HMT Week-12.pdf arranged by asad naeem shahh

HEAT & MASS TRANSFER
Arranged By
PROF. DR. ASAD NAEEM SHAH
MED, UET LAHORE
anaeems@uet.edu.pk

RADIATION HEAT
TRANSFER Cont.…
Arranged by Prof. Dr. Asad Naeem Shah

RADIATION SHIELDS
▪ A method to reduce radiant heat transfer between two
particular surfaces by introducing an additional surface
resistance in the heat flow path so that overall heat transfer
can be retarded is known as radiation shielding.
▪ These shields do not deliver or remove any heat from the
overall system.
▪ However, the shields only place another resistance in the
heat-flow path to retard the heat.
▪ Thin sheets of plastic coated with highly reflecting metallic
films on both sides serve as very effective radiation shields.

RADIATION SHIELDS Cont.
➢APPLICATIONS:
o These shields are used for the insulation of cryogenic storage tanks,
e.g., liquid hydrogen, liquid oxygen and LNG storage tanks, etc.
o A good application of radiation shield is also observed in the
measuring of fluid temperature by a thermometer or a
thermocouple which is shielded to reduce the effects of radiation.
➢HEAT EXCHANGE:
▪ Consider two parallel infinite planes as
shown in Fig. 1. Heat exchange between
these surfaces may be calculated as
follows:
Fig. 1

Fig. 2
ൗ
𝑞
𝐴 =
𝐸𝑏1 − 𝐸𝑏2
1 − 𝜖1
𝜖1
+
1
𝐹12
+
1 − 𝜖2
𝜖2
→ (1)
▪ Now considering the same two planes, but with a radiation
shield placed between them (Fig. 2).
▪ Under the S.S condition, heat
transfer between surface 1 and
the shield must be precisely the
same as that between the shield
and surface 2, i.e.,
𝑞
𝐴 1−3
=
𝑞
𝐴 3−2

▪ Therefore,
𝑞
𝐴 1−3
= ൗ
𝑞
𝐴 =
1 − 𝜖1
𝜖1
+
1
𝐹13
+
1 − 𝜖3
𝜖3
and
𝑞
𝐴 3−2
=
1 − 𝜖3
𝜖3
+
1
𝐹32
+
1 − 𝜖2
𝜖2
▪ Thus, overall heat transfer of the system is:
⇒ ൗ
𝑞
𝐴 =
1 − 𝜖1
𝜖1
+
1
𝐹13
+
1 − 𝜖3
𝜖3
+
1 − 𝜖3
𝜖3
+
1
𝐹32
+
1 − 𝜖2
𝜖2
→ (2)

▪ The radiation network corresponding to the situation of the
latter case (shown in Fig. 2) is depicted in Fig. 3:
▪ As the number of resistances are increased by inserting the 3rd
plate between the two plates, heat exchange rate is reduced.
This additional 3rd plate is called the radiation shield.
Fig. 3

▪ As 𝐹13 = 𝐹32 = 1, so Eqn. (2) becomes:
ൗ
𝑞
𝐴 =
1
𝜖1
+
1
𝜖2
+
2
𝜖3
− 2
⇒ ൗ
𝑞
𝐴 =
𝜎 𝑇1
4
− 𝑇2
4
1
𝜖1
+
1
𝜖2
+
2
𝜖3
− 2
→→ (4)
▪ If there are more than one shields, number of resistances can
be found for the subsequent heat transfer analysis as follows:
No. of resistances = 3(n+1); where n = number of shields

PROBLEM
A cryogenic fluid flows through a long tube of 20-mm diameter,
the outer surface of which is diffuse and gray with 𝜖1 = 0.020
and 𝑇1 = 77𝐾. This tube is concentric with a larger tube of 50-
mm diameter, the inner surface of which is diffuse and gray with
𝜖2 = 0.05 and 𝑇2 = 300𝐾. The space between the surfaces is
evacuated. Calculate the heat gain by the cryogenic fluid per unit
length of tubes. If a thin radiation shield of 35-mm diameter and
𝜖3 = 0.02 (both sides) is inserted midway between the inner and
outer surfaces, calculate the change (percentage) in heat gain per
unit length of the tubes.

