Page 1 of 35
`
`BOREALIS EXHIBIT 1060
`
`Page 1 of 35
`
`BOREALIS EXHIBIT 1060
`
`

`
`Page 2 of 35
`
`

`
`Principles of
`HEAT TRANSFER
`third edition
`
`Frank Kreith
`••
`University of Colorado
`
`IN TEXT EDUCATIONAL PUBLISHERS
`New York and London
`
`Page 3 of 35
`
`

`
`Copyright© 1958, 1965, by International Textbook Company
`
`Copyright© 1973, by Intext Press, Inc.
`
`All rights reserved. No part of this book may be reprinted, reproduced,
`or utilized in any form or by any electronic, mechanical, or other means,
`now known or hereafter invented, including photocopying and recording,
`or in any information storage and retrieval system, without permission in
`writing from the Publisher.
`
`Library of Congress Cataloging in Publication Data
`
`Kreith, Frank.
`Principles of heat transfer.
`
`(Series in mechanical engineering)
`Includes bibliographical references.
`1. Heat- Transmission. I. Title.
`QC320.K7 1973
`536'.2
`73-1784
`ISBN 0-7002-2422-X
`
`Intext Educational Publishers
`257 Park Avenue South
`New York, New York 10010
`
`C1P
`
`7 . ~.~fly 19i73
`' . ' ( -\
`
`•.
`
`Page 4 of 35
`
`

`
`Direction of Heat Flow
`
`Direction of Heat Flow
`
`T(x)
`
`_,-+ c...;x_T ____ 1• 4T
`
`+tlx
`
`+T l
`
`- - --+x
`
`+T l
`
`- ----+X
`
`Fig. 1·1. Sketch illustrating sign convention for conduction heot flow.
`
`ex pressed in Btufhr, the arCH A in sq rt, and the Lempcralurc gradienl
`dTj dx in F / ft. The th erm al conduclivity k i a property or the ma terial
`and indi atcs the qua nti ty o f hea L that will flow across a unit area if the
`temperature gradient is unity. The uni Ls for k used in this tex t a re British
`therm a l units per bo ur per square fool per unit temp~ raturc gradient in
`degrees Fahrenheit per foot, i.e.,
`
`Btu/hr sq ft
`F /ft
`
`or
`
`Btu - - -
`hr ft F
`
`In the SI system (Systeme International d'Unites), the units of
`thermal conductivity are watts per square meter per unit temperature
`gradient in degree centigrade (or Kelvin) per meter, i.e.,
`
`and
`
`watt j m 2
`---'--=
`K/m
`
`watt
`m K
`
`1 wjm K = 0.578 Btujhr ft F
`
`Thermal conductivities of engineering materials at atmospheric pressure
`range from about 4 X 10 - 3 for gases through about 1 X
`lQ - 1 for liquids
`to 2.4 x 10 2 Btujhr ft F for copper. Orders of magnitudes of the thermal
`conductivity of various classes of material are shown in Table 1-1 and Fig.
`1-2. Materials having a high thermal conductivity are called conductors,
`while materials of low thermal conductivity are referred to as insulators.
`
`1 Throughout the remainder of this book units of physica l qu antities will be given .n
`conventionally abbrevia ted form without expanding them .
`
`8
`
`INTRODUCTION
`
`Page 5 of 35
`
`

`
`Tobie 1-2 Order of magnitude of convective heat transfer coefficients he.
`
`Btu/hr sq ft F
`
`1-
`
`5
`
`6-
`
`30
`
`5-
`50
`300
`10-
`50- 2,000
`500- 10,000
`1 ,000- 20,000
`
`300
`30-
`60-
`1,800
`6,000
`300 -
`3,000 - 60,000
`6,000 - 120.000
`
`Air, frt!c convection
`Superheated steam or air,
`forced convection
`Oil, forced convccti n
`Wutcr, forced convection
`Water, boiling
`Steam, condensing
`
`vective heat transfer as
`
`and the thermal resistance to convective heat transfer R"' which is equal to
`the reciprocal of the conductance, as
`
`( 1· 16)
`
`( 1-17)
`
`1-4. Combined heat-transfer mechanisms
`
`ln the preceding section the three mechanisms of heat transfer have
`been con. idered separately. In practice, however, heal is usually tran(cid:173)
`ferred in steps through a number of different serie -connected section .
`the tran fer frequently occurring by two mechanisms in parallel for a given
`section in Lhe system. The transfer of heat from the products of corn(cid:173)
`bu Lion in the combustion chamber of a rocket motor thr ugh a thin wall
`Loa coolant nowing in an annulus over t.he out ide of the wall will illu.(cid:173)
`trale such a case(Fig. 1-4).
`Products of combustion contain gases. uch as CO C02 • and H 20
`which emit and absorb radiation . In the first ·ection or this system. heal
`f the wall of
`is therefore transferred fr m the hot gas to the inner surface
`the rockel motor by the mechanisms of convection and radiation acting in
`parallel. The Lotal rate of heat flow q to the surface of the wall some dis·
`tance from the nozzle is
`
`q = qc + q,
`= hcA (Tg - Tsg) + h,A (Tg - Tsg)
`
`( 1-18)
`
`14
`
`INTRODUCTION
`
`Page 6 of 35
`
`

