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Heat & Mass Transfer - acharya ng ranga agricultural university
Heat & Mass Transfer - acharya ng ranga agricultural university
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ACHARYA N.G. RANGA AGRICULTURAL UNIVERSITY<br />
<strong>Course</strong> <strong>No</strong>. FDEN- 221<br />
<strong>Course</strong> HEAT AND MASS TRANSFER<br />
<strong>Title</strong>:<br />
<strong>Credits</strong>: 2 (1 + 1)<br />
Prepared by<br />
Er. B. SREENIVASULA REDDY<br />
Assistant Professor (Food Engineering)<br />
College of Food Science and Technology<br />
Chinnarangapuram, Pulivendula – 516390<br />
YSR (KADAPA) District, Andhra Pradesh
DEPARTMENT OF FOOD ENGINEERING<br />
1 <strong>Course</strong> <strong>No</strong> : FDEN - 221<br />
2 <strong>Title</strong> : Heat and Mass Transfer<br />
3 Credit hours : 2 (1+1)<br />
4 General Objective : To impart knowledge to students on different<br />
modes of heat transfer through extended surfaces,<br />
study of heat exchanges and principles of mass<br />
transfer<br />
5 Specific Objectives :<br />
a) Theory : By the end of the course, the students will acquire<br />
knowledge from different modes of heat transfer,<br />
extended surfaces, boiling and condensation<br />
process and principles of heat exchangers which<br />
are very essential in dairy and food industries<br />
b) Practical By the end of the course, the students will learn<br />
efficient design of heat exchangers on the basis of<br />
overall heat transfer coefficient and LMTD<br />
A) Theory Lecture Outlines<br />
1 Introduction to Heat Transfer- Basic Mechanisms of Heat Transfer -<br />
Conduction Heat transfer - Fourier's Law of Heat Conduction- Convection<br />
Heat Transfer- Radiation Heat Transfer<br />
2 The basic equation that governs the transfer of heat in a solid - Thermal<br />
conductivity<br />
3 One-dimensional steady-state conduction of heat through some simple<br />
geometries - Conduction Through a Flat Slab or Wall - Conduction Through<br />
a Hollow Cylinder - Conduction Through a Hollow Sphere<br />
4 Conduction Heat Transfer Through A Composite Plane Wall - Conduction<br />
Heat Flow Through A Composite Cylinder<br />
5 The Overall Heat-Transfer Coefficient - Critical Thickness of Insulation<br />
6 Heat Source Systems: One-dimensional steady state heat conduction with<br />
heat generation : Heat Flow through slab / Plane Wall<br />
7 Steady state heat conduction with heat dissipation to environment-<br />
Introduction to extended surfaces (FINS) of uniform area of cross section -<br />
Different fin configurations - General Conduction Analysis Equation<br />
8 Equation of temperature distribution with different boundary conditions, Fin<br />
Performance and Overall surface efficiency of FINS<br />
9 Principles Of Unsteady-State Heat Transfer: Derivation of Basic Equation;
Simplified Case For Systems With Negligible Internal Resistance ; Total<br />
Amount of Heat Transferred ; Dimensional Analysis in Momentum Transfer<br />
10 Some important empirical relations used for determination of heat transfer<br />
coefficient: Nusselt’s number, Prandtl number, Reynold’s number, Grashoff<br />
number<br />
11 Radiation - heat transfer, Radiation Properties, radiation through black and<br />
grey surfaces, determination of shape factors<br />
12 Introduction to condensing and boiling heat transfer, Condensation Heat-<br />
Transfer Phenomena, Film Condensation Inside Horizontal Tubes , Boiling<br />
Heat Transfer<br />
13 Heat exchangers- general introduction; Double-pipe heat exchanger; Shelland-tube<br />
heat exchanger; Cross-flow exchanger; fouling factors, LMTD<br />
14 Design problems on heat exchangers: Calculation of heat exchanger size<br />
from known temperatures, Problem on Shell-and-tube heat exchanger ,<br />
Design of shell-and-tube heat exchanger<br />
15 Introduction To Mass Transfer: A Similarity of Mass, Heat, and Momentum<br />
Transfer Processes; Fick's Law for Molecular Diffusion<br />
16 Molecular Diffusion In Gases : Equimolar Counter diffusion in Gases<br />
B) Practical Class Outlines<br />
1 Determination of thermal conductivity of milk and dairy products<br />
2 Tutorials on heat conduction through slab, cylinder and sphere<br />
3 Tutorials on heat conduction through slab, cylinder and sphere<br />
4 Tutorials on heat conduction through slab, cylinder and sphere<br />
5 Tutorials on extended surfaces (FINS)<br />
6 Tutorials on unsteady state heat conduction<br />
7 Determination of specific heat of food materials<br />
8 Tutorials on determination of Nusselt’s number by dimensional analysis<br />
9 Tutorials on LMTD and NTU method of analysis of heat exchangers<br />
10 Study of shell and tube heat exchanger<br />
11 Study of plate heat exchanger<br />
12 Study on temperature distribution and heat transfer in HTST pasteurizer<br />
13 Study on temperature distribution and heat transfer in HTST pasteurizer<br />
14 Design problems on heat exchangers - I<br />
15 Design problems on heat exchangers - II<br />
16 Design problems on heat exchangers - III<br />
References<br />
1 Geankoplis, C.J. 1978. Transport Processes and Unit Operations. Allyn and<br />
Bacon Inc., Newton, Massachusetts.<br />
2 Holman, J. P. 1989. Heat Transfer. McGraw Hill Book Co., New Delhi.<br />
3 Incropera, F. P. and De Witt, D .P. 1980. Fundamentals of Heat and Mass
Transfer. John Wiley and Sons, New York.<br />
4 Gupta, C. P. and Prakash, R. 1994. Engineering Heat Transfer. Nem Chand<br />
and Bros., Roorkee
LECTURE NO.1<br />
INTRODUCTION TO HEAT TRANSFER- BASIC MECHANISMS OF HEAT<br />
TRANSFER - CONDUCTION HEAT TRANSFER - FOURIER'S LAW OF<br />
HEAT CONDUCTION- CONVECTION HEAT TRANSFER- RADIATION<br />
HEAT TRANSFER<br />
Introduction<br />
Heat transfer is the science that seeks to predict the energy transfer<br />
that may take place between material bodies as a result of a temperature<br />
difference. Thermodynamics deals with systems in equilibrium; it may be used<br />
to predict the amount of energy required to change a system from one<br />
equilibrium state to another; it may not be used to predict how fast a change<br />
will take place since the system is not in equilibrium during the process. Heat<br />
transfer supplements the first and second principles of thermodynamics by<br />
providing additional experimental rules which may be used to establish<br />
energy-transfer rates.<br />
As an example of the different kinds of problems that are treated by<br />
thermodynamics and heat transfer, consider the cooling of a hot steel bar that<br />
is placed in a pail of water. Thermodynamics may be used to predict the final<br />
equilibrium temperature of the steel bar-water combination. Thermodynamics<br />
will not tell us how long it takes to reach this equilibrium condition or what the<br />
temperature of the bar will be after a certain length of time before the<br />
equilibrium condition is attained. Heat transfer may be used to predict the<br />
temperature of both the bar and the water as a function of time.<br />
Basic Mechanisms of Heat Transfer<br />
Heat transfer may occur by anyone or more of the three basic<br />
mechanisms of heat transfer: conduction, convection, and radiation.<br />
1. Conduction Heat transfer. In conduction, heat can be conducted through<br />
solids, liquids, and gases. The heat is conducted by the transfer of the energy<br />
of motion between adjacent molecules. In a gas the "hotter" molecules, which<br />
have greater energy and motions, impart energy to the adjacent molecules at<br />
lower energy levels. This type of transfer is present to some extent in all<br />
solids, gases, or liquids in which a temperature gradient exists. In conduction,
energy can also be transferred by "free" electrons, which is quite important in<br />
metallic solids. Examples of heat transfer mainly by conduction are heat<br />
transfer through walls of exchangers or a refrigerator, heat treatment of steel<br />
forgings, freezing of the ground during the winter, and so on.<br />
Fourier's Law of Heat Conduction<br />
The basic rate transfer process equation for processes such as<br />
momentum transfer, heat transfer, mass transfer and electric current is as<br />
follows:<br />
driving force<br />
rate of a transfer process = ----------- (1)<br />
resis tan ce<br />
This equation states what we know intuitively: that in order to transfer a<br />
property such as heat or mass, we need a driving force to overcome a<br />
resistance.<br />
The transfer of heat by conduction also follows this basic equation and<br />
is written as Fourier's law for heat conduction in fluids or solids:<br />
where<br />
q x<br />
A<br />
dT = − k<br />
--------------- (2)<br />
dx<br />
qx<br />
is the heat-transfer rate in the x direction in watts (W), A is<br />
the cross-sectional area normal to the direction of flow of heat in m 2 , T is<br />
temperature in K, x is distance in m, and k is the thermal conductivity in<br />
W/m·K in the SI system. The quantity<br />
The quantity<br />
dT / dx<br />
q x<br />
/ A is called the heat flux in W/m 2 .<br />
is the temperature gradient in the x direction. The minus<br />
sign in eq. (2) is required because if the heat flow is positive in a given<br />
direction, the temperature decreases in this direction.<br />
Fourier's law, eq.(2), can be integrated for the case of steady-state<br />
heat transfer through a flat wall of constant cross-sectional area A, where the<br />
inside temperature is T 1 at point 1 and T 2 at point 2, a distance of x2 −x<br />
1 m<br />
away. Rearranging eq. (2),<br />
qx<br />
A<br />
x2<br />
∫<br />
x1<br />
dx<br />
= −<br />
k<br />
∫<br />
T2<br />
T1<br />
dT<br />
-------(3)<br />
Integrating, assuming that k is constant and does not vary with<br />
temperature and dropping the subscript x on<br />
qx<br />
for convenience,
q<br />
A<br />
k<br />
= ( T ) 1<br />
−T<br />
2 -------- (4)<br />
x −x<br />
2<br />
1<br />
2. Convection Heat Transfer. The transfer of heat by convection implies the<br />
transfer of heat by bulk transport and mixing of macroscopic elements of<br />
warmer portions with cooler portions of a gas or liquid. It also often refers to<br />
the energy exchange between a solid surface and a fluid. A distinction must<br />
be made between forced-convection heat transfer, where a fluid is forced<br />
to flow past a solid surface by a pump, fan, or other mechanical means, and<br />
natural or free convection, where warmer or cooler fluid next to the solid<br />
surface causes a circulation because of a density difference resulting from<br />
the temperature differences in the fluid. Examples of heat transfer by<br />
convection are loss of heat from a car radiator where the air is being<br />
circulated by a fan, cooking of foods in a vessel being stirred, cooling of a hot<br />
cup of coffee by blowing over the surface, and so on.<br />
Convective Heat-Transfer Coefficient<br />
It is well known that a hot piece of material will cool faster when air is<br />
blown or forced past the object. When the fluid outside the solid surface is in<br />
forced or natural convective motion, we express the rate of heat transfer from<br />
the solid to the fluid or vice versa, by the following equation:<br />
q= hA ( T w<br />
−T<br />
f<br />
) -------------- (5)<br />
where q is the heat-transfer rate in W, A is the area in m 2 , T w is the<br />
temperature of the solid surface in K, T f is the average or bulk temperature of<br />
the fluid flowing past in K, and h is the convective heat-transfer coefficient in<br />
W/m 2 .K.<br />
The coefficient h is a function of the system geometry, fluid properties,<br />
flow velocity, and temperature difference. In many cases, empirical<br />
correlations are available to predict this coefficient, since it often cannot be<br />
predicted theoretically. Since we know that when a fluid flows past a surface<br />
there is a thin, almost stationary layer or film of fluid adjacent to the wall which<br />
presents most of the resistance to heat transfer, we often call the coefficient h<br />
a film coefficient.<br />
3. Radiation Heat Transfer:- Radiation differs from heat transfer by<br />
conduction and convection in that no physical medium is needed for its
propagation. Radiation is the transfer of energy through space by means of<br />
electromagnetic waves in much the same way as electromagnetic light<br />
waves transfer light. The same laws that govern the transfer of light govern<br />
the radiant transfer of heat. Solids and liquids tend to absorb the radiation<br />
being transferred through them, so that radiation is important primarily in<br />
transfer through space or gases. The most important example of radiation is<br />
the transport of heat to the earth from the sun. Other examples are cooking of<br />
food when passed below red-hot electric heaters, heating of fluids in coils of<br />
tubing inside a combustion furnace, and so on.<br />
The rate of energy emitted by a black body is proportional to the fourth<br />
power of the absolute temperature of the body and directly proportional to its<br />
surface area. Thus<br />
q emitted<br />
= σAT<br />
4<br />
where σ is the proportionality constant and is called the Stefan-<br />
Boltzmann constant with the value of 5.669 x 10 -8 W/m 2 .K 4 . The equation is<br />
called the Stefan-Boltzmann law of thermal radiation, and it applies only to<br />
blackbodies.<br />
Problem Heat Loss Through an Insulating Wall<br />
Calculate the heat loss per m 2<br />
of surface area for an insulating wall<br />
composed of 25.4 -mm-thick fiber insulating board, where the inside<br />
temperature is 352.7 K and the outside temperature is 297.1 K. The thermal<br />
conductivity of fiber insulating board is 0.048 W/m.K<br />
Solution:<br />
The thickness x2 − x1<br />
= 0.0254 m.<br />
Substituting into the eq.<br />
q<br />
A<br />
= k<br />
0.048<br />
( T ) 1<br />
−<br />
2<br />
=<br />
x −x<br />
T 0.0254<br />
2<br />
1<br />
= 105.1 W/m 2<br />
(352 .7−297<br />
.1)
LECTURE NO.2<br />
THE BASIC EQUATION THAT GOVERNS THE TRANSFER OF HEAT IN A<br />
SOLID - THERMAL CONDUCTIVITY<br />
The basic equation that governs the transfer of heat in a solid<br />
Fig.2.1 Elemental volume for one-dimensional heat conduction analysis<br />
Consider (Fig. 2.1) the general case where the temperature may be<br />
changing with time and heat sources may be present within the body. For the<br />
element of thickness dx the following energy balance may be made:<br />
Energy conducted in left face + heat generated within element<br />
= change in internal energy + energy conducted out right face<br />
These energy quantities are given as follows:<br />
Energy in left face =<br />
q x<br />
∂T<br />
=−kA<br />
∂x<br />
Energy generated within element =<br />
q .<br />
Adx<br />
Change in internal energy =<br />
ρcA<br />
∂T<br />
dx ∂τ<br />
Energy out right face =<br />
q<br />
x+<br />
dx<br />
=− kA<br />
∂T<br />
∂x<br />
x+<br />
dx<br />
⎡ ∂T<br />
∂ ⎛ ∂T<br />
⎞ ⎤<br />
= − A⎢k<br />
+ ⎜k<br />
⎟dx<br />
⎥<br />
⎣ ∂x<br />
∂x<br />
⎝ ∂x<br />
⎠ ⎦
(In the derivations, the expression for the derivative at<br />
x + dx<br />
has been<br />
written in the form of a Taylor-Series expression with only the first two terms<br />
of the series employed for the development.)<br />
where<br />
.<br />
q = energy generated per unit volume, W /m 3<br />
c = specific heat of material, J /kg .°C<br />
ρ = density, kg/m 3<br />
Combining the relations above gives<br />
or<br />
heat source.<br />
∂<br />
− kA T<br />
+<br />
∂x<br />
q .<br />
Adx =<br />
∂T<br />
ρcA<br />
dx ∂τ<br />
⎡ ∂T<br />
− A⎢k<br />
⎣ ∂x<br />
.<br />
∂ ⎛ ∂T<br />
⎞ ∂T<br />
⎜k<br />
⎟dx<br />
+ q=ρc<br />
∂x<br />
⎝ ∂x<br />
⎠ ∂ τ<br />
∂ ⎛ ∂T<br />
⎞ ⎤<br />
+ ⎜k<br />
⎟dx<br />
∂x<br />
⎥<br />
⎝ ∂x<br />
⎠ ⎦<br />
This is the one-dimensional steady state heat-conduction equation with<br />
Figure 2.2 Elemental volume for three-dimensional heat-conduction analysis:<br />
(a) cartesian coordinates; (b) cylindrical coordinates;<br />
(c) spherical coordinates.
