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# Heat Transfer and Heat Generation

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## Abstract

Thermal conductivity is a decisive parameter for the sizing of heat sinks for semiconductors or for other elements in electronic circuits. Heat production and conductivity may also be an important topic in elastomer friction. In this chapter, it will be shown that also problems of heat and electrical conductivity can be exactly solved within the framework of the method of dimensionality reduction. The mappability is not only limited to the thermal/electrical conductivity or resistance, but also includes local parameters, such as the temperature distribution on the surface. This Chapter can be used as a reference work for thermal contacts and frictional heat generation..
115
8.1 Thermal Conductivity and Resistance
Thermal conductivity is a decisive parameter for the sizing of heat sinks for semi-
conductors or for other elements in electronic circuits. It is deﬁned as
where
Q
is the heat ﬂux through the element and
δT
is the difference in tempera-
ture between both ends. Alternately, the thermal resistance
RW
is used, which is
simply the inverse of the thermal conductivity:
The heat ﬂux density
q
in an isotropic continuum is proportional to the temperature
where
is the speciﬁc thermal conductivity.
The change in temperature in a homogenous medium is described by the heat
equation
in which
ρ
is the density and
c
is the speciﬁc heat capacity of the medium. Using
the thermal diffusivity
α=c,
Eq. (8.4) can also be written in the form
(8.1)
W
(8.2)
R
W
=1
W=δT
Q.
(8.3)
q=−
T,
(8.4)
ρ
c
T
t
=−div q=T
,
(8.5)
T
t=
α�T
.
Chapter 8
Heat Transfer and Heat Generation
V.L. Popov and M. Heß, Method of Dimensionality Reduction
in Contact Mechanics and Friction, DOI 10.1007/978-3-642-53876-6_8
Markus Heß and Valentin L. Popov
v.popov@tu-berlin.de
116 8 Heat Transfer and Heat Generation
In the steady-state case, the temperature distribution must satisfy the Laplace
equation
the solution of which is the next topic of discussion for various boundary condi-
tions. The results will directly show that also heat conductivity problems can be
exactly solved within the framework of the method of dimensionality reduction.
The mappability is not only limited to the thermal conductivity or resistance, but
rather includes also local parameters, such as the temperature distribution on the
surface.
8.2 Temperature Distribution for a Point Heat Source
on a Conductive Half-Space
We consider a point heat source Q on an isotropic half-space, as shown in Fig. 8.1.
With the exception of the location of the point source, let the entire surface be ide-
ally insolated (adiabatic) and at an inﬁnite distance, the temperature
T0
is reached.
With these thermal boundary conditions, the solution to the steady-state conduc-
tion problem (see, for example [1]) is
On the surface of the half-space
(z=0)
, the resulting temperature distribution is
A relationship equivalent to (8.8) appears also in the elastic problem, which was
shown by Francis [2], among others. The normal surface displacement of an elas-
tic half-space caused by a normal force at the origin is [3]
(8.6)
T=0,
(8.7)
δ
T(R)
:=
T(R)
T0
=Q
2πR
with R
:=
x2
+
y2
+
z2
.
(8.8)
δ
T(r)
:=
T(r)
T0
=Q
2πr
with r
:=
x2
+
y2
.
(8.9)
¯u
z(r)
=
1ν
2
πE
FN
r.
Fig. 8.1 Point heat source Q
on a homogeneous half-space
with the thermal diffusivity
α
v.popov@tu-berlin.de
117
Following this analogy and the interpretation of (8.8) and (8.9) as Green’s func-
tions of the corresponding problem, arbitrary heat ﬂux density distributions
q(x,y)
on the surface of the half-space present no difﬁculties. In place of the explicit cal-
culation of the integral
we can call on the solution of the (equivalent) elastic problem and transfer this
directly to the heat transfer problem. For this, we need only undertake the follow-
ing reassignments:
where
q(x,y)
is the component of the heat ﬂux density that is normal to the sur-
face. Figure 8.2 shows an example of a constant heat ﬂux density (isoﬂux) on a
circular area with the radius a. Determining the corresponding temperature distri-
bution on the surface is the goal of Problem 5.
