Heat Transfer
CH21004
Figure
5.1,ofthis
requirement
takesApplying
theby
form
change
the
internalisenergy.
Equation an
1.11c
to the
contro
temperature
response
determined
formulating
overall
energy
b
ches T!. This reduction is due to convection
heat
transfer
at
the
solid–liquid
interFigureThis
5.1,
this
requirement
takesthe
the rate
form
solid.
balance
must
of heat loss at the surface t
T!. Thisofreduction
is due
to convection
heat
transfer
at the relate
solid–liquid
intere.reaches
The essence
the lumped
capacitance
method
is
the
assumption
that
the
tem#Ė out "
Ė st
change
of the
internal
energy.
Applying
Equation
1.11c to the contr
face. The essence of the lumped
method
is
the assumption
that
the tem5.1
!capacitance
The
Lumped
Capacitance
Method
ature of the solid is spatially uniforFigure
m at any
instant
during
the
transient
process.
#form
Ė out " Ė st
5.1,instant
this requirement
takes the
perature of the solid is spatially uni
form at any
during the transient
process.
or
s assumption implies that temperature gradients within the solid are negligible.
This assumption implies that temperature
or gradients within the solid are negligible.
From
Fourier’s
law,
heat
conduction
of aa temperature
temperature
#gradient
Ė st"VcddTT
out!"
From Fourier’s law, heat conductioninin the
the absence
absence of
#hAs(Tgradient
#Ė T
)"
#hAis
Tclearly
#
T!) " "Vc dt
plies
the the
existence
of of
infinite
Such aa condition
condition
clearly
t<0
s(is
Ti conductivity.
implies
existence
infinitethermal
thermal
conductivity.
Such
dt
or
= Ti
Tapproximated
possible.
However,
the
condition
is
closely
if
the
resistance
to
conimpossible. However, the condition
is closely approximated
if the resistance
to conIntroducing
thethe
temperature
difference
dT
Introducing
temperature
difference
ction
within
the the
solid
is small
compared
toheat
heattransfer
transfer
duction
within
solid
is small
comparedwith
withthe
the resistance
resistance to
#hAs(between
Tbetween
# T!) " "Vc
dt
solid
andand
its surroundings.
For
is, in
infact,
fact,the
thecase.
case.
the solid
its surroundings.
Fornow
nowwe
weassume
assume that
that this is,
•
!
!
T
#
T
= qT
!
!consider
!Eout
T#
conv
Liquid
!
In neglecting
temperature
gradients
withinthe
thethe
solid,
we
longer
In neglecting
temperature
gradients
within
solid,
we can
canno
nodifference
longerconsider
Introducing
temperature
the problem
from
within
framework
the heat
heat equation.
equation.
Instead,
the
transient
problem
from
within
thethe
framework
ofof
the
Instead,
the
• transient
andand
recognizing
that
(
d
!/
dt
)
"
(
d
T
/dt
)) ifif T! is constant,
constant,ititfollows
followsthat
that
recognizing
that
(
d
!/
dt
)
"
(
d
T
/dt
E
T(t)
st on the
temperature
response
is determinedbybyformulating
formulating
an overall
overall energy
mperature
response
is determined
an
energybalance
balance
on
the
! ! T # T!
t ≥ 0 at the surface to the"Vrate
solid.
This
balance
must
relate
the
rate
of
heat
loss
of of
crate
id. This balance must relate
the
rate
of
heat
loss
at
the
surface
to
the
"
V
c
d
!
d
T∞ <Applying
Ti
T =1.11c
T(t) to the control volume of" # !!
change
of the
internal
energy.
Equation
and Equation
recognizing
thatto(dthe
!/dtcontrol
) " (dT/dt
nge
of the
internal
energy.
Applying
1.11c
volume
hAA)ss ifdt
dtT!ofis constant, it follows that
h
Figure 5.1, this requirement takes the form
ure 5.1, this requirement takes the form
c d! the initial condition, for whi
FIGURESeparating
5.1
Cooling
of a hot metal
forging.
Separating
variables
andintegrating
integrating"V
from
variables
and
from
condition, for w
" #initial
!