The general three-dimensional heat-conduction equation is<br />
∂ ⎛ ∂T<br />
⎜k<br />
∂x<br />
⎝ ∂x<br />
⎞<br />
⎟+<br />
⎠<br />
∂ ⎛ ∂T<br />
⎜k<br />
∂y<br />
⎝ ∂y<br />
⎞<br />
.<br />
∂ ⎛ ∂T<br />
⎞ ∂T<br />
⎟+ ⎜k<br />
⎟+ q=ρc<br />
⎠ ∂z<br />
⎝ ∂z<br />
⎠ ∂ τ<br />
For constant thermal conductivity equation can be written as<br />
where the quantity<br />
2 2 2<br />
∂ T ∂ T ∂ T q 1 ∂T<br />
+ + + =<br />
2 2 2<br />
∂x<br />
∂x<br />
∂x<br />
k α ∂τ<br />
.<br />
α = k / ρc<br />
is called the thermal diffusivity of the material.<br />
The larger the value of α , the faster heat will diffuse through the material. A<br />
high value of α could result either from a high value of thermal conductivity,<br />
which would indicate a rapid energy-transfer rate, or from a low value of the<br />
thermal heat capacity ρ c . A low value of the heat capacity would mean that<br />
less of the energy moving through the material would be absorbed and used<br />
to raise the temperature of the material; thus more energy would be available<br />
for further transfer. α has the units of m 2 /s.<br />
The three-dimensional heat-conduction equation with heat generation<br />
for cylindrical or spherical co-ordinates is<br />
Cylindrical coordinates:<br />
Spherical coordinates:<br />
2<br />
2 2<br />
∂ T 1 ∂T<br />
1 ∂ T ∂ T q 1 ∂T<br />
+ + + + =<br />
2<br />
2 2 2<br />
∂r<br />
r ∂r<br />
r ∂φ<br />
∂z<br />
k α ∂τ<br />
.<br />
1 ∂<br />
r ∂r<br />
2<br />
2<br />
( rT )<br />
+<br />
r<br />
2<br />
1 ∂ ⎛ ∂T<br />
⎞<br />
⎜sin<br />
θ ⎟+<br />
sin θ ∂θ<br />
⎝ ∂θ<br />
⎠ r<br />
2<br />
1<br />
sin<br />
2<br />
2<br />
∂ T q 1 ∂T<br />
+ =<br />
2<br />
θ ∂φ<br />
k α ∂τ<br />
The reduced form of the general equations for several cases of practical<br />
interest.<br />
Steady-state one-dimensional heat flow (no heat generation):<br />
2<br />
d T<br />
2<br />
dx<br />
= 0<br />
Steady-state one-dimensional heat flow in cylindrical coordinates (<strong>No</strong><br />
heat generation):<br />
.<br />
2<br />
d T<br />
2<br />
dr<br />
1 dT<br />
+<br />
r dr<br />
= 0
Steady-state one-dimensional heat flow with heat sources:<br />
2<br />
d T<br />
2<br />
dx<br />
.<br />
q<br />
+ = 0 k<br />
Thermal conductivity<br />
Thermal conductivity, k, is the property of a material that indicates its<br />
ability to conduct heat. Thermal conductivity is measured in W/m·K. The<br />
thermal conductivity predicts the rate of energy loss (in watts, W) through a<br />
piece of material.<br />
Thermal conductivity, k, also defined as the quantity of heat Q that flows<br />
per unit time through a food of unit thickness and unit area having unit<br />
temperature difference between faces.<br />
The reciprocal of thermal conductivity is thermal resistivity. In general,<br />
the thermal conductivity is strongly temperature-dependent.<br />
Thermal energy may be conducted in solids by two modes: lattice<br />
vibration and transport by free electrons. In good electrical conductors a<br />
rather large number of free electrons move about in the lattice structure of the<br />
material. Just as these electrons may transport electric charge, they may also<br />
carry thermal energy from a high-temperature region to a low-temperature<br />
region, as in the case of gases. In fact, these electrons are frequently referred<br />
to as the electron gas. Energy may also be transmitted as vibrational energy<br />
in the lattice structure of the material. In general, however, this latter mode of<br />
energy transfer is not as large as the electron transport, and for this reason<br />
good electrical conductors are almost always good heat conductors, viz.,<br />
copper, aluminum, and silver, and electrical insulators are usually good heat<br />
insulators. A notable exception is diamond, which is an electrical insulator, but<br />
which can have a thermal conductivity five times as high as silver or copper. It<br />
is this fact that enables a jeweler to distinguish between genuine diamonds<br />
and fake stones. A small instrument is available that measures the response<br />
of the stones to a thermal heat pulse. A true diamond will exhibit a far more<br />
rapid response than the non genuine stone.
LECTURE NO.3<br />
ONE-DIMENSIONAL STEADY-STATE CONDUCTION OF HEAT THROUGH<br />
SOME SIMPLE GEOMETRIES - CONDUCTION THROUGH A FLAT SLAB<br />
OR WALL - CONDUCTION THROUGH A HOLLOW CYLINDER -<br />
CONDUCTION THROUGH A HOLLOW SPHERE<br />
CONDUCTION HEAT TRANSFER<br />
One-dimensional steady-state conduction of heat through some simple<br />
geometries<br />
Fig. 3.1 Heat conduction in a flat wall: (a) geometry of wall, (b) temperature<br />
plot.<br />
Conduction Through a Flat Slab or Wall<br />
Consider a flat slab or wall (Fig. 3.1) where the cross-sectional area A<br />
and k in are constant,<br />
q k<br />
The eq. = ( T1<br />
−T<br />
2)<br />
can be rewrite as<br />
A x −x<br />
2<br />
1<br />
q<br />
A<br />
=<br />
k<br />
∆x<br />
where ∆ x=<br />
x 2<br />
−x1<br />
.<br />
( T 1<br />
−T<br />
2)<br />
The above indicates that if T is substituted for T 2 and x for x<br />
2 , the<br />
temperature varies linearly with distance, as shown in Fig.3.1(b).<br />
If the thermal conductivity is not constant but varies linearly with<br />
temperature, then substituting<br />
integrating,<br />
k = a=<br />
bT into the above equation and
q<br />
A<br />
T 2 .<br />
where<br />
T1<br />
+ T<br />
q 2<br />
a + b<br />
2<br />
k<br />
=<br />
( T1<br />
−T2<br />
) =<br />
m ( T1<br />
T2<br />
)<br />
A ∆x<br />
∆x<br />
−<br />
b T 1<br />
+ T<br />
k a<br />
2<br />
m<br />
= +<br />
2<br />
This means that the mean value of k (i.e., k m ) to use in<br />
k<br />
= ( T ) 1<br />
−T<br />
2 is the value of k evaluated at the linear average of T 1 and<br />
∆x<br />
The rate of a transfer process equals the driving force over the<br />
q k<br />
resistance and the equation = ( T 1<br />
−T<br />
2)<br />
can be rewritten in that form<br />
A ∆x<br />
as:<br />
where<br />
R ∆x<br />
/ kA<br />
T −T2<br />
T1<br />
−T2<br />
q = =<br />
∆x<br />
/ kA R<br />
1 =<br />
driving force<br />
resis tan ce<br />
= A and is the resistance in K/W.<br />
Fig. 3.2 Heat conduction in a cylinder<br />
Conduction Through a Hollow Cylinder<br />
In many instances in the process industries, heat is being transferred<br />
through the walls of a thick-walled cylinder, such as a pipe that may or may<br />
not be insulated. Consider the hollow cylinder in Fig.3.2 with an inside radius<br />
of r 1 , where the temperature is T 1 , an outside radius of r 2 having a temperature<br />
of T 2 , and a length of L m. Heat is flowing radially from the inside surface to<br />
the outside. Rewriting Fourier's law, with distance dr instead of dx,<br />
q x<br />
A<br />
= −<br />
k<br />
dT<br />
dr
The cross-sectional area normal to the heat flow is<br />
A =2πrL<br />
Substituting A value and rearranging, and integrating,<br />
q<br />
2πL<br />
r2<br />
∫<br />
r1<br />
dr<br />
r<br />
= −<br />
k<br />
∫<br />
T2<br />
T1<br />
dT<br />
q=<br />
k<br />
2πL<br />
ln( r<br />
2<br />
/ r1<br />
)<br />
( T −T<br />
)<br />
1<br />
2<br />
Conduction Through a Hollow Sphere<br />
Heat conduction through a hollow sphere is another case of onedimensional<br />
conduction. Using Fourier's law for constant thermal conductivity<br />
with distance dr, where r is the radius of the sphere,<br />
the radius.<br />
q x<br />
A<br />
= −<br />
k<br />
dT<br />
dr<br />
The cross-sectional area normal to the heat flow is<br />
2<br />
A = 2πr<br />
Substituting A value and rearranging, and integrating,<br />
q<br />
4π<br />
r2<br />
∫<br />
r1<br />
dr<br />
2<br />
r<br />
= −<br />
k<br />
∫<br />
T2<br />
T1<br />
( T −T<br />
)<br />
4πk<br />
1 2<br />
q =<br />
1/ r −1/<br />
r<br />
1<br />
2<br />
=<br />
dT<br />
T −T<br />
( 1/ r −1/<br />
r ) / 4πk<br />
It can easily be shown that the temperature varies hyperbolically with<br />
1<br />
1<br />
2<br />
2
LECTURE NO.4<br />
CONDUCTION HEAT TRANSFER THROUGH A COMPOSITE PLANE WALL<br />
- CONDUCTION HEAT FLOW THROUGH A COMPOSITE CYLINDER<br />
CONDUCTION HEAT TRANSFER THROUGH A COMPOSITE PLANE<br />
WALL<br />
Fig. 4.1 Heat flow through multilayer wall<br />
Consider the heat flow through composite wall made of several<br />
materials of different thermal conductivities and thicknesses. An example is a<br />
wall of a cold storage, constructed of different layers of materials of different<br />
insulating properties. All materials are arranged in series in the direction of<br />
heat transfer, as shown in the above Figure.<br />
The thickness of the walls are x 1, x 2 , and x 3 and the thermal<br />
conductivites of the walls are K 1 , K 2 , and K 3 , respectively. The temperatures at<br />
the contact surfaces are T 2 , T 3 , and T 4.<br />
From Fourier’s Law,<br />
q<br />
=−K A<br />
A<br />
This may be written as<br />
dT<br />
dx<br />
∆T<br />
= T<br />
x1<br />
T<br />
1<br />
−T<br />
2<br />
= . q<br />
K A<br />
x2<br />
T<br />
2<br />
−T<br />
3<br />
= . q<br />
K A<br />
1<br />
2<br />
2<br />
−T<br />
x3<br />
T<br />
4<br />
−T<br />
3<br />
= . q<br />
K A<br />
3<br />
1<br />
q.<br />
∆x<br />
=−<br />
KA
Total temperature difference, ∆T<br />
=∆T<br />
1<br />
+∆T<br />
2<br />
+ ∆T<br />
3<br />
⎡ x1<br />
x2<br />
x3<br />
T1 −T4<br />
= q.<br />
⎢ + +<br />
⎣K1<br />
A K2<br />
A K31<br />
where, T1 − T4<br />
= thermal potential responsible for heat flow. The<br />
⎤<br />
⎥<br />
A⎦<br />
⎡ x1<br />
x2<br />
x<br />
⎢ + +<br />
⎣K1<br />
A K2<br />
A K31<br />
3<br />
⎤<br />
⎥<br />
A⎦<br />
is known as the total thermal resistance of the<br />
composite ass. It is similar to the electrical resistance in series.<br />
The thermal circuit for multilayer rectangular system is shown in the<br />
following figure.<br />
Fig 4.2 Electrical analog of one dimensional heat transfer through composite<br />
wall<br />
T1<br />
−T4<br />
q =<br />
R + R + R<br />
CONDUCTION HEAT FLOW THROUGH A COMPOSITE CYLINDER<br />
1<br />
2<br />
3<br />
Fig.4.3 One-dimensional heat flow through multiple cylindrical sections and<br />
electrical analog<br />
Consider a long cylinder of inside radius r i , outside radius r o , and length<br />
L, such as the one shown in above Figure 4.3. We expose this cylinder to a
temperature differential T i - T o and determine what the heat flow will be. For a<br />
cylinder with length very large compared to diameter, it may be assumed that<br />
the heat flows only in a radial direction, so that the only space coordinate<br />
needed to specify the system is r. Again, Fourier's law is used by inserting the<br />
proper area relation. The area for heat flow in the cylindrical system is<br />
A =2πrL<br />
so that Fourier's law is written<br />
with the boundary conditions<br />
q<br />
r<br />
dT<br />
= −k<br />
Ar<br />
or<br />
dr<br />
q r<br />
=−2πk rL<br />
T = T i at r = r i ,<br />
dT<br />
dr<br />
T = T o at r = r o<br />
The solution to equation<br />
q<br />
r<br />
dT<br />
= −k<br />
Ar<br />
is<br />
dr<br />
q=<br />
k<br />
2πL<br />
ln( r<br />
2<br />
/ r1<br />
)<br />
( T −T<br />
)<br />
1<br />
2<br />
and the thermal resistance in this case is<br />
R<br />
th<br />
ln( ro<br />
/ ri<br />
)<br />
=<br />
2πkL<br />
The thermal-resistance concept may be used for multiple-layer<br />
cylindrical walls just as it was used for plane walls. For the three-layer system<br />
shown in Figure the solution is<br />
2π<br />
L ( T1<br />
−T4<br />
)<br />
q =<br />
ln( r2<br />
/ r1<br />
) ln( r3<br />
/ r2<br />
) ln( r4<br />
/ r2<br />
)<br />
+ +<br />
K K K<br />
a<br />
The thermal circuit is also shown in Figure.<br />
B<br />
B
LECTURE NO.5<br />
THE OVERALL HEAT-TRANSFER COEFFICIENT - CRITICAL THICKNESS<br />
OF INSULATION<br />
5.1 THE OVERALL HEAT-TRANSFER COEFFICIENT<br />
Fig. 5.1 Heat flow with convective boundaries in plane wall<br />
Fig. 5.2 Heat flow with convective boundaries in plane wall- Electrical<br />
Analog<br />
In many practical situations the surface temperatures (or boundary<br />
conditions at the surface) are not known, but there is a fluid on both sides of<br />
the solid surfaces. Consider the plane wall in the Figure 5.1, with a hot fluid at<br />
temperature T 1 on the inside surface and a cold fluid at T 4 on the outside<br />
surface. The convective coefficient on the outside is h o W/m 2 .K and h i on the<br />
inside.<br />
The heat transfer is expressed by<br />
q=<br />
h A<br />
i<br />
K<br />
A<br />
A<br />
( T −T<br />
) = ( T −T<br />
) = h A T − )<br />
1 2<br />
2 3 o<br />
(<br />
3<br />
T4<br />
∆x<br />
A<br />
The heat-transfer process may be represented by the resistance<br />
network in electrical analog Figure, and the overall heat transfer is calculated
as the ratio of the overall temperature difference to the sum of the thermal<br />
resistances:<br />
T1<br />
−T4<br />
T1<br />
−T<br />
q=<br />
=<br />
1 ∆x<br />
A<br />
1<br />
+ +<br />
∑R<br />
h A K A h A<br />
∴<br />
4<br />
i<br />
A<br />
The overall heat transfer by combined conduction and convection is<br />
frequently expressed in terms of an overall heat-transfer coefficient U, defined<br />
by the relation<br />
q=<br />
U A∆<br />
T overall<br />
o<br />
where ∆ T overall<br />
= T 1<br />
−T<br />
4 and U is<br />
U<br />
1<br />
∆x<br />
1<br />
A<br />
= + + W/m<br />
hi<br />
K<br />
A<br />
h<br />
2 .K<br />
o<br />
value as<br />
The overall heat-transfer coefficient is also related to the R<br />
U =<br />
1<br />
R value<br />
A more important application is heat transfer from a fluid outside a<br />
cylinder, through a metal wall, to a fluid inside the tube, as often occurs in<br />
heat exchangers.<br />
boundaries.<br />
Figure 5.3 Resistance analogy for hollow cylinder with convection