Let it be mentioned that the equivalence is limited to the surface and is not
valid for the ﬁeld within the media. This does not, however, affect the heat ﬂux Q
through the surface, which is calculated by integrating the heat ﬂux density over
the surface:
In the elastic problem, this is the role of the normal force, which is similarly
deﬁned as the integral of the normal stress.
It is known from Chap. 3 that every axially-symmetric elastic contact problem
can be mapped exactly to a one-dimensional model. Due to the existing equiva-
lence between the heat transfer and the elastic contact, characterized by the reas-
signments in (8.11), the dimensionality reduction must also be valid for these
problems.
(8.10)
δ
T(x,y)=
1
2π

A
q(˜x,˜y)
(x
−˜
x)2
+
(y
−˜
y)2
d˜xd˜y
,
(8.11)
p
(x,y)
�→
q(x,y),
¯
uz(x,y)
�→
δT(x,y), and E/
1
ν2
�→
2
,
(8.12)
Q
:=
A
q(x,y)dA
.
Fig. 8.2 Constant heat ﬂux density from a circular area of radius a into the half-space; cross-
sectional view in the xz plane (left), top view (right)
8.2 Temperature Distribution for a Point Heat Source...
v.popov@tu-berlin.de
118 8 Heat Transfer and Heat Generation
8.3 The Universal Dependence of Thermal Conductivity
and Contact Stiffness
If two half-spaces are in an ideal thermal contact by means of a circular area with
a
and the temperature difference between the two is
δT
at an inﬁnite dis-
tance, then the entire steady-state heat ﬂux through the contact area is
and the conductivity of the contact is [1]
Here,
1
and
2
denote the speciﬁc thermal conductivity of the two half-spaces
and we can summarize
as a type of effective speciﬁc thermal conductivity.
Comparing this to the contact stiffness of a circular contact with the radius
a
,
shows that there exists the following relationship between the thermal conductivity
and contact stiffness:
Both properties are proportional to the characteristic length of the contact. Interestingly,
the validity of Eq. (8.16) goes much beyond the circular contact. It is, likewise, valid for
individual contacts with arbitrarily formed isothermal contact areas and even remains
unchanged for the contact between rough surfaces (Sevostianov and Kachanov [4],
Barber [5]). This universal relation has a very important meaning, because with its help,
one must not investigate both the thermal and elastic behavior of a contact separately.
Contact stiffness and thermal conductivity are connected in a simple way.
It is generally known that thermal conduction and electrical conduction are
equivalent problems. If a constant electric potential difference U is applied at a
sufﬁciently large separation distance over the contact between two half-spaces,
then a steady-state electric current ﬂows through the contact area. If we once again
assume a circular ideal contact (without impurities), then the entire current must
ﬂow through this constriction, which is characterized by the so-called constriction
resistance
RE
and can be interpreted as the contact resistance. The entire electrical
current I through the equipotential contact area is
and the corresponding constriction resistance is
(8.13)
Q=
4
a
δT
(8.14)
W
:= Q
δT=
4awith
1
=1
1+1
2
.
(8.15)
k
z
:= dF
N
dδ=
2aE
,
(8.16)
W
=
2
E
kz
.
(8.17)
I
=4a
ρ1+ρ2
U
(8.18)
R
E
:= U
I=ρ
1
+ρ
2
4a,
v.popov@tu-berlin.de
119
where
ρ1
and
ρ2
are the speciﬁc resistances of the two bodies. If instead of the
resistances in Eq. (8.18), we use the inverse of the (speciﬁc) electrical conductivi-
ties, this leads to the electrical contact conductivity
Completely identically to the thermal contact, the electrical conductivity is pro-
portional to the contact length. Except for the form factor, the proportionality is
also valid for contact areas of other forms as well as multiple micro-contacts suf-
ﬁciently far from one another. For the latter, the contact length is the sum of the
characteristic diameters for the so-called a-spots [6].
