Connecting Transient and Steady-State Analyses
Using Heat Transfer and Fluids Examples
Washington Braga
Mechanical Engineering Department
Pontifical Catholic University of Rio de Janeiro, PUC-Rio
Rio de Janeiro, RJ, Brazil
wbraga@mec.puc-rio.br
ABSTRACT. In many undergraduate courses, emphasis is given to the
analysis of steady state situations, in spite of the fact that unsteady situations
are quite common in engineering problems. For instance, while discussing
heat transfer, unsteady state topics are introduced to students without
connection to steady state situations, as nature could handle unsteadiness
separately. This paper presents a few common situations, some of them often
treated only at their steady counterpart, that have been used to offer students an
interesting and pedagogically rich unsteady and steady analysis. The
methodology proposed herein, bridging unsteady to steady state situations,
helps subject integration, presents some criteria for model simplification and
allows further discussion on transient topics. The current paper is mainly
focused on Heat Conduction. However, similar analysis may be made for
situations involving Radiation and Convective Heat transfer.
Nomenclature
A
area, m2
coefficient, W/m2K
AR
aspect ratio, dimensionless
k
thermal conductivity, W/m.K
B
parameter, defined in Eq. (5), s-1
L
length, m
Bi
Biot number, dimensionless
m
fin parameter, m-1
C
parameter, defined in Eq. (5), K/s
P
Ct
Constant, defined in Eq. (19)
rate of energy generation inside
the sphere, W
c
specific heat, J/kg.K
Pe
fin perimeter, m-1
D
parameter, defined in Eq. (5), s-1
Q
rate of transfer of energy, W
E
parameter, defined in Eq. (5), K/s
q"
heat flux, W/m2
F
dimensionless parameter, Eq. (15)
h
Fo
Fourier number, dimensionless
convection
heat
transfer
coefficient inside the recipient,
W/m2.K; height of the channel, m
H
height, m
V
volume, m3
h
convective
R
radius of a cylindrical rod, m
heat
transfer
S
dimensionless parameter, Eq. (50)
Subscripts
T
temperature, K or C
b
base of an extended surface
t
time, s
c
characteristic
U
wall velocity, m/s
f
fluid
u
velocity component
H
relative to the height
x
coordinate, m
i
initial condition
L
relative to the length
R
radiation
thermal diffusivity, m /s
SS
steady-state
dimensionless length
s
surface; sphere
temperature difference, K
sup
superficial
mass density, kg/m3
t
transversal or cross sectional
∞
free stream conditions
Greek Letters
α
η
θ
ρ
φ
2
dimensionless temperature
Superscript
*
steady state condition
1. INTRODUCTION
Engineering problems occur both during unsteady and steady state situations. In
fact, experience indicates that many problems only occur or are more demanding during
transient situations. Nonetheless, in mechanical engineering courses, quite often, emphasis
is given to steady state analyses. Furthermore, most undergraduate heat transfer books
introduce steady and unsteady problems as two separate topics, perhaps because they
require different analytical methods: a simple one for steady situations and a quite
demanding one for unsteady problems. As it may be seen, most unsteady state examples
discussed leads to the same steady state solution: thermal equilibrium with the ambient, a
situation not always found.
Pedagogically speaking, the classical approach does not allow knowledge
construction, and does not help students to build up connections among the new material
and the many transient situations they met in their lives. To some students, this procedure
may indicate separate routes, solutions and physics and often, they end up believing that
there is no unsteadiness behind a steady state. Consequently, it may be concluded that a
more detailed analysis should be made linking both situations.
This paper intends to present a few classroom-type situations in which a simple and
yet interesting analysis may be made to introduce appropriately the path from unsteady to
steady situations: the lumped formulation cooling problem, the heat transfer problem in a 1D slab, a fin (extended surface), a couple of 2-D situations and the flow of fluid inside a
channel, the Couette problem.
In all such problems, the unsteady and steady parts are
analyzed in order to obtain the time necessary to reach steady state as function of the
relevant physical parameters, an important piece of information not only for many
2
industrial problems but also to an adequate understanding of the physical situation. It is
expected that following a similar analysis, students may start to visualize correctly that
steady state may occur in some situations, not always, but only at the end of an unsteady
situation. Sure, the transient phase may or may not be fast, depending on fluid thermal
properties and physical geometry, but now students may understand why.
