Chapter 12
Principles of Steady-State Heat
Transfer
Nehal I. Abu-Lail
Thank you
Jordan University of Science and Technology
• Dr. Sameer Al-Asheh
• Dr. Mohammad Al-Saleh
• Dr. Fahmi Abu Al-Rub
• Google images
Heat Transfer
The basic requirement for heat transfer is the
presence of a temperature difference.


The rate of heat transfer in a certain direction depends
on the magnitude of the temperature gradient in
that direction.
The larger the temperature gradient, the higher the
rate of heat transfer.
Application Areas of Heat Transfer
More Applications of Heat Transfer
Drying of fruits
Fuel burning
Drying of wood
Evaporation of water &
weather control
Cooking
Distillation
Heat Transfer Mechanisms

Heat can be transferred in three basic modes:



Conduction,
Convection,
Radiation.

All modes of heat transfer require the existence of a
temperature difference.

All modes are from the high-temperature medium to a lowertemperature one.
12.3A Unsteady State Heat Balance

Making an unsteady-state heat balance for the x
direction only on the element of volume or control
volume in Fig. 12.3-1 by using Eqs. (12.3-1) and (12.32), with the cross-sectional area being A m2 and the
consumption term ignored
𝐴𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = 𝐼𝑛 − 𝑂𝑢𝑡 + 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 − 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛
𝜕𝐸
= 𝑞𝑥 𝑥 − 𝑞𝑥 𝑥+∆𝑥 + 𝑞 ∆𝑥. 𝐴
𝜕𝑡
12.3-4
12.3-1
12.3A Unsteady State Heat Balance
When only considering energy changes caused by a difference in temperature (ΔE = ΔH =
mcpΔT), thus ignoring potential and kinetic energy changes,
𝜕𝐸
𝜕𝑇
= 𝜌𝐶𝑃
∆𝑥. 𝐴
𝜕𝑡
𝜕𝑡
12.3-5
Substituting back into Eq. (12.3-4),
𝜌𝐶𝑃
𝜕𝑇
∆𝑥. 𝐴
𝜕𝑡
= 𝑞𝑥 𝑥 − 𝑞𝑥 𝑥+∆𝑥 + 𝑞 ∆𝑥. 𝐴
12.3-6
where 𝑞 is rate of heat generated per unit volume. Assuming no heat generation and also assuming
steady-state heat transfer, where the rate of accumulation is zero, Eq. (12.3-6) becomes
𝑞𝑥 𝑥 = 𝑞𝑥 𝑥+∆𝑥
12.3-7
This means the rate of heat input by conduction = the rate of heat output by conduction; or qx is a
constant with time for steady-state heat transfer.
12.3A Unsteady State Heat Balance
When only considering energy changes caused by a difference in temperature (ΔE = ΔH =
mcpΔT), thus ignoring potential and kinetic energy changes,
𝜕𝐸
𝜕𝑇
= 𝜌𝐶𝑃
∆𝑥. 𝐴
𝜕𝑡
𝜕𝑡
12.3-5
Substituting back into Eq. (12.3-4),
𝜌𝐶𝑃
𝜕𝑇
∆𝑥. 𝐴
𝜕𝑡
= 𝑞𝑥 𝑥 − 𝑞𝑥 𝑥+∆𝑥 + 𝑞 ∆𝑥. 𝐴
12.3-6
where q is rate of heat generated per unit volume. Assuming no heat generation and also assuming
steady-state heat transfer, where the rate of accumulation is zero, Eq. (12.3-6) becomes
𝑞𝑥 𝑥 = 𝑞𝑥 𝑥+∆𝑥
12.3-7
This means the rate of heat input by conduction = the rate of heat output by conduction; or qx is a
constant with time for steady-state heat transfer.

In this chapter, we are concerned with a control volume where the
rate of accumulation of heat is zero and we have steady-state heat
transfer.

The rate of heat transfer is then constant with time, and the
temperatures at various points in the system do not change with
time.