<strong>No</strong>te that the area for convection is not the same for both fluids in this<br />
case, these areas depend on the inside tube diameter and wall thickness. In<br />
this case the overall heat transfer would be expressed by<br />
q =<br />
1<br />
h A<br />
i<br />
i<br />
T1<br />
−T<br />
ln<br />
o<br />
+<br />
2πKL<br />
4<br />
( r / r )<br />
i<br />
1<br />
+<br />
h A<br />
o<br />
o<br />
The terms A i and A o represent the inside and outside surface areas of<br />
the inner tube. The overall heat-transfer coefficient may be based on either<br />
the inside or the outside area of the tube. Accordingly,<br />
Ui<br />
=<br />
1<br />
h<br />
i<br />
1<br />
Ai<br />
ln<br />
o<br />
+<br />
2πKL<br />
( r / r )<br />
i<br />
Ai<br />
+<br />
A<br />
o<br />
1<br />
h<br />
o<br />
U<br />
o<br />
=<br />
Ao<br />
A<br />
i<br />
1<br />
h<br />
i<br />
+<br />
1<br />
Ao<br />
ln<br />
o<br />
2πKL<br />
( r / r )<br />
i<br />
1<br />
+<br />
h<br />
o<br />
The general notion, for either the plane wall or cylindrical coordinate<br />
system, is that<br />
U<br />
A<br />
1<br />
=<br />
∑R<br />
th<br />
=<br />
R<br />
1<br />
th , Overall<br />
5.2 CRITICAL THICKNESS OF INSULATION<br />
Fig. 5.4 Critical radius for insulation of cylinder or pipe<br />
Consider a layer of insulation is installed around the outside of a<br />
cylinder whose radius r 1 , is fixed and with a length L. The cylinder has a high<br />
thermal conductivity and the inner temperature T 1 at point r 1 outside the<br />
cylinder is fixed. An example is the case where the cylinder is a metal pipe
with saturated steam inside. The outer surface of the insulation at T 2 is<br />
exposed to an environment at T o where convective heat transfer occurs. It is<br />
not obvious if adding more insulation with a thermal conductivity of k will<br />
decrease the heat-transfer rate.<br />
At steady state the heat-transfer rate q through the cylinder and the<br />
insulation equals the rate of convection from the surface:<br />
( T − )<br />
q = h A<br />
2 -------------- (1)<br />
o<br />
T o<br />
T1<br />
−T4<br />
q =<br />
1 ln<br />
o i<br />
+<br />
h A 2πKL<br />
i<br />
i<br />
( r / r )<br />
1<br />
+<br />
h A<br />
As insulation is added, the outside area, which is<br />
o<br />
o<br />
--------- (2)<br />
A<br />
2<br />
= 2πr<br />
L , increases,<br />
but T 2 decreases. However, it is not apparent whether q increases or<br />
decreases. To determine this, an equation similar to Eq. (2) with the<br />
resistance of the insulation represented by Eq.(3) is written using the two<br />
resistances:<br />
r2<br />
−r<br />
R =<br />
K A<br />
1<br />
lm<br />
⎛r<br />
⎞<br />
2<br />
ln<br />
⎜<br />
r<br />
⎟<br />
1<br />
=<br />
⎝ ⎠<br />
2πKL<br />
A<br />
lm = log mean area<br />
2<br />
q =<br />
ln<br />
2<br />
K<br />
πL<br />
( T1<br />
−To<br />
)<br />
( r / r ) 1<br />
1<br />
+<br />
r h<br />
2<br />
o<br />
As the outside radius, r 2 , increases, then in the denominator, the first<br />
term increases but the second term decreases.<br />
Thus, there must be a critical radius, r c , that will allow maximum rate of<br />
heat transfer, q.<br />
To determine the effect of the thickness of insulation on q, we take the<br />
derivative of q with respect to r 2 , equate this result to zero, and obtain the<br />
following for maximum heat flow. The maximization condition is<br />
d<br />
q<br />
dr<br />
=<br />
−2πL<br />
( T −T<br />
⎡<br />
⎢<br />
⎣<br />
1<br />
( r / r )<br />
⎛ 1<br />
)<br />
⎜<br />
⎝ r2<br />
K<br />
2 ln<br />
2 1<br />
1<br />
K<br />
o<br />
+<br />
r h<br />
2<br />
1<br />
−<br />
r h<br />
0<br />
⎤<br />
⎥<br />
⎦<br />
2<br />
2<br />
2<br />
o<br />
⎞<br />
⎟<br />
⎠
T 1 ,T 0 , K, L, r o , r i are constant terms.<br />
Therefore,<br />
⎛ 1<br />
⎜<br />
⎝r2<br />
K<br />
1<br />
−<br />
2<br />
r<br />
1 1 =<br />
r K r<br />
2<br />
2<br />
h o<br />
2<br />
2<br />
h o<br />
⎞<br />
⎟=<br />
0<br />
⎠<br />
When outside radius becomes equal to critical radius, or r 2 = r c , we get<br />
( r 2<br />
) =<br />
cr<br />
k<br />
h<br />
o<br />
Where (r 2 ) cr is the value of the critical radius when the heat transfer rate<br />
is a maximum. Hence, if the outer radius r 2 is less than the critical value,<br />
adding more insulation will actually increase the heat- transfer rate q. Also, if<br />
the outer radius is greater than the critical, adding more insulation will<br />
decrease the heat transfer rate. Using values of K and h o typically<br />
encountered, the critical radius is only a few mm. As a result, adding<br />
insulation on small electrical wires could increase the heat loss. Adding<br />
insulation to large pipes decreases the heat transfer rate.<br />
radius of r 2 ,<br />
When no insulation is provided then for a metal pipe with an outside<br />
q<br />
bare<br />
= 2πr<br />
2<br />
L ho<br />
2<br />
( T −T<br />
)<br />
The rate of heat transfer from an insulated pipe, where the annular<br />
insulating shell has an inside radius of r 2 and an outer radius of r 3 ,<br />
o<br />
q<br />
insulated<br />
Then,<br />
2 r3<br />
L ho<br />
= π<br />
r3<br />
h<br />
1+<br />
K<br />
( T −T<br />
)<br />
o<br />
2<br />
r<br />
ln<br />
r<br />
3<br />
2<br />
o<br />
q<br />
q<br />
insulated<br />
bare<br />
r<br />
=<br />
r<br />
3<br />
2<br />
⎡<br />
⎢ 1<br />
⎢<br />
⎢ r3<br />
ho<br />
1+<br />
⎢<br />
⎣ K<br />
ln<br />
⎤<br />
⎥<br />
⎥<br />
r3<br />
⎥<br />
r ⎥<br />
2 ⎦
LECTURE NO.6<br />
HEAT SOURCE SYSTEMS: ONE-DIMENSIONAL STEADY STATE HEAT<br />
CONDUCTION WITH HEAT GENERATION: HEAT FLOW THROUGH<br />
SLAB / PLANE WALL<br />
HEAT-SOURCE SYSTEMS<br />
A number of interesting applications of the principles of heat transfer<br />
are concerned with systems in which heat may be generated internally.<br />
Nuclear reactors are one example; electrical conductors and chemically<br />
reacting systems are others.<br />
Fig.6.1 Sketch illustrating one- dimensional conduction problem with<br />
heat generation.<br />
One-dimensional steady state heat conduction with heat generation:<br />
Heat Flow through slab / Plane Wall<br />
Consider the plane wall with uniformly distributed heat sources shown<br />
in the above Figure. The thickness of the wall in the x direction is 2L, and it is<br />
assumed that, the dimensions in the other directions are sufficiently large that<br />
the heat flow may be considered as one-dimensional. The heat generated per<br />
unit volume is q • , and we assume that the thermal conductivity does not vary<br />
with temperature. (This situation might be produced in a practical situation by<br />
passing a current through an electrically conducting material)<br />
The differential equation which governs the heat flow is
2<br />
d T<br />
2<br />
dx<br />
.<br />
q<br />
+ = 0 k<br />
2<br />
d T q<br />
=−<br />
2<br />
dx k<br />
Integrating twice with respect to x results in<br />
dT<br />
dx<br />
.<br />
.<br />
q<br />
=− x+<br />
c<br />
k<br />
.<br />
q 2<br />
T = − x + c1x<br />
+ c -------------- (1)<br />
2<br />
2k<br />
For the boundary conditions we specify the temperatures<br />
on either side of the wall, i.e.,<br />
T = T w at x = ± L<br />
1<br />
Since the temperature must be the same on each side of the wall, c 1<br />
must be zero.<br />
dT<br />
= 0,<br />
q<br />
• = 0⇒c<br />
1<br />
= 0<br />
The temperature at the mid plane (x = 0) is denoted by T o and from<br />
Equation (1)<br />
At mid plane = x=0 and T= T 0<br />
T o = c 2<br />
The temperature distribution equation (1) becomes<br />
2<br />
T =− x +<br />
.<br />
q<br />
2k<br />
T<br />
0<br />
T<br />
.<br />
q<br />
2k<br />
x<br />
2<br />
−T<br />
0<br />
= − --------------- (2)<br />
Assumed T=T w at x= L<br />
.<br />
q<br />
2k<br />
L<br />
2<br />
T w<br />
−T<br />
0<br />
= − ----------------- (3)<br />
(2)<br />
(3)<br />
T −T0<br />
⇒<br />
T −T<br />
w<br />
0<br />
2<br />
⎛ x ⎞<br />
= ⎜ ⎟<br />
⎝ L ⎠<br />
------------ (4)
An expression for the midplane temperature T o may be obtained through an<br />
energy balance. At steady-state conditions the total heat generated must<br />
equal the heat lost at the faces. Thus<br />
⎛<br />
2⎜−<br />
KA<br />
⎝<br />
dT<br />
dx<br />
⎤<br />
⎥<br />
⎦<br />
x=<br />
L<br />
⎞ •<br />
⎟=<br />
q.<br />
A.2L<br />
⎠<br />
where A is the cross-sectional area of the plate. The temperature gradient at<br />
the wall is obtained by differentiating Equation (4):<br />
T −T0<br />
T −T<br />
w<br />
0<br />
⎛<br />
= ⎜<br />
⎝<br />
x<br />
L<br />
2<br />
⎞<br />
⎟<br />
⎠<br />
T −T<br />
x<br />
⎝L<br />
⎛ ⎞<br />
0<br />
= ( Tw<br />
−T0<br />
) ⎜ ⎟ ⎠<br />
2x<br />
dT =<br />
w 0<br />
.<br />
2<br />
L<br />
( T −T<br />
) dx<br />
dT<br />
⎤<br />
⎦<br />
at x= L ( T T )<br />
dx ⎥<br />
=<br />
w−<br />
o<br />
. L<br />
x=<br />
L<br />
2<br />
2<br />
Since,<br />
dT<br />
−K.<br />
dx<br />
= q<br />
•<br />
L<br />
dT<br />
dx<br />
•<br />
q L<br />
= − K<br />
•<br />
qL<br />
T w<br />
− 2 0<br />
=−<br />
L K<br />
( T )<br />
•<br />
2<br />
qL<br />
T w<br />
−T<br />
0<br />
=−<br />
2K<br />
•<br />
2<br />
qL<br />
T<br />
0<br />
= + T w<br />
2K<br />
----- (5)
LECTURE NO.7<br />
STEADY STATE HEAT CONDUCTION WITH HEAT DISSIPATION TO<br />
ENVIRONMENT- INTRODUCTION TO EXTENDED SURFACES<br />
(FINS) OF UNIFORM AREA OF CROSS SECTION - DIFFERENT FIN<br />
CONFIGURATIONS - GENERAL CONDUCTION ANALYSIS<br />
EQUATION<br />
Steady State heat conduction with heat dissipation to environment:<br />
Introduction to extended surfaces (FINS) of Uniform area of cross<br />
section<br />
The term extended surface is commonly used to depict an important<br />
special case involving heat transfer by conduction within a solid and heat<br />
transfer by convection (and/or radiation) from the boundaries of the solid.<br />
Heat transfer from the boundaries of a solid to be in the same direction as<br />
heat transfer by conduction in the solid. In contrast, for an extended surface,<br />
the direction of heat transfer from the boundaries is perpendicular to the<br />
principal direction of heat transfer in the solid.<br />
Fig. 7.1 Combined conduction and convection in a structural element.<br />
Consider a strut that connects two walls at different temperatures and<br />
across which there is fluid flow (Figure 7.1). With T 1 >T 2 , temperature gradients<br />
in the x-direction sustain heat transfer by conduction in the strut. However,<br />
with T 1 >T 2 > T∞there is concurrent heat transfer by convection to the fluid,<br />
causing q x , and hence the magnitude of the temperature gradient,<br />
decrease with increasing x.<br />
dT / dx<br />
An extended surface is used specifically to enhance heat transfer<br />
between a solid and adjoining fluid. Such an extended surface is termed a fin.<br />
, to
The heat transfer rate may be increased by increasing the surface area<br />
across which the convection occurs. This is done by employing fins that<br />
extend from the wall into the surrounding fluid. The thermal conductivity of the<br />
fin material can have a strong effect on the temperature distribution along the<br />
fin and therefore influences the degree to which the heat transfer rate is<br />
enhanced. Ideally, the fin material should have a large thermal conductivity to<br />
minimize temperature variations from its base to its tip. In the limit of infinite<br />
thermal conductivity, the entire fin would be at the temperature of the base<br />
surface, thereby providing the maximum possible heat transfer enhancement.<br />
Fig. 7.2 Use of fins to enhance heat transfer from a plane wall (a) Bare<br />
surface (b) Finned surface.<br />
Examples of fin applications are easy to find. Consider the<br />
arrangement for cooling engine heads on motorcycles and lawn mowers or for<br />
cooling electric power transformers. Consider also the tubes with attached fins<br />
used to promote heat exchange between air and the working fluid of an air<br />
conditioner. Two common finned tube arrangements are shown in Figure 7.3<br />
Fig.7.3 Schematic of typical finned-tube heat exchangers.
Different fin configurations<br />
Different fin configurations are illustrated in Figure 7.4. A straight fin is<br />
any extended surface that is attached to a plane wall. It may be of uniform<br />
cross-sectional area, or its cross-sectional area may vary with the distance x<br />
from the wall. An annular fin is one that is circumferentially attached to a<br />
cylinder, and its cross section varies with radius from the wall of the cylinder.<br />
The foregoing fin types have rectangular cross sections, whose area may be<br />
expressed as a product of the fin thickness t and the width w for straight fins<br />
or the circumference<br />
2 πr<br />
for annular fins. In contrast a pin fin, or spine, is an<br />
extended surface of circular cross section. Pin fins may also be of uniform or<br />
non-uniform cross section. In any application, selection of a particular fin<br />
configuration may depend on space, weight, manufacturing, and cost<br />
considerations, as well as on the extent to which the fins reduce the surface<br />
convection coefficient and increase the pressure drop associated with flow<br />
over the fins.<br />
FIGURE 7.4 Fin configurations. (a) Straight fin of uniform cross section. (b)<br />
Straight fin of non-uniform cross section. (c) Annular fin. (d) Pin fin / spine.<br />
General Conduction Analysis Equation<br />
As engineers we are primarily interested in knowing the extent to which<br />
particular extended surfaces or fin arrangements could improve heat transfer<br />
from a surface to the surrounding fluid. To determine the heat transfer rate<br />
associated with a fin, we must first obtain the temperature distribution along<br />
the fin.