#Ė out " Ė st
(5.1)the
h
A
dt
T
(0)
"
T
,
we
then
obtain
s (5.1)
Ė out""TĖ, stwe
i
T#(0)
then obtain
Transient Heat Conduction
or
i
!!
!!
!
! heatthe
t
reaches TSeparating
is due
convection
transfer
atcondition,
the solid–liquid
!. This reduction
variables
andtointegrating
initial
for wh
"Vc from
d
!
d
T
!
t
"
V
c
"
#
dt
(5.2)
#
h
A
(
T
#
T
)
"
"
V
c
d
!
s Tessence
face. The
the
is the
assumption that the
(0) !" T of
, we
obtaincapacitance
dthen
T lumped
hAs method
"#
! !(5.2)
0 dt
#hAs(T # T!) " "i Vc dt
perature of the solid isdtspatially uniformhAats any
the transient pr
!i !instant during
0
!
t
Introducing the temperature difference
where implies that temperature"gradients
Vc d! within the solid are negligib
This assumption
" # dt
where
roducing the temperature difference
!
h
A
From
Fourier’s
!the
0 of a temperature gr
s! (5.3)
i
Tabsence
!!
T # T! law, heat conduction in
i!
i # T!
implies! the
of infinite thermal conductivity.
!i !(5.3)
Ti # T! Such a condition is c
!
Texistence
# T!
where
and recognizing that (d!/dtimpossible.
) " (dT/dt) ifHowever,
T! is constant,
follows that
the itcondition
is closely approximated if the resistance to
!i !
# T!
d recognizing that (d!/dt) "
(dT/dt"within
)Vifc dT!! is
it follows
that with
duction
theconstant,
solid is small
compared
theTiresistance
to heat transfer be
!
"#!
i
Chapter 5
!
" #
#t !
(!Vc) ! R tCt
hAs
Transient Conduction
hAs
the integrals,
it follows
to convection
heatthat
transfer and Ct is the lumped thermal
where REvaluating
t is the resistanceor
capacitance of the solid. Any increase in Rt or Ct will cause a solid to respond more
Evaluating
the integrals,
it follows
that This behavior is analogous to the
slowly
to changes
in its thermal
environment.
!Vc "i "
hAs (5.5)
T " T!
ln ! t!through
voltage decay that occurs when a capacitor is
discharged
a
resistor
in
an
!
exp
"
t
hAs " "
T
"
T
!
V
c
i
!
i
electrical RC circuit.
!Vc "i
(5.5)
ln ! tup to some time t, we simTo determine the total energy transfer hQAoccurring
"
s
or
Equation 5.5 may be used to determine the time required for the solid
ply write
Lumped Capacitance Method
$ " #%
!
! $ " #%
t
t
temperature
T, or, conversely,
Equation 5.6 may be used to compu
hAs
T!
" hTA!s " dt
Q ! " q dt
(5.6)
! by the !
expat"some time
t t.
ture reached
solid
!Vc
"i 0 Ti " T! 0
The foregoing results indicate
hAs that the difference between the
" T " T!
(5.6) infini
!
exp "
t
must!decay
exponentially
to zero as t approaches
1
Equation
5.5 maytemperatures
be used
" T! the time required
!Vc for the solid to reach some
"i toTidetermine
iorconversely,
is shown Equation
in Figure5.65.2.
Equation
5.6 it
also evident th
temperature T, or,
mayFrom
be used
to compute
theistemperaVc
ρ
___
τt =
= RtCt
may time
be interpreted
as a thermal time constant expressed as
(!Vsolid
A )some
hAc/h
ture
reached
by
the
t. time required
s
Equation 5.5 may be used toatsdetermine
the
for the solid to reach some
The foregoing results indicate that the difference between the solid and fluid
temperature T, or, conversely, Equation 5.6 may be used to compute the temperatemperatures must decay exponentially to zero as t approaches infinity. This behavture reached by the solid at some time t.
