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 theanalysis of steady state situations, in spite of the fact that unsteady situationsare quite common in engineering problems. For instance, while discussingheat transfer, unsteady state topics are introduced to students withoutconnection to steady state situations, as nature could handle unsteadinessseparately. This paper presents a few common situations, some of them oftentreated only at their steady counterpart, that have been used to offer students aninteresting and pedagogically rich unsteady and steady analysis. Themethodology proposed herein, bridging unsteady to steady state situations,helps subject integration, presents some criteria for model simplification andallows further discussion on transient topics. The current paper is mainlyfocused on Heat Conduction. However, similar analysis may be made forsituations involving Radiation and Convective Heat transfer.

Nomenclature

A area, m2

AR aspect ratio, dimensionless

B parameter, defined in Eq. (5), s-1

Bi Biot number, dimensionless

C parameter, defined in Eq. (5), K/s

Ct Constant, defined in Eq. (19)

c specific heat, J/kg.K

D parameter, defined in Eq. (5), s-1

E parameter, defined in Eq. (5), K/s

F dimensionless parameter, Eq. (15)

Fo Fourier number, dimensionless

H height, m

h convective heat transfer

coefficient, W/m2K

k thermal conductivity, W/m.K

L length, m

m fin parameter, m-1

P rate of energy generation insidethe sphere, W

Pe fin perimeter, m-1

Q rate of transfer of energy, W

q" heat flux, W/m2

h convection heat transfercoefficient inside the recipient,W/m2.K; height of the channel, m

V volume, m3

R radius of a cylindrical rod, m

2

S dimensionless parameter, Eq. (50)

T temperature, K or C

t time, s

U wall velocity, m/s

u velocity component

x coordinate, m

Greek Letters

αα thermal diffusivity, m2/s

ηη dimensionless length

θθ temperature difference, K

ρρ mass density, kg/m3

φφ dimensionless temperature

Subscripts

b base of an extended surface

c characteristic

f fluid

H relative to the height

i initial condition

L relative to the length

R radiation

SS steady-state

s surface; sphere

sup superficial

t transversal or cross sectional

∞∞ free stream conditions

Superscript

* steady state condition

1. INTRODUCTION

Engineering problems occur both during unsteady and steady state situations. Infact, experience indicates that many problems only occur or are more demanding duringtransient situations. Nonetheless, in mechanical engineering courses, quite often, emphasisis given to steady state analyses. Furthermore, most undergraduate heat transfer booksintroduce steady and unsteady problems as two separate topics, perhaps because theyrequire different analytical methods: a simple one for steady situations and a quitedemanding one for unsteady problems. As it may be seen, most unsteady state examplesdiscussed leads to the same steady state solution: thermal equilibrium with the ambient, asituation not always found.

Pedagogically speaking, the classical approach does not allow knowledgeconstruction, and does not help students to build up connections among the new materialand the many transient situations they met in their lives. To some students, this proceduremay indicate separate routes, solutions and physics and often, they end up believing thatthere is no unsteadiness behind a steady state. Consequently, it may be concluded that amore detailed analysis should be made linking both situations.

This paper intends to present a few classroom-type situations in which a simple andyet interesting analysis may be made to introduce appropriately the path from unsteady tosteady situations: the lumped formulation cooling problem, the heat transfer problem in a 1-D slab, a fin (extended surface), a couple of 2-D situations and the flow of fluid inside achannel, the Couette problem. In all such problems, the unsteady and steady parts areanalyzed in order to obtain the time necessary to reach steady state as function of therelevant physical parameters, an important piece of information not only for many

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industrial problems but also to an adequate understanding of the physical situation. It isexpected that following a similar analysis, students may start to visualize correctly thatsteady state may occur in some situations, not always, but only at the end of an unsteadysituation. Sure, the transient phase may or may not be fast, depending on fluid thermalproperties and physical geometry, but now students may understand why.

2. SIMPLE COOLING PROBLEM

Usually discussed in an introductory heat transfer course, this problem describes thecooling (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-identifiedfluid, that far from the hot sphere is maintained at some uniform temperature. The heattransfer coefficient, h, assumed to be uniform, takes care of convection and/or radiationbetween the sphere and the inner walls of the recipient containing both the fluid and thesphere.