Fig.7.5 Energy balance for an extended surface<br />
A general form of the energy equation for an extended surface is as<br />
follows:<br />
2<br />
d T<br />
2<br />
dx<br />
⎛ 1<br />
⎜<br />
⎝ Ac<br />
dA<br />
dx<br />
⎞dT<br />
⎟<br />
⎠ dx<br />
⎛ 1<br />
−<br />
⎜<br />
⎝ Ac<br />
h dA<br />
k dx<br />
⎞<br />
⎟<br />
⎠<br />
( T −T<br />
) = 0<br />
c<br />
s<br />
+<br />
∞<br />
Its solution for appropriate boundary conditions provides the<br />
temperature distribution, which may be used with Fourier’s equation (<br />
dT<br />
q x<br />
= −kA ) to calculate the conduction rate at any x.<br />
dx
LECTURE NO.8<br />
EQUATION OF TEMPERATURE DISTRIBUTION WITH DIFFERENT<br />
BOUNDARY CONDITIONS, FIN PERFORMANCE AND OVERALL<br />
SURFACE EFFICIENCY OF FINS<br />
Fins of Uniform Cross-Sectional Area<br />
To solve general form of fin energy equation it is necessary to be more<br />
specific about the geometry. We begin with the simplest case of straight<br />
rectangular and pin fins of uniform cross section (Figure 8.1). Each fin is<br />
attached to a base surface of temperature T(O) = T b and extends into a fluid<br />
of temperature<br />
T<br />
∞.<br />
Fig. 8.1 Straight fins of uniform cross section (a) Rectangular fin (b) Pin<br />
fin<br />
For the prescribed fins, A c is a constant and A s = P x , where A s is the<br />
surface area measured from the base to x and P is the fin perimeter.<br />
Accordingly, with dA c /dx = 0 and dA s /dx = P, general form of the energy<br />
equation for an extended surface reduces to<br />
2<br />
d T<br />
2<br />
dx<br />
hP<br />
−<br />
kA<br />
c<br />
( T − T ) = 0<br />
∞<br />
--------(1)<br />
To simplify the form of this equation, we transform the dependent<br />
variable by defining an excess temperature θ as<br />
where, since<br />
into<br />
∞<br />
Equation (3), we then obtain<br />
θ ( x ) = T ( x)<br />
−T<br />
-------------(2)<br />
∞<br />
T is a constant, d θ / dx = dT / dx . . Substituting Equation (2)
where<br />
2<br />
d θ 2<br />
− θ= 0<br />
2 m<br />
dx<br />
hP<br />
m 2 =<br />
-------------(4)<br />
kA c<br />
--------(3)<br />
Equation (3) is a linear, homogeneous, second-order differential equation with<br />
constant coefficients. Its general solution is of the form<br />
mx −mx<br />
θ ( x)<br />
= C1<br />
e + C2e<br />
---------------- (5)<br />
By substitution it may readily be verified that Equation (5) is indeed a<br />
solution to Equation (3).<br />
To evaluate the constants C 1 and C 2 of Equation (5), it is necessary to<br />
specify appropriate boundary conditions. One such condition may be specified<br />
in terms of the temperature at the base of the fin (x = 0)<br />
θ =<br />
( 0)= Tb<br />
−T<br />
∞<br />
θb<br />
----------------- (6)<br />
The second condition, specified at the fin tip (x = L), may correspond to<br />
one of four different physical situations.<br />
The first condition, case A, considers convection heat transfer from the<br />
fin tip. Applying an energy balance to a control surface about this tip (Figure<br />
8.2), we obtain<br />
hA<br />
c<br />
[ T ( L)−T<br />
]<br />
∞<br />
= −kA<br />
dT<br />
dx<br />
x=<br />
L<br />
or<br />
h<br />
dT<br />
( L)=−<br />
kA<br />
dx<br />
θ --------------- (7)<br />
x=<br />
L<br />
Figure 8.2 Conduction and convection in a fin of uniform cross section.
That is, the rate at which energy is transferred to the fluid by<br />
convection from the tip must equal the rate at which energy reaches the tip by<br />
conduction through the fin. Substituting Equation (5) into Equations (6) and<br />
(7), we obtain, respectively,<br />
And<br />
θb = C 1<br />
+ C 2 ----------- (8)<br />
mL −mL<br />
−mL<br />
h( C1e<br />
+ C2e<br />
) = km(<br />
C2e<br />
−C1<br />
e<br />
Solving for C 1 , and C 2 , it may be shown, after some manipulation, that<br />
θ cosh m(<br />
L − x)<br />
+ ( h / mk )sinh m(<br />
L − x)<br />
=<br />
θ cosh mL + ( h / mk ) sinh mL<br />
b<br />
mL<br />
)<br />
------- (9)<br />
The form of this temperature distribution is shown schematically in<br />
Figure 8.2. <strong>No</strong>te that the magnitude of the temperature gradient decreases<br />
with increasing x. This trend is a consequence of the reduction in the<br />
conduction heat transfer q x<br />
(x)<br />
with increasing x due to continuous<br />
convection losses from the fin surface.<br />
We are particularly interested in the amount of heat transferred from<br />
the entire fin. From Figure 8.2 it is evident that the fin heat transfer rate<br />
may be evaluated in two alternative ways, both of which involve use of the<br />
temperature distribution. The simpler procedure, and the one that we will use,<br />
involves applying Fourier's law at the fin base. That is,<br />
q<br />
f<br />
= q<br />
b<br />
=− kA<br />
c<br />
dT<br />
dx<br />
=− kA<br />
dθ<br />
c<br />
x=0 dx<br />
x=0<br />
q<br />
f<br />
----- (10)<br />
Hence, knowing the temperature distribution, θ (x)<br />
, q<br />
f may be<br />
evaluated, giving<br />
q<br />
f<br />
sinh mL + ( h / mk )cosh mL<br />
= hPkA<br />
cθ b<br />
------- (11)<br />
cosh mL + ( h / mk )sinh mL<br />
However, conservation of energy dictates that the rate at which heat is<br />
transferred by convection from the fin must equal the rate at which it is<br />
conducted through the base of the fin. Accordingly, the alternative formulation<br />
for<br />
q<br />
f is<br />
q<br />
q<br />
f<br />
f<br />
∫<br />
A f<br />
[ ( x T∞] dA<br />
s<br />
= h T )−<br />
∫A<br />
f<br />
= h )<br />
θ ( x dA<br />
s -------------- (12)
where<br />
Af<br />
Af is the total, including the tip, fin surface area. Substitution of<br />
Equation (9) into Equation (12) would yield Equation (11):<br />
The second tip condition, case B, corresponds to the assumption that<br />
the convective heat loss from the fin tip is negligible, in which case the tip may<br />
be treated as adiabatic and<br />
dθ<br />
dx<br />
x =L<br />
=0<br />
-------- (13)<br />
Substituting from Equation (5) and dividing by m, we then obtain<br />
C<br />
mL −mL<br />
1<br />
e −C2e<br />
=<br />
Using this expression with Equation (8) to solve for C 1 and C 2 and<br />
substituting the results into Equation (5), we obtain<br />
θ cosh m(<br />
L −x)<br />
=<br />
θ cosh mL<br />
b<br />
0<br />
--------- (14)<br />
Using this temperature distribution with Equation (10), the fin heat<br />
transfer rate is then<br />
q<br />
f<br />
= hPkA<br />
cθb<br />
tanh mL ----- (15)<br />
In the same manner, we can obtain the fin temperature distribution<br />
and, heat transfer rate for case C, where the temperature is prescribed at the<br />
fin tip. That is, the second boundary condition is θ ( L)<br />
= θ , and the resulting<br />
expressions are of the form<br />
θ<br />
=<br />
θ<br />
b<br />
( θ θ )<br />
L<br />
/<br />
b<br />
sinh mx + sinh m(<br />
L − x)<br />
sinh mL<br />
L<br />
------- (16)<br />
q<br />
f<br />
=<br />
cosh mL −θ<br />
L<br />
/ θb<br />
hPkA<br />
cθb<br />
sinh mL<br />
The very long fin, case D, is an interesting extension of these results.<br />
In particular, as L →∞, θ →0<br />
and it is easily verified that<br />
L<br />
θ<br />
=e −<br />
θ<br />
b<br />
mx<br />
q<br />
f<br />
=<br />
hPkA<br />
c<br />
θ<br />
b<br />
Fin Performance<br />
Fins are used to increase the heat transfer from a surface by<br />
increasing the effective surface area. However, the fin itself represents a<br />
conduction resistance to heat transfer from the original surface. For this
eason, there is no assurance that the heat transfer rate will be increased<br />
through the use of fins. An assessment of this matter may be made by<br />
evaluating the fin effectiveness ε<br />
f . It is defined as the ratio of the fin heat<br />
transfer rate to the heat transfer rate that would exist without the fin. Therefore<br />
q<br />
f<br />
εf<br />
=<br />
hA θ<br />
c,<br />
b<br />
b<br />
where<br />
A<br />
c , b<br />
is the fin cross-sectional area at the base. In any rational design<br />
the value of εf<br />
should be as large as possible, and in general, the use of fins<br />
may rarely be justified unless ε<br />
f<br />
≥ 2.<br />
Fin effectiveness is enhanced by the choice of a material of high<br />
thermal conductivity. Aluminum alloys and copper come to mind. However,<br />
although copper is superior from the stand point of thermal conductivity.<br />
aluminum alloys are the more common choice because of additional benefits<br />
related to lower cost and weight. Fin effectiveness is also enhanced by<br />
increasing the ratio of the perimeter to the cross-sectional area. For this<br />
reason, the use of thin, but closely spaced fins, is preferred.<br />
Another measure of fin thermal performance is provided by the fin<br />
efficiencyη f . The maximum driving potential for convection is the<br />
temperature difference between the base (x = 0) and the fluid, θ b<br />
= T b<br />
−T<br />
∞.<br />
Hence the maximum rate at which a fin could dissipate energy is the rate that<br />
would exist if the entire fin surface were at the base temperature. However,<br />
since any fin is characterized by a finite conduction resistance, a temperature<br />
gradient must exist along the fin and the above condition is an idealization. A<br />
logical definition of fin efficiency is therefore<br />
q<br />
f<br />
η<br />
f<br />
= =<br />
qmax<br />
q<br />
hA<br />
f<br />
f<br />
θ<br />
b<br />
where A f is the surface area of the fin.<br />
Overall Surface Efficiency<br />
In contrast to the fin efficiency<br />
η<br />
f , which characterizes the<br />
performance of a single fin, the overall surface efficiency η<br />
o characterizes an<br />
array of fins and the base surface to which they are attached. Representative<br />
arrays are shown in Figure 8.3, where S designates the fin pitch. In each case<br />
the overall efficiency is defined as
where<br />
q<br />
f<br />
η<br />
f<br />
= =<br />
qmax<br />
qt<br />
hA θ<br />
t<br />
b<br />
q<br />
t is the total heat rate from the surface area A t associated with both<br />
the fins and the exposed portion of the base (often termed the prime surface).<br />
If there are N fins in the array, each of surface area A f , and the area of the<br />
prime surface is designated as A b , the total surface area is<br />
A = NA + A<br />
t<br />
f<br />
b<br />
The maximum possible heat rate would result if the entire fin surface,<br />
as well as the exposed base, were maintained at<br />
T<br />
b .<br />
Figure 8.3 Representative fin arrays. (a) Rectangular fins. (b) Annular fins.
LECTURE NO.9<br />
PRINCIPLES OF UNSTEADY-STATE HEAT TRANSFER: DERIVATION OF<br />
BASIC EQUATION; SIMPLIFIED CASE FOR SYSTEMS WITH<br />
NEGLIGIBLE INTERNAL RESISTANCE; TOTAL AMOUNT OF HEAT<br />
TRANSFERRED ; DIMENSIONAL ANALYSIS IN MOMENTUM TRANSFER<br />
PRINCIPLES OF UNSTEADY-STATE HEAT TRANSFER: DERIVATION OF<br />
BASIC EQUATION<br />
Introduction<br />
In steady state heat-transfer systems the temperature at any given<br />
point and the heat flux were always constant over time. In unsteady state or<br />
transient processes the temperature at any given point in the system changes<br />
with time. Before steady-state conditions can be reached in a process, some<br />
time must elapse after the heat-transfer process is initiated to allow the<br />
unsteady-state conditions to disappear.<br />
Unsteady-state heat transfer is important because of the large<br />
number of heating and cooling problems occurring industrially. In metallurgical<br />
processes it is necessary to predict cooling and heating rates for various<br />
geometries of metals in order to predict the time required to reach certain<br />
temperatures. In food processing, for example, the canning industry,<br />
perishable canned foods are heated by immersion in steam baths or chilled by<br />
immersion in cold water. In the paper industry wood logs are immersed in<br />
steam baths before processing. In most of these processes the material is<br />
suddenly immersed in a fluid of higher or lower temperature.<br />
Fig. 9.1 Unsteady State conduction in one direction
Derivation of Unsteady-State Conduction Equation<br />
To derive the equation for unsteady-state conduction in one direction in<br />
a solid, refer to Fig. 9.1. Heat is being conducted in the x direction in the<br />
∆ x, ∆y<br />
, ∆z<br />
in size. For conduction in the x direction, write<br />
q x<br />
∂T<br />
= −kA<br />
------- (1)<br />
∂x<br />
∂T<br />
The term means the partial or derivative of T with respect to x,<br />
∂x<br />
with the other variables, y, z, and time t, being held constant. Next, making a<br />
heat balance on the cube, we can write<br />
rate of heat input + rate of generation = rate of heat output + rate of heat accumulati on<br />
---------- (2)<br />
The rate of heat input to the cube is<br />
rate of heat input =<br />
q<br />
∂T<br />
x / x<br />
= −k<br />
( ∆y<br />
∆z<br />
) ----------(3)<br />
∂x<br />
x<br />
is<br />
Also, rate of heat output =<br />
The rate of accumulation of heat in the volume<br />
q<br />
∂T<br />
x / x +∆ x<br />
= −k<br />
( ∆y<br />
∆z<br />
)<br />
------- (4)<br />
∂x<br />
x+∆<br />
x<br />
, in time ∂ t<br />
∆ x ∆y<br />
, ∆z<br />
∂T<br />
rate of heat accumulation = ( ∆x , ∆y<br />
, ∆z<br />
) ρc<br />
------(5)<br />
p<br />
∂t<br />
The rate of heat generation in volume ∆ x, ∆y<br />
, ∆z<br />
is<br />
rate of heat generation =<br />
( ∆x<br />
∆y<br />
∆z<br />
) q ------(6)<br />
.<br />
Substituting Eqs. (3)-(6) into (2) and dividing by<br />
Letting<br />
respect to x or<br />
⎛∂T<br />
−k<br />
⎜<br />
x<br />
q<br />
⎝ ∂<br />
+<br />
.<br />
x<br />
∂T<br />
−<br />
∂x<br />
∆x<br />
x+∆x<br />
⎞<br />
⎟<br />
⎠<br />
= ρc<br />
p<br />
∂T<br />
∂t<br />
∆ x,<br />
∆y<br />
, ∆z<br />
------- (7)<br />
∆ x approach zero, we have the second partial of T with<br />
2<br />
∂T / ∂x<br />
2<br />
on the left side. Then, rearranging,<br />
∂T<br />
∂t<br />
k<br />
=<br />
ρc<br />
p<br />
2<br />
∂ T<br />
2<br />
∂x<br />
.<br />
q<br />
+<br />
ρc<br />
p<br />
2<br />
∂ T<br />
= α<br />
2<br />
∂x<br />
.<br />
q<br />
+<br />
ρc<br />
p<br />
--------(8)
where α is<br />
k / ρc<br />
p , thermal diffusivity. This derivation assumes constant<br />
k ,and<br />
,ρ c p . In SI units, α = m 2 /s, T = K, t = s, k = W/m.K, ρ = kg/m 3 , q . =<br />