1 (!Vcthat
ior is shown in Figure 5.2. From Equation 5.6 it#tis!also evident
) ! the
R tCquantity
t
h
A
The
foregoing
results
indicate
that
the
difference
between
the
solid
and
fluid
s
(!Vc/hAs) may be interpreted as a thermal time constant expressed as
temperatures
must decay exponentially to zero as t approaches infinity. This behav0.368
resistance
heat that
transfer
and Ct is the l
where
ior is shown in Figure
5.2.RFrom
Equation
5.6toit convection
is also evident
the quantity
t is the
(5.7) a solid t
capacitance
in Rt orasCt will cause
as a#tof
the
rm1asolid.
l ti(!mVecAny
const
atCntt expressed
(!Vc/hAs) may be interpreted
! the
) ! Rincrease
hAs
slowly to changes in its thermal environment. This behavior is an
0
voltage
decay
that occurs
when aand
capacitor
discharged
τt, 2 τt, to
τt, 4 1
convection
heat transfer
Ct is theislum
ped thermathrough
l
where
Rt is theτtresistance
,1
3
(5.7)
#
!
(!
V
c
)
!
R
C
t
electrical
Rt Cincrease
circuit.
t t cause a solid to respond more
capacitance of the
solid. Any
hA in Rt or Ct will
or
$ " #%
θ ______
T – T∞
__
=
θ i Ti – T∞
" #
" #
" #
voltage
decay the
thatintegrals,
occurs when
a capacitor
is discharged through a resistor in an
Evaluating
it follows
that
electrical RC circuit.
To determine the total energy transfer Q occurring up to some time t, we sim!Vc "i
(5.5)
ply write
ln ! t
"
h
A
s
t
t
Q ! q dt ! hAs " dt
0
0
or
Lumped Capacitance Method
!
5.2
!
!
259
Validity of the Lumped Capacitance Method
$ " #%
hAs
" T " T!
!
! exp "
t
!Vc
"i Ti " T!
1
(5.6)
Substituting for ! from τEquation
5.6 and integrating, we obtain
Vc
ρ
___
=
=
R
C
t hA
t t
Equation 5.5 may be used tos determine the time required for the solid to reach some
temperature T, or, conversely, Equation 5.6 may be used
to compute the tempera(5.8a)
Q # ("Vc)!i 1 " exp " #t
ture reached by the solid at some time t.
t
The foregoing results indicate that the difference between the solid and fluid
temperatures
exponentially
to zero
as t in
approaches
infinity.
This
The
quantity Qmust
is, ofdecay
course,
related to the
change
the internal
energy
of behavthe solid,
ior0.368
is shown
in Figure
and
from
Equation
1.11b5.2. From Equation 5.6 it is also evident that the quantity
(!Vc/hAs) may be interpreted as a thermal time constant expressed as
"Q # $E st
(5.8b)
θ ______
T – T∞
__
=
θ i Ti – T∞
!
" #$
" #
(5.7)
For quenching Q is positive and
Equa#t !the 1solid
(!Vexperiences
c) ! R tCt a decrease in energy.
h
A
0 5.6, and 5.8a also apply to situations
s
tions 5.5,
where the solid is heated (! ! 0), in
τt, 2 τt, 3
τt, 4
τ t, 1
t internal energy of the solid increases.
which
is negative
the
resistanceand
to convection
heat transfer and C is the lumped thermal
wherecase
R isQthe
t
t
capa5.2
citance
of the solid.
Any increase
will cause asolids
solid for
to different
respond more
FIGURE
Transient
temperature
responseinof Rlumped
t or Ct capacitance
thermal
timetoconstants
slowly
changes# .in its thermal environment. This behavior is analogous to the
tions, this criterion is readily extended to transient processes. One su
kA (T T"s,1Tand
tained at a temperature
the other surface is exposed to a fluid
s,1
s,2) # hA(Ts,2 " T!)
