Radiation Heat Transfer

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Chapter 8

• Radiation Heat Transfer

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8-1 Introduction
Three modes of heat transfer conduction,
convection radiation We have learnt conduction
and convection
Radiation
Heat transferred by electromagnetic radiation
Electromagnetic radiation which is propagated as
a result of temperature difference
Thermal radiation
Examples Sun, boiler
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8-2 Physical mechanism

• Thermal radiation is one kinds of the
  electromagnetic radiation, so it is propagated at
  the speed of light, 310m/s. The following
  relation holds

wherec speed of light ? wavelength
? frequency

• The propagation of thermal radiation takes place
  in the form of discrete quanta, each quantum
  having an energy of

h6.625 10-34J.s is Plancks constant

• relativistic relation between mass and energy

• Stefan-Boltzmann Law

Eb is called the emissive power (??? ) of a
blackbody, the energy radiated per unit time and
per unit area by a black body T is the surface
absolute temperature
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• Electromagnetic spectrum

Cosmic rays up to 4?10-7
?m Gamma rays 4?10-7 to 1.4
?10-4 ?m X-rays 1?10-5
to 2 ?10-2 ?m Ultraviolet rays 1?10-2
to 3.9 ?10-4 ?m Visible light
0.38 to 0.76 ?m Infrared rays
0.76 to 1000 ?m Radio and
1000 to 2?1010 ?m
Hertzian waves Heat rays 0.1
to 100 ?m
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• Thermal radiation 0.1100µm
• Visible light 0.35-0.75(0.380.76 )µm
• most body on the earth lt2000K
  0.38-100 µm
• most of
  them 0.76-20 µm
• The sun a black body at 5762K 0.2---2µm
• Infrared 0.76---20 µm
• Near infrared (????)
• Far infrared (????)

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8-3 Radiation properties
Absorbed Qa ReflectedQ? TransmittedQt
Q
Heat balance
or
absorptivity???(?)
Set
reflectivity???(?)
transmissivity???(?)
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• Solid body and liquid
• Most solid body and liquids do not transmit
  thermal radiation, so for many applied problems
  the transmissivity may be taken as zero t0

• two types of reflections
• specular reflection(????)incident
  anglereflection angle
• diffuse reflection(???) the incident beam is
  distributed ,uniformly in all directions after
  reflection
• gasesGases do not reflect thermal radiation, so
  ?0 ??1

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• For solids and liquids the absorption and
  reflection takes place on the surface, the
  surface properties are important to thermal
  radiation, glass and water?
• Radiation and absorption take place inside gas,
  the surface properties do not matter.

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• Ideal bodies

perfect black body (??)
perfect reflector (??????) specular (??)
diffuse perfect reflector (??)
transparent media (???)

• There is no ideal body.
• Constructing a blackbody
• (a large cavity with very small opening
• the opening is 6 the area of the
  cavitya0.996)
• blackbody is reference standard
• incompressible fluid
• windows

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• Emissive power
• defined as the energy emitted by the
  body per unit area and per unit time.

• Kirchhoffs law
• A body is inside a black enclosure. At
  equilibrium

The body is replaced with a blackbody with the
same size and shape, and at equilibriumT1T2
The 1st eq. is divided by 2nd eq.

• Emissivity (??)
• Hemispherical or Total Emissivity

• The emissivity of a body is equal to its
  absorptivity at the same temperature.
  Kirchhoffs Law Kirchhoffs Identity

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• It is known from Kirchhoffs Law
• (1). For a given T,E?,? ?, ? ? E
• (2).? EltEb,E/?Eb Eb/?b
• ? ?lt1 (?b1)

• Total properties
• the integrated behavior of the material
  over all wavelengths.
• emissivity and absorptivity
• In reality the properties vary with temperature,
  wavelength and
• surface condition
• Monochromatic Emissivity or Spectral
  Emissivity(????)

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• Monochromatic or Spectral Emissive Power (?????)
• the energy radiated per unit time and per unit
  area per unit wavelength by a body

Fig.
The total area under the curve
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• Gray body

• Kirchhoffs Law works for monochromatic
  radiation,then
• ? ? ? ?
• An application of Kirchhoffs Law to gray body
• (1). ?? (?)const., ? ? (?)const.
• (2). ??
• ? depends on its surface properties
• so does ?

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• Planks Law (1900)
• blackbody radiation experiment
• previous work
• Ultraviolet catastrophe
• Planks Law

Eb? spectral emissive power, W/m3
? wave length, m T --temperature, K
C1 --- the first radiation constant,
3.74210-16 W m2 C2 --- the second
radiation constant, 1.438 10-2m K?
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• From Plancks law, we know
• E?f(?,T )
• E? has maximum
• T ?? max ?