2. SIMPLE COOLING PROBLEM
Usually discussed in an introductory heat transfer course, this problem describes the
cooling (or heating) of a small diameter, high thermal conductivity sphere (or cylinder),
initially at temperature Ti, that is dropped inside a pool, containing some non-identified
fluid, that far from the hot sphere is maintained at some uniform temperature. The heat
transfer coefficient, h, assumed to be uniform, takes care of convection and/or radiation
between the sphere and the inner walls of the recipient containing both the fluid and the
sphere.
A more interesting situation occurs when it is considered that the temperature of the
fluid may also change according to heat transfer not only to the sphere but also to an
external environment [1]. Considering the presence of some internal source of thermal
energy inside the sphere, Joule heating, for instance, say P , and taking into account some
thermal energy lost to that external environment1, QR , the lumped formulation balance of
energy for both fluid and sphere, neglecting the participation of the recipient walls, may be
written as:
( ρcV )f
•
fluid:
•
sphere:
dTf
= hAsup [ Ts − Tf ] − QR
dt
(1)
dTs
= P − hA sup [ Ts − Tf ]
dt
(2)
( ρcV )s
This system of equations must be solved considering as initial conditions:
•
Ts (t = 0) = Ts,i
(3)
•
Tf (t = 0) = Tf , i
(4)
To avoid solving the above system, one usually neglects the energy equation for the
fluid and considers only the thermal profile for the sphere. Naturally, such approximation
simplifies the problem but, on my account, it also reduces the chances of a better
understanding of its physics. Therefore, solving the system is recommended. Its solution
[2] is:
1
Actually, this term depends on the temperature difference between the recipient and the environment. It’s
taken herein as a fixed entity for simplicity.
3
B(Ts,i − Tf , i ) − ( B + D)t

CD − BE
e
=
− 1 +
T
(t)
t + Ts,i
 s
B+D
B+D

 T (t) = ( T − T ) D 1 − e − ( B + D)t  + CD − BE t + T
f
s,i
f ,i
f,i


B+D
B+D
(5)
where the following definitions apply:
•
B=
•
D=
hAsup
(ρcV )s
hA sup
(ρcV )f
•
C=
P
(ρcV)s
•
E=
QR
(ρcv )f
Several interesting results may be found:
1. After some time, Ts (t) → Tf (t) , that is, whenever the non-linear drops to zero, the
temperature difference between sphere and fluid drops to zero, as it may be readily
verified.
2. The multiplier constant CD − BE , that appears on the linear term on the right hand
side of both equations, is in fact a relation between the internal source of thermal energy
inside the sphere, P , and the heat lost to the external environment, QR . Whenever
both terms are equal, we may have a steady state.
3. Whenever P = QR , the final, i.e. the steady state temperature may be obtained from
simple Thermodynamics arguments, and it is:
T* =
(ρcV )s Ts,i + (ρcV )f Tf ,i
(ρcV )s + (ρcV )f
(6)
or
T* =
BTf ,i + DTs,i
B+D
(7)
4. We often assume that the fluid has a much larger thermal inertia (or capacitance), that
is, a larger (ρ cV) than the corresponding value for the solid sphere. In this case, any
temperature variation for the fluid may be neglected to obtain:
•
D = 0 , therefore, T* = Tf , i = Tf
(8)
•
Ts (t) = ( Tf − Ts,i ) 1 − e − Bt  + Ts,i
(9)
4
Critical to the present analysis is the evaluation of the time needed to allow body
and fluid to reach a common temperature that may or may not be, the final steady state
temperature (that is, provided P = QR ). Through observation of Equation (5), we may
conclude that such situation happens whenever the exponential term drops to zero. Noticing
the characteristic of the exponential function, a sufficient accurate estimate of the time
necessary to attain the desired situation is obtained assuming a large number for the
exponent. I usually consider 8, but any other suitable number may be obtained (see a more
thorough discussion on this below). Therefore, it may be concluded that steady state is
