radiation. (He, H2, Ar, O2, N2).
Dipole gases and higher polyatomic gases emit and absorb radiant energy significantly. (CO2, CO, H2O, SO2, NH3).
Absorption of thermal radiant by gas depend on pressure and temperature.
i.e. partial pressure of gases, amount of absorption.
If the gas is heated, it radiates energy to surrounding.
Characteristic Mean Beam Length of Absorbing Gas
The absorption of energy by gas T1, P and characteristic length, L (mean beam length). Mean beam length, L specific geometry.
The mean beam length has been evaluated for various geometries in Table 4.11-1 (pg. 321).
For other shapes,
where,
L = m, V = m3 of gases A = surface area of enclosure m2.
A
VL 6.3
εG, αG and radiation of a gas
q = εG TG
4 - (rate of radiation emitted) from the gas.
q = α αG T14 - (rate of absorption of energy as it is
radiated back to the gas from midpoint of surface element.)
The net rate of radiant transfer between a gas at TG and a black surface of finite area A1 at T1.
41
4 TTAq GGG
Figure 4.11-10 gives the gas emissivity, εG of CO2 at a total pressure of the system = 1.0 atm abs using
at TG.
pG = partial pressure of CO2 in atm.
L = mean beam length in m (Table 4.11-1 ). αG of CO2 is determined also from Figure 4.11-10 but at T1.
(T1 =temperature of midpoint of surface element.)
But instead of using pG L, need to use:
• The value obtained from the chart is then multiplied by
correction factor .
GG T
TLp 1
65.0
1
TTG
LpG
For the case when wall of enclosure are not black. An approximation when the emissivity is greater than 0.7, effective emissivity, ε’ can be used
Hence the radiant transfer between a gas at TG and not black surface at T1
41
4' TTAq GGG
2
0.1'
Example 4.11.7 Page 321
A furnace is in the form of a cube 0.3 m on a side inside and these interior walls can be approximated as black surfaces. The gas inside at 1.0 atm total pressure at 1100K contains 10 mol% CO2 and the rest is O2 and N2. The small amount of water vapor present will be neglected. The walls of the furnace are maintained at 600K by external cooling. Calculate the total heat transfer to the walls neglecting heat transfer by convection.
Ans: 2.6 kW