A HEAT TRANSFER TEXTBOOK - THIRD EDITION Episode 2 Part 1 doc

We deduce the temperature for the point of interest, A, by a simple proportionality: T point A = 2.1 5.6 40 − 0 = 15 ◦C The shape factor A heat conduction shape factor S may be deﬁned fo

Trang 1

§5.7 Steady multidimensional heat conduction 239

• If you have doubts about whether any large, ill-shaped regions are

correct, ﬁll them in with an extra isotherm and adiabatic line to

be sure that they resolve into appropriate squares (see the dashed

lines in Fig.5.21)

• Fill in the ﬁnal grid, when you are sure of it, either in hard pencil or

pen, and erase any lingering background sketch lines

• Your ﬂow channels need not come out even Notice that there is an

extra 1/7 of a channel in Fig.5.21 This is simply counted as 1/7 of

a square in eqn (5.65)

• Never allow isotherms or adiabatic lines to intersect themselves.

When the sketch is complete, we can return to eqn (5.65) to compute

the heat ﬂux In this case

Q = N

I k ∆T = 2(6.14)

4 k ∆T = 3.07 k∆T

When the authors of [5.3] did this problem, they obtained N/I = 3.00—a

value only 2% below ours This kind of agreement is typical when ﬂux

plotting is done with care

Figure 5.22 A ﬂux plot with no axis of symmetry to guide

construction

Trang 2

240 Transient and multidimensional heat conduction §5.7

One must be careful not to grasp at a false axis of symmetry Figure5.22shows a shape similar to the one that we just treated, but with un-

equal legs In this case, no lines must enter (or leave) the corners A and

B The reason is that since there is no symmetry, we have no guidance

as to the direction of the lines at these corners In particular, we know

that a line leaving A will no longer arrive at B.

Example 5.8

A structure consists of metal walls, 8 cm apart, with insulating

ma-terial (k = 0.12 W/m·K) between Ribs 4 cm long protrude from one

wall every 14 cm They can be assumed to stay at the temperature ofthat wall Find the heat ﬂux through the wall if the ﬁrst wall is at 40◦Cand the one with ribs is at 0◦C Find the temperature in the middle ofthe wall, 2 cm from a rib, as well

Figure 5.23 Heat transfer through a wall with isothermal ribs.

Trang 3

Solution. The ﬂux plot for this conﬁguration is shown in Fig.5.23

For a typical section, there are approximately 5.6 isothermal

incre-ments and 6.15 heat ﬂow channels, so

Q = N

I k ∆T = 2(6.15)

5.6 (0.12)(40 − 0) = 10.54 W/m

where the factor of 2 accounts for the fact that there are two halves

in the section We deduce the temperature for the point of interest,

A, by a simple proportionality:

T point A = 2.1

5.6 (40 − 0) = 15 ◦C

The shape factor

A heat conduction shape factor S may be deﬁned for steady problems

involving two isothermal surfaces as follows:

Thus far, every steady heat conduction problem we have done has taken

this form For these situations, the heat ﬂow always equals a function of

the geometric shape of the body multiplied by k ∆T

The shape factor can be obtained analytically, numerically, or through

ﬂux plotting For example, let us compare eqn (5.65) and eqn (5.66):

This shows S to be dimensionless in a two-dimensional problem, but in

three dimensions S has units of meters:

For a three-dimensional body, eqn (5.69) is unchanged except that the

dimensions of Q and R t diﬀer.8

8 Recall that we noted after eqn ( 2.22) that the dimensions of R tchanged, depending

on whether or not Q was expressed in a unit-length basis.

Trang 4

Figure 5.24 The shape factor for two similar bodies of

diﬀer-ent size

The virtue of the shape factor is that it summarizes a heat conduction

solution in a given conﬁguration Once S is known, it can be used again and again That S is nondimensional in two-dimensional conﬁgurations means that Q is independent of the size of the body Thus, in Fig.5.21, S

is always 3.07—regardless of the size of the ﬁgure—and in Example5.8, S

is 2(6.15)/5.6 = 2.196, whether or not the wall is made larger or smaller When a body’s breadth is increased so as to increase Q, its thickness in the direction of heat ﬂow is also increased so as to decrease Q by the

same factor

Example 5.9

Calculate the shape factor for a one-quarter section of a thick cylinder

Solution. We already know R t for a thick cylinder It is given byeqn (2.22) From it we compute

Scyl= 1kRt = 2π

ln(r o/ri)

so on the case of a quarter-cylinder,

2 ln(r o/r i )

The quarter-cylinder is pictured in Fig.5.24for a radius ratio, r o /r i =

3, but for two diﬀerent sizes In both cases S = 1.43 (Note that the same S is also given by the ﬂux plot shown.)