• Wiens displacement Law (1893)
• Performing derivative of Planks Law with
  respect to ?,and setting

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Example Calculate the wavelengths of maximum
points in the radiation curves of blackbodies at
2000K, 5800K Solutionfrom Wiens displacement
law when T2000K ? max2.9?10-3/20001.
45 ?m when T5800K ?
max2.9?10-3/58000.50 ?m the wavelength
at which the spectral emissive power is maximum
is in infrared region for most surfaces on the
earth that of the sun is in visible light
region
real surfaces do not follow Wiens
displacement law, but the trend of change is
similar. Steel when Tlt500ºC,no visible
light,color keeps unchanged T goes up,color
?dark red, bright red, bright yellow and white
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• Stefan-Boltzman Law
• Stefan obtained based on experiments in 1879
• Boltzman based on thermodynamics in 1884
• We get in by integrating Planks Law

An alternative form is
Where C0 is 5.67W/(m2K4)?
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• Radiation function
• We are frequently interested in the amount of
  energy radiated from a blackbody in a certain
  specified wavelength range.

That is the area under the curve between ? 1
and ? 2 .
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The fraction of the total energy emitted between
? 1 and ? 2 is
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The fraction can be expressed in terms of the
single variable ? T

• f(? T) is called Radiation Functions or
  Fractional functions (??????)
• The values of Radiation Functions is given in
  table 8-1 on page 374
• The energy emitted between wavelengths ? 1 and ?
  2 .

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Example Calculate the percents of the total
emitted energy that lie in visible and infrared
wavelength when the blackbody is at temperatures
1000K, 1400K, 3000K, 6000K
Radiation and color white clothing is
comfortable in summer,since we receive the
radiation from sun (0.2-2 ?m), the fraction of
visible light is 0.455 the radiation under
2000K is independent of color
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Glass is essentially transparent for visible
light, but almost totally opaque for thermal
radiation emitted at ordinary room temperature
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• Solid angle (???)

• plane angle ,s-arc length,r- radius, ?s/r
  (rad)
• An angle formed by three or more planes
  intersecting at a common point.

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Then
Steradian (???)
semicircle
hemisphere
If dAc is an element of the spherical surface,the
corresponding elemental solid angle
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• Intensity of radiation (??????)

the energy emitted per unit time per unit
projected surface area normal to the p direction
into a unit elemental solid angle centered around
the p direction
is projected surface area
where
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Importance of projected surface area
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• Lamberts Law

• The blackbody intensity of radiation is
  independent of angular
• that is

Problems Is a radiator against Lamberts Law?
From the intensity the energy emitted is
The distance between a person and the radiator is
const. d?const. dAconst.
I(?)const. cos ??const. At ? 0, dq(?)
reaches maximum
Lamberts Law is also called Cosine Law.

• A surface obeys Lamberts Law is called diffuse
  surface
• blackbody and gray body are diffuse

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• Emissive Power and Intensity of Radiation
• dq(?) /dA is the energy emitted per unit
  surface area into the ? direction

• If Lamberts law holds emissive power is

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8-4 Radiation shape factor

• Definition

The fraction of energy leaving surface m which
reaches surface n is called radiation shape
factor, or view factor, angle factor,
configuration factor, Fm-n.
receiving surface
emitting surface

• from the definition F1-2 ?F 1 , F1-2 ?1
• generally F1-2 ? F2-1

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• Heat exchange between blackbodies
• The energy leaving surface 1 is

The energy leaving surface 1 and arriving surface
2 is
Surface 2 absorbs all the energy since it is
blackbody.
In a similar way, surface 1 absorbs the energy
from surface 2 is
The net heat exchange is
Reciprocal relation
space resistance ??????
analogy
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If the two surfaces come into temperature
equilibrium That is Eb1Eb2 and F1-20
Does this relation works for all surfaces ?
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• Calculation of shape factors
• From element of area dA1 to element of area dA2

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The energy leaving dA1
If dA1 is diffuse surface (obeying Lamberts
Law)
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In a similar way
obviously
From element of area dA1 to area A2
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In a similar way
From area A1 to element of area dA2
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If surface 1 is diffuse, and I is independent of
position
And
From area to area
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In a similar way
By comparison

• preconditiondiffuse surface, I uniform
  distribution
• blackbody and gray body satisfy the precondition
• real bodies are approximately taken as gray
  bodies
• radiation shape factor depends on only
  geometries of bodies

• Some special radiation shape factors

Two surfaces can not view each other F1.2F2.10
Infinite parallel planes
Two surfaces on a plane F1.2F2.10
F1.2F2.11
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Real-surface behavior

• 1. Behavior
• E ? f(?,T ) selective
• E ?lt Eb?
• Do not follow Lamberts Law
• Directional preferences
• No shift in the peak of the curve

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• 2. Directional or angular Emissivity (????)
• To illustrate the directional properties, we
  define
• directional emissivity

? is azimuth angle. Actually I(?, ?) is often
independent of ? ,then
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• conductor ? ? ? (?) ? , then ?