A more interesting situation occurs when it is considered that the temperature of thefluid may also change according to heat transfer not only to the sphere but also to anexternal environment [1]. Considering the presence of some internal source of thermalenergy inside the sphere, Joule heating, for instance, say P , and taking into account some

thermal energy lost to that external environment1, RQ , the lumped formulation balance of

energy for both fluid and sphere, neglecting the participation of the recipient walls, may bewritten as:

• fluid: (( )) [[ ]]fsup s f Rf

dTcV hA T T Q

dtρ = − −ρ = − − (1)

• sphere: (( )) [[ ]]ssup s fs

dTcV P hA T T

dtρ = − −ρ = − − (2)

This system of equations must be solved considering as initial conditions:

• 0s s,iT (t ) T= == = (3)

• 0f f , iT (t ) T= == = (4)

To avoid solving the above system, one usually neglects the energy equation for thefluid and considers only the thermal profile for the sphere. Naturally, such approximationsimplifies the problem but, on my account, it also reduces the chances of a betterunderstanding 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’staken herein as a fixed entity for simplicity.

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(( ))

1

1

s,i f , i (B D)ts s,i

(B D)tf s,i f ,i f , i

B(T T ) CD BET (t) e t T

B D B DD CD BE

T (t) T T e t TB D B D

− +− +

− +− +

−− −− = − + += − + + + ++ +

−− = − − + += − − + + + ++ +

(5)

where the following definitions apply:

• (( ))

sup

s

hAB

cV==

ρρ

• (( ))

sup

f

hAD

cV==

ρρ

• (( ))

s

PC

cV==

ρρ

• (( ))

R

f

QE

cv==

ρρ

Several interesting results may be found:

1. After some time, s fT (t) T (t)→→ , that is, whenever the non-linear drops to zero, the

temperature difference between sphere and fluid drops to zero, as it may be readilyverified.

2. The multiplier constant CD BE−− , that appears on the linear term on the right handside 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, RQ . Whenever

both terms are equal, we may have a steady state.

3. Whenever RP Q== , the final, i.e. the steady state temperature may be obtained from

simple Thermodynamics arguments, and it is:

(( )) (( ))(( )) (( ))

s,i f ,i* s f

s f

cV T cV TT

cV cV

ρ + ρρ + ρ==

ρ + ρρ + ρ (6)

or

f ,i s,i* BT DTT

B D

++==

++ (7)

4. We often assume that the fluid has a much larger thermal inertia (or capacitance), thatis, a larger ( cV)ρρ than the corresponding value for the solid sphere. In this case, anytemperature variation for the fluid may be neglected to obtain:

• 0D == , therefore, *

f , i fT T T= == = (8)

• (( )) 1 Bts f s,i s,iT (t) T T e T−− = − − += − − + (9)

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Critical to the present analysis is the evaluation of the time needed to allow bodyand fluid to reach a common temperature that may or may not be, the final steady state

temperature (that is, provided RP Q== ). Through observation of Equation (5), we may

conclude that such situation happens whenever the exponential term drops to zero. Noticingthe characteristic of the exponential function, a sufficient accurate estimate of the timenecessary to attain the desired situation is obtained assuming a large number for theexponent. I usually consider 8, but any other suitable number may be obtained (see a morethorough discussion on this below). Therefore, it may be concluded that steady state isobtained whenever:

8*(B D)t+ =+ = (10)

or

(( )) 181* s

sup

cV Dt

hA B

−−ρρ = += + that may be written as:

(( ))(( ))

8 1

1

*

s

f

FocVBi

cV

== ρρ

++ ρρ

(11)

In the above equation, the Fourier Number, defined as s2c

tFo

L

αα== , and the Biot number,

defined as c

s

hLBi

k== , are introduced in terms of a characteristic length, cL , defined as

the ratio of the sphere’s volume to surface area. Considering the standard situation inwhich the fluid is handled as having an infinite heat capacity, we obtain a simpler result:

(( )) 8 88* s s

sup sup

cV kt Fo*

hA h(V / A ) Bi

ρρ= → = == → = = (12)