`
`,II
`
`2
`Steady one-dimensional
`heat conduction
`
`2-1. Walls of simple geometrical configuration
`
`In this section we shall consider steady-state heat conduction through
`simple systems in which the temperature and the heat flow are functions
`of a single coordinate.
`
`Plane wall. The simplest case of one-dimensional heat flow, namely
`heat conduction through a plane wall, was treated in Sec. 1-3. We found
`that, for uniform temperatures over the hot and the cold surfaces, the
`rate of heat flow by conduction through a homogeneous material is given
`by
`
`Ak
`qk = L (Tho! - Tcold)
`
`( 2- 1)
`
`EXAMPLE 2-1. T he interior surfaces of the wulls in a large building arc to be
`maintained a1 70 F while the outer surface temperature is - 10 F. The walls are
`10 in. thick and constructed from a brick material having a tbcnna l conductivity
`of0.4 Btujhr ft F. Calculate the hea t loss for each square foot of wall surface per
`hour.
`
`Solwion: 1f we neglect the effect of the corners where the walls meet and
`the effect of mortared brick joints, Eq. l-2 applie . Substituting the thermal con(cid:173)
`ductivity and the pertinent dimensions in their proper units (e.g., L = ~ ft) we
`obtain
`
`29
`
`Page 7 of 35
`
`

`
`( - 10)1 = 38.4 Btufhr sq ft
`!_ ~ (0.4)[70 -
`10/\2
`A
`
`Thus, 38.4 Btu will be lost from the buHding ?er hour through each square foot of
`
`wall surface area.
`
`Ans.
`
`Hollow cylinders. Radial heat \l oW by conduction thr ugh a hollow
`circular cylinder is another one-dimensi nal conduction pr blem of con(cid:173)
`siderable practical im?ortance. Typical examples arc conduction t hrough
`
`pipes and through pipe insulati n.
`If the cylinder i. homogeneous a nd sufficiently tong that end effects
`1
`while
`may be neglected and the inner surface temperature is constant at T
`, the rate o f
`the utcr surface temperature is maintained uniformly at T
`
`0
`
`heat c nduction is. fr m Eq. l-1,
`
`- kA dT
`dr
`where dT / dr ~ temperature gradient in the radial direction.
`For the hollow cylinder( Fig. 2-l), the area is a [unction of the radius
`
`(2-2)
`
`and
`
`(2-3)
`
`where r is the radius and I the length of the cylinder. The rate of heal flow
`
`\
`
`illustrating nomendoture
`Fig. 2-l. Sket(h
`through o hollow (y\inder.
`
`for
`
`conduction
`
`30
`
`STEADY ONE-DIMENSIONAL HEAT CONDUCTION
`
`Page 8 of 35
`
`

`
`by conduction can thus be expressed as
`
`dT
`qk = -k21rr!-
`dr
`
`(2-4)
`
`Separating the variables and integrating between T0 at r0 and T, at r;
`yields
`
`T1· -T=~In'0
`21r kl
`r,
`
`o
`
`Solving Eq. 2-5 for qk yields
`
`T;- To
`In (r0 jr;)
`27rkl
`
`(2-5)
`
`(2-6)
`
`Lhc equntion for calculating the ra le of heat conduction Lhrough a hollow
`circular cylinder such as a pipe. An inspection of Eq. 2-6 shows that the
`mte of radial heat flow varies directly with the cylinder length /, the
`Lhermal conductiv ity k, the lemp~ralure difference belween lhc.:: inner and
`outer. urface T, - T,1 , and inversely a Lhc nMumllogaritllm 1 of the ratio
`r the ouLsidc and inside radii r,.jr1 or the corresponding diameter ratio
`D,) D,. By ana logy to Lhe cuse of a plane wall and Ohm's law. tlH!
`thermal resistance of the hollow cylinder is
`
`{2-7)
`
`The temperature distribution in the curved wall is obtained by inte(cid:173)
`grating Eq. 2-4 from the inner radius r, and the corresponding tempera(cid:173)
`ture T; to an arbitrary radius rand the corresponding temperature T, or
`
`!, qk
`
`,, k (27r/)
`
`dr
`
`r
`
`-
`
`l T(r)
`
`T,
`
`dT
`
`Which gives
`
`T(r)
`
`1The natural logarithm ora number, In, is 2.3026 times the logarithm to base 10.
`
`2-1. WALLS OF SIMPLE GEOMETRICAL CONFIGURATION
`
`31
`
`brr
`
`Page 9 of 35
`
`