obtained whenever:
(B + D)t * = 8
(10)
or
8 ( ρcV )s  D  −1
t =
1+
hA sup  B 
*
that may be written as:


8
1
Fo* = 
Bi 
( ρcV )s
1 +
( ρcV )f







(11)
In the above equation, the Fourier Number, defined as
defined as
Bi =
Fo =
αs t
, and the Biot number,
L2c
hL c
, are introduced in terms of a characteristic length, Lc , defined as
ks
the ratio of the sphere’s volume to surface area. Considering the standard situation in
which the fluid is handled as having an infinite heat capacity, we obtain a simpler result:
t* = 8
( ρcV )s
hA sup
→ Fo* =
8k s
8
=
h(V / A sup ) Bi
(12)
As expected, the larger the thermal inertia (or the convective thermal resistance), the
longer it will take to reach steady state. In any event, with such expression, students may
observe clearly the influence of a large heat transfer coefficient or a small mass on the
attainability of a steady state situation, either for the cooling of thermal systems but also for
effective temperature measurement using a termocouple. Figures 1, 2 and 3 illustrate some
of the analysis that may be done. Figure 1 describes the temperature profiles for the fluid
and the sphere for the case in which the thermal capacitances of both bodies are of the same
order of magnitude (considered as equal, for simplicity). Figure 2 is obtained considering a
5
much larger thermal capacitance (taken as infinity) for the fluid than for the sphere. In such
situation, no significant temperature drop is observed for the fluid. Finally, Fig. 3 indicates
the situation in which the temperature difference between fluid and sphere drops to zero
after a while but no steady state is achievable. In such case, the heat generated inside the
sphere is larger than the heat lost to the external environment. Mathematically, this
indicates that the exponential term in Equation (5) drops to zero but not the linear one.
Fig.1. Temperature Profiles for same order thermal capacitance bodies.
Steady state is attainable,
Fig. 2 Temperature Profiles for Standard Cooling Problem. Fluid has
infinite thermal capacity. Steady state is attainable.
6
Fig. 3. Temperature Profiles for a situation in which no steady state is attainable
Temperature difference between bodies drops to zero.
Another way to pick a reasonable number involves the attainable precision on
temperature measurements. That is, assuming that it is reasonable to measure unsteady
temperature differences in the order of some small number, say 0.5% of the total variation,
but nothing less, it is sufficient to consider steady state whenever:
e − Bt = 0, 005 ≈ e −5
As it is seen, there is no significant difference between such approaches.
3. FLAT PLATE, LUMPED FORMULATION
Consider a flat plate, having a (relatively) small thickness
heat flux on the left surface,
q
"
R,
L , subjected to a radiant
and a convective heat transfer to an ambient fluid at
temperature T∞ and having a convective heat transfer coefficient, h , on the other surface.
At time t = 0, the initial temperature is Ti , considered constant. Assuming constant
thermodynamical properties, the First Law of Thermodynamics on such situation, may be
written as:
ρcV
∂T
=  q"R − h(T − T∞ )  A sup
∂t
T = Ti
at
t=0≤x≤L
(13)
(14)
that has an exact solution, the following temperature profile:
7
φ(Fo) =
1 
1
+  F −  exp {− Bi Fo}
Bi 
Bi 
In such equation, Bi
=
(15)
hL
αt
T − T∞
T − T∞
, Fo = 2 , φ = "
and F = "i
. As it
k
L
qRL / k
qR L / k
may be seen, the assymptotic characteristic exponential function precludes a sharp
definition for the final condition, that is the steady state. To overcome such difficulty, one
may use the Ritz Integral Method [3], in which a trial profile such as:
φ% (Fo) = φSS f(Fo)
φ%
is an approximate profile, φSS is the steady state solution and f(Fo) is
arbitrary (usually polinomial) function such as a quadratic profile,
is used, where
some
(16)
f(Fo) = a1 + a 2Fo + a 3 Fo 2 ,
or
f(Fo) = a1 + a 2Fo + a 3Fo2 + a4 Fo3 .
