Computational Fluid Mechanics and Heat Transfer Third Edition_6 doc

184 Chapter 4: Analysis of heat conduction and some steady one-dimensional problemsthe dimensionless functional equation for the local velocity cf.Section8.5.. The inside heat transfer c

184 Chapter 4: Analysis of heat conduction and some steady one-dimensional problems

the dimensionless functional equation for the local velocity (cf.Section8.5)

4.7 A steam preheater consists of a thick, electrically

conduct-ing, cylindrical shell insulated on the outside, with wet streamflowing down the middle The inside heat transfer coefficient

is highly variable, depending on the velocity, quality, and so

on, but the flow temperature is constant Heat is released at

˙

q J/m3s within the cylinder wall Evaluate the temperaturewithin the cylinder as a function of position Plot Θ against

ρ, where Θ is an appropriate dimensionless temperature and

ρ = r /ro Use ρ i = 2/3 and note that Bi will be the parameter

of a family of solutions On the basis of this plot, recommend

criteria (in terms of Bi) for (a) replacing the convective ary condition on the inside with a constant temperature condi-tion; (b) neglecting temperature variations within the cylinder

bound-4.8 Steam condenses on the inside of a small pipe, keeping it at

a specified temperature, T i The pipe is heated by electricalresistance at a rate ˙q W/m3 The outside temperature is T ∞andthere is a natural convection heat transfer coefficient,h around

the outside (a) Derive an expression for the dimensionlessexpression temperature distribution, Θ = (T − T ∞ )/(Ti − T∞ ),

as a function of the radius ratios, ρ = r /ro and ρ i = ri/ro;

a heat generation number, Γ = ˙ qr o2/k(T i − T∞ ); and the Biot

number (b) Plot this result for the case ρ i = 2/3, Bi = 1, and

for several values of Γ (c) Discuss any interesting aspects ofyour result

4.9 Solve Problem 2.5 if you have not already done so, putting

it in dimensionless form before you begin Then let the Biotnumbers approach infinity in the solution You should get thesame solution we got in Example 2.5, using b.c.’s of the firstkind Do you?

4.10 Complete the algebra that is missing between eqns (4.30) and

eqn (4.31b) and eqn (4.41)

4.11 Complete the algebra that is missing between eqns (4.30) and

eqn (4.31a) and eqn (4.48)

Problems 185

4.12 Obtain eqn (4.50) from the general solution for a fin [eqn (4.35)],

using the b.c.’s T (x = 0) = T0 and T (x = L) = T∞ Comment

on the significance of the computation

4.13 What is the minimum length, l, of a thermometer well

neces-sary to ensure an error less than 0.5% of the difference between

the pipe wall temperature and the temperature of fluid flowing

in a pipe? The well consists of a tube with the end closed It

has a 2 cm O.D and a 1.88 cm I.D The material is type 304

stainless steel Assume that the fluid is steam at 260◦C and

that the heat transfer coefficient between the steam and the

tube wall is 300 W/m2K [3.44 cm.]

4.14 Thin fins with a 0.002 m by 0.02 m rectangular cross section

and a thermal conductivity of 50 W/m·K protrude from a wall

and haveh  600 W/m2K and T0 = 170 ◦C What is the heat

flow rate into each fin and what is the effectiveness? T ∞ =

20◦C

4.15 A thin rod is anchored at a wall at T = T0 on one end and is

insulated at the other end Plot the dimensionless temperature

distribution in the rod as a function of dimensionless length:

(a) if the rod is exposed to an environment at T ∞ through a

heat transfer coefficient; (b) if the rod is insulated but heat is

removed from the fin material at the unform rate−˙ q = hP(T0−

T ∞ )/A Comment on the implications of the comparison.