`
`'WI."· ~~ ~ -
`
`-
`
`- -
`
`-·
`
`\
`
`fig. 2-2. Temperature distribution in o hollow cylinder.
`
`Thus the temperature in a hollow circular cylinder is a logarithmic func(cid:173)
`tion of the radius r (Fig. 2-2), while for a plane waH the temperature
`
`distribution is linear.
`For some applications it i helpful to huvc the equation for heal con-
`duction Lhrough a curved wall in the arne form a~ Eq. 2- 1 fo r a plane
`wall. To obtain this r 1rm of equali<>n we equale Lhc right-hand sides of
`Eqs. 2- 1 and 2-6 using, however, L ~ (r, -
`r 1), the thickness through
`which heat i conducted, and A === if in Eq. 2-1. This yields
`
`from which if is
`
`AI}- AI
`ln(A 0 /A ,)
`
`(2-8)
`
`The area A defined by Eq. 2-8 is called the logarithmic mean area. The
`rate of heat conduction through a hollow circular cylinder can then be ex-
`
`pressed as
`
`(2-9)
`
`For values of Ao/A
`
`1
`
`< 2 (i.e., rofr1 < 2) the arithmetic mean area.
`
`32
`
`STEADY ONE-DIMENSIONAL HEAT CONDUCTION
`
`Page 10 of 35
`
`

`
`(A"+ A1)/2 is within 4 percent of the logarithmic mean area and may be
`used with satisfuctory accuracy. For thicker walls this approximation is
`generally not acccpt;:lble.
`
`EXAMPLE 2-2. Calculate the heat loss from 10 ft of 3-in. nominal-diameter
`pipe covered with It in. of a n insulaling material having a thermal conductivity
`of 0.040 Btufhr ft F. Assum e thal the inner and outer surface temperatures
`of the insulation are 400 and 80 F, respectively.
`
`Solution: The outside diameter of a nominal 3-in. pipe is 3.50 in. This is
`also the inside diameter of the insulation. The outside diameter of the insulation
`is 6.50 in. The logarithmic mean area is
`
`- A;
`A 0
`In (A 0 /A;)
`
`101!'(6.50 - 3.50)/12
`ln(6.50/3.50)
`
`=
`
`7·85 = 12.70 sq ft
`0.62
`
`/ r; < 2, the arithmetic mean area would be an acceptable approximation
`
`Since r0
`and
`
`10,11'(6.50 + 3.50)
`(2)( 12)
`
`13.10 sq ft
`
`Applying Eq. 2-9, the rate of heat loss is
`
`400 - 80
`qi(. = - - -- - -- -
`(1.5/12)/(0.04)(12.7)
`
`1300 Btu/hr
`
`Ans.
`
`Sphericol and parallelepiped shells. A sphere has lhe largest vol(cid:173)
`ume per ou tside surface area of any geometrical con figuration. For this
`reaso n a hollow pbere is sometimes used in the chem i.cal industry for
`low-lem peralure work when heal lo ses are to be kcpl at a minimum .
`Conduction th rough a spherica l shell i also a one-dimensional .leady(cid:173)
`slate problem if the interior and ex terior surface tempera lures are uniform
`and constant. Tbe rate of heal conductio n for lh is ca ·e (Fig. 2-3) is
`
`(2-1 0)
`
`if the material is homogeneous.
`Equation 2- 10 can also be u ed as an approximnllon for paraJiele(cid:173)
`piped shells which have a small inner cavity surrounded by a thick wall.
`An example o f such a system would be a small furnace urroundcd by a
`large th ick ness or insulating material. For this type of geometry the heat
`
`2-1. WALLS OF SIMPLE GEOMETRICAL CONFIGURATION
`
`33
`
`Page 11 of 35
`
`

`
`illustrating nomenclature
`Fig. 2-3. Sketch
`for conduction through spherical shell.
`
`11ow. especial ly in thc.: corners. is nol perpendicular lU the bounding sur(cid:173)
`face ·, and h~.:nce cunn ot :;triclly be c n:;idcrcd one-dimensio nal. fl o-.: ev~r.
`" hen th e cavity i:; roughly cubic and the surround in g wal ls ure thick
`(A .. / A, ~ 2 ). the ra l< of heal \low ca n he e<timu ted a ceo rd i ng to Sch u·
`mann I) by mulliplyin g th~; geomt:tric mean arcu in Eq. 2-10,
`A 11A ,,
`by the semi-em pi rica I oorr<Cl ion i'actor 0. 725. More accu nile cor reel ion
`ttl. (2) and an; . umnwrized
`rac.:t<HS h<lVe been determ ini.!d b Langmuir d
`
`in Ref. 8.
`EXAMPlE 2-3. Tht.: working. chambc:.r or ;m eb:tricall y· h~ated laburutory
`i'urn:n;c is 6 by~ by I 2 ln . unci th ..: wu\1~. 6 in. thick n all sidt.:s, are made or a
`rel'r<l~Lory br1ck (k = 0.- Blu/ hr n r ). I I" the ten1p~r:\tUre ul the interior surl\\C~
`is to bt.: rn :.linLai ncd a l 2000 F whtlc the outsioe :;urface tcmpcr<lture is 300 F.
`c. ll!11<1lc lht: power con. uJnpuon in kilm' atls (kw).
`
`Solution: Under steady-state conditions the power consumption will equal
`
`the heat loss. The inner surface area A; is
`
`A = 2 (6 X 8) + (6 X 12) + (8 X 12)
`\44
`
`I
`
`3 sq ft
`
`The outer surface area Ao is
`
`(\8 X 20) + (\8 X 24) + (20 X 24)
`144
`
`2
`
`I 7.7 sq ft
`
`Since A,/ A, > 2, we can usc Eq. 2-10 with the em pirica I correction factor O.J25.
`
`34
`
`STEADY ONE-DIMENSIONAL HEAT CONDUCTION
`
`Page 12 of 35
`
`