a
The constants,
cubic
a1 ,a 2 ,a3
one,
and
a4
are
determined according to the conditions:
f(Fo = 0) = 0
f(Fo = Fo*) = 1
(17)
f '(Fo = Fo*) = 0
f ''(Fo = Fo*) = 0
In such equations, Fo * (or t* ) indicates the necessary time to reach steady state, the
only unknown remaining in Equation (16) subjected to Equation (17). According to the
% , written in terms of the unknown Fo * , must be
integral method, the trial profile φ
introduced in the dimensionless form of the integral equation obtained from the energy
equation:
Fo*
∫
0
∂φ
d ( Fo ) =
∂Fo
Fo*
∫ [1 − Biφ] d(Fo)
(18)
0
Depending on the order of the approximation, different results may be obtained, such as:
•
•
quadratic profile:
cubic profile:
3
3 L2
Fo* =
→ t* =
Bi
Bi α
4
4 L2
Fo* =
→ t* =
Bi
Bi α
Previous experience with integral methods does not allow us to conclude which
approximation gives better results. Other options may be chosen: for instance, considering
8
that the thickness of the boundary layer over a flat plate is arbitrarily defined as the height
in which the velocity reaches 99% of the external uniform velocity, we may use a similar
description to obtain a value such as:
•
5
5 L2
Fo* =
→ t* =
Bi
Bi α
As a matter of fact, the important fact is that all previous options indicate that the
correct answer may be generically represented as:
Ct
Ct L2
Fo* =
→ t* =
Bi
Bi α
in which the constant,
(19)
Ct , is to be determined following any suitable criteria.
In short,
higher Fo* indicates longer time to reach steady state. According to the previous analysis
given herein, I’ve been using a number such as 8.
4. FLAT PLATE, DISTRIBUTED FORMULATION
In the literature, the study of unsteady state in situations in which the internal
(conductive) resistance is not negligible compared to the external (convective) resistance is
usually introduced using a 1-D flat plate as model. The physical situation is such that the
initial condition is such that the plate has an uniform temperature, say Ti , and suddenly it
is dropped inside a non specified medium having uniform temperature
T∞ . The convective
heat transfer coefficient, h , is considered constant. In such situation, it is expected that
after some time, the wall reaches thermal equilibrium with the medium, having the same
final temperature.
The (classical) solution is obtained using the method of separation of variables, a
lengthy procedure that involves eigenvalues and eigenfunctions (sine or a cosine series
expansion - Fourier’s series - for a Cartesian slab). I have been using another problem,
with a better result, perhaps because of the students’ greater familiarity. Recognizing that
on the initial stages of a junior level heat transfer course, a steady state 1-D flat plate is
thoroughly studied, I have been discussing the unsteady state associated to it, trying to infer
how long it takes to develop. According to a balance of energy and Fourier’s law of heat
conduction, students learn fundamentals about a steady profile such as:
T(x) = c1x + c2
in which constants c1 and c2 will be determined following some boundary conditions. The
simplest case involves specified surface temperatures, at x = 0, T = T1 and
x = L, T = T2 2.
2
To analyze the unsteadiness of such problem, we may consider the
Most interesting cases involving convective or radiative heat flux boundary conditions are easily handled.
9
following model:
1 ∂T ∂ 2 T
=
α ∂t ∂x 2
(20)
subject to:
•
t = 0, T(x, 0) = Ti
(21a)
•
T(x = 0, t) = T1
(21b)
•
T(x = L, t) = T2
(21c)
Applying the superposition method, we may write that a tentative solution may be
found according to the temperature profile given by:
T(x, t) = TSS (x) + Tt (x, t)
(22)
For the case under analysis, the first term is simply the steady profile:
T2 − T1
x + T1
(23)
L
and the term Tt (x,t) indicates the unsteady part of our problem, that is expected to drop
TSS (x) =
to zero eventually. Doing so, our problem is reduced to:
1 ∂Tt ∂ 2 Tt
=
α ∂t
∂x 2
(24)
and the boundary conditions are: at
x = 0:
T(x = 0, t) = T1 → TSS (x = 0) + Tt (x = 0) = T1
However, by definition:
TSS (x = 0) = T1 , so we may say that
Tt (x = 0) = 0
and at x = L :
(25a)
T(x = L, t) = T2 → TSS (x = L) + Tt (x = L) = T2
Similarly,
TSS (x = L) = T2 , therefore, we have that:
Tt (x = L) = 0
The initial condition deserves a similar analysis. At
(25b)
t = 0,T(x, 0) = Ti . Therefore,
T(x, t = 0) = To → TSS (x) + Tt (x, t = 0) = Ti
(25c)
Consequently,
10
Tt (x, t = 0) = Ti − TSS (x) = (Ti − T1 ) −
T2 − T1
x
L
(26)
that may be written as:
t = 0, Tt (x, 0) = a + bx
Using the standard procedure for solving unsteady 1D problems [4,5], we obtain:
Tt (x, t) = exp( −αλ 2n t)  d cos ( λ nx ) + e sin ( λ n x ) 
(27)
in which the eigenvalues are given by
λ n L = nπ
(28)
Applying the non-homogeneous initial condition yields
Tt (x, 0) = a + bx = e n sin ( λ n x )
d=0
and:
(29)
Using the standard procedure [3,4,5], we obtain that:
en =
2
 −a + ( −1) n ( a + bL ) 
nπ
(30)
Consequently, the complete temperature profile, including both the unsteady and steady
solutions, is written as
T(x, t) = a + bx + ∑ e n sin ( λ n x ) e −αλ n t
2
(31)
n =1
This result may be presented graphically, as indicated in Fig. 4. The results were
obtained for the following set of conditions:
Ti = 25 C ; T1 = 100 C ; T2 = 30 C
and are displayed as function of the Fourier number, previously defined.