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• (a) Wet ice, (b) Wood, (c) Glass, (d) Paper,
• (e) Clay, (f) Copper oxide, (g) Aluminum oxide
• nonconductor ? ? ? (?) ? ? (90)0
• different from conductor

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Which is conductor?
conductor
nonconductor
(a) Electrical conductor (b) Electrical
nonconductor
3. Spectral Directional Emissivity (??????)
4. Total Emissivity
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• 5. e depends on
• materialnonconductor about 0.9,conductor
  lt0.2
• surface conditionaluminum about 0.2,high
  polish 0.05
• temperatureee(T),EbEb(T)?
• e is a function only of the surface itself ,
• independent of the surroundings
• 6. Determination of emissivity of a real
  surface
• normal total emissivity

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Highly polished metal surfaces
Smooth surfaces
rough surface
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7. Total Absorptivity (???)
8. Monochromatic Absorptivity (??(??)???)
material Surface condition temperature
Absorbing surface
? depends on
Incident radiation
?(?) measured
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For black source
?(?) and Eb? are known, ? can be obtained by
numerical integration, as shown in the right
figure.
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8-5 Relations between shape factors 1. Properties
of shape factors ? The reciprocal property (???)
? The enclosure property (???) energy
balance divided by q1
? Decomposability (???)
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If surface 1 is split into n surfaces, we have
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From the reciprocal relation
Substitute into the top eq.
That is
Keep them in mind
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2. Calculation of shape factors 1). Algebra
method (???) Based on the properties of shape
factors. Example Calculate all the shape factors
of an enclosure formed by
three convex or plane surfaces
Solution There are 9 shape factors, but
F11F22F330
The enclosure property
The reciprocal property
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2). Cross-string method ExampleTwo surfaces are
2-D, infinite in extend in one direction and
characterized by all cross sections normal to the
infinite direction being identical. Calculate the
shape factor F1-2. From the enclosure property
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3) Integration Integrating the definition
equation.
4) Charts and literatures Fig. 8-12,
---Fig. 8-16 Handbook of radiation shape
factors
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8-6 Heat exchange between nonblackbodies

• Real surfaces are regarded as gray body in
  engineering
• The assumption is justified by practice utility
• A assumptions

1. Diffuse surface
2. Uniform in temperature
3. Properties are constant over all the surface

• G irradiation (????)total radiation incident
  upon a
• surface per unit time and per unit area

• J radiaosity (????)total radiation which
  leaves
• a surface per unit time and per unit
  area

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From the figure
The energy leaving the surface
Eliminating G
Solving for J
surface resistance ??????
Analogy
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• The exchange of radiant energy by two surfaces

The energy leaving surface 1 and arriving surface
2 is
The energy leaving surface 2 and arriving surface
1 is
The net interchange between the two surfaces
Reciprocal relation
space resistance ??????
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Problem If there are other surfaces, what is
the situation?
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• Essentials of the radiation network method

surface resistance between the radiosity
potentials
space resistance
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• Three body problem
• 1. network

2. Nodal equations 3. Kirchhoffs current law to
the circuit
The sum of the current entering a node is zero.
3. Solving for radiosity Ji 4. Net heat exchange
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• Two special cases
• 1. Insulated surfaces (refractory surfaces)
  (????)

F30, J3Eb3
Temperature depends on condition Dynamic
equilibrium
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2. Black surfaces and surfaces with large areas
Black surface 3, ? 31
Surface 3 is large enough compared to the size
of other surfaces R30 J3Eb3 as shown in
fig.
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8-7 Simple two bodies systems
1. Surface 1 is not concave,F1,21
tube
2. Surface 1 is not concave,F1,21,and A1/A2?1
Vacuum flask
3. Surface 1 is not concave,F1,21,andA1/A2?0
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• Convex object in large enclosure

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• Apparent emissivity (????) of a cavity

Ai is concave Imaginary surface Ao covering the
opening
From reciprocity
Definition of apparent emissivity
Two limiting cases
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• System emissivity (????)

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8-8 Radiation shields (???)

• 1. Enhancement of radiation heat transfer
• radiation shape factor ?
• emissivity ?

• 2. Reducing radiation heat transfer
• emissivity?
• shield

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• Radiation shields

If the emissivities of all the surfaces are
equal, That is ?1?2 ?3? The overall heat
transfer
Combining the two eq. And noting

• conduction resistance of
• surface 3 is neglected
• heat flow is one half
• if ?30.05, ?1?20.8
• the heat flow is 1/27

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• application of
• radiation shields

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8-9 Gas radiation
1. Introduction

• There are usually gaseous media between solid
  surfaces
• Gas may emit and absorb thermal radiation
• at low temperature
• Diatomic gases with nonpolar symmetrical
  molecular
• structure, such as O2, N2, H2, air, are
  essentially transparent
• Gases with more complex molecules, such as CO,
  CO2, H2O,
• CH4, SO2,NH3,CmHn, do emit and absorb
  thermal
• radiation___ at least in certain wavelength
  bands(??)