4.16 A tube of outside diameter d o and inside diameter d i carries

fluid at T = T1 from one wall at temperature T1 to another

wall a distance L away, at T r Outside the tube h ois negligible,

and inside the tube h i is substantial Treat the tube as a fin

and plot the dimensionless temperature distribution in it as a

function of dimensionless length

4.17 (If you have had some applied mathematics beyond the usual

two years of calculus, this problem will not be difficult.) The

shape of the fin in Fig.4.12is changed so that A(x) = 2δ(x/L)2b

instead of 2δ(x/L)b Calculate the temperature distribution

and the heat flux at the base Plot the temperature distribution

and fin thickness against x/L Derive an expression for ηf

186 Chapter 4: Analysis of heat conduction and some steady one-dimensional problems

4.18 Work Problem 2.21, if you have not already done so,

nondi-mensionalizing the problem before you attempt to solve it Itshould now be much simpler

4.19 One end of a copper rod 30 cm long is held at 200◦C, and the

other end is held at 93◦C The heat transfer coefficient in tween is 17 W/m2K (including both convection and radiation)

be-If T ∞ = 38◦C and the diameter of the rod is 1.25 cm, what isthe net heat removed by the air around the rod? [19.13 W.]

4.20 How much error will the insulated-tip assumption give rise to

in the calculation of the heat flow into the fin in Example4.8?

4.21 A straight cylindrical fin 0.6 cm in diameter and 6 cm long

protrudes from a magnesium block at 300◦C Air at 35◦C isforced past the fin so thath is 130 W/m2K Calculate the heatremoved by the fin, considering the temperature depression ofthe root

4.22 Work Problem4.19considering the temperature depression in

both roots To do this, find mL for the two fins with insulated

tips that would give the same temperature gradient at each

wall Base the correction on these values of mL.

4.23 A fin of triangular axial section (cf Fig 4.12) 0.1 m in length

and 0.02 m wide at its base is used to extend the surface area

of a 0.5% carbon steel wall If the wall is at 40◦C and heatedgas flows past at 200◦ C (h = 230 W/m2K), compute the heat

removed by the fin per meter of breadth, b, of the fin Neglect

temperature distortion at the root

4.24 Consider the concrete slab in Example 2.1 Suppose that the

heat generation were to cease abruptly at time t = 0 and the

slab were to start cooling back toward T w Predict T = Twas afunction of time, noting that the initial parabolic temperatureprofile can be nicely approximated as a sine function (Withoutthe sine approximation, this problem would require the seriesmethods of Chapter5

4.25 Steam condenses in a 2 cm I.D thin-walled tube of 99%

alu-minum at 10 atm pressure There are circular fins of constantthickness, 3.5 cm in diameter, every 0.5 cm on the outside The

Problems 187

fins are 0.8 mm thick and the heat transfer coefficient from

themh = 6 W/m2K (including both convection and radiation)

What is the mass rate of condensation if the pipe is 1.5 m in

length, the ambient temperature is 18◦ C, and h for

condensa-tion is very large? [ ˙mcond= 0.802 kg/hr.]

4.26 How long must a copper fin, 0.4 cm in diameter, be if the

tem-perature of its insulated tip is to exceed the surrounding air

temperature by 20% of (T0− T∞ )? Tair = 20 ◦ C and h = 28

W/m2K (including both convection and radiation)

4.27 A 2 cm ice cube sits on a shelf of aluminum rods, 3 mm in

diam-eter, in a refrigerator at 10◦C How rapidly, in mm/min, does

the ice cube melt through the wires if h at the surface of the

wires is 10 W/m2K (including both convection and radiation)

Be sure that you understand the physical mechanism before

you make the calculation Check your result experimentally

h sf = 333, 300 J/kg.

4.28 The highest heat flux that can be achieved in nucleate

boil-ing (called qmax—see the qualitative discussion in Section9.1)

depends upon ρ g , the saturated vapor density; h fg, the

la-tent heat vaporization; σ , the surface tension; a characteristic

length, l; and the gravity force per unit volume, g(ρ f − ρg),

where ρ f is the saturated liquid density Develop the

dimen-sionless functional equation for qmax in terms of

dimension-less length

4.29 You want to rig a handle for a door in the wall of a furnace The

door is at 160◦C You consider bending a 16 in length of ¼ in

0.5% carbon steel rod into a U-shape and welding the ends to

the door Surrounding air at 24◦ C will cool the handle (h = 12

W/m2K including both convection and radiation) What is the

coolest temperature of the handle? How close to the door can

you grasp it without being burned? How might you improve

the handle?