To solve problems in steady-state heat transfer, various
mechanistic expressions in the form of differential equations for
the different modes of heat transfer such as Fourier’s law are
integrated. Expressions for the temperature profile and heat flux
are then obtained in this chapter.

In Chapter 14, the conservation-of-energy equations (4.2-2) and
(12.3-6) will be used again when the rate of accumulation is not
zero and unsteady-state heat transfer occurs.

The mechanistic expression for Fourier’s law in the form of a
partial differential equation will be used where the temperature at
various points and the rate of heat transfer change with time.

In Section 14.6, a general differential equation of energy change
will be derived and integrated for various specific cases to
determine the temperature profile and heat flux.
12.3 Conduction as a Basic Mechanism of
Heat Transfer
Conduction is the transfer of energy from the more energetic
particles of a substance to the adjacent less energetic ones as a
result of interactions between the particles.
Conduction can take place in solids,
liquids, or gases.



Consider a gas occupy
the two surfaces shown
in the drawing with no
bulk motion at any
point:
T1>T2
Energy  T
  T  Kinetic energy  Molecular energy
  collisions of neighbouring molecules
 Transfer of energy from more energetic to less energetic
https://www.youtube.com
/watch?v=vuBPgTMKUc0
 Energy is related to the random translation of motion of https://www.youtube.c
om/watch?v=9joLYfaye
internal energy + vibrational motion of the molecules
 Thus, energy transfers in the direction of decreasing temperature e8
• Examples of conduction:
• Walls of homes
• Heat exchangers
• Freezing of ground in winter
• Heat treatment of steel forgings
• Keeping constant skin temperature
Frostbite is a disease of skin when
exposed to freezing temperatures
12.3C Fourier’s Law of Heat Conduction
qx
dT
 k
A
dx
W
W K

2
m
m.K m
Assumptions:
1. Homogeneous media
2. 1Dimensional flow
12.3-3
btu
btu 0 F

2
hr. ft
hr. ft.0 F ft
qx: Heat transfer rate by conduction,
A: Area normal to flow, K: thermal
conductivity
dT/dx= Temperature gradient
• For steady state heat transfer, constant cross sectional area of a large plate wall, qx is constant
and we can integrate Fourier’s law for transfer between two points (x1, T1) and (x2 and T2),
separation of variables:
x2