SOLUTION
SYSTEM WITHOUT SHIELD:
ൗ
𝑞
𝐿 =
𝜋𝐷1𝜎 𝑇1
4
− 𝑇2
4
1
𝜖1
+
𝐷1
𝐷2
1
𝜖2
− 1
⇒ ൗ
𝑞
𝐿 = −0.50 𝑊/𝑚

SYSTEM WITH SHIELD:
⇒ ൗ
𝑞
𝐿 = −0.25 𝑊/𝑚
PERCENTAGE HEAT GAINED:
⇒
𝑞𝑤/𝑜 − 𝑞𝑤
𝑞𝑤/𝑜
= 50%
1
2

GAS RADIATION
▪ Radiation exchange between a gas and a heat-transfer surface
is considerably more complex than the situations described in
the preceding sections. Unlike most solid bodies, gases are in
many cases transparent to radiation.
▪ When gases absorb and emit radiation, they usually do so only
in certain narrow wavelength bands.
▪ Some gases, such as 𝑁2, 𝑂2 , and others of nonpolar
symmetrical molecular structure (e.g., propane), are
essentially transparent at low temperatures, while 𝐶𝑂2, 𝐻2𝑂
and various hydrocarbon (HC) gases radiate to an appreciable
extent.

GAS RADIATION Cont.
▪ The absorption of radiation in gas layers may be described
analytically considering the system shown in Fig. 1. A
monochromatic beam of radiation having an intensity 𝐼𝜆
impinges on the gas layer of thickness 𝑑𝑥.
Fig. 1: Absorption in a
gas layer.
▪ The decrease in intensity resulting from
absorption in the layers is assumed to be
proportional to the thickness of the layer
𝑑𝑥 and the intensity of radiation 𝐼𝜆 at
that point, thus:
𝑑𝐼𝜆 ∝ 𝐼𝜆 𝑑𝑥
⇒ 𝑑𝐼𝜆 = −𝑎𝜆𝐼𝜆𝑑𝑥 →→ (1)
where the proportionality constant 𝒂𝝀 is called
the monochromatic absorption coefficient.

GAS RADIATION Cont.
▪ Integrating the Eqn. (1) gives:
න
𝐼𝜆0
𝐼𝜆𝑥 𝑑𝐼𝜆
𝐼𝜆
= න
0
𝑥
−𝑎𝜆𝑑𝑥
⇒
𝐼𝜆𝑥
𝐼𝜆0
= 𝑒−𝑎𝜆𝑥
→→ (2)
▪ Equation (2) is called Beer’s law and represents the
exponential-decay formula experienced in many types of
radiation analyses dealing with absorption.
▪ In accordance with the definition of monochromatic
emissivity (i.e., 𝜖𝜆 =
𝐸𝜆
𝐸𝑏𝜆
), the monochromatic transmissivity
is also defined as:
𝐼𝜆𝑥
𝐼𝜆0
= 𝜏𝜆

GAS RADIATION Cont.
⇒ 𝜏𝜆 = 𝑒−𝑎𝜆𝑥 →→ 3 𝑢𝑠𝑖𝑛𝑔 𝐸𝑞𝑛. (2)
▪ If the gas is non-reflecting, then:
𝜏𝜆 + 𝛼𝜆 = 1
⇒ 𝛼𝜆 = 1 − 𝑒−𝑎𝜆𝑥 →→ 4 𝑢𝑠𝑖𝑛𝑔 𝐸𝑞𝑛. (3)
▪ It is obvious from Eqn. (4) that the absorptivity is a function of
the thickness of the gas layer 𝑥 and disports the temperature
dependence as well.
▪ Gases frequently absorb only in narrow wavelength bands e.g.,
water vapor has an absorptivity of about 0.7 between 1.4 and
1.5 𝜇𝑚, about 0.8 between 1.6 and 1.8 𝜇𝑚, about 1.0 between
2.6 and 2.8 𝜇𝑚, and about 1.0 between 5.5 and 7.0 𝜇𝑚.