Of course, the conductivity for arbitrary contacts can also be determined from
the incremental contact stiffness, because Eq. (8.16) remains absolutely valid
when replacing the thermal properties by the analogous electrical properties.
8.4 The Implementation of the Steady-State Current Flow
Within the Framework of the Reduction Method
The contact stiffness of arbitrary axially-symmetric bodies and rough contact is
correctly mapped using the method of dimensionality reduction. A simple way for
calculating the thermal and electrical conductivity of a (rough) contact consists of
ﬁrst determining the contact stiffness using the method of dimensionality reduc-
tion and subsequently calculating the conductivity using Eq. (8.16). Alternatively,
we can look at every element of the linearly elastic foundation as having a (spe-
ciﬁc) conductivity of
The latter is imperative, when mapping contacts with arbitrary thermal or electri-
cal boundary conditions.1 Due to the analogy with the elastic problem in the form
of the reassignments in Eq. (8.11), both the global relations and the local parame-
ters on the surface can be correctly mapped. According to Eq. (8.11), the thermal
ﬂow density
q(r)
takes over the role of the normal stress
σzz
and temperature, the
role of the normal surface displacement.
As an example, we want to investigate the thermal contact between two half-
spaces. At an inﬁnite distance, there exists a temperature difference of
δT
. The
non-contacting surface is adiabatic and the contact area has a radius of a. We
would like to determine the heat ﬂux Q, the thermal resistance
RW,
and the distri-
bution of the heat ﬂux density q within the contact area. In the three-dimensional
(8.19)
E
:= I
δV
=
4a
Ewith
1
E=1
E1+1
E2
.
(8.20)
∆Λ =2·x.
1 In the following, we constrict ourselves to the mapping of thermal contacts, because these can
be directly transferred to electrical contacts.
8.3 The Universal Dependence of Thermal Conductivity and Contact Stiffness
v.popov@tu-berlin.de
120 8 Heat Transfer and Heat Generation
problem, there is a so-called isothermal contact area. This means that every point
on the contact area has the same temperature. The equivalent elastic problem is
the indentation of a ﬂat cylinder, the equivalent proﬁle of which remains the same.
This leads to the fact that the temperature in all of the elements of the foundation
is the same and also that the heat ﬂux through every element
Q
is independent of
the coordinate:
The ﬂux density
j
(per unit length in the one-dimensional system) is equal to
and the entire ﬂux is found by integration of the one-dimensional ﬂux density over
the contact area:
which corresponds to the three-dimensional result (8.13). The same is true for the
thermal resistance
Analogously to the elastic contact, we can calculate the three-dimensional heat
ﬂux density
q(r)
by using the Abel transformation (3.37) of the one-dimensional
ﬂux density
j(x)
:
In the present case of a constant, one-dimensional ﬂux density according to (8.22),
the integral on the right-hand side of (8.25) disappears so that only the three-
dimensional ﬂux density remains:
Also this result corresponds exactly to the three-dimensional distribution. In the
thermal contact considered, we assume an isothermal contact surface. In the case
of an axially-symmetric, spatial temperature distribution, we must transfer the
three-dimensional to a one-dimensional temperature distribution. The respective
transformation takes place in the familiar way (3.27):
(8.21)
Q(x)=∆Λ ·δT(x)=
2
·x·δT.
(8.22)
j(x)
=Q(x)
x=
2
·
δT
(8.23)
Q
:=
a
a
j(x)dx=2
a
0
2δTdx=4aδT
,
(8.24)
R
W
:= δT
Q=1
4
a
.
(8.25)
q
(r):= − 1
π
1
r
d
dr
a
r
x·j(x)
x2
r2dx=−1
π
a
r
j(x)
x2
r2dx+1
π
j(a)
a2
r2
.
(8.26)
q
(r)
=1
π
2δT
a
2
r
2
.
