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Summary of HeatTransfer by Radiation
Chapters 12 and 13
CH EN 3453 – Heat Transfer
Reminders…• Homework #12 due today (last one!)
– Turn in by 4:00 PM to ChE main office– Scores on web site are updated, so you should be able to complete #1(a)– Sorry ‘bout that part (b) on problem #1
• Consider it a trick question...
• Final project report due Wednesday by 8:00 PM– Email the file to report@chen3453.com– Check the rubric one last time to make sure you have done everything
required
• Final exam Wednesday, December 17 from 8:00 AM to 10:00 AM– 50% is review of conduction and convection, mostly multiple choice– 50% is three calculation problems relating to radiation
• Wednesday: Conduction review• Friday: Convection review
Radiation with Participating Media(Gaseous Emission and Absorption)
• Gas radiation– Nonpolar gases (O2, N2) neither emit nor
absorb radiation– Polar gases (CO2, H2O, hydrocarbons) do
• In most cases, contribution of gas to radiation can be safely neglected
• Notable exception: H2O and CO2 at high temperatures (e.g. in combustion gases)
General Considerations• The medium separating surfaces of an enclosure may affect radiation
at each surface through its ability to absorb, emit and/or scatter (redirect) radiation.
• Participating media may involve semitransparent solids and liquids, as well as polar gases such as CO2, H2O, CH4, and O3.
• Radiation transport in participating media is a volumetric phenomenon, and for polar gases is confined to discrete wavelength bands.
• Beer’s law: A simple relation for predicting the exponential decay of radiation propagating through an absorbing medium.
...where !! is the spectral absorption coefficient (m–1)
• Transmissivity and absorptivity of medium of thickness L
Emissivity of Water Vapor
Emissivity of Carbon Dioxide
Pressure Correction
H2O
CO2
H2O + CO2 Correctionεg = εw + εc − Δε
Gas Radiation - Geometries
Example - Problem 13.126
A gas turbine combustion chamber may be approximated as a long tube of 0.4-m diameter. The combustion gas is at a pressure and temperature of 1 atm and 1000°C, while the chamber surface temperature is 500°C. If the combustion gas contains 0.15 mol fraction each of carbon dioxide and water vapor, what is the net radiative heat flux between the gas and chamber surface, which may be approximated as a blackbody?
Example: Problem 13.129Products of combustion (2000 K, 1 atm) flow through a long, 0.25-m-diameter pipe whose outer surface is black. Combustion gas contains CO2 and H2O, each at 0.1 atm. Gas may be treated as air in fully developed flow at 0.25 kg/s. Pipe is cooled by water in cross flow at 0.3 m/s and 300 K. Determine the pipe wall temperature and heat flux. Emission from the pipe wall may be neglected.
Review of Radiation
Radiation Spectrum
Intensity vs. Wavelength and Direction
The Solid Angle
Solid Angle Geometry
+
θ2
A2
ω =A2 cosθ2
r2 A2 cosθ2
r
Projected Area
Radiation Heat Transfer• Energy transfer between two elements A1
and A2
q1− j = I × A1 cosθ1 ×ω j−1
= I × A1A2 cosθ1 cosθ2r2
From Example 12.1...
ω 3−1 = ω 4−1 =A3r2
=10−3m2
0.5m( )2= 4.00 ×10−4 sr
ω2−1 =A2 cosθ2
r2=10−3m2 × cos30°
0.5m( )2= 3.46 ×10−3sr
Blackbody• Hypothetical perfect radiative surface
• Absorbs all incident radiation, regardless of wavelength and direction
• Emits maximum theoretical energy
• Diffuse emitter – Radiation emitted evenly in all directions
The Planck Distribution• Emissive power of a blackbody depends
on temperature and wavelength
• Planck figured out this relation
• Plot of E vs. λ looks like this:
NOTES:
• Total power increases with temperature
• At any given wavelength the magnitude of emitted radiation increases with temperature
• Wavelength of radiation decreases with temperature
• Sun is approximated by blackbody at 5800 K
• At T < 800 K, most radiation in infrared
Wien’s Displacement Law• For a given temperature, spectral emission
goes through a maximum at a given wavelength.
• Wien figured this one out:
• This maximum is indicated by the dashed line in Figure 12.12
Stefan-Boltzmann Law• If one were to integrate any of the curves
shown in Figure 12.12 over the entire range of wavelengths, one would get the total emissive power for a blackbody:
• The Stefan-Boltzmann constant σ is:
Eb =C1
λ5 exp C2 / λT( ) −1⎡⎣ ⎤⎦dλ
0
∞
∫= σT 4
Text
σ = 5.670 × 10–8 W/m2·K4
Band Emission• Amount of total emitted radiation depends on
range of wavelengths of emission
• Effective emissivity determined by integrating over wavelengths
• Table 12.1, column “F” provides fraction of total integrated area to a given wavelength
Example: 12.29• The spectral, hemispherical emissivity of
tungsten may be approximated by the distribution given below. What is the total hemispherical emissivity when the filament temperature is 2900 K.
Radiation Transfer Types• Emission (E)
– Associated with energy transfer due to surface temperature
• Irradiation (G)– Radiation incident onto a surface– Irradiation can have three fates:
• Absorption by the surface(α = absorptivity = fraction of G absorbed)
• Reflection by the surface(ρ = reflectivity = fraction of
• Transmission through the material(τ = transmissivity = fraction transmitted)
Irradiation onto a Surface• Irradiation can have three fates:
– Absorption by the surface (α = absorptivity = fraction of G absorbed)
– Reflection by the surface (ρ = reflectivity = fraction of G reflected)
– Transmission through the material (τ = transmissivity = fraction of G transmitted)
• Sum of α + ρ + τ = 1
Radiosity (J)• Total radiation leaving a surface.• Sum of emission plus reflected portion of
irradiation.
Example - Problem 12.52Consider an opaque, diffuse surface for which the spectral absorptivity and irradation are shown below. What is the absorptivity of the surface for the prescribed irradiation. If the surface is at 1250 K, what is its emissive power?
View Factors• Fraction of radiation from surface i that is
captured by surface j
• Summation rule:
• Reciprocity:
Fij = 1j=1
N
∑
AiFij = AjFji
Spaceresistance
Surfaceresistance
Review: Radiation between Surfaces
Review: Two-Surface Enclosure
Spaceresistance
Surfaceresistance
Surfaceresistance
Radiation Shield
Reradiating Surface
“Direct Method” for Solving Networks
• Useful for systems with >2 surfaces• Balance radiant energy around each surface
node i :
• Solve systemof equations
Multimode Heat Transfer
Example - Problem 13.66Two parallel, aligned disks 0.4 m diameter and 0.1 m apart are located in a large room with walls at 300 K. One of the disks is at 500K with emissivity of 0.6 while the backside of the second disk is well insulated. What is the temperature of the insulated disk?
Radiation with Participating Media(Gaseous Emission and Absorption)
• Gas radiation– Nonpolar gases (O2, N2) neither emit nor
absorb radiation– Polar gases (CO2, H2O, hydrocarbons) do
• In most cases, contribution of gas to radiation can be safely neglected
• Exception:
Emissivity of Water Vapor
Emissivity of Carbon Dioxide
H2O + CO2 Correctionεg = εw + εc − Δε
Gas Radiation - Geometries
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