W/m 3 , and<br />
c<br />
p = J/kg.K.<br />
For conduction in three dimensions, a similar derivation gives<br />
2<br />
∂T<br />
⎛∂<br />
T<br />
= α<br />
2<br />
t<br />
⎜<br />
∂ ⎝ ∂x<br />
2<br />
∂ T<br />
+<br />
2<br />
∂y<br />
2<br />
∂ T ⎞ q<br />
+<br />
2<br />
z<br />
⎟<br />
∂<br />
+<br />
⎠ ρ<br />
.<br />
c p<br />
--------- (9)<br />
In many cases, unsteady-state heat conduction is occurring but the<br />
rate of heat generation is zero. Then Eqs. (8) and (9) become<br />
∂T<br />
∂t<br />
2<br />
∂ T<br />
= α ------(10)<br />
2<br />
∂x<br />
2 2 2<br />
∂T ⎛∂<br />
T ∂ T ∂ T ⎞<br />
= α ⎜ + +<br />
⎟<br />
2 2 2 ----- (11)<br />
∂t<br />
⎝ ∂x<br />
∂y<br />
∂z<br />
⎠<br />
z and time t.<br />
Equations (10) and (11) relate the temperature T with position x, y, and<br />
SIMPLIFIED CASE FOR SYSTEMS WITH NEGLIGIBLE INTERNAL<br />
RESISTANCE<br />
Basic Equation<br />
Consider a solid which has a very high thermal conductivity or very low<br />
internal conductive resistance compared to the external surface resistance,<br />
where convection occurs from the external fluid to the surface of the solid.<br />
Since the internal resistance is very small, the temperature within the solid is<br />
essentially uniform at any given time.<br />
An example would be a small, hot cube of steel at T o K at time t = 0,<br />
suddenly immersed in a large bath of cold water at<br />
T ∞which is held constant<br />
with time. Assume that the heat transfer coefficient h in W/m 2 .K is constant<br />
with time. Making a heat balance on the solid object for a small time interval of<br />
time dt s, the heat transfer from the bath to the object must equal the change<br />
in internal energy of the object:<br />
hA<br />
( T T ) dt = c V dT<br />
∞ −<br />
p<br />
ρ<br />
where A is the surface area of the object in m 2 , T the average<br />
temperature of the object at time t in s, ρ the density of the object in kg/m 3 ,
and V the volume in m 3 . Rearranging the equation and integrating between<br />
the limits of T = T o when t = 0 and T = T when t = t,<br />
T = T<br />
∫<br />
dT<br />
T −T<br />
T = T0<br />
∞<br />
hA<br />
=<br />
c ρV<br />
p<br />
∫t<br />
t = t<br />
= 0<br />
dt<br />
⎛ hA<br />
t<br />
c p V ⎟ ⎞<br />
−⎜<br />
⎝ ρ ⎠<br />
T −T<br />
∞<br />
= e<br />
T0<br />
−T<br />
∞<br />
This equation describes the time-temperature history of the solid<br />
object. The term<br />
c p<br />
ρ V is often called the lumped thermal capacitance of the<br />
system. This type of analysis is often called the lumped capacity method or<br />
Newtonian heating or cooling method.<br />
Equation for Different Geometries<br />
In using the above equation the surface / volume ratio of the object<br />
must be known. The basic assumption of negligible internal resistance was<br />
made in the derivation. This assumption is reasonably accurate when<br />
N Bi<br />
hx<br />
= 1<br />
calculate the temperature of the ball after 1 h (3600 s). The average physical<br />
properties are k = 43.3 W/m.K, ρ = 7849 kg/m3 , and c p = 0.4606 kJ/kg.K.<br />
Solution: For a sphere characteristic dimension<br />
V 25 .4<br />
x1 = =<br />
r = = 8.47 × 10<br />
A 3 1000 × 3<br />
The Biot Number<br />
hx 11 .36 (8.47 × 10<br />
N Bi<br />
= =<br />
k 43.3<br />
−3<br />
m<br />
−3<br />
1 =<br />
)<br />
0.00222<br />
Then,<br />
This value is
q( t)<br />
= hA ( T0<br />
−T<br />
∞<br />
) e<br />
⎛ hA ⎞<br />
−⎜<br />
⎟t<br />
c p V<br />
⎝ ρ ⎠<br />
To determine the total amount of heat Q in W.s or J transferred from<br />
the solid from time t = 0 to t = t, we can integrate the above equation:<br />
Q=<br />
Q=<br />
c<br />
t<br />
∫<br />
= t<br />
t = 0<br />
p<br />
q(<br />
t)<br />
d(<br />
t)<br />
=<br />
t<br />
∫<br />
= t<br />
t = 0<br />
hA ( T<br />
⎛<br />
⎜<br />
hA<br />
0<br />
−T<br />
⎞<br />
⎟<br />
− t<br />
⎜c<br />
V ⎟<br />
⎝ pρ<br />
⎠<br />
ρV<br />
( T 0<br />
−T<br />
∞<br />
)[1 −e<br />
]<br />
∞<br />
) e<br />
⎛ ⎞<br />
⎜<br />
hA<br />
− ⎟t<br />
⎝<br />
cpρV<br />
⎠<br />
dt<br />
Problem 9.2 Total Amount of Heat in Cooling<br />
For the conditions in problem 9.1, calculate the total amount of heat<br />
removed up to time t = 3600 s.<br />
Solution:<br />
From Problem 9.1,<br />
hA<br />
cpρV<br />
11 .36<br />
−4<br />
−1<br />
= 3.71 × 10<br />
− 3<br />
= (0.4606 × 1000 )(7849 )(8.47 × 10 )<br />
s<br />
Q=<br />
c<br />
3<br />
3<br />
−5<br />
3<br />
Also, V = 4πr<br />
/3=<br />
4( π)(0.0254<br />
) / 3=<br />
6.864 × 10 m<br />
Substituting the values into<br />
p<br />
⎛<br />
⎜<br />
hA<br />
⎞<br />
⎟<br />
− t<br />
⎜c<br />
V ⎟<br />
⎝ pρ<br />
⎠<br />
ρV<br />
( T 0<br />
−T<br />
∞<br />
)[1 −e<br />
]<br />
−4<br />
−5<br />
−(3.71<br />
× 10 )( 3600 )<br />
( 0.4606 × 1000 )( 7849 )( 6.864 × 10 )(699<br />
.9−394 .3) × [1 −<br />
]<br />
Q = e<br />
= 5.589X10 4 J<br />
DIMENSIONAL ANA LYSIS IN MOMENTUM TRANSFER<br />
Dimensional Analysis of Differential Equations<br />
Dimensional homogeneity requires that every term in a given equation<br />
have the same units. Then, the ratio of one term in the equation to another<br />
term is dimensionless. Knowing the physical meaning of each term in the<br />
equation, we are then able to give a physical interpretation to each of the<br />
dimensionless parameters or numbers formed. These dimensionless<br />
numbers, such as the Reynolds number and others, are useful in correlating<br />
and predicting transport phenomena in laminar and turbulent flow.<br />
Often it is not possible to integrate the differential equation describing a<br />
flow situation. However, we can use the equation to find out which
dimensionless numbers can be used in correlating experimental data for this<br />
physical situation.<br />
Systems that are geometrically similar are said to be dynamically<br />
similar if the parameters representing ratios of forces pertinent to the situation<br />
are equal. This means that the Reynolds, Euler, or Froude numbers must be<br />
equal between the two systems.<br />
This dynamic similarity is an important requirement in obtaining<br />
experimental data for a small model and extending these data to scale up to<br />
the large prototype. Since experiments with full-scale prototypes would often<br />
be difficult and / or expensive, it is customary to study small models. This is<br />
done in the scale-up of chemical process equipment and in the design of<br />
ships and airplanes.<br />
Dimensional Analysis Using the Buckingham Method<br />
The method of obtaining the important dimensionless numbers from<br />
the basic differential equations is generally the preferred method. In many<br />
cases, however, we are not able to formulate a differential equation which<br />
clearly applies. Then a more general procedure is required, known as the<br />
Buckingham method. In this method the listing of the important variables in<br />
the particular physical problem is done first. Then we determine the number of<br />
dimensionless parameters into which the variables may be combined by using<br />
the Buckingham pi theorem.<br />
The Buckingham theorem states that the functional relationship among<br />
‘q’ quantities or variables whose units may be given in terms of ‘u’<br />
fundamental units or dimensions may be written as (q - u) independent<br />
dimensionless groups, often called<br />
π ' s . [This quantity u is actually the<br />
maximum number of these variables that will not form a dimensionless group.]<br />
Let us consider the following example, to illustrate the use of this<br />
method. An incompressible fluid is flowing inside a circular tube of inside<br />
diameter D. The significant variables are pressure drop∆ p , velocity v,<br />
diameter D, tube length L, viscosity µ , and density ρ . The total number of<br />
variables is q = 6.
The fundamental units or dimensions are u = 3 and are mass M, length L, and<br />
2<br />
time t. The units of the variables are as follows: ∆pin M / Lt , v in L/t, D in L,<br />
µ in M/Lt, and ρ in M/L 3 . The number of dimensionless groups or π ' s is<br />
q− u , or 6-3 = 3. Thus,<br />
)<br />
π1 = f ( π2<br />
, π3<br />
Next, we must select a core group of u (or 3) variables which will<br />
appear in each π group and among them contain all the fundamental<br />
dimensions. Also, no two of the variables selected for the core can have the<br />
same dimensions. In choosing the core, the variable whose effect one desires<br />
to isolate is often excluded (for example,<br />
∆ p ). This leaves us with the<br />
variables v, D, µ and ρ to be used. (L and D have the same dimensions.)<br />
We will select D, v, and p to be the core variables common to all three<br />
groups. Then the three dimensionless groups are<br />
π<br />
a b c 1<br />
1<br />
= D v ρ ∆p<br />
π =<br />
π<br />
d e f 1<br />
2<br />
D v ρ L<br />
i 1<br />
3<br />
=D<br />
g v h ρ µ<br />
To be dimensionless, the variables must be raised to certain exponents<br />
a, b, c, and so forth.<br />
First we consider the π1<br />
group:<br />
π<br />
a b c 1<br />
1<br />
= D v ρ ∆p<br />
To evaluate these exponents, we write the above equation dimensionally by<br />
substituting the dimensions for each variable:<br />
M<br />
o<br />
o<br />
L t<br />
o<br />
= 1=<br />
L<br />
a<br />
b<br />
c<br />
⎛ L ⎞ ⎛ M ⎞<br />
⎜ ⎟ ⎜ 3 ⎟<br />
⎝ t ⎠ ⎝ L ⎠<br />
Next we equate the exponents of L on both sides of this equation, of M,<br />
and finally of t:<br />
(L) 0 = a + b - 3c - 1<br />
(M) 0 = c + 1<br />
(t) 0 = -b - 2<br />
Solving these equations, a = 0, b = -2, and c = -1.<br />
a b c 1<br />
Substituting these values into Eq. π = D v ρ ∆ ,<br />
1<br />
p<br />
M<br />
Lt<br />
2<br />
= ∆p =<br />
v ρ<br />
. π 1 2<br />
N<br />
Eu
Repeating this procedure for π<br />
2 and π<br />
3 ,<br />
L<br />
π 2<br />
=<br />
D<br />
Dv ρ<br />
π<br />
3<br />
= = N Re<br />
µ<br />
Finally, substituting π<br />
1 , π<br />
2 and π<br />
3 into equation π1 = f ( π2<br />
, π3<br />
)<br />
∆p<br />
⎛<br />
= f ⎜<br />
2<br />
v ρ ⎝<br />
L<br />
D<br />
Dvp ⎞<br />
, ⎟<br />
µ ⎠<br />
This type of analysis is useful in empirical correlations of data.<br />
However, it does not tell us the importance of each dimensionless group,<br />
which must be determined by experimentation, nor does it select the variables<br />
to be used.
LECTURE NO.10<br />
SOME IMPORTANT EMPIRICAL RELATIONS USED FOR DETERMINATION<br />
OF HEAT TRANSFER COEFFICIENT: NUSSELT’S NUMBER, PRANDTL<br />
NUMBER, REYNOLD’S NUMBER, GRASHOFF NUMBER<br />
The Nusselt Number<br />
For forced convection of a single-phase fluid with moderate<br />
temperature differences, the heat flux per unit area<br />
q<br />
"<br />
w<br />
is nearly<br />
proportional to the temperature difference ∆ T = Tw −T<br />
* . From the Newton’s<br />
law of cooling:<br />
q<br />
"<br />
w<br />
=<br />
h<br />
( T −T* )<br />
w<br />
where h is called the heat transfer coefficient, with units of W/m 2 .<br />
But h is dimensional and thus its value depends on the units used. The<br />
traditional dimensionless from of h is the Nusselt number, Nu, which may be<br />
defined as the ratio of convection heat transfer to fluid conduction heat<br />
transfer under the same conditions. Consider a layer of fluid of width L and<br />
temperature difference ( T w<br />
− T*<br />
) . Assuming that the layer is moving so that<br />
convection occurs, the heat flux would be,<br />
q<br />
"<br />
w<br />
=<br />
h<br />
( T −T* )<br />
w<br />
If, on the other hand, the layer were stagnant, the heat flux would be<br />
entirely due to fluid conduction through the layer:
q<br />
k ( Tw<br />
T* =<br />
)<br />
L<br />
" −<br />
w<br />
The Nusselt number is defined as the ratio of these two:<br />
N<br />
"<br />
qw(<br />
convection<br />
=<br />
"<br />
q ( conduction<br />
hL<br />
k<br />
uL<br />
=<br />
w<br />
)<br />
)<br />
A Nusselt number of order unity would indicate a sluggish motion little<br />
more effective than pure fluid conduction: for example, laminar flow in a long<br />
pipe. A large Nusselt number means very efficient convection: For example,<br />
turbulent pipe flow yields Nu of order 100 to 1000.<br />
Prandtl Number(Pr)<br />
The Prandtl number Pr is a dimensionless number approximating the<br />
ratio of momentum diffusivity (kinematic viscosity) and thermal diffusivity. It is<br />
named after the German physicist Ludwig Prandtl. It can be expressed as<br />
Pr<br />
υ<br />
=<br />
α<br />
where<br />
Pr = Prandtl's number<br />
υ = momentum diffusivity (m 2 /s)<br />
α = thermal diffusivity (m 2 /s)<br />
The Prandtl number can alternatively be expressed as<br />
µc<br />
Pr =<br />
p<br />
k<br />
where
μ = absolute or dynamic viscosity (kg/m. s, cP)<br />
c p = specific heat capacity (J/kg K,)<br />
k = thermal conductivity (W/m K)<br />
The Prandtl Number is often used in heat transfer and free and forced<br />
convection calculations.<br />
<strong>No</strong>te that whereas the Reynolds number and Grashof number are<br />
subscripted with a length scale variable, Prandtl number contains no such<br />
length scale in its definition and is dependent only on the fluid and the fluid<br />
state. As such, Prandtl number is often found in property tables alongside<br />
other properties such as viscosity and thermal conductivity.<br />
Typical values for Pr are:<br />
• around 0.015 for mercury<br />
• around 0.16-0.7 for mixtures of noble gases or noble gases with<br />
hydrogen<br />
• around 0.7-0.8 for air and many other gases,<br />
• between 4 and 5 for R-12 refrigerant<br />
• around 7 for water (At 20 degrees Celsius)<br />
• between 100 and 40,000 for engine oil<br />
• around 1 × 10 25 for Earth's mantle.<br />
For mercury, heat conduction is very effective compared to convection:<br />
thermal diffusivity is dominant. For engine oil, convection is very effective in<br />
transferring energy from an area, compared to pure conduction: momentum<br />
diffusivity is dominant.<br />
In heat transfer problems, the Prandtl number controls the relative<br />
thickness of the momentum and thermal boundary layers. When Pr is small, it<br />
means that the heat diffuses very quickly compared to the velocity<br />
(momentum). This means that for liquid metals the thickness of the thermal<br />
boundary layer is much bigger than the velocity boundary layer.