x1
T
2
qx
dx   k  dT ,12.3-13
A
T1
qx
T T
k 1 2
A
x2  x1
12.3-14
Note that T is a linear profile in X (similar to velocity profile we have seen for a moving plate).
The rate of transport of heat can be thought of as:
Rate of a transfer process 
Driving force
Re sis tan ce
12.3-8
This equation states what we know
intuitively: that in order to transfer a
property such as heat or mass, we need a
driving force to overcome a resistance.
qx
dT
 k
A
dx
•
•
•
12.3-3
The units in Eq. (12.3-3) may also be expressed in the cgs
system, with qx in cal/s, A in cm2, k in cal/s · °C · cm, T in
°C, and x in cm.
In the English system, qx is in btu/h, A in ft2, T in °F, x in
ft, k in btu/h · °F · ft, and qx/A in btu/h · ft2.
From Appendix A.1, the conversion factors for thermal
conductivity are:
𝑏𝑡𝑢
𝑐𝑎𝑙
3
12.3-9
1
= 4.1365 × 10
2
ℎ𝑟. 𝑓𝑡 . ℉
𝑠. 𝑐𝑚. ℃
𝑏𝑡𝑢
𝑊
1
= 1.73073
2
ℎ𝑟. 𝑓𝑡 . ℉
𝑚. 𝐾
12.3-10
Units of Heat Flux
For heat flux and power,
𝑏𝑡𝑢
𝑊
1
= 3.1546 2
2
ℎ𝑟. 𝑓𝑡
𝑚
𝑏𝑡𝑢
1
= 0.29307 𝑊
ℎ𝑟
12.3-11
12.3-12
Question?
What will happen to the rate of heat transfer, q,
through the wall if:
 T across the wall doubles?
 A normal to the direction of heat transfer is
doubled?
 Wall thickness is doubled?
T
q x kA
x
Example:
The wall of an industrial furnace is constructed from
0.15 m thick fireclay brick having a thermal
conductivity of 1.7 W/m.K. Measurements made
during steady state operation reveal temperatures of
1400 and 1150 K at the inner and outer surfaces,
respectively. What is the rate of heat loss through a
wall which is 0.5 m by 3 m on a side?
Solution
Schematic: Always try to provide it
Knowns: Steady state; L = 0.15 m; A = 0.5 × 3 m2;
k = 1.7 W/m.K; T1 = 1400 L; T2 = 1150 K.
Assumptions: 1. Steady state, 2. One dimensional
through the wall, 3. Constant properties.
Analysis: heat transfer by conduction  Use Fouriers’law
q/ A k
(1400  1150) K
T1  T2
 1.7 W/m.K
L
0.15 m
= 2833 W/m2 (Flux)
Heat transfer rate: q  (q / A) A = (0.5 m × 3 m) 2833 W/m2 = 4250 W
Example 12.3-1. Heat Loss through an
Insulating Wall
Calculate the heat loss per m2 of surface area for an
insulating wall composed of 25.4-mm-thick fiber
insulating board, where the inside temperature is 352.7
K and the outside temperature is 297.1 K.
Solution of Example 12.3-1. Heat Loss
Through an Insulating Wall
Calculate the heat loss per m2 of surface area for an
insulating wall composed of 25.4-mm-thick fiber
insulating board, where the inside temperature is 352.7
K and the outside temperature is 297.1 K.
From Appendix A.3, the thermal conductivity of fiber insulating
board is 0.048 W/m·K. The thickness x2 – x1 = 0.0254 m.
Substituting into Eq. (12.3-14):
𝑞
𝑘
0.048
𝑊
=
𝑇1 − 𝑇2 =
352.7 − 297.1 = 105.1 2
𝐴 𝑥2 − 𝑥1
0.0254
𝑚
12.3D Thermal Conductivity-k


The thermal conductivity of a material is a measure of the
ability of the material to conduct heat.
High value for thermal conductivity
Good heat conductor