11
Fig. 4. Unsteadiness of 1-D temperature profile
The question remaining to be answered is again the time span necessary for the
steady state. Noticing the transient temperature profile, we may conclude that steady state
is reached whenever the exponential term drops significantly to zero. As we are aware, the
most critical eigenvalue is the first one [4,5]. Consequently:
exp( −αλ12 t) = exp[− Fo∗ ( λ1L) 2 ] ≈ exp[ −8]
(32)
In the present situation, λ1L = π , see Equation (28), and therefore, we may
conclude that steady state happens whenever:
Fo∗ ≈
8
8
8
= 2≈
= 0, 8
2
10
π
( λ1L)
in which it was considered that
(33)
ð 2 ≈ 10 .
It may be noticed in Fig. 4, that for Fo ≥ 0, 5 , the steady state (linear) profile is
visually obtained, indicating that this simple analysis is convenient. However, for smaller
Fourier numbers, the transient temperature profiles are far from the steady state profile,
clearly indicating the reason why the Fourier’s law of heat conduction may not be taken as:
q = − kA
∂T
∆T
= kA
∂x
L
(34)
before the steady state is achieved, a situation usually not clearly understood. Naturally,
the actual profiles depend on the data used as boundary and initial conditions and other
situations may be easily studied.
For students, it may also be interesting to compare transient times for different
material, to indicate the influence of the thermal diffusivity (or the length) as shown in
12
Table 1.
Table 1: Time for Steady State, considering a 1D flat plate
with thickness = 0,2 m
5. UNSTEADY PROFILE IN EXTENDED SURFACES
Extended surfaces is one of those topics that display an interesting unsteady profile,
often not discussed among students. An energy balance for a constant transversal area fin,
with constant thermal properties and convective heat transfer coefficient is given by:
∂2θ
1 ∂θ
2
−
θ
=
m
∂x 2
α ∂t
in which
θ(x) = T(x) − T∞ ,
(35)
is the fin excess temperature, m
2
=
hPe
kA T
is the fin
parameter and α is the thermal diffusivity of the material used on the fin. The initial and
the boundary conditions, chosen for the sake of simplicity, are expressed by:
•
t = 0, θ(x) = 0
(36a)
•
x = 0, θ(x = 0) = θb , the temperature at the root of fin
(36b)
•
x = L, θ(x = L) = 0
(36c)
A straight forward analysis similar to the one made for the previously discussed example,
uses:
θ(x, t) = θSS (x) + θt (x, t)
where
θSS (x)
is the steady state temperature profile for this type of fin (very long fin)3:
θSS (x) = θbe − mx
3
(37)
(38)
Other situations may be handled similarly.