(8.27)
δ
T1D(x)=δT3D(0)+|x|
|x|
0
δT
3D(r)
x2
r2dr
.
v.popov@tu-berlin.de
121
The constant term on the right-hand side disappeared in the equivalent elastic
problem by choosing the appropriate coordinates,2 the second term expresses the
same relationship as that in Eq. (3.27). As it will be seen in the next section, the
inverse question is also interesting: How can we determine the three-dimensional
temperature distribution from the one-dimensional distribution? Referring to [7],
the inverse transformation is
In the mentioned literature, the transformation is given as well as the physical
interpretation that allows for the calculation of the one-dimensional ﬂux density
distribution from the three-dimensional distribution:
We would like to clarify its application using a simple example. In this example,
we assume that a stable constant thermal ﬂux density is given on the surface of the
half-space within a circle of radius a (see Fig. 8.2) of
and the rest of the surface is adiabatic. We want to ﬁnd the one-dimensional ﬂux
density and the one-dimensional and three-dimensional temperature distribution.
When taking Eq. (8.30) into account, Eq. (8.29) provides the one-dimensional ﬂux
density
which of course leads to the entire ﬂux of the original contact after integrating
over the contact length:
In the one-dimensional model, the temperature of the element is proportional to
the ﬂux density at that point (Eq. 8.22). For this example, it is
2 The point of the indenter is the origin of the coordinate system used for the indenter proﬁle.
(8.28)
δ
T3D(r)=2
π
r
0
δT1D(x)
r2
x2dx
.
(8.29)
j(x)=2
a
x
r·q(r)
r2
x2dr
.
(8.30)
q(r)
=
q0for 0 <r<a,
(8.31)
j(x)=2
a
x
rq0
r2
x2dr=2q0
a2x2
,
(8.32)
Q
=
a
a
j(x)dx=4q0
a
0
a2x2dx=4q0a2
π/2
0
cos2ϕdϕ=q0πa2
.
(8.33)
δ
T1D(x)
=1
2
j(x)
=q
0
a2
x2
.
8.4 The Implementation of the Steady-State Current…
v.popov@tu-berlin.de
122 8 Heat Transfer and Heat Generation
With the help of Eq. (8.28), it follows that
for which the complete elliptical integral of the second kind is shortened to
E
.
Comparing this to the expressions found in literature [8] veriﬁes it to be correct.
Further applications of the transformation formulas are handled in the problems at
the end of this chapter.
8.5 Heat Generation and Temperature in the Contact
of Elastic Bodies
Until now, we have only investigated cases with no relative motion between the
bodies. Furthermore, steady-state thermal states have been assumed. We would
like to continue to respect the latter, but now allow for relative motion between the
bodies. For this, we consider a stationary point source Q under which a half-space
moves with a constant speed of v in the x-direction; this is sketched in Fig. 8.3.
While the x, y, z coordinate system is stationary, the
˜x,˜y,˜z
system moves with
the body. To describe the temperature distribution (measured in the stationary sys-
tem), the Laplace Eq. (8.6) must be supplemented by a convective term:
the steady-state solution of which is [1]
(8.34)
δ
T3D(r)=2q0
π
r
0
a2x2
r2x2dx=2q0a
π
π/2
0
1(r/a)2sin2ϕdϕ=2q0a
πEr
a
,
(8.35)
T
=v
α
T
x,
(8.36)
δ
T(x,y,z)
=
T(x,y,z)
T0
=Q
2πR
ev(Rx)
2αwith R
:=
x2
+
y2
+
z2
.
Fig. 8.3 Stationary point
source under which a half-
space moves at a constant
speed of v in the x-direction
v.popov@tu-berlin.de
123
In order to calculate the temperature distribution for a distributed thermal ﬂux den-
sity of the surface, Eq. (8.36) must be used as Green’s function. This is especially
essential for the investigation of frictional contacts, for which the (entire) frictional
energy is transformed into heat. However, this is only necessary for one part of the
solution. For the body on which the stationary frictional position is located, we
can simply use the solution for the stationary case (8.7). Only in the special case
of very low speeds or very small Péclet numbers
can we add the approximation for the other body and, therefore, take advantage of
all equivalencies for the entirety of both surfaces (a is the contact radius). We will
constrict ourselves in the following to such cases.