Trang 5

Figure 5.25 Heat transfer through a

thick, hollow sphere

Example 5.10

Calculate S for a thick hollow sphere, as shown in Fig.5.25

Solution. The general solution of the heat diﬀusion equation in

spherical coordinates for purely radial heat ﬂow is:

Trang 6

Table5.4includes a number of analytically derived shape factors foruse in calculating the heat flux in different configurations Notice thatthese results will not give local temperatures To obtain that information,one must solve the Laplace equation,∇2T = 0, by one of the methods

listed at the beginning of this section Notice, too, that this table is stricted to bodies with isothermal and insulated boundaries

re-In the two-dimensional cases, both a hot and a cold surface must bepresent in order to have a steady-state solution; if only a single hot (orcold) body is present, steady state is never reached For example, a hotisothermal cylinder in a cooler, inﬁnite medium never reaches steadystate with that medium Likewise, in situations 5, 6, and 7 in the table,the medium far from the isothermal plane must also be at temperature

T2in order for steady state to occur; otherwise the isothermal plane andthe medium below it would behave as an unsteady, semi-inﬁnite body Ofcourse, since no real medium is truly inﬁnite, what this means in practice

is that steady state only occurs after the medium “at inﬁnity” comes to

a temperature T2 Conversely, in three-dimensional situations (such as

4, 8, 12, and 13), a body can come to steady state with a surrounding

infinite or semi-infinite medium at a different temperature

Example 5.11

A spherical heat source of 6 cm in diameter is buried 30 cm below thesurface of a very large box of soil and kept at 35◦C The surface ofthe soil is kept at 21◦C If the steady heat transfer rate is 14 W, what

is the thermal conductivity of this sample of soil?

k = 14 W(35 − 21)K

1− (0.06/2) 2(0.3) 4π (0.06/2) m = 2.545 W/m·K

Readers who desire a broader catalogue of shape factors should refer

to [5.16], [5.18], or [5.19]

Trang 7

Table 5.4 Conduction shape factors: Q = S k∆T

2 Conduction through wall of a long

4 The boundary of a spherical hole of

radius R conducting into an inﬁnite

medium

and2.15

5 Cylinder of radius R and length L,

transferring heat to a parallel

7 An isothermal sphere of radius R

transfers heat to an isothermal

plane; R/h < 0.8 (see item 4)

4π R

245

Trang 8

Table 5.4 Conduction shape factors: Q = S k∆T (con’t).

8 An isothermal sphere of radius R,

near an insulated plane, transfers

heat to a semi-inﬁnite medium at

9 Parallel cylinders exchange heat in

an inﬁnite conducting medium

2R1R2

10 Same as 9, but with cylinders

widely spaced; L R1and R2

2π

cosh−1

L 2R1

+ cosh −1

L 2R2

11 Cylinder of radius R i surrounded

by eccentric cylinder of radius

12 Isothermal disc of radius R on an

otherwise insulated plane conducts

heat into a semi-inﬁnite medium at

T ∞below it

13 Isothermal ellipsoid of semimajor

axis b and semiminor axes a

conducts heat into an inﬁnite

medium at T ∞ ; b > a (see 4)

4π b

4

1− a2 b2tanh−14

1− a2 b2 meter [5.16]

246

Trang 9

§5.8 Transient multidimensional heat conduction 247

Figure 5.26 Resistance vanishes where

two isothermal boundaries intersect

The problem of locally vanishing resistance

Suppose that two diﬀerent temperatures are speciﬁed on adjacent sides

of a square, as shown in Fig.5.26 The shape factor in this case is

S = N

I = ∞

4 = ∞ (It is futile to try and count channels beyond N 10, but it is clear that

they multiply without limit in the lower left corner.) The problem is that

we have violated our rule that isotherms cannot intersect and have

cre-ated a 1/r singularity If we actually tried to sustain such a situation,

the ﬁgure would be correct at some distance from the corner However,

where the isotherms are close to one another, they will necessarily

inﬂu-ence and distort one another in such a way as to avoid intersecting And

S will never really be inﬁnite, as it appears to be in the ﬁgure.