13
and indicates that the eigenvalues are given by:
 nπ 
λ =
+ m2

 L
2
2
(39)
Therefore, the complete temperature profile is given by:
θ(x) = θSS (x) + ∑ c ne −αλ n t sin ( λ n x )
2
(40)
in which the integration constants are given by:
−2
cn =
φSS sin ( nπx )dx
L ∫0
L
(41)
Repeating the previous analysis, we obtain that steady state is reached when
exp( −αλ12 t) = exp[− Fo∗ ( λ1L) 2 ] ≈ exp[ −8]
that is:
Fo∗ ≈
8
8
8
=
≈
( λ1L) 2 ( π 2 + mL2 ) (10 + mL2 )
(42)
It is now simpler to understand how the cooling rate is affected by a higher heat
transfer coefficient, obtained for instance increasing the velocity of the cooling fluid, but
also by the thermal conductivity, the cross section and the perimeter of the fin. See Fig. 5
for a graphical display of the transient behavior. It may be noticed the quick time evolution
of the dimensionless temperature profile, comparing for instance the solutions for Fo = 0,03
and Fo = 0,08 and then the solutions for Fo = 0,08 and Fo* = 0,42, that corresponds to the
steady state.
Fig. 5. Unsteady Temperature Profiles in Fins for m = 3,0.
14
6. A TWO DIMENSIONAL FLAT PLATE
2L , initially at a uniform temperature Ti . At
time t = 0, the wall is placed in a medium that is at some temperature T∞ , far from the
Consider a plane wall of thickness
wall. Heat transfer occurs by convection with a uniform and constant heat transfer
coefficient h. Mathematically, the problem may be defined according to the following
energy balance:
∂ 2 T ∂ 2 T 1 ∂T
+
=
∂x 2 ∂y 2 α ∂t
(43)
The standard procedure to obtain an analytical solution to this problem starts with
the proposition that variables can be separated:
T(x, y, t) = X(x, t)Y(y, t)
(44)
Generalizing previous results, we may write that the critical Fourier Number is
given by:
αt ∗
8
Fo = 2 =
L
(λ 1L)2 +  (µ1H)2 / AR 2 
∗
in which
AR =
(45)
2H
is the (geometrical) aspect ratio of the 2D plate and the two sets of
2L
eigenvalues are given by:
tan(λ1L) =
Bi L
λ1L
and
tan(µ1H) =
Bi H
µ1H
Naturally, it is interesting to have a general criteria to justify the usage of a 2D
problem instead of a much simpler 1D problem, as before. A simple one may be proposed
using Equation (45). For instance, it is obvious that whenever:
( λ1L) 2 >>  (µ1H) 2 / AR 2 
one may neglect the influence of the heat transferred along the horizontal surfaces and treat
the problem as a standard 1D vertical flat plate. For the present purpose, a ratio of 10 will
be considered reasonable for that. Therefore, whenever:
(µ H )
( λ1L ) = 10 1 2
( AR )
2
2
To make things easier, we may consider as before that
π ≈ 10 , resulting in:
15
λ 1L = π
µ 1H
AR
To generalize those concepts, we may define a physical aspect ratio (as it takes into
account the influence of the boundary conditions through the eigenvalues) as:
AR* =
( µ 1H )
( λ 1L )
(46)
Therefore:
αt ∗
8
Fo = 2 =
2
L
( λ1L)2 1 + ( AR*) / AR 2 


∗
Consequently, whenever
(47)
AR > π ( AR*) , that is, the geometrical aspect ratio is
greater than π times the physical aspect ratio, the 2D problem may be reduced to a 1D
vertical plate problem, possessing a much simpler solution. Similarly, if AR* > πAR ,
the 2D problem may be treated as a 1D horizontal plate problem. For other values, the 2D
model becomes relevant.
To illustrate such effect, some results are shown in Table 2, in which the nondimensional temperature differences between the 2D and the 1D approximations are
displayed. The chosen geometrical aspect ratio is 2,0. The ambient temperature is 40 C
and the initial uniform temperature is 450 C. The thermophysical properties are the thermal
conductivity (= 1,3 W/mK) and the thermal diffusivity (= 1,1E-6 m2/s). The convective
heat transfer coefficient is assumed to be 100 W/m2K along the vertical surface. Along the
horizontal surface, the corresponding coefficient is treated as parameter and handled as the
Biot number defined as Bi H = h H H / k . The results are displayed as function of time.