4.30 A 14 cm long by 1 cm square brass rod is supplied with 25 W at

its base The other end is insulated It is cooled by air at 20◦C,

withh = 68 W/m2K Develop a dimensionless expression for

Θ as a function of εf and other known information Calculate

the base temperature

188 Chapter 4: Analysis of heat conduction and some steady one-dimensional problems

4.31 A cylindrical fin has a constant imposed heat flux of q1 at one

end and q2at the other end, and it is cooled convectively alongits length Develop the dimensionless temperature distribu-

tion in the fin Specialize this result for q2= 0 and L → ∞, and

compare it with eqn (4.50)

4.32 A thin metal cylinder of radius r o serves as an electrical

re-sistance heater The temperature along an axial line in one

side is kept at T1 Another line, θ2 radians away, is kept at

T2 Develop dimensionless expressions for the temperaturedistributions in the two sections

4.33 Heat transfer is augmented, in a particular heat exchanger,

with a field of 0.007 m diameter fins protruding 0.02 m into aflow The fins are arranged in a hexagonal array, with a mini-mum spacing of 1.8 cm The fins are bronze, and h f aroundthe fins is 168 W/m2K On the wall itself, h wis only 54 W/m2K.Calculatehefffor the wall with its fins (heff= Qwalldivided by

Awalland [Twall− T∞].)

4.34 Evaluate d(tanh x)/dx.

4.35 An engineer seeks to study the effect of temperature on the

curing of concrete by controlling the temperature of curing in

the following way A sample slab of thickness L is subjected

to a heat flux, q w, on one side, and it is cooled to temperature

T1 on the other Derive a dimensionless expression for thesteady temperature in the slab Plot the expression and offer

a criterion for neglecting the internal heat generation in theslab

4.36 Develop the dimensionless temperature distribution in a

spher-ical shell with the inside wall kept at one temperature and theoutside wall at a second temperature Reduce your solution

to the limiting cases in which routside rinside and in which

routsideis very close to rinside Discuss these limits

4.37 Does the temperature distribution during steady heat transfer

in an object with b.c.’s of only the first kind depend on k?

Explain

4.38 A long, 0.005 m diameter duralumin rod is wrapped with an

electrical resistor over 3 cm of its length The resistor imparts

Problems 189

a surface flux of 40 kW/m2 Evaluate the temperature of the

rod in either side of the heated section if h = 150 W/m2K

around the unheated rod, and Tambient= 27 ◦C.

4.39 The heat transfer coefficient between a cool surface and a

satu-rated vapor, when the vapor condenses in a film on the surface,

depends on the liquid density and specific heat, the

tempera-ture difference, the buoyant force per unit volume (g[ρ f −ρg ]),

the latent heat, the liquid conductivity and the kinematic

vis-cosity, and the position (x) on the cooler Develop the

dimen-sionless functional equation forh.

4.40 A duralumin pipe through a cold room has a 4 cm I.D and a

5 cm O.D It carries water that sometimes sits stationary It

is proposed to put electric heating rings around the pipe to

protect it against freezing during cold periods of −7 ◦C The

heat transfer coefficient outside the pipe is 9 W/m2K (including

both convection and radiation) Neglect the presence of the

water in the conduction calculation, and determine how far

apart the heaters would have to be if they brought the pipe

temperature to 40◦C locally How much heat do they require?

4.41 The specific entropy of an ideal gas depends on its specific

heat at constant pressure, its temperature and pressure, the

ideal gas constant and reference values of the temperature and

pressure Obtain the dimensionless functional equation for

the specific entropy and compare it with the known equation

4.42 A large freezer’s door has a 2.5 cm thick layer of insulation

(kin= 0.04 W/m2K) covered on the inside, outside, and edges

with a continuous aluminum skin 3.2 mm thick (kAl = 165

W/m2K) The door closes against a nonconducting seal 1 cm

wide Heat gain through the door can result from conduction

straight through the insulation and skins (normal to the plane

of the door) and from conduction in the aluminum skin only,

going from the skin outside, around the edge skin, and to the

inside skin The heat transfer coefficients to the inside, h i,

and outside,h o, are each 12 W/m2K, accounting for both

con-vection and radiation The temperature outside the freezer is

25◦C, and the temperature inside is−15 ◦C.