RADIATION PROPERTIES OF
THE ENVIRONMENT
▪ The major portion of solar energy is concentrated in the short-
wavelength region as revealed by the radiation spectrum of the sun.
Consequently, the real surfaces may exhibit different absorption
properties for solar radiation than for long-wavelength
“earthbound” radiation.
▪ Meteorologists and hydrologists use the term insolation to describe
the intensity of direct solar radiation incident on a horizontal surface
per unit area and per unit time, designated with the symbol ‘I’.
▪ Insolation is analogous to the term irradiation employed for incident
radiation. The meteorological literature also uses the following unit:
1 𝑙𝑎𝑛𝑔𝑙𝑒𝑦 𝐿𝑦 = 1 𝑐𝑎𝑙/𝑐𝑚2
41.86 𝑘𝐽/𝑚2

THE ENVIRONMENT Cont.
▪ Insolation and radiation intensity are frequently expressed in
langley per unit time e.g., the Stefan-Boltzmann constant may be
given as:
𝜎 = 0.826 × 10−10 𝐿𝑦/𝑚𝑖𝑛 ∙ 𝐾4
▪ Radiation heat transfer in the environment is governed by the
absorption, scattering, and reflection properties of the atmosphere
and natural surfaces.
➢ SCATTERING OF RADIATION:
▪ Following two types of scattering phenomena occur in the
atmosphere:
1. Molecular scattering
2. Particulate scattering

RADIATION PROPERTIES OF THE
ENVIRONMENT Cont.
o Molecular scattering is observed because of the interaction of
radiation with individual molecules. The blue color of the sky results
from the scattering of the violet (short) wavelengths by the air
molecules.
o Particulate scattering results from the interaction of radiation with
the many types of particles that may be suspended in the air e.g.,
dust, smog, and water droplets are all particulate scattering centers.
The scattering process is governed mainly by the size of the particle
in comparison with the wavelength of radiation. Maximum
scattering occurs when wavelength and particle size are equal and
decreases progressively for longer wavelengths. For wavelengths
smaller than the particle size, the radiation tends to be reflected.

▪ The term albedo (𝐴) is used to describe the reflective
properties of surfaces and is defined by:
𝐴 =
𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑒𝑑 𝑒𝑛𝑒𝑟𝑔𝑦
𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑒𝑛𝑒𝑟𝑔𝑦
→→ (1)
▪ The albedo of the surface is
dependent on the angle the incoming
solar rays make with the surface. The
albedo is also termed as reflectivity.
The albedos of some natural surfaces
are given in Table 1.
Table 1

▪ The atmosphere absorbs radiation quite selectively in narrow-
wavelength bands. The absorption for solar radiation occurs in
entirely different bands relative to the absorption of the radiation
from the earth because of the different spectrums for the two types
of radiation. The approximate spectrums for solar and earth
radiation, is shown in Fig. 1.
▪ A quick inspection of these curves (Fig. 1) will show that the
atmosphere transmits most of the short-wavelength radiation while
absorbing most of the back radiation from the earth. Therefore, the
atmosphere acts like a greenhouse, trapping the incoming radiation
to provide energy and warmth for humans on earth.

RADIATION PROPERTIES OF THE ENVIRONMENT Cont.
Fig.1: Thermal-radiation spectrums for the sun and earth with
primary absorption bands indicated by shaded areas.
It is important
to note the scale
differential on
the two curves.
Arranged by Prof. Dr. Asad N Shah

BEER’S LAW FOR ABSORPTION AND
SCATTERING OF RADIATION
▪ The absorption and scattering of radiation may be described with
Beer’s law assuming that both processes are superimposed on each
other:
𝐼𝜆𝑥
𝐼𝜆0
= 𝑒−𝑎𝜆𝑥
→→ (2)
where 𝑎𝜆 is called the monochromatic absorption coefficient. For a
scattering process, however, 𝑎𝜆 may be replaced by a scattering
coefficient 𝑘𝜆.
▪ Assuming the superimposition, the appropriate coefficients based on
over all wavelengths are then defined as:
𝑎𝑚𝑠 = average molecular scattering coefficient over all wavelengths
𝑎𝑝𝑠 = average particulate scattering coefficient over all wavelengths
𝑎 = average absorption coefficient over all wavelengths
𝑎𝑡 = 𝑎𝑚𝑠 + 𝑎𝑝𝑠 + 𝑎 = total attenuation coefficient over all wavelengths.