The mass transfer analog of the Prandtl number is the Schmidt number.<br />
Boundary Layer<br />
In physics and fluid mechanics, a boundary layer is that layer of fluid<br />
in the immediate vicinity of a bounding surface where effects of viscosity of<br />
the fluid are considered in detail. In the Earth's atmosphere, the planetary<br />
boundary layer is the air layer near the ground affected by diurnal heat,<br />
moisture or momentum transfer to or from the surface. On an aircraft wing the<br />
boundary layer is the part of the flow close to the wing. The boundary layer<br />
effect occurs at the field region in which all changes occur in the flow pattern.<br />
The boundary layer distorts surrounding non-viscous flow. It is a phenomenon<br />
of viscous forces. This effect is related to the Reynolds number.<br />
Fig.10.1 Boundary layer visualization, showing transition from laminar to<br />
turbulent condition<br />
Reynolds number<br />
In fluid mechanics, the Reynolds number Re is a dimensionless<br />
number that gives a measure of the ratio of inertial forces ρV 2 /L to viscous<br />
forces μV/L 2 and consequently quantifies the relative importance of these two<br />
types of forces for given flow conditions. The concept was introduced by<br />
George Gabriel Stokes in 1851, but the Reynolds number is named after<br />
Osborne Reynolds (1842–1912), who popularized its use in 1883.<br />
Reynolds numbers frequently arise when performing dimensional<br />
analysis of fluid dynamics problems, and as such can be used to determine
dynamic similitude between different experimental cases. They are also used<br />
to characterize different flow regimes, such as laminar or turbulent flow:<br />
laminar flow occurs at low Reynolds numbers, where viscous forces are<br />
dominant, and is characterized by smooth, constant fluid motion, while<br />
turbulent flow occurs at high Reynolds numbers and is dominated by inertial<br />
forces, which tend to produce chaotic eddies, vortices and other flow<br />
instabilities.<br />
Grashof number Gr<br />
The Grashof number Gr is a dimensionless number in fluid dynamics<br />
and heat transfer which approximates the ratio of the buoyancy to viscous<br />
force acting on a fluid. It frequently arises in the study of situations involving<br />
natural convection. It is named after the German engineer Franz Grashof.<br />
Gr<br />
L<br />
3<br />
g β(<br />
T −T∞<br />
) L<br />
= for vertical flat plates<br />
s<br />
2<br />
υ<br />
Gr<br />
L<br />
3<br />
g β(<br />
T −T∞<br />
) D<br />
= for pipes<br />
s<br />
2<br />
υ<br />
where the L and D subscripts indicates the length scale basis for the Grashof<br />
Number.<br />
g = acceleration due to Earth's gravity<br />
β = volumetric thermal expansion coefficient<br />
(equal to approximately 1/T, for ideal fluids, where T is absolute<br />
temperature)<br />
T s = surface temperature<br />
T ∞ = bulk temperature<br />
L = length<br />
D = diameter<br />
υ = kinematic viscosity
The transition to turbulent flow occurs in the range 10 8 < Gr L < 10 9 for<br />
natural convection from vertical flat plates. At higher Grashof numbers, the<br />
boundary layer is turbulent; at lower Grashof numbers, the boundary layer is<br />
laminar.<br />
LECTURE NO.11<br />
RADIATION - HEAT TRANSFER, RADIATION PROPERTIES, RADIATION<br />
THROUGH BLACK AND GREY SURFACES, DETERMINATION OF<br />
SHAPE FACTORS<br />
Introduction<br />
Thermal radiation is that electromagnetic radiation emitted by a body<br />
as a result of its temperature.<br />
PHYSICAL MECHANISM<br />
There are many types of electromagnetic radiation; thermal radiation is<br />
only one. Regardless of the type of radiation, we say that it is propagated at<br />
the speed of light, 3 x 10 8<br />
wavelength and frequency of the radiation,<br />
where<br />
c = speed of light<br />
λ= wavelength<br />
υ = frequency<br />
m/s. This speed is equal to the product of the<br />
c=λυ<br />
The unit for A may be centimeters, angstroms (1<br />
ο<br />
A = 10 -8 cm), or<br />
micrometers (1 µ m = 10<br />
-6<br />
m). A portion of the electromagnetic spectrum is<br />
shown in Figure 11.1. Thermal radiation lies in the range from about 0.1 to<br />
100 µ m, while the visible-light portion of the spectrum is very narrow,<br />
extending from about 0.35 to 0.75 µ m.
Fig.11.1 Electromagnetic spectrum<br />
The propagation of thermal radiation takes place in the form of discrete<br />
quanta, each quantum having an energy of<br />
E=hυ<br />
where h is Planck's constant and has the value<br />
h = 6.625 X 10 -34 J.s<br />
The total radiation energy emitted is proportional to absolute<br />
temperature to the fourth power:<br />
4<br />
E b<br />
= σT<br />
The above equation is called the Stefan-Boltzmann law, E b is the<br />
energy radiated per unit time and per unit area by the ideal radiator, and σ is<br />
the Stefan-Boltzmann constant, which has the value<br />
σ = 5.669 x 10 -8 W/m 2 . K 4<br />
where E b , is in watts per square meter and T is in degrees Kelvin.<br />
We are interested in radiant exchange with surfaces-hence the reason<br />
for the expression of radiation from a surface in terms of its temperature. The<br />
subscript b in Equation denotes that this is the radiation from a blackbody. We<br />
call this blackbody radiation because materials which obey this law appear<br />
black to the eye; they appear black because they do not reflect any radiation.<br />
Thus a blackbody is also considered as one which absorbs all radiation<br />
incident upon it. E b is called the emissive power of a blackbody.<br />
It is important to note at this point that the "blackness" of a surface to<br />
thermal radiation can be quite deceiving in so far as visual observations are<br />
concerned. A surface coated with lampblack appears black to the eye and<br />
turns out to be black for the thermal-radiation spectrum. On the other hand,<br />
snow and ice appear quite bright to the eye but are essentially "black" for
long-wavelength thermal radiation. Many white paints are also essentially<br />
black for long-wavelength radiation.<br />
Fig. 11.2 Sketch showing effects of incident radiation.<br />
RADIATION PROPERTIES<br />
When radiant energy strikes a material surface, part of the radiation is<br />
reflected, part is absorbed, and part is transmitted, as shown in Figure 11.2.<br />
We define the reflectivity ρ as the fraction reflected, the absorptivity α as<br />
the fraction absorbed, and the transmissivity τ as the fraction transmitted.<br />
Thus<br />
ρ+ α+<br />
τ=1<br />
Most solid bodies do not transmit thermal radiation, so that for many<br />
applied problems the transmissivity may be taken as zero. Then<br />
ρ +α=1 ( τ=0<br />
)<br />
Two types of reflection phenomena may be observed when radiation<br />
strikes a surface. If the angle of incidence is equal to the angle of reflection,<br />
the reflection is called specular. On the other hand, when an incident beam is<br />
distributed uniformly in all directions after reflection, the reflection is called<br />
diffuse. These two types of reflection are depicted in Figure 11.3.<br />
Figure 11.3 (a) Specular ( φ=<br />
1<br />
φ2<br />
) and (b) diffuse reflection.
<strong>No</strong>te that a specular reflection presents a mirror image of the source to<br />
the observer. <strong>No</strong> real surface is either specular or diffuse. An ordinary mirror<br />
is quite specular for visible light, but would not necessarily be specular over<br />
the entire wavelength range of thermal radiation. Ordinarily, a rough surface<br />
exhibits diffuse behavior better than a highly polished surface. Similarly, a<br />
polished surface is more specular than a rough surface. The influence of<br />
surface roughness on thermal-radiation properties of materials is a matter of<br />
serious concern and remains a subject for continuing research.<br />
The emissive power of a body E is defined as the energy emitted by<br />
the body per unit area and per unit time.<br />
The ratio of the emissive power of a body to the emissive power of a<br />
blackbody at the same temperature is equal to the absorptivity of the body.<br />
This ratio is defined as the emissivity ε of the body,<br />
E<br />
ε=<br />
E b<br />
so that<br />
ε=α<br />
The above equation is called Kirchhoff's identity.<br />
The Gray Body<br />
A gray body is defined such that the monochromatic emissivity ε λ of<br />
the body is independent of wavelength. The monochromatic emissivity is<br />
defined as the ratio of the monochromatic emissive power of the body to the<br />
monochromatic emissive power of a blackbody at the same wavelength and<br />
temperature. Thus<br />
ε λ<br />
E<br />
=<br />
λ<br />
E b λ<br />
The glass, which is essentially transparent for visible light, is almost<br />
totally opaque for thermal radiation emitted at ordinary room temperatures.
Figure 11.4 Method of constructing a blackbody enclosure.<br />
Construction of a Blackbody<br />
The concept of a blackbody is an idealization; i.e., a perfect blackbody<br />
does not exist. All surfaces reflect radiation to some extent, however slight. A<br />
blackbody may be approximated very accurately, however, in the following<br />
way. A cavity is constructed, as shown in Figure 11.4, so that it is very large<br />
compared with the size of the opening in the side. An incident ray of energy is<br />
reflected many times on the inside before finally escaping from the side<br />
opening. With each reflection there is a fraction of the energy absorbed<br />
corresponding to the absorptivity of the inside of the cavity. After the many<br />
absorptions, practically all the incident radiation at the side opening is<br />
absorbed. It should be noted that the cavity of Figure 11.4 behaves<br />
approximately as a blackbody emitter as well as an absorber.<br />
Fig.11.5 Sketch showing area elements used in deriving radiation shape<br />
factor<br />
RADIATION SHAPE FACTOR
Consider two black surfaces A 1 and A 2 , as shown in Figure 11.5. We<br />
wish to obtain a general expression for the energy exchange between these<br />
surfaces when they are maintained at different temperatures. The problem<br />
becomes essentially one of determining the amount of energy which leaves<br />
one surface and reaches the other. To solve this problem the radiation shape<br />
factors are defined as<br />
F 1-2 = fraction of energy leaving surface 1 which reaches surface 2<br />
F 2-1 = fraction of energy leaving surface 2 which reaches surface 1<br />
F m-n = fraction of energy leaving surface m which reaches surface n<br />
Other names for the radiation shape factor are view factor, angle<br />
factor, and configuration factor. The energy leaving surface 1 and arriving at<br />
surface 2 is<br />
E b1 A 1 F 12<br />
and the energy leaving surface 2 and arriving at surface 1 is<br />
E b2 A 2 F 21<br />
Since the surfaces are black, all the incident radiation will be absorbed, and<br />
the net energy exchange is<br />
E b1 A 1 F 12 - E b2 A 2 F 21 = Q 1-2<br />
If both surfaces are at the same temperature, there can be no heat<br />
exchange, that is, Q 1-2 = 0. Also,<br />
E b1 = E b2<br />
so that<br />
A 1 F 12 = A 2 F 21<br />
The net heat exchange is therefore<br />
Q 1-2 = A 1 F 12 (E b1 - E b2 ) = A 2 F 21 (E b1 - E b2 )<br />
The above equation is known as a reciprocity relation, and it applies in<br />
a general way for any two surfaces m and n:<br />
A m F mn = A n F nm<br />
Although the relation is derived for black surfaces, it holds for other<br />
surfaces also as long as diffuse radiation is involved.
LECTURE NO.12<br />
INTRODUCTION TO CONDENSING AND BOILING HEAT TRANSFER,<br />
CONDENSATION HEAT-TRANSFER PHENOMENA, FILM<br />
CONDENSATION INSIDE HORIZONTAL TUBES, BOILING HEAT<br />
TRANSFER<br />
Introduction<br />
The convection processes are associated with a change of phase of a<br />
fluid. The two most important examples are condensation and boiling<br />
phenomena.<br />
In many types of power or refrigeration cycles one is interested in<br />
changing a vapor to a liquid, or a liquid to a vapor, depending on the particular<br />
part of the cycle under study. These changes are accomplished by boiling or<br />
condensation, and the engineer must understand the processes involved in<br />
order to design the appropriate heat-transfer equipment. High heat-transfer<br />
rates are usually involved in boiling and condensation, and this fact has also<br />
led designers of compact heat exchangers to utilize the phenomena for<br />
heating or cooling purposes not necessarily associated with power cycles.<br />
CONDENSATION HEAT-TRANSFER PHENOMENA<br />
Consider a vertical flat plate exposed to a condensable vapor. If the<br />
temperature of the plate is below the saturation temperature of the vapor,<br />
condensate will form on the surface and under the action of gravity will flow<br />
down the plate. If the liquid wets the surface, a smooth film is formed, and the<br />
process is called film condensation. If the liquid does not wet the surface,<br />
droplets are formed which fall down the surface in some random fashion. This<br />
process is called dropwise condensation.<br />
In the film-condensation process the surface is blanketed by the film,<br />
which grows in thickness as it moves down the plate. A temperature gradient<br />
exists in the film, and the film represents a thermal resistance to heat transfer.<br />
In dropwise condensation a large portion of the area of the plate is<br />
directly exposed to the vapor; there is no film barrier to heat flow, and higher<br />
heat-transfer rates are experienced. In fact, heat-transfer rates in dropwise
condensation may be as much as 10 times higher than in film<br />
condensation.<br />
Because of the higher heat-transfer rates, dropwise condensation<br />
would be preferred to film condensation, but it is extremely difficult to maintain<br />
since most surfaces become wetted after exposure to a condensing vapor<br />
over an extended period of time. Various surface coatings and vapor additives<br />
have been used in attempts to maintain dropwise condensation, but these<br />
methods have not met with general success. Measurements indicate that the<br />
drop conduction is the main resistance to heat flow for atmospheric pressure<br />
and above. Nucleation site density on smooth surfaces can be of the order of<br />
10 8 sites per square centimeter, and heat-transfer coefficients in the range of<br />
170 to 290 kW/m 2 .°C.<br />
FILM CONDENSATION INSIDE HORIZONTAL TUBES<br />
Condensation inside tubes is of considerable practical interest because<br />
of applications to condensers in refrigeration and air-conditioning systems, but<br />
unfortunately these phenomena are quite complicated and not amenable to a<br />
simple analytical treatment. The overall flow rate of vapor strongly influences<br />
the heat transfer rate in the forced convection condensation system, and this<br />
in turn is influenced by the rate of liquid accumulation on the walls.<br />
BOILING HEAT TRANSFER<br />
When a surface is exposed to a liquid and is maintained at a<br />
temperature above the saturation temperature of the liquid, boiling may occur,<br />
and the heat flux will depend on the difference in temperature between the<br />
surface and the saturation temperature. When the heated surface is<br />
submerged below a free surface of liquid, the process is referred to as pool<br />
boiling. If the temperature of the liquid is below the saturation temperature,<br />
the process is called sub-cooled, or local, boiling. If the liquid is maintained<br />
at saturation temperature, the process is known as saturated, or bulk,<br />
boiling.<br />
The different regimes of boiling are indicated in Figure 12.1, where<br />
heat-flux data from an electrically heated platinum wire submerged in water
are plotted against temperature excess T w – T sat . In region I free-convection<br />
currents are responsible for motion of the fluid near the surface. In this region<br />
the liquid near the heated surface is superheated slightly, and it subsequently<br />
evaporates when it rises to the surface. The heat transfer in this region can be<br />
calculated with the free-convection relations. In region II bubbles begin to form<br />
on the surface of the wire and are dissipated in the liquid after breaking away<br />
from the surface. This region indicates the beginning of nucleate boiling. As<br />
the temperature excess is increased further, bubbles form more rapidly and<br />
rise to the surface of the liquid, where they are dissipated. This is indicated in<br />
region III. Eventually, bubbles are formed so rapidly that they blanket the<br />
heating surface and prevent the inflow of fresh liquid from taking their place.<br />
At this point the bubbles coalesce and form a vapor film which covers the<br />
surface. The heat must be conducted through this film before it can reach the<br />
liquid and effect the boiling process. The thermal resistance of this film causes<br />
a reduction in heat flux, and this phenomenon is illustrated in region IV, the<br />
film-boiling region. This region represents a transition from nucleate boiling to<br />
film boiling and is unstable. Stable film boiling is eventually encountered in<br />
region V. The surface temperatures required to maintain stable film boiling are<br />
high, and once this condition is attained, a significant portion of the heat lost<br />
by the surface may be the result of thermal radiation, as indicated in region VI.<br />
Figure 12.1 Heat-flux data from an electrically heated platinum wire
in<br />
An electrically heated wire is unstable at point a since a small increase<br />
∆ Tx<br />
, at this point results in a decrease in the boiling heat flux. But the wire<br />
still must dissipate the same heat flux, or its temperature will rise, resulting in<br />
operation farther down to the boiling curve. Eventually, equilibrium may be<br />
reestablished only at point b in the film-boiling region. This temperature<br />
usually exceeds the melting temperature of the wire, so that burnout results. If<br />
the electric-energy input is quickly reduced when the system attains point a, it<br />
may be possible to observe the partial nucleate boiling and unstable film<br />
region.<br />
In nucleate boiling, bubbles are created by the expansion of entrapped<br />
gas or vapor at small cavities in the surface. The bubbles grow to a certain<br />
size, depending on the surface tension at the liquid-vapor interface and the<br />
temperature and pressure. Depending on the temperature excess, the<br />
bubbles may collapse on the surface, may expand and detach from the<br />
surface to be dissipated in the body of the liquid, or at sufficiently high<br />
temperatures may rise to the surface of the liquid before being dissipated.<br />
When local boiling conditions are observed, the primary mechanism of heat<br />
transfer is thought to be the intense agitation at the heat-transfer surface,<br />
which creates the high heat-transfer rates observed in boiling. In saturated, or<br />
bulk, boiling the bubbles may break away from the surface because of the<br />
buoyancy action and move into the body of the liquid. In this case the heattransfer<br />
rate is influenced by both the agitation caused by the bubbles and the<br />
vapor transport of energy into the body of the liquid.