a If the door is 1 m wide, estimate the one-dimensional heat

gain through the door, neglecting any conduction around

190 Chapter 4: Analysis of heat conduction and some steady one-dimensional problems

the edges of the skin Your answer will be in watts permeter of door height

b Now estimate the heat gain by conduction around the

edges of the door, assuming that the insulation is fectly adiabatic so that all heat flows through the skin.This answer will also be per meter of door height

per-4.43 A thermocouple epoxied onto a high conductivity surface is

in-tended to measure the surface temperature The ple consists of two each bare, 0.51 mm diameter wires One

thermocou-wire is made of Chromel (Ni-10% Cr with kcr= 17 W/m·K) and

the other of constantan (Ni-45% Cu with kcn= 23 W/m·K) The

ends of the wires are welded together to create a measuring

junction having has dimensions of D w by 2D w The wires tend perpendicularly away from the surface and do not touch

ex-one another A layer of epoxy (kep = 0.5 W/m·K separates

the thermocouple junction from the surface by 0.2 mm Air

at 20◦C surrounds the wires The heat transfer coefficient tween each wire and the surroundings is h = 28 W/m2K, in-cluding both convection and radiation If the thermocouple

be-reads Ttc = 40 ◦ C, estimate the actual temperature T s of the

surface and suggest a better arrangement of the wires

4.44 The resistor leads in Example 4.10 were assumed to be

“in-finitely long” fins What is the minimum length they each musthave if they are to be modelled this way? What are the effec-

tiveness, εf, and efficiency, ηf, of the wires?

References

[4.1] V L Streeter and E.B Wylie Fluid Mechanics McGraw-Hill Book

Company, New York, 7th edition, 1979 Chapter 4

[4.2] E Buckingham Phy Rev., 4:345, 1914.

[4.3] E Buckingham Model experiments and the forms of empirical

equa-tions Trans ASME, 37:263–296, 1915.

[4.4] Lord Rayleigh, John Wm Strutt The principle of similitude Nature,

95:66–68, 1915

References 191

[4.5] J O Farlow, C V Thompson, and D E Rosner Plates of the dinosaur

stegosaurus: Forced convection heat loss fins? Science, 192(4244):

1123–1125 and cover, 1976

[4.6] D K Hennecke and E M Sparrow Local heat sink on a convectively

cooled surface—application to temperature measurement error Int.

J Heat Mass Transfer, 13:287–304, 1970.

[4.7] P J Schneider Conduction Heat Transfer Addison-Wesley

Publish-ing Co., Inc., ReadPublish-ing, Mass., 1955

[4.8] A D Kraus, A Aziz, and J.R Welty Extended Surface Heat Transfer.

John Wiley & Sons, Inc., New York, 2001

5 Transient and multidimensional

heat conduction

When I was a lad, winter was really cold It would get so cold that if you went outside with a cup of hot coffee it would freeze I mean it would freeze fast That cup of hot coffee would freeze so fast that it would still be hot after it froze Now that’s cold! Old North-woods tall-tale

James Watt, of course, did not invent the steam engine What he did do

was to eliminate a destructive transient heating and cooling process that

wasted a great amount of energy By 1763, the great puffing engines of

Savery and Newcomen had been used for over half a century to pump the

water out of Cornish mines and to do other tasks In that year the young

instrument maker, Watt, was called upon to renovate the Newcomen

en-gine model at the University of Glasgow The Glasgow enen-gine was then

being used as a demonstration in the course on natural philosophy Watt

did much more than just renovate the machine—he first recognized, and

eventually eliminated, its major shortcoming

The cylinder of Newcomen’s engine was cold when steam entered it

and nudged the piston outward A great deal of steam was wastefully

condensed on the cylinder walls until they were warm enough to

accom-modate it When the cylinder was filled, the steam valve was closed and

jets of water were activated inside the cylinder to cool it again and

con-dense the steam This created a powerful vacuum, which sucked the

piston back in on its working stroke First, Watt tried to eliminate the

wasteful initial condensation of steam by insulating the cylinder But

that simply reduced the vacuum and cut the power of the working stroke

193

194 Transient and multidimensional heat conduction §5.2

Then he realized that, if he led the steam outside to a separate condenser,

the cylinder could stay hot while the vacuum was created

The separate condenser was the main issue in Watt’s first patent(1769), and it immediately doubled the thermal efficiency of steam en-gines from a maximum of 1.1% to 2.2% By the time Watt died in 1819, hisinvention had led to efficiencies of 5.7%, and his engine had altered theface of the world by powering the Industrial Revolution And from 1769until today, the steam power cycles that engineers study in their ther-modynamics courses are accurately represented as steady flow—ratherthan transient—processes