We will now consider a frictional contact with the frictional coefﬁcient
µ,
for which the contact partners move with a relative speed of v with respect to
one another. For the heat generated on the contact surface, the following is
valid:
for which
p(x,y)
denotes the normal stress distribution and
FN,
the normal force
distribution. The heat ﬂows into both half-spaces respectively according to
The distribution between the two sometimes causes difﬁculties, because the
weighted function
β
is generally dependent on x and y in order not to violate the
continuity of the temperature within the contact area [9]. We circumvent the prob-
lem by assuming that one of the contacts is non-conductive, so that the entire heat
ﬂows into the other body. We would now like to determine the temperature dis-
tribution on the surface of this body by using the reduction method; its speciﬁc
thermal conductivity is
. We consider an element of the linearly elastic founda-
tion with the coordinate
x
that is indented by
uz(x)
. The known force acting on
this element is then
fN=Ex·uz(x)
. The frictional power of the element is
Q(x)=µνfN(x)=µvE x·uz(x),
for which the resulting temperature differ-
ence of the element is
The temperature difference for the three-dimensional model at the point r on the
surface within the contact area can be obtained using Eq. (8.28). The temperature
can be calculated even outside of the contact surface. For this, we must simply
change the upper boundary of the integral in Eq. (8.28):
(8.37)
Pe := va
2α1,
(8.38)
q(x,y)=µvp(x,y)Q=µvFN,
(8.39)
q1(x,y)
=
β
·
µvp(x,y)and q2(x,y)
=
(1
β)
·
µvp(x,y).
(8.40)
δ
T1D(x)
=Q(x)
2·x=E
2
µ
·
v
·
uz(x)
.
(8.41)
δ
T3D(r)=2
π
a
0
δT1D(x)
r2
x2dxfor r>a
.
8.5 Heat Generation and Temperature in the Contact of Elastic Bodies
v.popov@tu-berlin.de
124 8 Heat Transfer and Heat Generation
Applying this classical transformation to the classical example of a parabolic frictional
contact is the topic of Problem 1. It is possible that the reader may not see the ben-
eﬁts of the method of dimensionality reduction compared to other methods because of
the complicated transformations. Therefore, we would like to emphasize the fact that
the reduction method maps global parameters such as normal force, indentation depth,
contact area/length, total heat ﬂow rate, and maximum surface temperature as well the
contact stiffness and resistance seemingly effortlessly and exactly. These relationships
are at the forefront of the investigation of rough contacts. If only information about
local parameters is of interest, then this can also be reconstructed using the transfor-
mation rules from the one-dimensional model.
8.6 Heat Generation and Temperature in the Contact
of Viscoelastic Bodies
Heat can not only be generated on the surface, but also directly in the material of
the contacting bodies, assuming that they exhibit viscoelastic properties. One can
qualitatively approximate the temperature distribution as follows. Let us consider an
element in a viscoelastic foundation at the point x and assume that it is deformed in
the vertical direction with the speed
˙uz(x,t)
. Thereby, the force produced is given by
for which an incompressible material is assumed (see Chap. 7).
The heat generation in the element is
If we interpret this heat generation as that produced in the frictional contact, then
we obtain the temperature in the element according to (8.40):
As an example, we consider a simple viscoelastic medium (Kelvin body). In this
case, the normal force is given by
and the temperature by
(8.42)
fN(x,t)=4x
t
0
G(tt)˙uz(x,t)dt
,
(8.43)
Q(x,t)=fN(x,t)·˙uz(x,t)uz(x,t)·4x
t
0
G(tt)˙uz(x,t)dt
.
(8.44)
δ
T1D(x,t)=Q(x,t)
2
·
x=2
˙uz(x,t)
t
0
G(tt)˙uz(x,t)dt
.