SCATTERING OF RADIATION Cont.
▪ If the insolations are defined as:
𝐼𝑐 = direct, cloudless sky insolation at earth’s surface
𝐼𝑜 = insolation at outer limits of earth’s atmosphere
▪ Then radiation insolation at the earth’s surface is expressed as:
𝐼𝑐
𝐼0
= 𝑒−𝑎𝑡𝑚 = 𝑒−𝒏𝒂𝒎𝒔𝑚 →→ (3)
where ‘𝑛’ is the turbidity factor of the air defined as:
𝑛 =
𝑎𝑡
𝑎𝑚𝑠
→ (𝑖)
and ‘𝑚’ is the relative thickness of the air mass or the air-mass-ratio,
defined as:
𝑚 =
𝑚𝑎
𝑚𝑧
=
actual optical mass
the mass if sun is at its zenith
→ (𝑖𝑖)

▪ The molecular scattering coefficient 𝑎𝑚𝑠 for air at atmospheric
pressure is given by Eagleson as:
𝑎𝑚𝑠 = 0.128 − 0.054 log 𝑚 →→ (4)
▪ The relative thickness of the air mass ‘𝑚’ is calculated as the
cosecant of the solar altitude 𝛼 following the Fig. 2:
𝜶
Fig. 2
𝑚𝑧
𝑚𝑎
= cos 𝜃𝑧 = cos 90 − 𝛼
⇒
𝑚𝑧
𝑚𝑎
= sin 𝛼
𝑚𝑎
𝑚𝑧
= csc 𝛼 = 𝑚 →→ (5)

▪ Moreover, the turbidity factor ‘𝑛’ involved in Eqn. (3) is a
convenient means of specifying atmospheric purity and clarity.
Its value ranges from about 2.0 for very clear air to 4 or 5 for
very smoggy, industrial environment.
▪ The insolation at the outer edge of the atmosphere may be
expressed in terms of the solar constant 𝐸𝑏𝑜 by:
𝐼𝑜 = 𝐸𝑏𝑜 sin 𝛼 →→ (5)
where α is again the angle the rays make with the horizontal.
▪ The value of solar constant is given as:
𝐸𝑏𝑜 = 442.4 𝐵𝑡𝑢/ℎ ∙ 𝑓𝑡2
= 2.00 𝑐𝑎𝑙/𝑐𝑚2 ∙ 𝑚𝑖𝑛
= 1395 𝑊/𝑚2

▪ An average variation of incident solar radiation for cloudy and
cloudless situations as a function of solar altitude angle is given in
Table 2.
Table 2

PROBLEM
A certain smoggy atmosphere has a turbidity of 4.0. Calculate the
direct, cloudless-sky insolation for a solar altitude angle of 75˚. How
much is this reduced from that of a clear sky?
𝐼𝑜 = 𝐸𝑏𝑜 sin 𝛼 = 1347.47 𝑊/𝑚2
𝑚 = csc 𝛼 = 1.035
𝑎𝑚𝑠 = 0.128 − 0.054 log 𝑚 = 0.1272
𝐼𝑐 = 𝐼0 𝑒−𝑛𝑎𝑚𝑠𝑚 = 1035 𝑊/𝑚2
⇒ 𝐼𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 = 23%
SOLUTION

HOME ASSIGNMENT
▪ Solve the following:
➢Examples: 8.2, 8.6, 8.7,8.8, 8.10
➢Problems: 8.23, 8.24, 8.25, 8.27, 8.31, 8.52,
8.54, 8.67, 8.68, 8.72 + Related Problems (by
Y. A. Cengel, 2nd Edition)

HMT Week-12.pdf arranged by asad naeem shahh

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