`
`and the heat loss is
`
`1700)
`qk = (0.2)(0. 725) v'3 x 17.7 - - = 3600 Btujhr
`(
`6/12
`
`Since 1 Btujhr = 2.93 x 10-4 kw, the power consumption is about 1.05 kw. Ans.
`
`It has already been men(cid:173)
`Effect of nonuniform thermal conductivity.
`tioned that the thermal conductivity varies with temperature. The varia(cid:173)
`tion of thermal conductivity with temperature may be neglected when the
`tempera ture ra nge under co nsiderat io n is not too la rge o r the tempera ture
`depe ndence of th e conduc tiv ity is no t too severe. On the o the r ba nd,
`whe n th e lempe ruture difTcren ce in ~ syste m causes substa nti,ll va riaLio n.
`in th e therm a l co nduc tivity, the tem p era ture depe nde nce mus t be taken
`into account.
`For numerous materials, especially within a limited temperature
`range, the variation of the thermal conductivity with the temperature can
`be represented by the linear function
`
`k ( T) = k 0 ( 1 + {3 k T)
`
`(2-11)
`
`where k0 is the thermal conductivity at T = 0 and {Jk is a constant called
`the temperature coefficient of thermal conductivity. When the variation of
`thermal conductivity is available in the form of a curve showing how k
`varies with T, the temperature coefficient can be obtained approximately
`by drawing a straight line between the temperatures of interest and
`measuring its slope. Then k0 is a hypothetical value of the thermal con(cid:173)
`du c tivity equ a l to the ordi nate intercept a t zero temperature. It is de ter(cid:173)
`mined gra phicall y by contin uing the stra ig ht line representing the ::~ctua l
`the rm a l co nduc t ivity over a limited tem pera ture range through the axis
`of conductivity at zero temperature (Fig. 2-4).
`With a linear approximation to the temperature variation of the
`thermal conductivity, the rate of heat flow by conduction through a
`plane wall is, from Eq. 2-1,
`
`Integration gives
`
`which can be written more conveniently as
`
`2- J. WALLS OF SIMPLE GEOMETRICAL CONFIGURATION
`
`35
`
`Page 13 of 35
`
`

`
`k ( T) = o.031 + o.oooo31T
`= o.o31 ( 1 + o.oonl
`
`dk
`?
`~Slope- = 0.000031
`dT
`
`X Experimental points
`
`0
`
`Temperature F
`
`Fig. 2-4. Graphical determination of the temperature coefficient of thermal conductivity.
`
`_ A ( Thol - Tcold) k (l
`
`qk -
`
`L
`
`0
`
`{3 TnoL + Tcold) _
`b. T
`- L/ Akm
`2
`k
`
`+
`
`(2
`
`12
`_
`
`)
`
`111
`
`= ko ll + f3k(.7j,
`01 + Tcolu)/21 represent a mean value
`f the
`where k
`thermal conductivity. For a linear variation of k wi I h T, the I hermal
`conducLivity in Eq. 1-2 should therefore be evaluated at the ariLhmetic
`meRn Lemperalure (T,,.,l + Tco1ct)f2.
`
`EXAMPLE 2-4. The conductivity of an HS percent nu1gncsiu insutating male(cid:173)
`(u) Determine l3k and ko
`rial is shown ns a function of tcmpertt ture in Fig. 2-4.
`for a linear Hpproximi.ltion between tOO and 300 F. (b)£ t:lmate the rate of heat
`now per unit area between these tcmpcrallircs for a slab or 3-in. thickness.
`
`(a) By means of the graphical mclhod illustrated in Fig. 2-4, the
`Solution:
`slope of the straight line connecting the thermal-conductivity curve between 100
`and 300 F is round to be + 0.000031 . The ordinate intercept al 0 degrees (deg) is
`
`0.03\. Thusweh<!YC
`
`k(T) = 0.031 (1 + 0.001 T)
`
`for 100 F < T < 300 F
`
`The mean temperature is 200 F and the mean value of the thermal conductivity is
`
`km = 0.031 (1 + 0.001 x 200) = 0.0372 Btufhr ft F
`
`Ans .
`
`....
`
`(b) The rate of heat flow per unit area is, from Eq. 2-4,
`
`200
`(3 1 12)/(0.0372)
`
`29.8 Btujhr sq ft
`
`Ans.
`
`36
`
`STEADY ONE-DIMENSIONAL HEAT CONDUCTION
`
`Page 14 of 35
`
`