The tactic of superposition

Consider the cooling of a stubby cylinder, such as the one shown in

Fig 5.27a The cylinder is initially at T = Ti, and it is suddenly

sub-jected to a common b.c on all sides It has a length 2L and a radius r o

Finding the temperature ﬁeld in this situation is inherently complicated

Trang 10

It requires solving the heat conduction equation for T = fn(r , z, t) with

b.c.’s of the ﬁrst, second, or third kind

However, Fig 5.27a suggests that this can somehow be viewed as acombination of an inﬁnite cylinder and an inﬁnite slab It turns out that

the problem can be analyzed from that point of view.

If the body is subject to uniform b.c.’s of the first, second, or thirdkind, and if it has a uniform initial temperature, then its temperatureresponse is simply the product of an infinite slab solution and an infinitecylinder solution each having the same boundary and initial conditions.For the case shown in Fig.5.27a, if the cylinder begins convective cool-

ing into a medium at temperature T ∞ at time t = 0, the dimensional

temperature response is

T (r , z, t) − T∞ =Tslab(z, t) − T∞×Tcyl(r , t) − T∞ (5.70a)

Observe that the slab has as a characteristic length L, its half thickness, while the cylinder has as its characteristic length R, its radius In dimen-

sionless form, we may write eqn (5.70a) as

Θ ≡ T (r , z, t) − T∞

T i − T∞ =

Θinf slab(ξ, Fos , Bis )

Θinf cyl(ρ, Foc , Bic)

(5.70b)For the cylindrical component of the solution,

Sec-Figure5.27b shows a point inside a one-eighth-infinite region, near thecorner This case may be regarded as the product of three semi-infinitebodies To find the temperature at this point we write

Θ ≡ T (x1, x2, x3, t) − T∞

Ti − T∞ = [Θsemi(ζ1, β)] [Θsemi(ζ2, β)] [Θsemi(ζ3, β)]

(5.71)

Trang 11

Figure 5.27 Various solid bodies whose transient cooling can

be treated as the product of one-dimensional solutions

249

Trang 12

in whichΘsemiis either the semi-inﬁnite body solution given by eqn (5.53)when convection is present at the boundary or the solution given byeqn (5.50) when the boundary temperature itself is changed at time zero.Several other geometries can also be represented by product solu-tions Note that for of these solutions, the value ofΘ at t = 0 is one for

each factor in the product

Example 5.12

A very long 4 cm square iron rod at T i = 100 ◦C is suddenly immersed

in a coolant at T ∞ = 20 ◦ C with h = 800 W/m2K What is the ature on a line 1 cm from one side and 2 cm from the adjoining side,after 10 s?

temper-Solution. With reference to Fig 5.27c, see that the bar may betreated as the product of two slabs, each 4 cm thick We ﬁrst evaluateFo1 = Fo2 = αt/L2 = (0.0000226 m2/s)(10 s) (0.04 m/2)2 = 0.565,

and Bi1 = Bi2 = hL k = 800(0.04/2)/76 = 0.2105, and we then

writeΘ

Trang 13

Transient multidimensional heat conduction 251

Product solutions can also be used to determine the mean

tempera-ture, Θ, and the total heat removal, Φ, from a multidimensional object

For example, when two or three solutions (Θ1, Θ2, and perhapsΘ3) are

multiplied to obtainΘ, the corresponding mean temperature of the

mul-tidimensional object is simply the product of the one-dimensional mean

temperatures from eqn (5.40)

Θ = Θ1 (Fo1, Bi1) × Θ2(Fo2, Bi2) for two factors (5.72a)

Θ = Θ1 (Fo1, Bi1) × Θ2(Fo2, Bi2) × Θ3(Fo3, Bi3) for three factors.

(5.72b)

Since Φ = 1 − Θ, a simple calculation shows that Φ can found from Φ1,

Φ2, andΦ3as follows:

Φ = Φ1 + Φ2(1 − Φ1) for two factors (5.73a)

Φ = Φ1 + Φ2(1 − Φ1) + Φ3(1 − Φ2) (1 − Φ1) for three factors (5.73b)

Example 5.13

For the bar described in Example5.12, what is the mean temperature

after 10 s and how much heat has been lost at that time?