Low value
Poor heat conductor or insulator
12.3D Thermal Conductivity - k
 It is a property of the material. It is a measure of the ability of the material to
conduct heat. Determined experimentally (empirical constant).
 k depends on microscopic structure of the substance;
• Generally: ksolids > kliq > kgas
Why??
•kAl=202 > kBeef=1.35> kWater=0.569 > kair=0.0242 W/m.K
•
•
•
In Table 12.3-1, thermal conductivities of a few materials are
given for the purpose of comparison.
More-detailed data are given in Appendix A.3 for inorganic
and organic materials and Appendix A.4 for food and
biological materials.
As seen in Table 12.3-1, gases have quite low values of
thermal conductivity, liquids have intermediate values, and
solid metals have very high values.
Table 12.3-1:
1. Gases
• Follows the kinetic theory of gases
• Molecules are always moving
• Energy transfers as molecules colloid from hotter to cooler
molecules
• Smaller molecules move faster and as such have higher
conductivity
k Tabsolute
• k is independent of pressure up to few atm’s.
• As P goes to vacuum pressures, k goes to 0.
• Since gas theory is pretty accurate, we can use it to predict k
for gases.
Prediction of k for Gases
•
•
Theories to predict thermal conductivities of gases are reasonably
accurate.
The predictive model developed by Chapman–Enskog for gases is
given below:
𝑇
0.0829
𝑀𝑊 12.3-15
𝑘=
𝜎 2 Ω𝑘
Where k is thermal conductivity in W/m.K, T is temperature (K), MW is
the molecular weight, s is the molecular diameter in (Å)(listed in A.6), and
Ω𝑘 is the Lenard-Jones Collision integral.
Ω𝑘 =1.16145 𝑇 ∗ −0.14874 +0.52487exp(−0.77329𝑇 ∗ )+2.16178exp(-2.43787𝑇 ∗ )
𝑘 𝑇
12.3-16
Where 𝑇 ∗ = 𝐵 12.3-17
𝜀
e/kB=59.7, kB is Boltzmann constant and e is a scaling parameter of length specific for
each molecule
Example 12.3-2. Chapman–Enskog for the
Prediction of Thermal Conductivity of H2
Use Eqs. 12.3-16, 12.3-17, and 12.3-18 to calculate the thermal
conductivity of H2 at 273 K and 1 atm, and then compare that
value to the one listed in Table 12.3-1.
Solution of Example 12.3-2. Chapman–Enskog
for the Prediction of Thermal Conductivity of
H2
• H2 (MW = 2 g/gmol), T=273 K, P= 1 atm, s=2.827 Å from
A.6. You are also given e/kB=59.7 K
0.0829
𝑘=
• We need Ω𝑘
Ω𝑘 =1.16145 𝑇 ∗
𝑇∗ =
𝑇
𝑀𝑊
𝜎 2 Ω𝑘
−0.14874 +0.52487exp(−0.77329𝑇 ∗ )+2.16178exp(-2.43787𝑇 ∗ )
𝑘𝐵 𝑇
=273/59.7=4.6
𝜀
Ω𝑘 =1.16145 4.6 −0.14874 +0.52487exp(−0.77329 ∗ 4.6)+2.16178exp(2.43787*4.6) = 0.941
Solution of Example 12.3-2. Chapman–Enskog
for the Prediction of Thermal Conductivity of
H2
𝑘=
0.0829
273
2
2.827 2 (0.941)
𝑊
=0.129𝑚.𝐾
• The value listed in the Table 12.3-1 at 273 K is 0.167 W/m.K
• The error between listed and calculated values is:
Error = (0.167-0.129)×100% /0.167 =22.8%
•
The correlation should be used with caution
2. Liquids
• Energy transfers as molecules colloid from hotter to cooler
molecules.
• Efficiency of energy transfer is better than in gases because
molecules are highly backed together.
• k’s are determined via empirical correlations.
• k varies largely linearly with T
k = a+bT with a and b being empirical constants 12.3-18
• For liquids, k is independent of P.
• Water has very high k compared to organic solvents.
• Dilute mixtures made largely with water have k values close to
that of pure water (skim milk & apple sauce).
3. Solids
• k for solids vary widely with homogeneous metals having very
high k values compared to industrial solids such as rock,
cardboard, wool or wood.
• Mechanisms of heat transfer in metals can be divided into two
categories:
• Metals: Free electrons carry heat while they move within the
metal lattice.
• Heat is conducted by transmission of energy vibrations
between adjacent atoms.
• k of insulating materials like rock wool approaches k for air
because they have lots of trapped air.
• Super insulators (layers of composite materials filled with liquid
H2) have lower k compared to air.
• k for ice is much higher than that of water. k for frozen food is
much higher than k for unfrozen food.
Thermal diffusivity
Heat conducted
k


Heat stored
cp
( m2 s )

The thermal diffusivity represents how fast heat diffuses
through a material compared to heat stored in it

Appears in the transient heat conduction analysis (Chapter 14).

A material that has a high thermal conductivity or a low heat
capacity will have a large thermal diffusivity.

The larger the thermal diffusivity, the faster the propagation of
heat into the medium.
12.4 Convection
12.4A Convection as a Basic Heat Transfer
Mechanism

The transfer of heat by convection implies the transfer of heat by
bulk transport and the mixing of macroscopic elements of warmer
portions with cooler portions of a gas or liquid.

It also often refers to the energy exchange between a solid surface
and a fluid.

Examples of heat transfer by convection are heat loss from a car
radiator where air is circulated by a fan, stirring foods being cooked
in a vessel, blowing across the surface of a hot cup of coffee to cool
it, and so on.
12.4A Convection as a Basic Heat Transfer
Mechanism

Convection is commonly classified into
two sub-modes:
 Forced convection: Where a fluid is
forced to flow past a solid surface by a
pump, fan, or other mechanical means,
and
 Natural (or free) convection: where
warmer or cooler fluid next to the
solid surface causes a circulation
because of a density difference
resulting from the temperature
differences in the fluid.