`
`Radi;ll heat conduction through hollow cylinders and spheres made of materials
`whose thcrmul conductivity varies linearly with temperature can be treated in a
`similar manner (see Problem 2-4).
`
`2-2. Composite structures
`
`The general method for analyzing problems of steady-state heat flow
`through composite structures has been presented in Sec. 1-4.
`In this
`section we shall consider some examples of composite structures in which
`In order
`the heat flow is one-dimensional, or at least approximately so.
`to make the treatment applicable to practical cases where the surface
`temperatures are generally not known, heat flow through thermal resis(cid:173)
`tances at the boundaries will be included in the treatment. We shall
`assume that the system is exposed to a high-temperature medium, i.e., a
`heat source, of known and constant temperature on one side and to a low(cid:173)
`temperature medium, i.e., a heat sink, of Known and constant temperature
`on the other side. The surface conductances between the medium and the
`surface will be taken as constant over a given surface.
`
`Composite walls. A composite wall, typical of the type used in a
`large furnace, is shown in Fig. 2-5. The inner layer, which is exposed to
`the high-temperature gases, is made of firebrick. The intermediate layer
`consists of an insulating brick and is followed by an outer layer of
`ordinary red brick. The temperature of the hot gases is 1j and the unit(cid:173)
`surface conductance over the interior surface is ii,.. The atmosphere
`surrounding the furnace is at a temperature To and the unit-surface con-
`
`Thermo I Circuit
`
`Fig. 2·5. Temperature distribution ond thermal cir<uit for
`heat flow through o series composite plane wall.
`
`2-2. COMPOSITE STRUCTURES
`
`37
`
`Page 15 of 35
`
`

`
`ductancc over the exterior su rfacc is h0 • Unde r th ese conditions there will
`be a continuo us heat flow from the hot guse. through the wall to the sur(cid:173)
`roundi ngs. Since the heat flow through a given area A is the same for any
`section, we obtain
`
`k,A (T, - T2)
`L,
`
`k 3A (TJ - T4)
`LJ
`
`(2-13)
`
`The symbols in Eq. 2-13 can be identified by inspection of Fig. 2-5.
`Equation 2- t 3 can be written in terms of the thermal resistances of the
`various sections as
`
`(2-14)
`
`where the resistances may be determined from Eqs. 1-3 and 1-13 or by
`comparison of corresponding terms in Eqs. 2-13 and 2-14. Solving for the
`various temperature differences in Eq. 2-14 we obtain
`
`T;- T, = qR,
`T, - T2 = qR2
`T2- TJ = qRJ
`T3- T4 = qR4
`T4- To= qRs
`
`Adding the left- and right-hand sides of these equations yields
`
`or
`
`q
`
`Ti- To
`
`11=5 L Rn
`
`n =I
`
`(2-15)
`
`(2-16)
`
`(2-17)
`
`,.
`
`The result expressed by Eq. 2-17, namely that the heat now through the
`five sections in series is equal to the over-all temperature potential divided
`by the sum of the thermal resistances in the palh of the heat flow, can a lso
`be obtained from Lhe thermal circuit hown in Fig. 2-5. Using t he analogy
`between the now of heaL a nd the flow of electric <>urrenl Eq . 2- 17 can be
`written directly.
`
`38
`
`STEADY ONE-DIMENSIONAL HEAT CONDUCTION
`
`Page 16 of 35
`
`

`
`EXAMPLE 2-5. A furnace wall c.:onc;ists of two layers, 9 in. of firebrick
`(k = 0. ~ Btu(hr ft F) and 5 in. of insulating brick (k = 0.1 Btujhr ft F). The
`temperature inside the furnace is 3000 f and the unit-surface conductance at the
`inside wu ll il:l 12 Btujhr sq ft F. The tcm r cr<tture of the surrounding atmosphere
`is 80 1- nnd the unit-surface conducwnce a l the outer wall is 2 Btujhr sq ft F.
`Neglecting the thermal resistance of t he mortur joint!., estimate (a) the rate of
`heat loss per square fool of wall and the temperatures at the (b) inner surface
`and (c) outer surface.
`
`Solution:
`
`(a) The rate of heat flow is obtained from Eq. 2-17 as
`
`q
`
`A
`
`3000 - 80
`
`rz + -&;o.s + -f2;o.l + ~
`2920
`- - = 513 Btujhr sq fl
`5.69
`
`2920
`0.083 + 0.94 + 4.17 + 0.50
`
`Ans.
`
`It is of interest to note that the insulating brick, while representing only about
`one-third of the wall thickness. accounts for three-quarters of the total thermal
`resista nee.
`(b ) Aprlying E.q. 2-14. the temperature drop be t~ecn the furn:u;c ~w. cs and
`71 = q N 1 = (:'i i J)(O.<HD) = 43 F. This reluuvcly
`the Interior sur i:H.:c i:. T, -
`small temperature dl!fercnc.:e is in tll.:l'on.l:tllL'e with pre tous cons ideration~ i n dic~t t­
`Jng that the thl!f11HI I rcsJStttnt:e of th~ fi rst . CCII )11 in the circuit '" negligibh:.
`Th us, heat can new without a large potcnt Jal nnd the lcmrenlturc at the in terior
`wa ll i~ Jh .. ~.~rly cq u<ll to that olt hc; f!Ul>~;s.l h at i~. T, = T1 -
`·D = 2957 F
`1l11.1.
`(c) The tempera Lure or the otner '\irrfaL·c. tlbta i iH~d in u like mu nner, is 336 F .
`..:fns.
`
`I n numerou. pr~l~tical app l i~ation~. combina t i~)ns of series- und
`par;c~ l h!l-connt:clcd heat-llow paths urc encount~:r~d . An c ample or ·uth a
`ca e is ill u tntted by the compo ·ite wa ll hown in Fig. 2-6. An approxi(cid:173)
`mate M lu tion can be obt<.~in~:d b) u~. uming thut the heat now IS es.e rHiully
`ne-dimensionul. The comro. ite ' all can then be divided into th ree sec(cid:173)
`tio ns. llle thcrmu l rc~i. li..lnce or each ~ection can bl.! dcl crminc::d wit h tht:
`aid r the therm:.ll circuit . ho'" n rn Fig. 2-6. The interml.!dia te layer
`c ln!>rsts of twl) ~cpa rate thermal puths in paru lld un d its thermal l: n(cid:173)
`duct:.rncc b lh<.: sum of th~.: rndi\ 1 du ~rl nndudanL·c. . Fm u ""aJI sectfon
`of height h 1 + h~ (Fig . 2-6) tht: L'Ondudancc is
`
`per unit length of wall. Using Eq. 1-24. the overall unit conductance U
`from surface to surface is
`
`2-2. COMPOSITE STRUCTURES
`
`39
`
`Page 17 of 35
`
`