L
T! ! Ts,1. The temperature
of this surface will be some intermediate
Validitywhich
of the
Lumped Capacitance Method
T! ! Ts,2 ! Ts,1. Hence under steady-state conditions the surfa
ance,
Equation
reduces
to
Chapter
5 ! 1.12,
Transient
Conduction
T
kA (T " T ) # hA(T " T )
s,1
s,2
s,2
!
qconv
L
where k is the thermal conductivity of the solid. Rearranging, we the
qcond
Bi << 1
Ts, 1
Bi ≈ 1
T
Bi >> 1
T s, 1
x
Ts, 2
Ts, 2
Ts,1 & Ts,2 (L/kA) R cond hL
$
$
$ ! Bi
Ts,2 & T! (1/hA) R conv
k
qcond
qconv
T
s, 2
Bi =
number
The
quantity
in Biot
Equation
5.9 is a dimensionles
Bi <<
1
T∞,(h
h L /k)Fappearing
T 2 mberIGURE
termedBthe
, and 5.3
it plays a fundamental role in condu
i ≈ 1 Biots, nu
Effect of Biot number on steady-state temperature
L
is
that involve surface convection effects. According to Equation 5.
distribution in a plane wall with surface convectio
T
s, 2 the Biot number provides a measure of the temp
trated in Figure 5.3,
the solid relativeT to the temperature difference between the surfac
s, 2
Bi >> 1
Note especially
the conditions
corresponding to Bi ! 1. The resul
T∞, h
FIGURE
5.3
for these conditions, it is reasonable
to assu
me a uniform temperat
of Biot number on steady-state
x a solid Lat any time during aEffect
within
transient
process. This result ma
distribution in a plane wall with surfa
ciated with interpretation of the Biot number as a ratio of therm
Equation 5.9. If Bi ! 1, the resistance to conduction within the sol
2
lem, the very first thing that one should do is calculate the Biot
lowing condition is satisfied
Bi !
T(x, 0) = Ti
hL c
" 0.1
k
T(x, 0) = Ti
T∞, h
Validity of the Lumped Capacitance
Method with using the lumped capacitance
261 method is
the
error associated
t
5.2 ! Validity of the Lumped
Capacitance
Method
261
nience, it is customary
to define the characteristic length
of Eq
!
ratio of the solid’s volume to surface area, Lc ! V/As. Such a d
nsient heating and cooling problems.
Hence,
confronted with such a probcalculation
of Lwhen
c for solids of complicated shape and reduces to t
hst thing that one should do is calculate the Biot number. If the folm,
veryT∞fi,rheating
entthe
Conduction
for aproblems.
of thickness
(FigureT∞5.4),
to rsuch
a long cy
T∞plane wallHence,
T∞ when2L
o /2 for
transient
and cooling
confronted
with
aT∞probwinglem,
condition
satisfied
for a sphere.
However,
if –one
wishes
to implement
–L
L
L
L
–L
L the criterion
the veris
–y
L firstLthing that one should do is calculate the Biot number. If the folBishould
<< 1
Bi ≈ 1 with the length
Bi >> 1scale correspo
fashion,
L
be
associated
x
c
lowing condition is satisfied
T ≈ T(t)
T = T(x, t)
T = T(x, t)
rals, it follows that
hL c temperature difference. Accordingly, for a symme
mum spatial
(5.10)
Bi ! plane
" 0.1
FIGURE 5.4 Transient temperature
Biot2numbers
in a plane
wall equal to th
cooled)
wall for
of different
thickness
L, Lc would
remain
kdistributions
h
L
c
symmetrically
by convection.
!Vc "cooled
However,Bifor
a long
or sphere, Lc would equal
the actu
i
(5.10)
!
" cylinder
0.1
(5.5)
ln ! t
k
" using thethan
hAswith
ro /2 or
ro /3.
e error associated
lumped
capacitance
method is small. For convenote
that, of
with
Lc ! V5.10
/As, the
exponent of Eq
ence, it is customary to define the Finally,
charactewe
ristic
length
Equation
as the
thethe
error
associated
using
the
method isfacilitates
small. For conveexpressed
asLc ! Vcapacitance
io of
solid’s
volumewith
to surface
area,lumped
/As. Such a definition
nience,
to define the
chaand
ractereduces
ristic length
of Equation 5.10
as the
culation
of it
Lc isforcustomary
solids of complicated
shape
to
the
half-thickness
L
t . Such
L c k t facilitates
hL c #t
ht !a hdefinition
hAs to surface area, Lc ! hVA/sA
T
"
T
"
ratio
of
the
solid’s
volume
!
s
!