Table 2. Temperature Differences between 1D and 2D
Approximations for a Flat Plate. Geometrical Aspect
Ratio (AR) = 2.0
As it can be seen, the temperature differences are reduced significantly whenever
16
AR* is small compared to AR (the geometrical one).
The temperature differences
increase with time but are reduced to zero as the steady state temperature is the same in
both cases.
7. Unsteady Profile for Short Cylinders
Following a similar analysis, we may conclude that, for a cylindrical rod of radius R
and height 2L, steady state is obtained whenever
αt ∗
8
Fo = 2 =
L
( λ1L)2 +  (µ1R) 2 / AR 2 
∗
L
and µ1 is the first eigenvalue obtained as root of the equation:
R
µ1J1 (µ1 ) + Bi J o (µ1 ) = 0
wherein
(48)
AR =
(49)
where J n indicates the Bessel function of n-th order. As before, a similar criteria may be
obtained to allow us to neglect the heat transfer through the horizontal surfaces (the infinite
cylinder case) or through the lateral surface (the infinite flat plate case).
8. Transient Couette Problem
A simple solution to the Navier-Stokes equation is obtained for the flow between
two parallel flat plates, one of which is at rest, the other is moving with constant velocity U.
This is the so-called Couette Problem [6]. The steady state solution is easily obtained as:
u
= η + Sη(1 − η)
U
(50)
in which the following definitions apply:
•
η=
y
, where h indicates the distance between the two plates, i.e. the channel
h
h 2  dP 
• S=−
2µU  dx 
Typically, for S > 0 , that is, for a pressure decreasing in the flow direction, the velocity is
positive over the whole channel. For negative values of P, however, the velocity may
become negative, indicating the existence of a back-flow near the stationary wall.
Following a similar procedure close to the ones already shown, we obtain the following
profile as transient solution:
17
u(y,t)
= η + Sη(1 − η) + ∑ d n exp{ −Foς 2n }sin ( ς n η)
U
(51)
where:
νt
h2
•
Fo =
•
ς n , a dimensionless eigenvalue, is given by λ n h = ς = nπ
 ( − 1) n
• dn = 2 
 ς n

S
2S 
2 
1
S
2
+
+
−
ς
−
n 

3 
ς 2n

 ς n 
(
)
For the sake of demonstration, Fig. 6 indicates the transient velocity profile for the
special case that S = −2 . As it may be seen, it is obvious that the slip condition is a
localized effect while the pressure gradient is a bulk effect. The developing of a boundary
layer4 type of flow along the moving wall is clearly displayed. It is also shown the final
steady state profile, for comparison purposes. A similar analysis may be done for the
Hagen-Poiseuilli flow, as done originally by Szymanski [7].
Fig. 6 Transient Couette Flow, indicating the development of both
a back-flow and a positive flow regions with time.
4
that is, a region in which the moving wall affects the flow.
18
9. Conclusions
This paper presents a few discussions on how to investigate steady state situations
as a final stage for heat transfer problems, having in mind undergraduate engineering
students. In all the situations treated here, it is shown how thermal and geometrical
parameters affect the time necessary to reach steady state, allowing a deeper understanding
of transient effects.
Considering the standard approach used in most textbooks, the
procedure discussed here has at least two advantages: the sequential analysis of the
problems and the link between concepts, allowing a clearer view of the evolution.
Although not shown here, there are many other situations that may be handled accordingly,
allowing, perhaps, a fuller integrated engineering course.
Acknowledgements
I would like to thank the reviewers for suggesting improvements and helpful
comments to the original manuscript.
9. References
[1] PITTS, D.R. & SISSOM, L.E., Heat Transfer, Schaum Outlines, 1977,
[2] KAPLAN, W. Advanced Mathematics for Engineers, Addison Wesley, 1981
[3] ARPACI, V.S., Conduction Heat Transfer, Addison Wesley, 1966
[4] BRAGA, W., Heat Transfer, Thomson Learning Pub. Co., 2003, in portuguese
[5] INCROPERA F.P. & DEWITT D.P. Fundamentals of Heat and Mass Transfer, Wiley,
N.York, 1996
[6] SCHLICHTING, H., McGraw-Hill Book Company, 6th Edition, 1968.
[7] SZYMANSKI, F., Proc. Intern. Congr. Appl. Mech. Stockholm, 1930.
19