`
`Thermal Circuit
`R4
`I
`I ----"""'""""""'
`I
`
`~ Brick
`
`Air
`
`0 Steel
`
`(l1 = 1 in.; L1 =
`Fig. 2-6. Thermal circuit for a parallel-series composite wall.
`1/32 in.; L3 = 1/4 in.; for Example 2-6, in which T1 is at the center.)
`
`u
`
`EXAMPLE 2-6. A layer of 2-in.-thick firebrick (kb = 1.0 Btuf hr ft F) is
`r taced between two t-i n.-t hick steel plates (k, = 30 Btu/ hr l't F). The face of
`the brick adjacent to the plates are rough, having solid-to-solid cormtcl only over
`30 percent or the total urea, wilh the average height of the aspen tics being :h in. If
`the outer steel-plate s urface temperatures are 200 and 800 F rt'.c;pccti vely, specify
`the rate of heat now per unit area.
`
`Solutiow The rea l system is tirst idealized by assumi ng that Lhe asperiLies
`or the s urface a re distributed, a. shown in Fig. 2-6. We not · th at the camp s'ite
`wall is sym melric<tl with respect to the center plane and therefore only consider
`one-half or the system. T he overall hea t-tra nsfer coefficien t for the composite
`wa ll is Lhcn
`
`...
`
`from an inspection of the thermal circuit.
`The thermal resistance of the steel plate R 3 is, on the basis of a unit area,
`equal to
`
`40
`
`STEADY ONE-DIMENSIONAL HEAT CONDUCTION
`
`Page 18 of 35
`
`

`
`~
`R 3 = - =
`k,
`
`~~
`( 12)(30)
`
`3
`= 0.694 x 10 - hrsq ft F/Btu
`
`The therma l resistance of lhe brick asperities R4 is, on the busis of a unit
`arcit, equal to
`
`32
`R4 = ~- l /
`( 12)(0.3)( 1.0)
`0.3 k"
`
`= 8.7 x J0 - 3 hrsq ft F/Btu
`
`Since the air is trapped in very small compartments, the em:cts or convection "lre
`small and it will be assumed that heut Bows through the air by cond uction. Ala
`temperature of 300 F, the cond uctivity of air k(l i$ 0.02 Btu/hr fl F. Then R5 ,
`the t.hcrmnl resistance of the air trapped between the asperities, is, on the ba sis
`of unit nrea, eq ual to
`
`L,
`R5 = -~-,.
`0.7 kQ
`
`1/32
`(J 2)(0.7)(0.02)
`
`J
`=> 187 x 10- hrsq fl F/ Btu
`
`The factors of 0.3 and 0.7 in R4 and R5 respect ively, represent the percent of the
`total area for the two sepa rate heat-flow paths.
`The total thermal resistance for the two paths, R4 and R; in parallel, is
`
`(8.7)(187) x I0-6
`.....:....__:....:....__;__ _ _
`(8.7 + 187) X JO-l
`
`=
`
`8 3
`. X
`
`IO - J I
`H Sq
`
`r F/ B
`t U
`l
`
`The therma l resistance of one half of lhe so lid brick, R1 , is
`
`2
`.!.
`2 ( 12)( 1.0)
`
`= 83.5xi0 - 3 hr sq ftF/ Btu
`
`and U, the overa ll heat-transfer coefficient, is
`
`U =
`
`1/2 X 10 3
`83.5 + 8.3 + 0.69
`
`= 5.4 Btuf hr sq fl F
`
`An inspection ofthc va lues fo r the various the rm al rcsistnnces show tha t the s teel
`olfcrs a negligible resistance, while the contact sccLion. although only ~ in. thick,
`comribuLes 10 percent to ~he total resistance. From Eq. 1-2 1. the r~te of heat fl o w
`per unit area is
`
`q - U 6 T = 5.4(800 - 200) = 3250 Btu/ h.r sq fl
`I f
`
`Ails.
`
`The thermal resistance between two surfaces is ca lled contact re(cid:173)
`sistance. T he ana lysis of the con tact resistance in the preceding problem
`
`2-2. COM.POSTTE STRUCTURES
`
`..
`
`Page 19 of 35
`
`