(8.45)
fN(x,t)=(4Guz(x,t)+4η˙uz(x,t))x
(8.46)
δ
T1D(x,t)
=2
(Guz(x,t)
+
η
˙
uz(x,t))
˙
uz(x,t)
.
v.popov@tu-berlin.de
125
The temperature is, therefore, dependent on the time and can generally either
8.7 Problems
Problem 1 A non-conducting, rigid body with a smooth surface slides over an
elastic half-space with a parabolically curved surface of radius R with a speed of
v0
. The modulus of elasticity E, Poisson’s ratio
ν
, and the thermal conductivity
of the half-space are given. Determine the temperature distribution of the surface of
the half-space using the reduction method and assuming steady-state conditions.
Solution We have already solved the purely elastic problem multiple times and
carry over several intermediate results. After converting the three-dimensional pro-
ﬁle to a one-dimensional equivalent proﬁle using the rule of Popov and calculating
the indentation depth into a linearly elastic foundation, we obtain the displacement
in the one-dimensional system:
for which the relationship between the indentation depth
d
and the normal force
FN
is given by the Hertzian relation
FN=4
3
E
R
1/2
d
3/2
. According to (8.40), this
leads to the one-dimensional temperature difference
Insertion of (8.48) into (8.28) results in the three-dimensional distribution of the
surface temperature within the contact area:
By using the Hertzian relationship between normal force and contact radius and
taking (8.38) into account, we obtain
(8.47)
u
z(x)=dx
2
R
with d=a
2
R,
(8.48)
δ
T1D(x)
=
E
2µ
·
vo
·
uz(x)
=
E
2µ
·
vo
·
d
x
2
R
with E
=
E
1
ν2
.
(8.49)
δ
T3D(r)=2
π
r
0
δT1D(x)
r2x2dx=µvoE
πR
r
0
a2x2
r2x2dx
=µvoE
πR
a2arcsin x
r
r
0+xr2x2
r
0r2
π/2
0
cos2ϕdϕ
.
=
µvoE
4R
2a2
r2
(8.50)
δ
T3D(r)
=3Q
16a
3
2a2
r2
for 0 <r<a
.
8.6 Heat Generation and Temperature in the Contact of Viscoelastic Bodies
v.popov@tu-berlin.de
126 8 Heat Transfer and Heat Generation
We obtain the distribution outside of the contact area from (8.41). Because the for-
mula differs from that valid in the contact area only by the upper bound of the
integral, we can simply carry over the antiderivative in (8.49). After rearrange-
ment, we obtain
The reader may be convinced of the validity of the results by the usage of equiva-
lency [2].
Problem 2 Determine the thermal resistance of the contact from Problem 1.
Assume a one-dimensional model.
Solution The thermal resistance, as deﬁned in (8.2), presumes an isothermal
contact that is not present here. Therefore, we refer here to the maximum surface
temperature that is present in the middle of the contact. This takes the role of the
indentation depth in the elastic contact, where the indentation depth is the same for
both the one-dimensional and three-dimensional models. Therefore, the maximum
temperature in the middle of the contact is also the same in both models. From
(8.48), we obtain
and with it, the thermal resistance
This result initially appears to conﬂict with the universal formula (8.16), because
this would result in
This is indeed the thermal resistance for a round contact, however, this relationship
is only (!) valid for isothermal contact areas. Even redeﬁning the thermal resist-
ance with respect to the average temperature instead of the maximum temperature
does not help. In this case, the thermal resistance is
although the deviation is not very large. The proportionality to the contact length
is of course always present.
Problem 3 A half-space with a parabolically curved surface having a radius
of curvature of R is pressed into a second half-space with a ﬂat surface. Before
(8.51)
δ
T3D(r)
=3Q
8πa
3

2a2
r2
arcsin
a
r+
a
r2
a2
for r>a
.
(8.52)
δ
Tmax
=
δT1D(0)
=
E
µvoa
2
2R=
3Q
8a
(8.53)
R
W
:= δT
max
Q=3
8
a.