LECTURE NO.13<br />
HEAT EXCHANGERS- GENERAL INTRODUCTION; DOUBLE-PIPE HEAT<br />
EXCHANGER; SHELL-AND-TUBE HEAT EXCHANGER; CROSS-FLOW<br />
EXCHANGER; FOULING FACTORS, LMTD<br />
Introduction<br />
In the process industries the transfer of heat between two fluids is<br />
generally done in heat exchangers. The most common type is one in which<br />
the hot and cold fluids do not come into direct contact with each other but are<br />
separated by a tube wall or a flat or curved surface. The transfer of heat from<br />
the hot fluid to the wall or tube surface is accomplished by convection,<br />
through the tube wall or plate by conduction, and then by convection to the<br />
cold fluid.<br />
Different Types<br />
Double-pipe heat exchanger<br />
The simplest exchanger is the double-pipe or concentric pipe<br />
exchanger. This is shown in Fig. 13.1, where one fluid flows inside one pipe<br />
and the other fluid flows in the annular space between the two pipes. The<br />
fluids can be in cocurrent or countercurrent flow. The exchanger can be made<br />
from a pair of single lengths of pipe with fittings at the ends or from a number<br />
of pairs interconnected in series. This type of exchanger is useful mainly for<br />
small flow rates.<br />
Fig.13.1 Flow in a double pipe heat exchanger<br />
Shell-and-tube exchanger<br />
If larger flows are involved, a shell-and-tube exchanger is used, which<br />
is the most important type of exchanger in use in the process industries. In
these exchangers the flows are continuous. Many tubes in parallel are used,<br />
where one fluid flows inside these tubes. The tubes, arranged in a bundle, are<br />
enclosed in a single shell and the other fluid flows outside the tubes in the<br />
shell side. The simplest shell-and-tube exchanger is shown in Fig. 13.2(a) for<br />
one shell pass and one tube pass, or a 1-1 counterflow exchanger. The cold<br />
fluid enters and flows inside through all the tubes in parallel in one pass. The<br />
hot fluid enters at the .other end and flows counterflow across the outside of<br />
the tubes. Cross baffles are used so that the fluid is forced to flow<br />
perpendicular across the tube bank rather than parallel with it. The added<br />
turbulence generated by this cross-flow increases the shell-side heat-transfer<br />
coefficient.<br />
Fig.13.2. Shell-and-tube heat exchangers:<br />
(a) 1 shell pass and 1 tube pass (1-1 exchanger);<br />
(b) 1 shell pass and 2 tube passes (1-2 exchanger).<br />
Fig. 13.2(b) a 1-2 parallel-counterflow exchanger is shown. The liquid<br />
on the tube side flows in two passes as shown and the shell-side liquid flows<br />
in one pass. In the first pass of the tube side, the cold fluid is flowing
counterflow to the hot shell-side fluid; in the second pass of the tube side, the<br />
cold fluid flows in parallel (cocurrent) with the hot fluid. Another type of<br />
exchanger has two shell-side passes and four tube passes. Other<br />
combinations of number of passes are also used sometimes, with the 1-2 and<br />
2-4 types being the most common.<br />
Cross-flow exchanger<br />
When a gas such as air is being heated or cooled, a common device<br />
used is the cross-flow heat exchanger shown in Fig. 13.3 (a). One of the<br />
fluids, which is a liquid, flows inside through the tubes, and the exterior gas<br />
flows across the tube bundle by forced or sometimes natural convection. The<br />
fluid inside the tubes is considered to be unmixed, since it is confined and<br />
cannot mix with any other stream. The gas flow outside the tubes is mixed,<br />
since it can move about freely between the tubes, and there will be a<br />
tendency for the gas temperature to equalize in the direction normal to the<br />
flow. For the unmixed fluid inside the tubes, there will be a temperature<br />
gradient both parallel and normal to the direction of flow.<br />
Fig. 13.3 Cross-flow heat exchanger<br />
A second type of cross-flow heat exchanger shown in Fig. 13.3 (b) is<br />
typically used in air-conditioning and space-heating applications. In this type<br />
the gas flows across a finned-tube bundle and is unmixed, since it is confined<br />
in separate flow channels between the fins as it passes over the tubes. The<br />
fluid in the tubes is unmixed.<br />
Fouling Factors and Typical Overall U Values<br />
In actual practice, heat-transfer surfaces do not remain clean. Dirt,<br />
soot, scale, and other deposits form on one or both sides of the tubes of an
exchanger and on other heat-transfer surfaces. These deposits offer<br />
additional resistance to the flow of heat and reduce the overall heat-transfer<br />
coefficient U. In petroleum processes coke and other substances can deposit.<br />
Silting and deposits of mud and other materials can occur. Corrosion products<br />
which could constitute a serious resistance to heat transfer may form on the<br />
surfaces. Biological growth such as algae can occur with cooling water and in<br />
the biological industries.<br />
To avoid or lessen these fouling problems, chemical inhibitors are often<br />
added to minimize corrosion, salt deposition, and algae growth. Water<br />
velocities above 1 m/s are generally used to help reduce fouling. Large<br />
temperature differences may cause excessive deposition of solids on surfaces<br />
and should be avoided if possible.<br />
The effect of such deposits and fouling is usually taken care of in<br />
design by adding a term for the resistance of the fouling on the inside and<br />
outside of the tube in Equation as follows:<br />
where,<br />
U<br />
i<br />
=<br />
1 1<br />
+<br />
h h<br />
i<br />
di<br />
1<br />
( r0<br />
−ri<br />
) Ai<br />
+<br />
k A<br />
A<br />
Alm<br />
Ai<br />
+<br />
A h<br />
o<br />
o<br />
Ai<br />
+<br />
A h<br />
hdi<br />
is the fouling coefficient for the inside and<br />
hdo<br />
the fouling coefficient for the outside of the tube in W/m 2 . K.<br />
THE LOG MEAN TEMPERATURE DIFFERENCE<br />
Consider the double-pipe heat exchanger shown in Figure 13-1. The<br />
fluids may flow in either parallel flow or counter-flow, and the temperature<br />
profiles for these two cases are indicated in Figure 13-4. The heat transfer in<br />
this double-pipe arrangement can be calculated with the following equation<br />
q<br />
= UA ∆T m ----- (13.1)<br />
where<br />
U = overall heat-transfer coefficient<br />
A = surface area for heat transfer consistent with definition of U<br />
∆ T m = suitable mean temperature difference across heat<br />
exchanger<br />
The above Fig.13.4 shows that the temperature difference between the<br />
hot and cold fluids varies between inlet and outlet, and the average value has<br />
o<br />
do
to be calculated in the above equation. For the parallel- flow heat exchanger<br />
shown in Figure 13.4 (a), the heat transferred through an element of area dA<br />
may be written<br />
dq =−m c<br />
h<br />
h<br />
.<br />
dT<br />
h<br />
.<br />
= m<br />
c<br />
c<br />
c<br />
dT<br />
c<br />
------ (13.2)<br />
Fig. 13.4 Temperature Profiles for (a) parallelflow and (b) counterflow<br />
in double-pipe heat exchanger<br />
where the subscripts h and c designate the hot and cold fluids, respectively.<br />
The heat transfer could also be expressed<br />
From Equation (13.2-6)<br />
dT<br />
h<br />
dq = U ( Th −T<br />
c)<br />
dA ------------ (13.3)<br />
.<br />
= −<br />
.<br />
.<br />
m<br />
dq<br />
h<br />
c<br />
h
where<br />
Thus<br />
dT<br />
c<br />
.<br />
= −<br />
.<br />
.<br />
m<br />
dq<br />
c<br />
c<br />
c<br />
ṁ represents the mass-flow rate and c is the specific heat of the fluid.<br />
1 1<br />
dT<br />
h<br />
− dT<br />
c<br />
= d( Th<br />
−Tc<br />
) =−dq<br />
( + )<br />
. . ---------- (13.4)<br />
m c m c<br />
h<br />
h<br />
c<br />
c<br />
Solving for dq from Equation (13.3) and substituting into Equation (13.4)<br />
gives<br />
⎛ ⎞<br />
d(<br />
Th<br />
−T<br />
c)<br />
u<br />
⎜ 1 1<br />
=−<br />
⎟<br />
dA<br />
Th<br />
T ⎜<br />
+<br />
.<br />
. ⎟ ------------ (13.5)<br />
−<br />
c<br />
⎝mh<br />
ch<br />
mc<br />
cc<br />
⎠<br />
This differential equation may now be integrated between conditions 1<br />
and 2 as indicated in Figure 13.4.<br />
The result is<br />
⎛ ⎞<br />
Th<br />
2<br />
−T<br />
c2<br />
=−<br />
⎜ 1 1<br />
ln uA<br />
⎟<br />
⎜<br />
+<br />
. .<br />
−<br />
⎟ --------------- (13.5)<br />
Th<br />
1<br />
Tc<br />
1<br />
⎝mh<br />
ch<br />
mc<br />
cc<br />
⎠<br />
Returning to Equation (13.2), the products<br />
ṁ cc c<br />
and<br />
.<br />
mhc h<br />
may be<br />
expressed in terms of the total heat transfer q and the overall temperature<br />
differences of the hot and cold fluids. Thus<br />
.<br />
m<br />
h<br />
c<br />
h<br />
=<br />
T<br />
h1<br />
q<br />
−T<br />
h2<br />
.<br />
q<br />
mc<br />
cc<br />
=<br />
Tc<br />
2<br />
−T<br />
c1<br />
Substituting these relations into Equation (13.5) gives<br />
( Th<br />
2<br />
−T<br />
c2)<br />
−(<br />
Th<br />
1−T<br />
c1)<br />
q= UA<br />
------- (13.6)<br />
ln [( T −T<br />
) / ( T −T<br />
)]<br />
h2<br />
c2<br />
h1<br />
c1<br />
Comparing Equation (13.6) with Equation (13.1), the mean temperature<br />
difference is the grouping of terms in the brackets. Thus<br />
( Th<br />
2<br />
−T<br />
c2)<br />
−(<br />
Th<br />
1−T<br />
c1)<br />
∆ Tm<br />
=<br />
--------<br />
ln[( Th<br />
2<br />
−T<br />
c2) / ( Th<br />
1−T<br />
c1)]<br />
(13.7)<br />
This temperature difference is called the log mean temperature<br />
difference (LMTD). It is the temperature difference at one end of the heat<br />
exchanger less the temperature difference at the other end of the exchanger
divided by the natural logarithm of the ratio of these two temperature<br />
differences.<br />
The above derivation for LMTD involves two important assumptions:<br />
(1) the fluid specific heats do not vary with temperature, and (2) the<br />
convection heat-transfer coefficients are constant throughout the heat<br />
exchanger.<br />
If a heat exchanger other than the double-pipe type is used, the heat<br />
transfer is calculated by using a correction factor applied to the LMTD for a<br />
counterflow double-pipe arrangement with the same hot and cold fluid<br />
temperatures. The heat-transfer equation then takes the form<br />
q= UAF ∆T m -------- (13.8)<br />
values of the correction factor F according are plotted in Figures 13.5 to 13.8<br />
for several different types of heat exchangers. When a phase change is<br />
involved, as in condensation or boiling (evaporation), the fluid normally<br />
remains at essentially constant temperature and the relations are simplified.<br />
For this condition, P or R becomes zero and we obtain<br />
F = 1.0 for boiling or condensation<br />
Figure 13.5 Correction-factor plot for exchanger with one shell pass and<br />
two, four or any multiple of tube passes.
Figure 13.6 Correction-factor plot for exchanger with two shell passes and<br />
four, eight, or any multiple of tube passes.
Figure 13.7 Correction-factor plot for single-pass cross-flow<br />
exchanger, both fluids unmixed<br />
Figure 13.8 Correction-factor plot for single-pass cross-flow<br />
exchanger one fluid mixed the other unmixed.
LECTURE NO.14<br />
DESIGN PROBLEMS ON HEAT EXCHANGERS: CALCULATION OF HEAT<br />
EXCHANGER SIZE FROM KNOWN TEMPERATURES, PROBLEM ON<br />
SHELL-AND-TUBE HEAT EXCHANGER, DESIGN OF SHELL-AND-TUBE<br />
HEAT EXCHANGER<br />
Problems on Design of Heat Exchangers<br />
Problem 14.1 Calculation of heat exchanger size from known<br />
temperatures<br />
Water at the rate of 68 kg/min is heated from 35 to 75 °C by an oil<br />
having a specific heat of 1.9 kJ /kg. °C. The fluids are used in a counterflow<br />
double-pipe heat exchanger, and the oil enters the exchanger at 110 °C and<br />
leaves at 75 °C. The overall heat-transfer coefficient is 320 W/m 2 .°C.<br />
Calculate the heat exchanger area.<br />
Solution<br />
The total heat transfer is determined from the energy absorbed by the<br />
water:<br />
.<br />
q = m c ∆T<br />
= (68) (4180) (75 - 35) = 11.37 MJ /min (a)<br />
w<br />
w<br />
w<br />
= 189.5 kW<br />
Since all the fluid temperatures are known, the LMTD can be calculated by<br />
using the temperature scheme in the following Figure.<br />
(110 −75)<br />
−(75<br />
−35)<br />
∆T m<br />
=<br />
= 37 .44 C<br />
ln[( 110 −75) / (75 −35)]
Then, Since q= UA ∆T<br />
m ,<br />
5<br />
1.895 × 10<br />
A = = 15 .82 m<br />
(320 )(37 .44 )<br />
Problem 14.2 Shell-and-tube heat exchanger<br />
Instead of the double-pipe heat exchanger of problem 14.1, it is desired to<br />
use a shell-and-tube exchanger with the water making one shell pass and the<br />
oil making two tube passes. Calculate the area required for this exchanger,<br />
assuming that the overall heat-transfer coefficient remains at 320 W/m 2 . °C.<br />
Solution<br />
To solve this problem, determine a correction factor from Figure 13.5 to be<br />
used with the LMTD calculated on the basis of a counterflow exchanger. The<br />
parameters according to the nomenclature of Figure 13.5 are<br />
T 1 = 35 °C T 2 = 75 °C t 1 = 110 °C t 2 = 75 °C<br />
t<br />
P=<br />
T<br />
1<br />
T<br />
R=<br />
t<br />
−t<br />
−t<br />
1<br />
75 −110<br />
=<br />
35 −110<br />
2 1<br />
=<br />
2<br />
−T<br />
−t<br />
1<br />
35 −75<br />
=<br />
75 −110<br />
1 2<br />
=<br />
0.467<br />
1.143<br />
2<br />
Hence, the correction factor is<br />
and the heat transfer is<br />
F = 0.81<br />
q=<br />
UAF<br />
∆<br />
T m<br />
so that<br />
5<br />
1.895 × 10<br />
A =<br />
= 19 .53 m<br />
(320 )(0.81)(37 .44 )<br />
2
Problem 14.3 Design of shell-and-tube heat exchanger<br />
Water at the rate of 3.8 kg/s is heated from 38 to 55 °C in a shell-andtube<br />
heat exchanger. On the shell side one pass is used with water as the<br />
heating fluid, 1.9 kg/s entering the exchanger at 93 °C. The overall heattransfer<br />
coefficient is 1419 W / m 2 . °C, and the average water velocity in the<br />
1.9 cm diameter tubes is 0.366 m/s. Because of space limitations the tube<br />
length must not be longer than 2.5 m. Calculate the number of tube passes,<br />
the number of tubes per pass, and the length of the tubes, consistent with this<br />
restriction.<br />
Solution<br />
First assume one tube pass and check to see if it satisfies the<br />
conditions of this problem. The exit temperature of the hot water is calculated<br />
from<br />
q=<br />
m c<br />
h<br />
h<br />
.<br />
dT<br />
h<br />
.<br />
= m<br />
c<br />
c<br />
c<br />
dT<br />
(a)<br />
(3.8)(4.18 )(55 −38 )<br />
∆T h<br />
=<br />
= 34 C<br />
(1.9)(4.18 )<br />
c<br />
so<br />
T<br />
h . exit = 93 - 94 = 59 °C<br />