As expected, the larger the thermal inertia (or the convective thermal resistance), thelonger it will take to reach steady state. In any event, with such expression, students mayobserve clearly the influence of a large heat transfer coefficient or a small mass on theattainability of a steady state situation, either for the cooling of thermal systems but also foreffective temperature measurement using a termocouple. Figures 1, 2 and 3 illustrate someof the analysis that may be done. Figure 1 describes the temperature profiles for the fluidand the sphere for the case in which the thermal capacitances of both bodies are of the sameorder of magnitude (considered as equal, for simplicity). Figure 2 is obtained considering a

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much larger thermal capacitance (taken as infinity) for the fluid than for the sphere. In suchsituation, no significant temperature drop is observed for the fluid. Finally, Fig. 3 indicatesthe situation in which the temperature difference between fluid and sphere drops to zeroafter a while but no steady state is achievable. In such case, the heat generated inside thesphere is larger than the heat lost to the external environment. Mathematically, thisindicates 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 hasinfinite thermal capacity. Steady state is attainable.

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Fig. 3. Temperature Profiles for a situation in which no steady state is attainableTemperature difference between bodies drops to zero.

Another way to pick a reasonable number involves the attainable precision ontemperature measurements. That is, assuming that it is reasonable to measure unsteadytemperature 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:

50 005Bte , e− −− −= ≈= ≈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 L , subjected to a radiant

heat flux on the left surface, "Rq , and a convective heat transfer to an ambient fluid at

temperatureT∞∞ and having a convective heat transfer coefficient,h , on the other surface.

At time t = 0, the initial temperature is iT , considered constant. Assuming constant

thermodynamical properties, the First Law of Thermodynamics on such situation, may bewritten as:

"R sup

TcV q h(T T ) A

t ∞∞

∂∂ ρ = − −ρ = − − ∂∂ (13)

iT T== at t 0 x L= ≤ ≤= ≤ ≤ (14)

that has an exact solution, the following temperature profile:

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{{ }}1 1(Fo) F exp Bi Fo

Bi Bi φ = + − −φ = + − −

(15)

In such equation,hL

Bik

== ,2

tFo

Lαα

== , "R

T Tq L / k

∞∞−−φ =φ = and i

"R

T TF

q L / k∞∞−−

== . As it

may be seen, the assymptotic characteristic exponential function precludes a sharpdefinition for the final condition, that is the steady state. To overcome such difficulty, onemay use the Ritz Integral Method [3], in which a trial profile such as:

SS(Fo) f(Fo)φ = φφ = φ%% (16)

is used, where φφ%% is an approximate profile, SSφφ is the steady state solution and f(Fo) is

some arbitrary (usually polinomial) function such as a quadratic profile,2

1 2 3f(Fo) a a Fo a Fo= + += + + , or a cubic one,2 3

1 2 3 4f(Fo) a a Fo a Fo a Fo= + + += + + + . The constants, 1 2 3a ,a ,a and 4a are

determined according to the conditions:

f(Fo 0) 0

f(Fo Fo*) 1

f '(Fo Fo*) 0

f ''(Fo Fo*) 0

= == == == == == == == =

(17)

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

integral method, the trial profile φφ%% , written in terms of the unknown Fo* , must be

introduced in the dimensionless form of the integral equation obtained from the energyequation:

(( )) [[ ]]0 0

1Fo* Fo*

d Fo Bi d(Fo)Fo∂φ∂φ

= − φ= − φ∂∂∫ ∫∫ ∫ (18)

Depending on the order of the approximation, different results may be obtained, such as:

• quadratic profile: 23 3 L

Fo* t*Bi Bi

= → == → =αα

• cubic profile: 24 4 L

Fo* t*Bi Bi

= → == → =αα

Previous experience with integral methods does not allow us to conclude whichapproximation gives better results. Other options may be chosen: for instance, considering

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that the thickness of the boundary layer over a flat plate is arbitrarily defined as the heightin which the velocity reaches 99% of the external uniform velocity, we may use a similardescription to obtain a value such as:

• 25 5 L

Fo* t*Bi Bi

= → == → =αα

As a matter of fact, the important fact is that all previous options indicate that thecorrect answer may be generically represented as:

2t tC C L

Fo* t*Bi Bi

= → == → =αα

(19)

in which the constant, tC , is to be determined following any suitable criteria. In short,

higher Fo* indicates longer time to reach steady state. According to the previous analysisgiven 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 isusually 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 iT , 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 thatafter some time, the wall reaches thermal equilibrium with the medium, having the samefinal temperature.