(8.54)
R
W
:=
E
2
·kz=
E
2
·
2
Ea=
1
4
a.
(8.55)
δ
T3D
Q=
9
32
a,
v.popov@tu-berlin.de
127
contact, the bodies exhibit the temperatures
T1
and
T2
. Upon bringing the bodies
together, a heat ﬂux ﬂows through the contact area. If the temperatures far from
the contact surface are held constant, then a steady-state ﬂow will occur after
some time. Let it be mentioned that the contact area is isothermal and temperature
related deformations are neglected. Calculate the dependence of the thermal resist-
ance on the normal force in the case of an elastic contact, which is qualitatively
shown in Fig. 8.4.
Solution We can solve the elastic problem and heat conduction problem sepa-
rately with the help of the method of dimensionality reduction. The solution of the
elastic problem can be found in Chap. 3. The dependence of the normal force on
For a round contact with an isothermal contact area Eq. (8.14) is valid with which
we further express the conductivity by means of the resistance:
We have also already derived this relationship with the reduction method. By
solving (8.56) with respect to the contact radius and inserting this into (8.57), we
obtain the desired dependence:
In conclusion, let it be noted that for the complete plastic contact, the result is
Problem 4 Determine the total current, the constriction resistance, and the current
density distribution for the electrical contact between two half-spaces with the spe-
ciﬁc resistances
ρ1
and
ρ2
within a circular area (radius a). It should be assumed
that far from the contact, there exist equipotential surfaces within the half-spaces
having a difference in potential of U. Furthermore, determine the radius b of the
partial contact area through which half of the total current ﬂows.
(8.56)
F
N(a)
=
4
3
Ea
3
R
with 1
E=
1ν
2
1
E1+
1ν
2
2
E2
.
(8.57)
R
W
=1
4
a
with
1
=1
1+1
2
.
(8.58)
R
W
=
(E
)
1/3
(48RFN)
1/3
F1/3
N
.
(8.59)
FN
a2
R
W
F
1/2
N.
Fig. 8.4 Qualitative
presentation of a Hertzian
conduction
8.7 Problems
v.popov@tu-berlin.de
128 8 Heat Transfer and Heat Generation
Solution We have already discussed the equivalent heat conduction problem in
Sect. 8.4. First, we reduce the electrical contact between two bodies to the steady-
state ﬂow through one body whose effective speciﬁc conductivity is
Between the circular equipotential surface, and another at inﬁnity (or at a sufﬁ-
ciently large distance), there exists the potential difference U. Because a con-
stant three-dimensional potential difference exists, no modiﬁcation whatsoever is
needed and it can be carried over to the one-dimensional system. Every element
in the linearly elastic foundation obtains the speciﬁc conductivity
∆Λ =2·x
and the following partial current ﬂows through each:
By summation of the partial currents through all of the elements in the foundation,
we obtain the total current:
and from this, the constriction resistance
The three-dimensional distribution of the ﬂux density within the contact area is
calculated using (8.25), which is trivial due to the constant one-dimensional
current density:
Of course, all results correspond to those in the three-dimensional problem. For
this solution of the supplemental problem, we may not assume a one-dimensional
current density, but must use the determined three-dimensional current density. For
this, we integrate (8.64) over the three-dimensional contact area with the upper
radial boundary b and require that the result corresponds to half of the current:
Elementary integration and a few rearrangements lead to
(8.60)
=1
ρ1+ρ2
.
(8.61)
I(x)=∆Λ ·δV(x)=2·x·U.
(8.62)
I
=
a
a
2δV(x)dx=4a
U
(8.63)
R
E
:= U
I=1
4a=ρ
1
+ρ
2
4a.
(8.64)
q
(r)
=1
π
2U
a
2
r
2
=I
2
πa
1
a
2
r
2
.
(8.65)
I
2
πa
b
0
r
a2
r22πdr!
=I
2
.