`
`fig. 2-7. Sketch illustrating nomenclature of composite cylinder wall.
`
`is only approximate because, in addition to roughness, the contact re(cid:173)
`sistance depends on the contact pressure. For more information on
`contact resistance, see Refs. 4, 5, 6, 17, and 18.
`
`Concentric cylinders. Radial heal flow through c nl;enLric cylinder,
`of different therm al conductivily is encountered in m any irH.Iu ·trill I instul(cid:173)
`\ations. An insulated pipe, with "' hal Ruid no, ing inside lind exposed to
`a colder medium on the outside, is typkul of such problems (Fig. 2-7).
`tr the pipe is relatively long, then the heal n
`through the walls will be
`in <t radhd direction . In tht.:: slC4ldy stale, Lhe rate of he:.tl Oow through
`each section is the same and is represented by
`
`q
`
`q
`
`q
`
`2Trr 1lfi;(T;- T,) =
`
`2Trk,/
`In (rz/r 1)
`
`21rk2/
`In (r 3 / r2)
`
`(T, - T2)
`
`(T
`2
`
`_ T3
`
`)
`
`q = 2Trr3/h 0 (T3 - To) =
`
`That- T,
`R,
`
`T,- T2
`R2
`
`T2- T3
`RJ
`
`Tcold
`T) -
`R4
`
`for the inner surface
`
`for the inner cylinder
`
`for the outer cylinder
`
`for the outer surface
`
`In most practical applications lhe Lemperamre of the fl uid inside and
`the temperature of the medium surrounding the insulation are k nown
`r
`specified. The intermediate tempera ture · ca n be elimi nutcd by add ilion of
`the temperature-difference terms :md trnn spo ilion . The resul ting ex pres-
`
`...
`
`42
`
`STEADY ONE-DIMENSIONAL HEAT CONDUCTION
`
`Page 20 of 35
`
`

`
`sion for the rate of heat flow through two concentric cylinders then
`becomes
`
`q
`
`(2-18)
`
`Tho! -
`
`Tcold
`
`n=4 L Rn
`
`n•l
`
`The overall heat-transfer coefficient V for this system can be based on
`any area, but its numerical value will depend on the area selected. Since
`the outer diameter is the easiest to measure in practice, Ao = 21rr3 / is
`usually chosen as the base area and the rate of heat flow is
`
`Then, by comparison with Eq. 2-18, the overall heat-transfer coefficient
`becomes
`
`(2-19)
`
`EXAMPLE 2-7. Calculate the heat loss per linear foot from a 3-in.-steel
`sched. 40 pipe (3.07 in. ID, 3.500 in. OD, k = 25 Btujhr ft F) covered with a
`Hn. thickness of asbestos insulation (k =- 0.11 Btujhr ft F). The pipe transports
`a fluid at 300 F with an inner unit-surface conductance of 40 Btujhr sq ft F and is
`exposed to ambient air at 80 F with an average outer unit-surface conductance
`of 4.0 Btu/ hr sq ft F.
`
`Solution: Using Eq. 2-18, the rate of heat transfer for a length I = 1 ft is
`
`1
`11"( .07/12)40
`
`220
`+ In (3 .5/3.07)
`- ---'--- - +
`27r(25)
`
`In (4.5/3.5)
`21r(O.ll)
`
`I
`-t- - - - - -
`7r(4.5/ 12) (4)
`
`220
`0.0312 + 0.00085 + 0.363 + 0.212
`
`= 362 Btujhr ft
`
`Ans.
`
`It is to be noted that the thermal resistance is concentrated in the insulation
`and in the low surface conductance at the outer Slirfnce, while the resistance of
`If the pipe were bare, the heat loss would be
`the metal wal l is negligible.
`1040 Bt u/ hr ft. or nearly thrice as large as with the insulution.
`
`2-2. COMPOSITE STRUCTURES
`
`43
`
`Page 21 of 35
`
`