A hot egg will
cool faster
when we blow
air on it.
Natural and Forced Convection:
The forces used to create convection currents in fluids are of two types:
 Forced convection (a): If the
currents are set in motion by action
of a mechanical device, such as a
pump or agitator blower. The flow is
independent of the density.
 Natural convection (b): If the
hot plate exposed to ambient air
without external sources of motion,
i.e. the movement of air is due to
density gradient near the plate (i.e.
difference in density causes the T
in the fluid).
Generally:
hfree < hforced
Convection
Examples:




Loss of heat from a car radiator when air is
circulated by a fan
Cooling of food in vessels being stirred
Cooling of a hot cup of coffee by blowing over it
Sweating
Hot, less dense
Water rises
Water cools,
become
more dense
sinks
12.4B Convective Heat Transfer Coefficient

The rate of convection heat transfer is expressed by
Newton’s law of cooling as:
q  hA(Tw  T f )


12.4-1
Tw: Wall temperature, Tf: Bulk Fluid temperature, A: Area normal to flow
h is the convection heat transfer coefficient in W/m2. °C. Often
called also as film coefficient as when fluid passes over a solid, there
will be a stationary thin film of fluid adjacent to the surface of solid
and this layer represents the most resistance to heat flow.
h depends on variables such as:

The surface geometry,
 h can be predicted empirically

The nature of fluid motion,
 Water has high h value.

The properties of the fluid,
 To convert h from English to
SI units: 1 btu/hr.ft2.oF=

The bulk fluid velocity,
5.6783 W/m2. K
 Temperature gradient.
Table 12.4-1
Forced convection
Natural convection
Forced convection
Example: Water at 300 K flows over both
sides of a plate of 1m × 2m in area,
maintained at 400 K. If the convective heat
transfer coefficient is 200 W/m2.K,
calculate the heat transfer.
q  2hA(Ts  T ) Why 2?
q= (2) (1) (2) m2 (200) (W/m2.K)(400 – 300) K
q = 80,000 W
Water 300 K
Solid plate 400 k
2m
1m
Example 12.4-1. Heat Loss
Through Convection
The forced convective heat-transfer coefficient of flowing hot
water (393 K) over a cold steel surface (283 K) is 1000.
Determine the heat transfer rate per unit surface area (W/m2)
from the water to the steel surface.
Solution of Example 12.4-1. Heat
Loss Through Convection
The forced convective heat-transfer coefficient of flowing hot
water (393 K) over a cold steel surface (283 K) is 1000.
Determine the heat transfer rate per unit surface area (W/m2)
from the water to the steel surface.
Given:
h=1000 W/m2.K
Tf= 393 K
Tw=283 K
𝑞
= ℎ(𝑇𝑓 − 𝑇𝑤 )=1000(393-283)=110,000 W/m2
𝐴
12.5 Radiation
12.5A Radiation as a Basic Heat Transfer
Mechanism
•
•
•
•
Radiation differs from heat transfer by conduction and
convection in that no physical medium is needed for its
propagation.
Radiation is the transfer of energy through space by means of
electromagnetic waves in much the same way as electromagnetic
light waves transfer light.
The same laws that govern the transfer of light govern the
radiant transfer of heat.
Solids and liquids tend to absorb the radiation being transferred
through them, so that radiation is important primarily in transfer
through space or gases.
Examples of Radiation