The (classical) solution is obtained using the method of separation of variables, alengthy procedure that involves eigenvalues and eigenfunctions (sine or a cosine seriesexpansion - 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 thaton the initial stages of a junior level heat transfer course, a steady state 1-D flat plate isthoroughly studied, I have been discussing the unsteady state associated to it, trying to inferhow long it takes to develop. According to a balance of energy and Fourier’s law of heatconduction, students learn fundamentals about a steady profile such as:

1 2T(x) c x c= += +

in which constants c1 and c2 will be determined following some boundary conditions. The

simplest case involves specified surface temperatures, at 10x ,T T= == = and

2x L,T T= == = 2. To analyze the unsteadiness of such problem, we may consider the

2 Most interesting cases involving convective or radiative heat flux boundary conditions are easily handled.

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following model:

2

2

1 T Tt x

∂ ∂∂ ∂==

α ∂ ∂α ∂ ∂ (20)

subject to:

• 0 0 it , T(x, ) T= == = (21a)

• 10T(x , t) T= == = (21b)

• 2T(x L, t) T= == = (21c)

Applying the superposition method, we may write that a tentative solution may befound according to the temperature profile given by:

SS tT(x, t) T (x) T (x, t)= += + (22)

For the case under analysis, the first term is simply the steady profile:

2 11SS

T TT (x) x T

L

−−= += + (23)

and the term tT (x,t) indicates the unsteady part of our problem, that is expected to drop

to zero eventually. Doing so, our problem is reduced to:

2

2

1 t tT Tt x

∂ ∂∂ ∂==

α ∂ ∂α ∂ ∂ (24)

and the boundary conditions are: at 0x == :

1 10 0 0SS tT(x , t) T T (x ) T (x ) T= = → = + = == = → = + = =

However, by definition: 10SST (x ) T= == = , so we may say that

0 0tT (x )= == = (25a)

and at x L== :

2 2SS tT(x L, t) T T (x L) T (x L) T= = → = + = == = → = + = =

Similarly, 2SST (x L) T= == = , therefore, we have that:

0tT (x L)= == = (25b)

The initial condition deserves a similar analysis. At 0 0 it ,T(x, ) T= == = . Therefore,

0 0o SS t iT(x, t ) T T (x) T (x, t ) T= = → + = == = → + = = (25c)

Consequently,

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2 110t i SS i

T TT (x, t ) T T (x) (T T ) x

L

−−= = − = − −= = − = − − (26)

that may be written as:

0 0tt , T (x, ) a bx= = += = +

Using the standard procedure for solving unsteady 1D problems [4,5], we obtain:

(( )) (( ))2t n n nT (x, t) exp( t) dcos x e sin x = −αλ λ + λ= −αλ λ + λ (27)

in which the eigenvalues are given by

nL nλ = πλ = π (28)

Applying the non-homogeneous initial condition yields 0d == and:

(( ))0t n nT (x, ) a bx e sin x= + = λ= + = λ (29)

Using the standard procedure [3,4,5], we obtain that:

(( ))21 n

ne a ( ) a bLn

= − + − += − + − + ππ (30)

Consequently, the complete temperature profile, including both the unsteady and steadysolutions, is written as

(( )) 2

1

ntn n

n

T(x, t) a bx e sin x e−αλ−αλ

==

= + + λ= + + λ∑∑ (31)

This result may be presented graphically, as indicated in Fig. 4. The results wereobtained for the following set of conditions:

25iT C== ; 1 100T C== ; 2 30T C==

and are displayed as function of the Fourier number, previously defined.

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Fig. 4. Unsteadiness of 1-D temperature profile

The question remaining to be answered is again the time span necessary for thesteady state. Noticing the transient temperature profile, we may conclude that steady stateis reached whenever the exponential term drops significantly to zero. As we are aware, themost critical eigenvalue is the first one [4,5]. Consequently:

2 21 1 8exp( t) exp[ Fo ( L) ] exp[ ]∗∗−αλ = − λ ≈ −−αλ = − λ ≈ − (32)

In the present situation, 1Lλ = πλ = π , see Equation (28), and therefore, we may

conclude that steady state happens whenever:

2 21

8 8 80 8

10Fo ,

( L)∗∗ ≈ = ≈ =≈ = ≈ =

λ πλ π (33)

in which it was considered that 2 10ð ≈≈ .