The total required heat transfer is obtained from Equation (a) for the cold fluid:<br />
q = (3.8)(4.18)(55 - 38) = 270 kW<br />
For a counterflow exchanger, with the required temperature<br />
LMTD<br />
=∆T<br />
m<br />
(93 −55)<br />
−(59<br />
−38)<br />
=<br />
= 28 .66<br />
ln[( 93 −55) / (59 −38)]<br />
<br />
C<br />
q=<br />
UA<br />
∆<br />
T m<br />
(b)<br />
3<br />
270 × 10<br />
A = = 6.639<br />
(1419 )(28 .66 )<br />
m<br />
2<br />
Using the average water velocity in the tubes and the flow rate, calculate the<br />
total flow area with<br />
.<br />
m c<br />
= ρAu
3.8<br />
A = = 0.0104<br />
(1000 )(0.366 )<br />
m<br />
2<br />
This area is the product of the number of tubes and the flow area per tube:<br />
0.0104<br />
=n πd<br />
4<br />
(0.0104 )(4)<br />
n= π (0.019 )<br />
2<br />
2<br />
=<br />
36 .7<br />
or n = 37 tubes. The surface area per tube per meter of length is<br />
2<br />
πd =π (0.019 ) = 0.0597 m / tube . m<br />
Recall that the total surface area required for a one-tube-pass exchanger was<br />
calculated in Equation (b) as 6.639 m 2 . Thus compute the length of tube for<br />
this type of exchanger from<br />
nπdL<br />
=6.639<br />
6.639<br />
L = = 3m<br />
(37 )(0.0597 )<br />
This length is greater than the allowable 2.438 m, so we must use<br />
more than one tube pass. When we increase the number of passes,<br />
correspondingly increase the total surface area required because of the<br />
reduction in LMTD caused by the correction factor F. Next try two tube<br />
passes. From Figure 13.5, F = 0.88, and thus<br />
A<br />
q<br />
2.70 × 10<br />
3<br />
total<br />
= =<br />
=<br />
UF ∆T<br />
m<br />
(1419 )(0.88)(28 .66 )<br />
7.54 m<br />
The number of tubes per pass is still 37 because of the velocity<br />
requirement. For the two-tube-pass exchanger the total surface area is now<br />
related to the length by<br />
so that<br />
A total<br />
=2 nπdL<br />
2<br />
7.54<br />
L = = 1. 707 m<br />
(2)(37 )(0.0597 )<br />
This length is within the 2.438 m requirement, so the final design choice is<br />
Number of tubes per pass = 37<br />
Number of passes = 2
Length of tube per pass = 1.707 m
LECTURE NO.15<br />
INTRODUCTION TO MASS TRANSFER: A SIMILARITY OF MASS, HEAT, AND<br />
MOMENTUM TRANSFER PROCESSES; FICK'S LAW FOR MOLECULAR<br />
DIFFUSION<br />
INTRODUCTION TO MASS TRANSFER<br />
A Similarity of Mass, Heat, and Momentum Transfer Processes<br />
The various separation processes have certain basic principles which<br />
can be classified into three fundamental transfer (or "transport") processes:<br />
momentum transfer, heat transfer, and mass transfer. The fundamental<br />
process of momentum transfer occurs in such operations as fluid flow,<br />
mixing, sedimentation, and filtration. Heat transfer occurs in conductive and<br />
convective transfer of heat, evaporation, distillation, and drying.<br />
The third fundamental transfer process, mass transfer, occurs in<br />
distillation, absorption, drying, liquid-liquid extraction, adsorption, ion<br />
exchange, crystallization, and membrane processes. When mass is being<br />
transferred from one distinct phase to another or through a single phase, the<br />
basic mechanisms are the same whether the phase is a gas, liquid, or solid.<br />
This was also shown for heat transfer, where the transfer of heat by<br />
conduction followed Fourier's law in a gas, solid, or liquid.<br />
General molecular transport equation. All three of the molecular transport<br />
processes - momentum, heat, and mass-are characterized by the general<br />
type of equation,<br />
rate of<br />
atransfer<br />
process<br />
driving force<br />
= --------(1)<br />
resis tan ce<br />
Molecular diffusion equations for momentum, heat, and mass transfer<br />
Newton's equation for momentum transfer for constant density can be<br />
written as follows<br />
µ d( v x<br />
ρ)<br />
τzx<br />
= −<br />
------- (2)<br />
ρ dz<br />
where<br />
τzx<br />
is momentum transferred / s.m 2 ,<br />
µ<br />
ρ is kinematic viscosity in m2 /s,
ρ and c p :<br />
z is distance in m, and<br />
v xρ is momentum/m 3 , where the momentum has units<br />
of kg.m/s.<br />
Fourier's law for heat conduction can be written as follows for constant<br />
where<br />
q d(ρc<br />
pT<br />
)<br />
z<br />
= −α<br />
--------- (3)<br />
A dz<br />
q z<br />
/ A is heat flux in W/m 2 ,<br />
α is the thermal diffusivity in m 2 /s, and<br />
ρc p<br />
T is J/m 3 .<br />
The equation for molecular diffusion of mass is Fick's law and can be<br />
written as follows for constant total concentration in a fluid:<br />
where,<br />
J<br />
*<br />
Az<br />
= −D<br />
AB<br />
dc<br />
dz<br />
A<br />
------------ (4)<br />
*<br />
J<br />
Az is the molar flux of component A in the z direction due to<br />
molecular diffusion in kg mol A/s·m 2 ,<br />
D<br />
AB the molecular diffusivity of the molecule A in B in m 2 /s,<br />
cA<br />
the concentration of A in kg mol / m 3 , and<br />
z the distance of diffusion in m.<br />
The similarity of Eqs. (2), (3), and (4) for momentum, heat, and mass<br />
transfer should be obvious. All the fluxes on the left-hand side of the three<br />
equations have as units transfer of a quantity of momentum, heat, or mass<br />
per unit time per unit area. The transport properties µ/ ρ, α and DAB<br />
have units of m 2 /s, and the concentrations are represented as momentum/m 3 ,<br />
J/m 3 , or kg mol/m 3 .<br />
Examples of Mass- Transfer Processes<br />
Mass transfer is important in many areas of science and engineering.<br />
Mass transfer occurs when a component in a mixture migrates in the same<br />
phase or from phase to phase because of a difference in concentration<br />
between two points. Many familiar phenomena involve mass transfer. Liquid<br />
in an open pail of water evaporates into still air because of the difference in<br />
all
concentration of water vapor at the water surface and the surrounding air.<br />
There is a "driving force" from the surface to the air. A piece of sugar added to<br />
a cup of coffee eventually dissolves by itself and diffuses to the surrounding<br />
solution. Many purification processes involve mass transfer. In uranium<br />
processing, a uranium salt in solution is extracted by an organic solvent.<br />
Fick's Law for Molecular Diffusion<br />
Molecular diffusion or molecular transport can be defined as the<br />
transfer or movement of individual molecules through a fluid by means of the<br />
random, individual movements of the molecules. Imagine the molecules<br />
traveling only in straight lines and changing direction by bouncing off other<br />
molecules after collision. Since the molecules travel in a random path,<br />
molecular diffusion is often called a random-walk process.<br />
Fig 15.1 Schematic diagram of molecular diffusion process<br />
In Fig. 15.1 the molecular diffusion process is shown schematically. A<br />
random path that molecule A might take in diffusing through B molecules from<br />
point (1) to (2) is shown. If there are a greater number of A molecules near<br />
point (1) than at (2), then, since molecules diffuse randomly in both directions,<br />
more A molecules will diffuse from (1) to (2) than from (2) to (1). The net<br />
diffusion of A is from high- to low-concentration regions.<br />
As another example, a drop of blue liquid dye is added to a cup of<br />
water. The dye molecules will diffuse slowly by molecular diffusion to all parts<br />
of the water. To increase this rate of mixing of the dye, the liquid can be<br />
mechanically agitated by a spoon and convective mass transfer will occur.
The two modes of heat transfer, conduction and convective heat transfer are<br />
analogous to molecular diffusion and convective mass transfer.<br />
First, we will consider the diffusion of molecules when the whole bulk<br />
fluid is not moving but is stationary. Diffusion of the molecules is due to a<br />
concentration gradient. The general Fick's law equation can be written as<br />
follows for a binary mixture of A and B:<br />
where<br />
J<br />
*<br />
Az<br />
= −cD<br />
AB<br />
dx<br />
dz<br />
A<br />
-------- (5)<br />
c is total concentration of A and B in kg mol A+B/m 3 , and<br />
xA<br />
is the mole fraction of A in the mixture of A and B.<br />
If c is constant, then since<br />
c = cx ,<br />
A<br />
A<br />
c dx = d( cx ) = dc -------------- (6)<br />
A<br />
A<br />
A<br />
Substituting into Eq. (5) obtain the following equation for constant total<br />
concentration:<br />
J<br />
*<br />
Az<br />
= −D<br />
AB<br />
dc<br />
dz<br />
A<br />
------------ (7)<br />
This equation is the one more commonly used in many molecular diffusion<br />
processes.<br />
Problem Molecular Diffusion of Helium in Nitrogen<br />
A mixture of He and N 2 gas is contained in a pipe at 298 K and 1 atm<br />
total pressure which is constant throughout. At one end of the pipe at point 1<br />
the partial pressure pA<br />
1 of He is 0.60 atm and at the other end 0.2 m (20 cm)<br />
p<br />
A2 = 0.20 atm. Calculate the flux of He at steady state if DAB<br />
of the He-N 2<br />
mixture is 0.687 X 10 -4 m 2 /s (0.687 cm 2 /s). Use SI units.<br />
Solution:<br />
Since total pressure P is constant, then c is constant, where c is as<br />
follows for a gas according to the perfect gas law:<br />
where<br />
PV = nRT ----------------------- (8)<br />
n<br />
V<br />
n is kg mol A+B,<br />
V is volume in m 3 ,<br />
P<br />
= = c<br />
-------------- (9)<br />
RT
T is temperature in K,<br />
R is 8314.3 m 3 Pa / kg mol.K or<br />
R is 82.057 x 10 -3 m 3 . atm / kg mol.K, and<br />
c is kg mol A+B/m 3 .<br />
For steady state the flux in Eq.(4) is constant. Also,<br />
constant. Rearranging Eq. (4) and integrating,<br />
DAB<br />
for a gas is<br />
J<br />
*<br />
Az<br />
z2<br />
∫<br />
z1<br />
dz<br />
= −D<br />
AB<br />
CA 2<br />
∫<br />
cA 1<br />
dc<br />
A<br />
J<br />
*<br />
Az<br />
( cA<br />
1<br />
−c<br />
A2)<br />
= DAB<br />
----------- (10)<br />
( z −z<br />
)<br />
2<br />
1<br />
Also, from the perfect gas law, pA V = nA<br />
RT , and<br />
pA<br />
nA<br />
cA = 1<br />
1<br />
=<br />
------------------- (11)<br />
RT V<br />
Substituting Eq. (11) into (10),<br />
J<br />
*<br />
Az<br />
( pA<br />
1<br />
−pA2)<br />
= DAB<br />
----------------- (12)<br />
RT ( z −z<br />
)<br />
2<br />
1<br />
This is the final equation to use, which is in a form easily used for<br />
gases. Partial pressures are p<br />
A1<br />
= 0.6 atm = 0.6 X 1.01325 X 10 5 = 6.08 X<br />
10 4 Pa and p<br />
A2<br />
using SI units,<br />
= 0.2 atm = 0.2 X 1.01325 X 10 5 = 2.027 X 10 4 Pa. Then,<br />
*<br />
J AZ<br />
(0.687<br />
=<br />
−4<br />
4<br />
× 10 )(6.08 × 10 −2.027<br />
8314 (298 )(0.20 −0)<br />
× 10<br />
4<br />
)<br />
= 5.63 X 10 -6 kg. mol A/s. m 2<br />
If pressures in atm are used with SI units,<br />
*<br />
J AZ<br />
(0.687 × 10<br />
(82.06 × 10<br />
=<br />
−3<br />
4<br />
)(0.60 −0.20)<br />
)(298 )(0.20 −0)<br />
−<br />
= 5.63 X 10 -6 kg. mol A/s. m 2<br />
Other driving forces (besides concentration differences) for diffusion<br />
also occur because of temperature, pressure, electrical potential, and other<br />
gradients.
LECTURE NO.16<br />
MOLECULAR DIFFUSION IN GASES: EQUIMOLAR COUNTER DIFFUSION<br />
IN GASES<br />
MOLECULAR DIFFUSION IN GASES: Equimolar Counter diffusion in<br />
Gases<br />
In Fig. 16.1 a diagram is given of two gases A and B at constant total<br />
pressure P in two large chambers connected by a tube where molecular<br />
diffusion at steady state is occurring. Stirring in each chamber keeps the<br />
concentrations in each chamber uniform. The partial pressure<br />
p > p and p > p . Molecules of A diffuse to the right and B to the left.<br />
A1 A2<br />
B2<br />
B1<br />
Since the total pressure P is constant throughout, the net moles of A diffusing<br />
to the right must equal the net moles of B to the left. If this is not so, the total<br />
pressure would not remain constant. This means that<br />
J<br />
= −<br />
------- (13)<br />
* *<br />
AZ<br />
J BZ<br />
FIGURE 16.1 Equimolar counter diffusion of gases A and B.<br />
The subscript z is often dropped when the direction is obvious. Writing<br />
Fick's law for B for constant c,<br />
<strong>No</strong>w since<br />
J<br />
*<br />
B<br />
B<br />
= −<br />
D<br />
BA<br />
dc<br />
dz<br />
P = p A<br />
+ p = constant, then<br />
Differentiating both sides,<br />
B<br />
c = c A<br />
+ c B ------- (15)<br />
dc<br />
A<br />
dc B<br />
------(14)<br />
= −<br />
---------(16)
Equating Eq. J<br />
J<br />
*<br />
Az<br />
= −D<br />
AB<br />
dc<br />
dz<br />
A<br />
to<br />
J<br />
*<br />
B<br />
= −<br />
D<br />
BA<br />
dc<br />
dz<br />
dc<br />
A * dc<br />
B<br />
= −DAB<br />
= − J<br />
B<br />
=−(<br />
− DBA<br />
------- (17)<br />
dz<br />
dz<br />
*<br />
A<br />
)<br />
Substituting Eq. (16) into (17) and canceling like terms,<br />
coefficient<br />
D<br />
AB<br />
= D BA<br />
This shows that for a binary gas mixture of A and B, the diffusivity<br />
DAB<br />
for A diffusing into B is the same as DBA<br />
for B diffusing into A.<br />
Problem Equimolar Counter diffusion<br />
Ammonia gas (A) is diffusing through a uniform tube 0.10 m long<br />
containing N 2 gas (B) at 1.0132 X 10 5 Pa pressure and 298 K. The diagram is<br />
similar to Fig. 16.2. At point 1, p<br />
A1<br />
= 1.013 X 10 4 Pa and at point 2, p<br />
A2<br />
=<br />
B<br />
,<br />
0.507 X 10 4 Pa. The diffusivity D<br />
AB<br />
= 0.230 X 10 -4 m 2 /s.<br />
(a) Calculate the flux<br />
*<br />
J<br />
A at steady state.<br />
(b) Repeat for<br />
*<br />
J B<br />
Solution:<br />
J<br />
*<br />
Az<br />
where<br />
= D<br />
AB<br />
( pA<br />
1<br />
−pA2)<br />
RT ( z −z<br />
)<br />
2<br />
1<br />
P = 1.0132 X 10 5 Pa,<br />
z 2 – z 1 = 0.10 m, and<br />
T = 298 K. Substituting into above equation<br />
J<br />
*<br />
A<br />
= D<br />
AB<br />
−4<br />
4<br />
( pA<br />
1−pA2)<br />
(0.23 × 10 )(1.013 × 10 −0.507<br />
=<br />
RT ( z −z<br />
) 8314 (298 )(0.10 −0)<br />
2<br />
1<br />
× 10<br />
4<br />
)<br />
= 4.70 X 10 -7 kg mol A/s.m 2<br />
Rewriting the above equation for component B for part (b) and noting that<br />
p<br />
p<br />
= P−p<br />
= 1.0132 X 10 5 - 1.013 X 10 4 = 9.119 X 10 4 Pa and<br />
B1 A1<br />
= P −p<br />
= 1.0132 X 10 5 - 0.507 X 10 4 = 9.625 X 10 4 Pa,<br />
B2 A2<br />
J<br />
*<br />
B<br />
= D<br />
AB<br />
−4<br />
4<br />
( pB<br />
1−pB2)<br />
(0.23 × 10 )(9.119 × 10 −9.625<br />
× 10<br />
=<br />
RT ( z −z<br />
) 8314 (298 )(0.10 −0)<br />
2<br />
1<br />
4<br />
)<br />
=- 4.70 X 10 -7 kg mol B/s.m 2
The negative value for<br />
*<br />
J<br />
B means the flux goes from point 2 to point 1.