!
a plane!wall of thickness
a long cylinder, and!to
! exp " 2L (Figure
t 5.4), to ro /2 for(5.6)
c Lr2o /3 k L 2
!
V
c
!
c
L
k
T
"
T
!
V
c
"
c
i
! L for solids of complicated shape and reduces to the half-thickness
i
c
c L
calculation
of
c if one wishes to implement the criterion in a conservative
a sphere.
However,
forLac plane
wall
of thickness
2L the
(Figure
5.4),
to corresponding
ro /2 for a longtocylinder,
and to ro /3
hion,
should
be
associated
with
length
scale
the
maxior
e used to determine the time required for the solid to reach some
$ " #%
ransient Conduction
or
where
integrals, it follows that
hAst
! Bi ! Fo
!hVAcst
!Vc
where
where
! Bi ! Fo
(5.11)
#t
Fo ! 2
Lc
(5.11)
!Vc "i
(5.5)
ln ! t
#
t
"
his
As termed
the FourierFnumber.
a dimensionless
o ! #2t It isFourier
(5.12) with
number time, which,
Fo !L cconduction
(5.12) Equa
ber, characterizes transient
problems. Substituting
2
Lc
5.6, we obtain
is termed the Fourier number. It is a dimensionless time, which, with the Biot numis termed the Fourier number. It is a dimensionless time, which, with the Biot number,ber,
characterizes
conduction
problems.
Substituting
Equation5.11
5.11 into
T!
hAconduction
" T"
T " T! transient
" characterizes
s
transient
problems.
Equation
! Substituting
! exp("
Bi ! Fo) into
(5.6)
!
! exp
"
t
Tobtain
!Vc
5.6,5.6,
we
obtain
"i we
"i Ti " T!
i " T!
$ " #%
" " TT"
"Tthe
T!!!
ay be used to determine the time required
solid
to reach
!! for
BBii !! some
F
(5.13)
!exp("
exp("
Fo
o))
(5.13)
T
"
T
"
iTi "
!!
i"i used
or, conversely, Equation 5.6 may be
to Tcompute
the temperathe solid at some timeEt.XAMPLE 5.1
oing results indicate that the difference between the solid and fluid
ust decay exponentially
zero as t approaches
infinity.
Thismay
behavA to
thermocouple
junction,
which
be approximated as a sphere, is
EXAMPLE
5.1
E
XAMPLE
5.1
Figure 5.2. From Equation
5.6 it ismeasurement
also evident that
quantity
temperature
in athegas
stream. The convection coefficien
be interpreted as a thermal time constant expressed as
2
junction
surface
and
the
gas
is
h
!
400
W/m
! K,
the
junction
A
thermocouple
junction,
which
may
be
approximated
as
a
sphere,
isisand
totobe
used
for
A thermocouple junction, which may be approximated as a sphere,
be
used
for t
properties in
are
! stream.
20
W/mThe
! K,convection
c ! 400 J/kg
! K, and
! ! 8500
kg/
temperature
measurement
ina akgas
gas
stream.
The
convection
coefficient
between
the
temperature
measurement
coefficient
between
the
1 and
junction
surface
the
gas
is h ! 400
W/m22for
! K,the
and
the junction thermophysical
the
junction
diameter
needed
thermocouple
to have a time con
(5.7)
#
!
(!
V
c
)
!
R
C
t
tist h ! 400 W/m ! K, and the junction thermophysical
junction surface
and
the
gas
Aks ! 20 W/m ! K, c ! 400 J/kg ! K, and ! ! 8500 kg/m3. Determine
properties arehthe
junction
is
atc 25°C
and
is !placed
in! a!gas
stream
that
is at 200°C,
3
properties
are
k
!
20
W/m
!
K,
!
400
J/kg
K,
and
8500
kg/m
.
Determine
the junction diameter
needed
for the thermocouple
to have a time constant of 1 s. If
it take
for
thefor
junction
199°C?
resistance
to convection
heat
transfer
and
is to
thereach
lumped
the
mal a time constant of 1 s. If
the junction
diameter
needed
theCthermocouple
to rhave
" #