The repeated transient heating and cooling that occurred in comen’s engine was the kind of process that today’s design engineermight still carelessly ignore, but the lesson that we learn from history

New-is that transient heat transfer can be of overwhelming importance day, for example, designers of food storage enclosures know that suchsystems need relatively little energy to keep food cold at steady condi-tions The real cost of operating them results from the consumption

To-of energy needed to bring the food down to a low temperature and thelosses resulting from people entering and leaving the system with food

The transient heat transfer processes are a dominant concern in the

de-sign of food storage units

We therefore turn our attention, first, to an analysis of unsteady heattransfer, beginning with a more detailed consideration of the lumped-capacity system that we looked at in Section1.3

We begin by looking briefly at the dimensional analysis of transient duction in general and of lumped-capacity systems in particular

con-Dimensional analysis of transient heat conduction

We first consider a fairly representative problem of one-dimensional sient heat conduction:

x =L = h (T − T1) x =L

§5.2 Lumped-capacity solutions 195

The solution of this problem must take the form of the following

dimen-sional functional equation:

T − T1= fn(T i − T1), x, L, t, α, h, kThere are eight variables in four dimensions (K, s, m, W), so we look for

8−4 = 4 pi-groups We anticipate, from Section4.3, that they will include

Fo



(5.3)

If the problem involved only b.c.’s of the first kind, the heat transfer

coefficient,h—and hence the Biot number—would go out of the problem.

Then the dimensionless function eqn (5.1) is

By the same token, if the b.c.’s had introduced different values of h at

x = 0 and x = L, two Biot numbers would appear in the solution.

The lumped-capacity problem is particularly interesting from the

stand-point of dimensional analysis [see eqns (1.19)–(1.22)] In this case,

nei-ther k nor x enters the problem because we do not retain any features

of the internal conduction problem Therefore, we have ρc rather than

α Furthermore, we do not have to separate ρ and c because they only

appear as a product Finally, we use the volume-to-external-area ratio,

V /A, as a characteristic length Thus, for the transient lumped-capacity

problem, the dimensional equation is

T − T∞ = fn(T i − T∞ ) , ρc, V /A, h, t

(5.5)

196 Transient and multidimensional heat conduction §5.2

This is exactly the form of the simple lumped-capacity solution, eqn (1.22)

Notice, too, that the group t/ T can be viewed as

t

hk(V /A)t ρc(V /A)2k = h(V /A) k · (V /A) αt 2 = Bi Fo (5.7)

Electrical and mechanical analogies to the lumped-thermal-capacity problem

The term capacitance is adapted from electrical circuit theory to the heat

transfer problem Therefore, we sketch a simple resistance-capacitancecircuit in Fig.5.1 The capacitor is initially charged to a voltage, E o Whenthe switch is suddenly opened, the capacitor discharges through the re-sistor and the voltage drops according to the relation

Normally, in a heat conduction problem the thermal capacitance,

ρcV , is distributed in space But when the Biot number is small, T (t)

§5.2 Lumped-capacity solutions 197

is uniform in the body and we can lump the capacitance into a single

circuit element The thermal resistance is 1/hA, and the temperature

difference (T − T∞ ) is analogous to E(t) Thus, the thermal response,

analogous to eqn (5.9), is [see eqn (1.22)]

T − T∞ = (Ti − T∞ ) exp



− hAt ρcV



Notice that the electrical time constant, analogous to ρcV /hA, is RC.

Now consider a slightly more complex system Figure 5.2 shows a

spring-mass-damper system The well-known response equation

(actu-ally, a force balance) for this system is

system with a forcing function

did not include it in the electrical circuit Mass has the effect of carrying

the system beyond its final equilibrium point Thus, in an underdamped

mechanical system, we might obtain the sort of response shown in Fig.5.3

if we specified the velocity at x = 0 and provided no forcing function.

Electrical inductance provides a similar effect But the Second Law of

Thermodynamics does not permit temperatures to overshoot their

equi-librium values spontaneously There are no physical elements analogous

to mass or inductance in thermal systems.