It may be noticed in Fig. 4, that for 0 5Fo ,≥≥ , the steady state (linear) profile isvisually obtained, indicating that this simple analysis is convenient. However, for smallerFourier 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:

T Tq kA 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 othersituations may be easily studied.

For students, it may also be interesting to compare transient times for differentmaterial, to indicate the influence of the thermal diffusivity (or the length) as shown in

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Table 1.

Table 1: Time for Steady State, considering a 1D flat platewith 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:

22

2

1m

x t∂ θ ∂θ∂ θ ∂θ

− θ =− θ =∂ α ∂∂ α ∂

(35)

in which (x) T(x) T∞∞θ = −θ = − , is the fin excess temperature,2 e

T

hPm

kA== is the fin

parameter and αα is the thermal diffusivity of the material used on the fin. The initial andthe boundary conditions, chosen for the sake of simplicity, are expressed by:

• 0 0t , (x)= θ == θ = (36a)

• 0 0 bx , (x )= θ = = θ= θ = = θ , the temperature at the root of fin (36b)

• 0x L, (x L)= θ = == θ = = (36c)

A straight forward analysis similar to the one made for the previously discussed example,uses:

SS t(x, t) (x) (x, t)θ = θ + θθ = θ + θ (37)

where SS(x)θθ is the steady state temperature profile for this type of fin (very long fin)3:

mxSS b(x) e−−θ = θθ = θ (38)

3 Other situations may be handled similarly.

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and indicates that the eigenvalues are given by:

2

2 2nm

Lππ λ = +λ = +

(39)

Therefore, the complete temperature profile is given by:

(( ))2n t

SS n n(x) (x) c e sin x−αλ−αλθ = θ + λθ = θ + λ∑∑ (40)

in which the integration constants are given by:

(( ))0

2 L

n SSc sin n x dxL−−

= φ π= φ π∫∫ (41)

Repeating the previous analysis, we obtain that steady state is reached when

2 21 1 8exp( t) exp[ Fo ( L) ] exp[ ]∗∗−αλ = − λ ≈ −−αλ = − λ ≈ −

that is:

2 2 2 21

8 8 8

10Fo

( L) ( mL ) ( mL )∗∗ ≈ = ≈≈ = ≈

λ π + +λ π + + (42)

It is now simpler to understand how the cooling rate is affected by a higher heattransfer coefficient, obtained for instance increasing the velocity of the cooling fluid, butalso by the thermal conductivity, the cross section and the perimeter of the fin. See Fig. 5for a graphical display of the transient behavior. It may be noticed the quick time evolutionof the dimensionless temperature profile, comparing for instance the solutions for Fo = 0,03and Fo = 0,08 and then the solutions for Fo = 0,08 and Fo* = 0,42, that corresponds to thesteady state.

Fig. 5. Unsteady Temperature Profiles in Fins for m = 3,0.

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6. A TWO DIMENSIONAL FLAT PLATE

Consider a plane wall of thickness 2L , initially at a uniform temperature iT . At

time t = 0, the wall is placed in a medium that is at some temperature T∞∞ , far from the

wall. Heat transfer occurs by convection with a uniform and constant heat transfercoefficient h. Mathematically, the problem may be defined according to the followingenergy balance:

2 2

2 2

T T 1 T

x y t

∂ ∂ ∂∂ ∂ ∂+ =+ =

∂ ∂ α ∂∂ ∂ α ∂ (43)

The standard procedure to obtain an analytical solution to this problem starts withthe 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 isgiven by:

2 2 2 21 1

8tFo

L ( L) ( H) / AR

∗∗∗∗ αα

= == = λ + µλ + µ

(45)

in which 2

2

HAR

L== is the (geometrical) aspect ratio of the 2D plate and the two sets of

eigenvalues are given by:

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LBitan( L)

Lλ =λ =

λλ and 1

1

HBitan( H)

Hµ =µ =

µµNaturally, 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 proposedusing Equation (45). For instance, it is obvious that whenever:

2 2 21 1( L) ( H) / AR λ >> µλ >> µ

one may neglect the influence of the heat transferred along the horizontal surfaces and treatthe problem as a standard 1D vertical flat plate. For the present purpose, a ratio of 10 willbe considered reasonable for that. Therefore, whenever:

(( )) (( ))(( ))

22 1

1 2

HL 10

AR

µµλ =λ =

To make things easier, we may consider as before that 10π ≈π ≈ , resulting in:

16

11

HL

AR

µµλ = πλ = π

To generalize those concepts, we may define a physical aspect ratio (as it takes intoaccount the influence of the boundary conditions through the eigenvalues) as:

(( ))(( ))

1

1

HAR*

L

µµ==

λλ (46)

Therefore:

(( ))2 22 21

8

1

tFo

L ( L) AR* / AR

∗∗∗∗ αα

= == = λ +λ +

(47)

Consequently, whenever (( ))AR AR*> π> π , that is, the geometrical aspect ratio is

greater than ππ times the physical aspect ratio, the 2D problem may be reduced to a 1Dvertical plate problem, possessing a much simpler solution. Similarly, if AR* AR> π> π ,the 2D problem may be treated as a 1D horizontal plate problem. For other values, the 2Dmodel becomes relevant.

To illustrate such effect, some results are shown in Table 2, in which the non-dimensional temperature differences between the 2D and the 1D approximations aredisplayed. The chosen geometrical aspect ratio is 2,0. The ambient temperature is 40 Cand the initial uniform temperature is 450 C. The thermophysical properties are the thermalconductivity (= 1,3 W/mK) and the thermal diffusivity (= 1,1E-6 m2/s). The convectiveheat transfer coefficient is assumed to be 100 W/m2K along the vertical surface. Along thehorizontal surface, the corresponding coefficient is treated as parameter and handled as the

Biot number defined as H HBi h H / k== . The results are displayed as function of time.

Table 2. Temperature Differences between 1D and 2DApproximations for a Flat Plate. Geometrical Aspect

Ratio (AR) = 2.0

As it can be seen, the temperature differences are reduced significantly whenever

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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 inboth cases.

7. Unsteady Profile for Short Cylinders

Following a similar analysis, we may conclude that, for a cylindrical rod of radius Rand height 2L, steady state is obtained whenever

2 2 2 21 1

8tFo

L ( L) ( R) / AR

∗∗∗∗ αα

= == = λ + µλ + µ

(48)

wherein L

ARR

== and 1µµ is the first eigenvalue obtained as root of the equation:

1 1 1 1 0oJ ( ) Bi J ( )µ µ + µ =µ µ + µ = (49)

where nJ 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 infinitecylinder 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 betweentwo 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:

uS (1 )

U= η + η − η= η + η − η (50)

in which the following definitions apply:

• yh

η =η = , where h indicates the distance between the two plates, i.e. the channel

• 2h dP

S2 U dx

= −= − µµ Typically, for S 0>> , that is, for a pressure decreasing in the flow direction, the velocity ispositive over the whole channel. For negative values of P, however, the velocity maybecome 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 followingprofile as transient solution:

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(( ))2n n n

u(y,t)S (1 ) d exp{ Fo }sin

U= η + η − η + − ς ς η= η + η − η + − ς ς η∑∑ (51)

where:

• 2

tFo

hνν

==

• nςς , a dimensionless eigenvalue, is given by nh nλ = ς = πλ = ς = π

• (( ))n

2n n2 3

n n n

( 1) S 2Sd 2 1 S 2

−− = + + − ς −= + + − ς − ς ς ςς ς ς 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 alocalized effect while the pressure gradient is a bulk effect. The developing of a boundarylayer4 type of flow along the moving wall is clearly displayed. It is also shown the finalsteady state profile, for comparison purposes. A similar analysis may be done for theHagen-Poiseuilli flow, as done originally by Szymanski [7].

Fig. 6 Transient Couette Flow, indicating the development of botha back-flow and a positive flow regions with time.

4 that is, a region in which the moving wall affects the flow.

19

9. Conclusions

This paper presents a few discussions on how to investigate steady state situationsas a final stage for heat transfer problems, having in mind undergraduate engineeringstudents. In all the situations treated here, it is shown how thermal and geometricalparameters affect the time necessary to reach steady state, allowing a deeper understandingof transient effects. Considering the standard approach used in most textbooks, theprocedure discussed here has at least two advantages: the sequential analysis of theproblems 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 helpfulcomments 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.