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2. Radiative properties of gases

• Here the gases are those with more complex
  molecules
• Gases frequently radiate and absorb only in
  narrow wavelength
• bands (??).
• Spectral emissive power of a blackbody obeys
  Plancks law
• E? for a solid body is relatively continuous for
  all wavelength

CO2 2.65-2.80 ?m 4.15-4.45 ?m
13.0-17.0 ?m H2O 2.55-2.84 ?m
5.60-7.60 ?m 12.0-30.0 ?m
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• greenhouse effect (????)
• Solar radiation

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• Most of solar radiation is in the wavelength
  range of 0.2-2 ?m
• But the radiation of low temperature is in long
  wavelength

• The glass is almost transparent for radiation
  with the wavelength
• less than 0.2 ?m, but essentially opaque to
  long-wavelength
• radiation above 3 or 4 ?m.
• The glass allows much more radiation to come
  in than can escape

• Greenhouse gas(????)

Gases with more complex molecules, such as
CO, CO2, H2O, CH4, SO2,NH3,CmHn
temperature ?,agricultural production?,ground
drown, ecological damage A research reports
all ground of USA is planted in trees, the
emitted CO2 can not be converted to O2
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2) The emission and absorption of gases take
place all over the volume of gases, the
shape of gas affects the properties
3) usually gas is nonreflecting
3. Beers Law
Beer believes that the decrease of spectral
intensity of radiation resulting from absorption
in the layer is to be proportional to the
thickness of the layer and the intensity at that
point.
a? monochromatic absorption coefficient
(??????) a?f(material, wavelength,thermodynamic
state)
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4. Spectral absorbtivity and spectral emissivity
for gases
When s??,
From Kirchhoffs law
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5. Mean beam length (??????)

• effect of the shape of gas
• for easy calculation

Mean beam length is thus the required radius of
s gas hemisphere such that it radiates a flux to
the center of its base equal to the average flux
radiated to the area of interest by actual volume
of gas
The mean beam length of various common gas shapes
are given in table8-
For the gas shapes other than listed in table
mean beam length may be approximated by
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8-10 Radiation network for an absorbing
and transmitting medium
Application of Kirchhoffs law
The energy leaving surface 1 which is transmitted
through the medium and arriving surface 2 is
The energy leaving surface 2 and arriving surface
1 is
The net exchange in the transmission process is
Network element
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The energy leaving the medium
The energy leaving medium and arriving surface 1
is
Of the energy leaving surface 1, the quantity
that reaches the medium (that is the absorbed by
the medium, since ??1) is
The net energy exchange is the difference between
the amount emitted by the medium toward the
surface 1 and that absorbed which emanated from
surface 1
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network
If no energy is delivered to the medium, the node
is floating.
Otherwise the node is grounded
If there is a temperature gradient in the medium,
the medium should be divided into layers.
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8-11 Radiation exchange with specular surfaces
8-12 Radiation exchange with transmitting,
reflecting, and absorbing media
Self-learning
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8-13 formulation for numerical solution Radiosity
of surface i is
Incident radiation of surface i
Substituting into J
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8-14 Solar Radiation
8-15 Radiation properties of the environment
Self-learning
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8-16 Effect of radiation on temperature
measurement
Example 8-23 A mercury-therometer having e0.9
hangs in a metal building and indicates a
temperature of Tt 20?. The wall of the building
are poorly insulated and have a temperature of Ts
5?. The value of h for the thermometer may be
taken as 8.3W/(m2.K). Calculate the true air
temperature. Solution A problem of enclosure
formed by two surfaces with At/A?0, the heat
balance
Measurement error is 8.6?. T ?, error ?
If e0.3, Tt 792?, Ts 600?,
h58.2W/(m2.K) Solving T8998.2 ?, error is
206.2 ?
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Example A thermocouple is used to measure the
temperature of flue gas. A shield is placed
outside of the thermocouple. If Ts600?, the
emissivities of the thermocouple and the shield
are 0.3, h116 W/(m2.K) , T81000?,calculate
the indicated temperature. SolutionThe shield
absorbs heat by two side convection is
The shield losses heat by radiation
At equilibrium q3q4,solving the two equ. for T3
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With the help of iteration or chart methods,
T3903?? The energy qt is transmitted by
convection to the thermocouple is
The energy which thermocouple dissipates energy
by radiation to the shield
At equilibrium ,qtq2,solving for T1951.2? ?
Measurement error