Radiation from sun to earth.
Cooking food when passed below red-hot electric heater.
Heating of fluids inside coils of a combustion furnace.
Radiation therapy in cancer treatment.
1. Nature of Radiant Heat Transfer
•
In conduction, heat is transferred from one part of a body to
another, and the intervening material is heated.
•
In convection, heat is transferred by the actual mixing of
materials and by conduction.
•
In radiant heat transfer, the medium through which the heat is
transferred usually is not heated.
•
Again, radiation heat transfer is the transfer of heat by
electromagnetic radiation.
1. Nature of Radiant Heat Transfer
•
Thermal radiation is a form of electromagnetic radiation similar
to X rays, light waves, gamma rays, and so on, differing only in
wavelength.
•
It obeys the same laws as light: it travels in straight lines, can be
transmitted through space and vacuum, and so on.
•
It is an important mode of heat transfer and is especially
important where large temperature differences occur, as, for
example, in a furnace with boiler tubes, in radiant dryers, or in
an oven baking food.
•
Radiation often occurs in combination with conduction and
convection. Ch. 17 will cover radiation in details.
1. Nature of Radiant Heat Transfer
In an elementary sense, the mechanism of radiant heat transfer is
composed of three distinct steps or phases:
•
The thermal energy of a hot source, such as the wall of a
furnace at T1, is converted into energy in the form of
electromagnetic-radiation waves.
•
These waves travel through the intervening space in straight
lines and strike a cold object at T2, such as a furnace tube
containing water to be heated.
•
The electromagnetic waves that strike the body are absorbed by
it and converted back to thermal energy or heat.
2. Absorptivity and Black Bodies
•
When thermal radiation (such as light waves) falls upon a body,
a portion is absorbed by the body in the form of heat, a
portion is reflected back into space, and a portion may actually
be transmitted through the body.
• For most cases in process engineering, bodies are opaque to
transmission, so this will be neglected. Hence, for opaque
bodies,
𝛼 + 𝜌 = 1 12.5-1
Where  is absorptivity or fraction absorbed and ρ is reflectivity or
fraction reflected.
2. Absorptivity and Black Bodies
•
•
•
•
•
•
•
•
A black body is defined as one that absorbs all radiant
energy and reflects none.
Hence, ρ = 0 and α = 1.0 for a black body.
Actually, in practice there are no perfect black bodies, but a
close approximation is a small hole in a hollow body, as
shown in Fig. 12.5-1.
The inside surface of the hollow body is blackened by
charcoal.
The radiation enters the hole and impinges on the rear
wall; a portion is absorbed there and a portion is reflected
in all directions.
The reflected rays impinge again, a portion is absorbed,
and the process continues.
Hence, essentially all of the energy entering is absorbed
and the area of the hole acts as a perfect black body.
The surface of the inside walls is “rough” and rays are
scattered in all directions, unlike a mirror, where they are
reflected at a definite angle.
2. Absorptivity and Black Bodies
•
As stated previously, a black body absorbs all radiant energy
falling on it and reflects none.
• Such a black body also emits radiation, depending on its
temperature, and does not reflect any.
• The ratio of the emissive power of a surface to that of a black
body is called emissivity ε and is 1.0 for a black body.
• Kirchhoff ’s law states that at the same temperature T1, α1 and ε1
of a given surface are the same, or
𝛼1 = 𝜀1 12.5-2
Eq. (12.5-2) holds for any black or nonblack solid surface.
3. Radiation From a Body and Emissivity.
•
The basic equation for heat transfer by radiation from a perfect
black body with an emissivity ε = 1.0 is
𝑞 = 𝐴𝜎𝑇 4
12.5-3
where q is heat flow in W, A is m2 surface area of the body, σ is a
constant 5.676 × 10–8 W/m2 · K4 (0.1714 × 10–8 btu/h·ft2·°R4),
