1D STEADY STATE HEAT
CONDUCTION (1)
Prabal Talukdar
Associate Professor
Department of Mechanical Engineering
IIT Delhi
E-mail: prabal@mech.iitd.ac.in
p
PTalukdar/Mech-IITD
Convection Boundary Condition
Heat conduction
at the surface in a
selected direction
=
Heat convection
at the surface in
the same direction
In writing the equations for convection
boundary conditions, we have selected
the direction of heat transfer to be the
positive x-direction at both surfaces. But
those expressions are equally applicable
when
h h
heatt ttransfer
f is
i iin th
the opposite
it
direction
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Radiative Boundary Condition
Heat conduction
Radiation exchange
at the surface in a = at the surface in the
selected direction
same direction
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Interface Boundary Conditions
The boundary conditions at an interface are
based on the requirements that
(1) two bodies in contact must have the
same temperature at the
h area off contact
and
(2) an interface (which is a surface) cannot
store anyy energy,
gy, and thus
the heat flux on the two sides of an
interface must be the same
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Generalized Boundary
Conditions
Heat transfer
to the surface
in all modes
in all modes
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=
Heat transfer
from the surface
in all modes
in all modes
Solution of steady heat
conduction equation
1D Cartesian
Differential Equation:
d
2
dx
T
2
Boundary Condition:
T (0 ) = T 1
= 0
Applying the boundary condition to the general solution:
Integrate:
T(x ) = C1x + C 2
dT
= C1
dx
0
0
Integrate again:
T (x ) = C 1 x + C 2
G
General
lS
Solution
l ti
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A bit
Arbitrary
C
Constants
t t
T1
Substituting:
T1 = C1.0 + C 2
C 2 = T1
It cannot involve x or T(x) after the
boundary condition is applied.
Cylindrical
Differential Equation:
-
Spherical
Differential Equation:
dT
d
) = 0
(r
dr
dr
d
dT
(r 2
) = 0
dr
dr
Integrate:
Integrate:
r
dT
= C1
dr
r2
Divide by r (r ≠ 0) :
Divide by r2 (r ≠ 0) :
C
dT
= 1
dr
r
C
dT
= 1
dr
r2
Integrate again:
T (r ) = C 1 ln r + C 2
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dT
= C1
dr
Integrate again:
T (r ) = −
which is the general solution.
C1
+ C2
r
During steady one-dimensional
heat conduction in a spherical (or
cylindrical) container, the total rate
of heat transfer remains constant,
but the heat flux decreases with
i
increasing
i radius.
di
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Heat Generation
Under steady conditions, the energy
balance for this solid can be expressed as
Rate of heat transfer
from solid
hAs((Ts‐T∞)
=
•
Ts = T ∞ +
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Vg
hA s
Rate of energy generation within the solid
g& V
A large plane wall of thickness 2L (As = 2Awall and V = 2LAwall),
)
A long solid cylinder of radius ro (As = 2πro L and V= πr2o L),
A solid sphere of radius r0 (As = 4πr2o L and V= 4/3πr3o )
•
Ts = T ∞ +
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Vg
hA s
Under steadyy conditions,, the
entire heat generated within the
medium is conducted through
the outer surface of the cylinder.
cylinder
The heat g
generated within this inner cylinder
y
must
be equal to the heat conducted through the outer
surface of this inner cylinder
Integrating from r = 0 where T(0) = T0 to r = ro where T(ro) = Ts yields
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• The maximum temperature
in a symmetrical solid with
uniform heat generation
occurs at its center
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1-D
1
D plane wall
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Energy balance
Rate of heat
transfer into the wall
Rate of heat
transfer out of the
wall
•
•
Q in − Q out =
dEwall
=0
dt
=
Rate of change of
energy
gy of the wall
dE wall
dt
for steady operation
Therefore, the rate of heat transfer into the wall must be equal to the rate
of heat transfer out of it. In other words, the rate of heat transfer through
the wall must be constant, Qcond, wall constant.
•
Fourier’s law of heat conduction for the wall
L •
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dT
dx
T2
∫ Qcond,wall dx
d = − ∫ kAdT
x=0
Q cond, wall = −kA
T=T1
constant
t t
Temp profile
•
Q cond , wall = kA
T1 − T2
L
(W)
The rate of heat conduction through a
plane wall is proportional to the
average thermal conductivity,
conductivity the
wall area, and the temperature
difference, but is inversely
proportional
i l to the
h wall
ll thickness
hi k
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Temp profile
1 D steady state heat conduction equation
Integrate the above equation twice
T(0) = Ts,1
Boundary conditions
Apply the condition at x = 0 and L
Ts,1 = C2
Ts,2 = C1L + C2 = C1L + Ts,1
Ts,2 − Ts,1
L
T(x) =
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= C1
Ts,2 − Ts,1
L
x + Ts,1
and
d
dT
(k
)=0
dx
dx
T (x ) = C 1 x + C 2
T ( L) = Ts, 2
Thermal Resistance Concept
Analogy between thermal and
electrical resistance concepts
T1 − T2
&
Q cond, wall =
R wall
R wall =
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L
kA
(W)
(oC/W)
Convection Resistance
•
= hA s ( T s − T ∞ )
Q convection
•
Q convection
i
=
Ts − T∞
R convection
1
R convection =
hA s
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(W)
(oC/W)
Radiation Resistance
•
4
Q rad = εσ A s ( T s4 − T surr
) = h rad A s ( T s − T surr ) =
R
rad
=
1
h rad A s
T s − T surr
R rad
(W)
(K/W)
Combined convection and radiation
•
h rad =
Q rad
2
= εσ ( T s2 + T surr
)( T s + T surr )
A s ( T s − T surr )
h combined
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= h conv + h rad
(W/m2K)
(W/m2K)
Possible when T∞ = Tsurr
The thermal resistance network for heat transfer through a plane wall subjected
to convection on both sides, and the electrical analogy
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Network subjected to convection on both sides
Rate of heat
convection into =
the wall
Rate of heat
conduction
through the wall
=
T1 − T2
= h2 A(T2 − T∞ 2 )
Q = h1 A(T∞1 − T1 ) = kA
L
•
Q=
T∞ 1 − T1 T1 − T2 T2 − T∞ 2
=
=
1 h1 A
L kA
1 h2 A
=
T∞ 1 − T1 T1 − T2 T2 − T∞ 2
=
=
R conv ,1
R wall
R conv , 2
•
Adding
g the numerators and denominators yields
y
•
Q=
T∞ 1 − T∞ 2
R total
R total = R conv ,1 + R wall + R conv , 2 =
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(W)
L
1
1
+
+
h1 A kA h2 A
Rate of heat
convection from the
wall
•
Q=
T∞1 − T∞ 2
Rtotal
(W)
The ratio of the temperature drop to the
thermal resistance across any layer is
constant, and thus the temperature drop
across any layer
l
iis proportional
ti
l tto th
the
thermal resistance of the layer. The larger
the resistance, the larger the temperature
drop.
p
•
ΔT = Q R
(oC)
This indicates that the temperature drop across
any layer is equal to the rate of heat transfer
times the thermal resistance across that layer
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It is sometimes convenient
to express heat transfer
through a medium in an
analogous manner to
Newton’s law of cooling as
•
Q = UAΔT
(W)
& = ΔT
Q
R total
UA =
The surface temperature of the wall can be
determined using the thermal resistance
concept, but by taking the surface at which the
temperature is to be determined as one of the
terminal surfaces.
R total
T∞1 − T1 T∞1 − T1
Q=
=
1
Rconv,1
h1 A
•
Known
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1