`
`0
`
`Critical thickness of insulation. The additio n of in ulali n to the
`outside or small pipt:.<s or wires d cs n L always reduce the heal transfer.
`We have previously noted that the n1dial ra:te of heat R w th ro ugh a hol(cid:173)
`I' the e uler radi us
`low cylinder is inversely proportion al to th e I garithm
`a nd lhe rule oJ heat dissipation from the outer surface is directly propor(cid:173)
`tional to this radius. Thus, f r a single-w:lll tube of fixed inner radius
`r, , an increase in outer radius ra (e.g .• th e insulalion thickness) increases
`the thermal resisumce due to c nduction logaritl!mically and at the same
`time reduces the th erm al resistance at the outer surface linearly with r
`Since the to tal therm al resistance is proportional to the sum of these two
`resistances. the ru le o r heat now may increa e as insu \at ion i added to a
`If the insulation thickness is then furth er increased,
`b(lre pipe or wire.
`the heal loss gr<.~dua\ly drops below the loss for a bare ·urfacc. This
`principle is widely utilized in electrical engineerin g where lagging is pro(cid:173)
`vided fo r current-carrying wires a nd cables not to reduce the heat loss,
`but to increase it. It is also of importa nce in refrigeration, where heat
`flow to the cold refrigerant should be kept at a minimum. In many such
`installations where small-diameter pipes are used, insulation on the out-
`side surface would increase the rate r heatl~ow.
`The relation between heul transfer and insulation thickne. s can be
`studied quantitati vely wi th the aid o r Eq. 2- 18. ln many practical
`itua(cid:173)
`tio ns the thermal resistance is concentrated in the insulatio n and at tbe
`outer urf;.ICC. We hall therefore . implify Eq . 2- 18 by assuming thai
`T; is the temperature al the inner surface o f the insulation. This boundary
`condition applies to an insulated electric wire whose outer surface tem(cid:173)
`perature T; is fixed by th e current density, lhe wire size, and the material.
`
`•
`
`Then,
`
`21rk/(T; - T0 )
`q == In (r 0 / r;) + k/horo
`
`(2·20)
`
`where r
`
`is the outer radius, r; the inner radius, and k the thermal con(cid:173)
`
`0
`
`ductivity of the insulation.
`For a fixed value of r;, the rate of heat flow is a function of r0
`), and will be a maximum at the value of r 0 for which
`q == q(r
`
`0
`
`dq
`-
`dro
`
`==
`
`-27rk/(T;- T0 )(1/ro- (k/hor/)1
`(ln (rofr;) + k/h1ol 2
`
`0
`
`, i.e.,
`
`(2·21)
`
`From Eq. 2-21 the radius for maximum heat transfer, called the critical
`
`radius, is roc = k/hn·
`EXAMPLE 2-8. An electrical cable, ! in. OD, is to be insulated with rubber
`(k ~ 0.09 Btu/hrft F). The cable is to be' located in air (il, ~ \.5 Btu/h< sq ft f)
`
`44
`
`STEADY ONE-DIMENSIONAL HEAT CONDUCTION
`
`Page 22 of 35
`
`

`
`•
`
`at 70 F. Investigate the effect of insulation thickness on the heat dissipation,
`assuming a cable surface temperature of 150 F.
`
`Solution: Applying Eq. 2-20, the rate of heat dissipation per unit length is
`
`q
`
`211"(0.09)(150 - 70)
`In (r0 /t) + (0.09)(12/l.5r0 )
`
`45
`2
`·
`In 4r0 + (0.72jr0 )
`
`Btujhr ft
`
`if r1
`, is in inches. The first term of the denominator is proportional lo the therma l
`resistance orLhe in ·u lat..ion, the second term to the surface resistance. l n Fig. 2-8
`each or these terms is plotted against the outer radius r 11 • The dotted line, repre(cid:173)
`sen ti ng the su m of both terms, has a minimum at roc = 0.72 in . Th is i
`the critical
`radius at which the rate of heat dissipation reaches a maximum value of
`
`2 = 22.8 Btujhr ft length
`q = 45
`·
`2.08
`
`If the wire were bare, the rate of heat dissipation would be 15.7 Btujhr ft
`length, a reduction of 45 percent.
`
`In a practical situation, the selection of the insulating thickness also
`requires a cost analysis, and usually a compromise between the desirabil(cid:173)
`ity of dissipating as much heat as possible and the necessity of keeping
`cost down must be . made. Such a compromise might be an insulation
`lhicknes of i in. (r0 = 0.5 in.), which requires only abou t 50 percent of
`lhe materi al, yet dissipates heat at a rate equal to 98 percent of the maxi(cid:173)
`mum rate.
`For cases when r; is larger than k/ho, addition of insulation will
`always reduce the rate of heat transfer, and the optimum insulation thick(cid:173)
`ness must be determined by a cost analysis that takes into account the cost
`
`3.0.-----.,-----.--~~---.---~----.
`
`0
`
`r0 , Inches
`
`Fig. 2·8. Variation of thermal resistance with outside radius for an insulated
`electric wire.
`
`2-2. COMPOSITE STRUCTURES
`
`45
`
`Page 23 of 35
`
`

`
`and depreciation of the insulation, the cost and depreciation of the equip(cid:173)
`ment required to make up for the energy lost as heat, and sometimes the
`space occupied by the insulation.
`
`2-3. Systems with heat sources
`
`Systems with heat sources (or sinks) are enco un tered in many
`branches of engineering. Typical examples are electric coils, resistance
`heaters, nuclear reactors, and the combustion of fuel in the fuel bed of
`a boiler furnace. The dissipation of heat