and T is temperature of the black body in K (°R).
•
•
For a body that is not a black body and has an emissivity ε < 1.0
the emissive power is reduced by ε, or
𝑞 = 𝐴𝜖𝜎𝑇 4 12.5-4
Substances that have emissivities of less than 1.0 are called gray
bodies when the emissivity is independent of the wavelength. All
real materials have an emissivity ε < 1.
Table 12.5-1. Total Emissivity, ε, of Various
Surfaces
Surface
T(K)
Polished aluminum
T(°F) Emissivity, ε
Polished iron
Oxidized iron
Polished copper
Asbestos board
Oil paints, all colors
500
850
450
373
353
296
373
440
1070
350
212
176
74
212
0.039
0.057
0.052
0.74
0.018
0.96
0.92–0.96
Water
273
32
0.95
3. Radiation From a Body and Emissivity.
•
Since the emissivity ε and absorptivity α of a body are equal at
the same temperature, the emissivity, like absorptivity, is low for
polished metal surfaces and high for oxidized metal surfaces.
•
Typical values are given in Table 12.5-1, but they do vary some
with temperature.
•
Most nonmetallic substances have high values. Additional data
are tabulated in Appendix A.3.
12.5B Radiation to a Small Object from Its
Surroundings
•
In the case of a small gray object of area A1 m2 at temperature
T1 in a large enclosure at a higher temperature T2, there is a net
radiation to the small object.
•
The small body emits an amount of radiation to the enclosure
given by Eq. (12.5-4) as 𝐴1 𝜀1 𝜎𝑇14 . The emissivity ε1 of this body
is taken at T1.
•
The small body also absorbs an amount of energy from its
surroundings at T2 given by 𝐴2 α12 𝜎𝑇24
•
The α12 is the absorptivity of body 1 for radiation from the
enclosure at T2. The value of α12 is approximately the same as
the emissivity of this body at T2.
12.5B Radiation to a Small Object from Its
Surroundings
•
The net heat of absorption is then, by the Stefan–Boltzmann
equation,
𝑞 = 𝐴1 𝜀1 𝜎𝑇14 − 𝐴1 𝛼12 𝜎𝑇24 = 𝐴1 𝜎 𝜀1 𝑇14 − 𝛼12 𝑇24 12.5-5
•
A further simplification of Eq. (12.5-5) is usually made for
engineering purposes by using only one emissivity for the small
body, at temperature T2. Thus,
𝑞 = 𝐴1 𝜎𝜀 𝑇14 − 𝑇24
12.5-6
Example 12.5-1. Radiation to a Metal Tube
A small oxidized horizontal metal tube with an OD of 0.0254 m
(1 in.), 0.61 m (2 ft) long, and with a surface temperature at 588 K
(600°F) is in a very large furnace enclosure with firebrick walls and
the surrounding air at 1088 K (1500°F). The emissivity of the metal
tube is 0.60 at 1088 K and 0.46 at 588 K. Calculate the heat transfer
to the tube by radiation using SI and English units.
Solution of Example 12.5-1. Radiation to a
Metal Tube
A small oxidized horizontal metal tube with an OD of 0.0254 m
(1 in.), 0.61 m (2 ft) long, and with a surface temperature at 588
K (600°F) is in a very large furnace enclosure with firebrick walls
and the surrounding air at 1088 K (1500°F). The emissivity of
the metal tube is 0.60 at 1088 K and 0.46 at 588 K. Calculate the
heat transfer to the tube by radiation using SI and English units.
•
Since the large-furnace surroundings are very large compared to
the small enclosed tube, the surroundings, when viewed from
the position of the small body, appear black even if they are
actually gray, and Eq. (12.5-6) is applicable.
Solution of Example 12.5-1. Radiation to a
Metal Tube
•
Substituting given values into Eq. (12.5-6) with an ε of 0.6 at
1088 K,
𝑞 = 𝐴1 𝜎𝜀 𝑇14 − 𝑇24
A1=pDL=p(0.0254 m)(0.61 m) = 0.049 m2
𝑞 = 𝐴1 𝜎𝜀 𝑇14 − 𝑇24
𝑊
q=0.049 m2(0.6)(5.676×108 2 4)[(588)4K4-(1088)4K4]=-2125 W
𝑚 .𝐾
•
You can do the same for English units
Other examples of small objects in large enclosures that occur in
the process industries are:
• A loaf of bread in an oven receiving radiation from the walls
around it,
• A package of meat or food radiating heat to the walls of a
freezing enclosure,
• A hot ingot of solid iron cooling and radiating heat in a large
room, and
• A thermometer measuring the temperature in a large duct.
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