Solution. For the Biot and Fourier numbers given in Example5.12,

Trang 14

252 Chapter 5: Transient and multidimensional heat conduction

Problems

5.1 Rework Example5.1, and replot the solution, with one change

This time, insert the thermometer at zero time, at an initial

temperature < (T i − bT ).

5.2 A body of known volume and surface area and temperature T i

is suddenly immersed in a bath whose temperature is rising

as Tbath = Ti + (T0− Ti)e t/τ Let us suppose that h is known, that τ = 10ρcV/hA, and that t is measured from the time of

immersion The Biot number of the body is small Find thetemperature response of the body Plot the response and the

bath temperature as a function of time up to t = 2τ (Do not

use Laplace transform methods except, perhaps, as a check.)

5.3 A body of known volume and surface area is immersed in

a bath whose temperature is varying sinusoidally with a

fre-quency ω about an average value The heat transfer coeﬃcient

is known and the Biot number is small Find the temperaturevariation of the body after a long time has passed, and plot italong with the bath temperature Comment on any interestingaspects of the solution

A suggested program for solving this problem:

• Write the diﬀerential equation of response.

• To get the particular integral of the complete equation, guess that T − Tmean = C1cos ωt + C2sin ωt Substitute this in the diﬀerential equation and ﬁnd C1 and C2valuesthat will make the resulting equation valid

• Write the general solution of the complete equation It

will have one unknown constant in it

• Write any initial condition you wish—the simplest one you

can think of—and use it to get rid of the constant

• Let the time be large and note which terms vanish from

the solution Throw them away

• Combine two trigonometric terms in the solution into a term involving sin(ωt − β), where β = fn(ωT ) is the

phase lag of the body temperature

5.4 A block of copper ﬂoats within a large region of well-stirred

mercury The system is initially at a uniform temperature, T i

Trang 15

Problems 253

There is a heat transfer coeﬃcient, h m, on the inside of the thin

metal container of the mercury and another one, hc, between

the copper block and the mercury The container is then

sud-denly subjected to a change in ambient temperature from T ito

Ts < Ti Predict the temperature response of the copper block,

neglecting the internal resistance of both the copper and the

mercury Check your result by seeing that it ﬁts both initial

conditions and that it gives the expected behavior at t → ∞.

5.5 Sketch the electrical circuit that is analogous to the

second-order lumped capacity system treated in the context of Fig.5.5

and explain it fully

5.6 A one-inch diameter copper sphere with a thermocouple in

its center is mounted as shown in Fig.5.28 and immersed in

water that is saturated at 211◦F The ﬁgure shows the

ther-mocouple reading as a function of time during the

quench-ing process If the Biot number is small, the center

temper-ature can be interpreted as the uniform tempertemper-ature of the

sphere during the quench First draw tangents to the curve,

and graphically diﬀerentiate it Then use the resulting values

of dT /dt to construct a graph of the heat transfer coeﬃcient

as a function of (Tsphere − Tsat) The result will give actual

values of h during boiling over the range of temperature

dif-ferences Check to see whether or not the largest value of the

Biot number is too great to permit the use of lumped-capacity

methods

5.7 A butt-welded 36-gage thermocouple is placed in a gas ﬂow

whose temperature rises at the rate 20◦C/s The

thermocou-ple steadily records a temperature 2.4◦C below the known gas

ﬂow temperature If ρc is 3800 kJ/m3K for the thermocouple

material, what ish on the thermocouple? [h = 1006 W/m2K.]

5.8 Check the point on Fig.5.7 at Fo = 0.2, Bi = 10, and x/L = 0

analytically

5.9 Prove that when Bi is large, eqn (5.34) reduces to eqn (5.33)

5.10 Check the point at Bi= 0.1 and Fo = 2.5 on the slab curve in

Fig.5.10analytically

Trang 14

25 2 Chapter 5: Transient and multidimensional heat conduction

Problems... Bi1) × Θ2(Fo2, Bi2) for two factors (5.7 2a)

Θ = ? ?1 (Fo1, Bi1)...

Trang 13

Transient multidimensional heat conduction 25 1

Product solutions can also be

Định dạng
Số trang	25
Dung lượng	246,66 KB