198 Transient and multidimensional heat conduction §5.2

Figure 5.3 Response of an unforced

spring-mass-damper system with an

initial velocity

Next, consider another mechanical element that does have a

ther-mal analogy—namely, the forcing function, F We consider a (massless) spring-damper system with a forcing function F that probably is time-

dependent, and we ask: “What might a thermal forcing function looklike?”

Lumped-capacity solution with a variable ambient temperature

To answer the preceding question, let us suddenly immerse an object at

a temperature T = Ti, with Bi 1, into a cool bath whose temperature is

rising as T ∞ (t) = Ti + bt, where Ti and b are constants Then eqn (1.20)becomes

To solve eqn (5.12) we must first recall that the general solution of

a linear ordinary differential equation with constant coefficients is equal

to the sum of any particular integral of the complete equation and thegeneral solution of the homogeneous equation We know the latter; it

is T − Ti = (constant) exp(−t/T ) A particular integral of the complete

equation can often be formed by guessing solutions and trying them inthe complete equation Here we discover that

T − Ti = bt − bT

§5.2 Lumped-capacity solutions 199

satisfies eqn (5.12) Thus, the general solution of eqn (5.12) is

T − Ti = C1e −t/T + b(t − T ) (5.13)

The solution for arbitrary variations of T ∞ (t) is given in Problem5.52

(see also Problems5.3,5.53, and5.54)

Example 5.1

The flow rates of hot and cold water are regulated into a mixing

cham-ber We measure the temperature of the water as it leaves, using a

thermometer with a time constant, T On a particular day, the

sys-tem started with cold water at T = Ti in the mixing chamber Then

hot water is added in such a way that the outflow temperature rises

linearly, as shown in Fig.5.4, with Texit flow = Ti + bt How will the

thermometer report the temperature variation?

Solution. The initial condition in eqn (5.13), which describes this

process, is T − Ti = 0 at t = 0 Substituting eqn (5.13) in the i.c., we

get

0= C1− bT so C1= bT

and the response equation is

T − (Ti + bt) = bTe −t/T − 1 (5.14)This result is graphically shown in Fig.5.4 Notice that the ther-

mometer reading reflects a transient portion, b T e −t/T, which decays

for a few time constants and then can be neglected, and a steady

portion, T i + b(t − T ), which persists thereafter When the steady

re-sponse is established, the thermometer follows the bath with a

tem-perature lag of b T This constant error is reduced when either T or

the rate of temperature increase, b, is reduced.

Second-order lumped-capacity systems

Now we look at situations in which two lumped-thermal-capacity systems

are connected in series Such an arrangement is shown in Fig.5.5 Heat is

transferred through two slabs with an interfacial resistance, h −1 c between

them We shall require that h c L1/k1, hcL2/k2, and hL2/k2 are all much

200 Transient and multidimensional heat conduction §5.2

Figure 5.4 Response of a thermometer to a linearly increasing

ambient temperature

less than unity so that it will be legitimate to lump the thermal tance of each slab The differential equations dictating the temperatureresponse of each slab are then

1Notice that we could also have used (ρcV )2/hc A for T2 since both h c and h act on

slab 2 The choice is arbitrary.

Thus we have reduced the pair of first-order equations, eqn (5.15) and

eqn (5.16), to a single second-order equation, eqn (5.19b)

The general solution of eqn (5.19b) is obtained by guessing a solution

of the form θ = C1e Dt Substitution of this guess into eqn (5.19b) gives

from which we find that D = −(b/2) ±3(b/2)2− c This gives us two

values of D, from which we can get two exponential solutions By adding

202 Transient and multidimensional heat conduction §5.2

them together, we form a general solution:

This is a pretty complicated result—all the more complicated when

we remember that b involves three algebraic terms [recall eqn (5.19a)].Yet there is nothing very sophisticated about it; it is easy to understand

A system involving three capacitances in series would similarly yield athird-order equation of correspondingly higher complexity, and so forth

A system involving three capacitances in series would similarly yield athird-order equation of correspondingly higher complexity, and so forth

are connected in series Such an arrangement is shown in Fig.5.5 Heat is

transferred through two slabs with an interfacial resistance, h −1 c... class="text_page_counter">Trang 17

200 Transient and multidimensional heat conduction §5.2

Figure 5.4 Response of a thermometer to a linearly