(8.66)
b=1
2
3a
0.866 a
.
v.popov@tu-berlin.de
129
Although the radius b divides the surface by the ratio 3:1, meaning that the outer
ring is only a quarter of the total area, half of the total current ﬂows through it.
Problem 5 Determine the temperature distribution on the surface as well as the
thermal resistance for the conduction problem (isoﬂux) shown in Fig. 8.2. Use the
analogy to the elastic problem, the solution of which is considered to be known.
According to this, the loading of an elastic half-space by a constant stress p over a
circular area with the radius a leads to the following normal surface displacements
(see, for example [3]):
where
K
and
E
are the complete elliptical integrals of the second kind:
Solution According to Eq. (8.11), we must only replace the displacement with the
temperature, the normal stress with the heat ﬂux density, and the effective modulus
of elasticity with double the conductivity:
In order to calculate the thermal resistance, we need the maximum surface tem-
perature. This is given in the center and has a value of
where we have taken
E(0)=π/2
into account. The resulting thermal resistance is
then
References
1. H.S. Carslaw, J.S. Jaeger, Conduction of Heat in Solids, 2nd edn. (Oxford University Press,
London, 1954)
2. H.A. Francis, Interfacial temperature distribution within a sliding Hertzian contact. ASLE
Trans. 14(1), 41–54 (1971)
3. K.L. Johnson, Contact Mechanics. (Cambridge University Press, Oxford, 1987)
4. I. Sevostianov, M. Kachanov, Incremental compliance and resistance of contacts and contact
clusters: Implications of the cross-property connection. Int. J. Eng. Sci. 47, 974–989 (2009)
(8.67)
¯u
z(r)=
41ν
2
pa
πEE(r/a)for r<a
41ν2pr
πE
E(r/a)
1a2
r2
K(a/r)
for r>a
,
(8.68)
K
(k):=
π/2
0
1
1
k2sin2ϕ
dϕand E(k):=
π/2
0
1k2sin2ϕdϕ
.
(8.69)
δ
T(r)=
2qa
πE(r/a)for r<a
2qr
π
E(r/a)
1
a2
r2
K(a/r)
for r>a
.
(8.70)
δ
Tmax
=
δT(0)
=qa
,
(8.71)
R
W
:= δT
max
Q=qa
πqa
2
=1
π
a.
8.7 Problems
v.popov@tu-berlin.de
130 8 Heat Transfer and Heat Generation
5. J.R. Barber, Bounds on the electrical resistance between contacting elastic rough bod-
ies, in Proceedings of the Royal Society of London. Series A: Mathematical, Physical and
Engineering Sciences, vol. 459, no. 2029 (2003), pp. 53–66
6. R. Holm, Die technische Physik der elektrischen Kontakte, Band 4, Edwards Bros. (1944)
7. M. Heß, Über die exakte Abbildung ausgewählter dreidimensionaler Kontakte auf Systeme mit
niedrigerer räumlicher Dimension (Cuvillier , Berlin, 2011), Kap. 2
8. M.M. Yovanovich, E.E. Marotta, Thermal spreading and contact resistances. Heat Transfer
Handbook, chap. 4 (2003), pp. 261–394
9. H. Blok, Theoretical study of temperature rise at surfaces of actual contact under oiliness
lubricating conditions, in General Discussion on Lubrication, vol. 2 (Institution of Mechanical
Engineers, 1937) , pp. 222–235
v.popov@tu-berlin.de
ResearchGate has not been able to resolve any citations for this publication.
Article
Full-text available
A method is developed for placing bounds on the electrical contact conductance between contacting elastic bodies with rough surfaces. An analogy is › rst estab- lished between contact conductance and the incremental sti¬ ness in the mechanical contact problem. Results from contact mechanics and the reciprocal theorem are then used to bracket the mechanical load{displacement curve between those for two related smooth contact problems. This enable bounds to be placed on the incremen- tal sti¬ ness and hence on the electrical conductance for the rough contact problem. The method is illustrated by two simple examples, but its greatest potential proba- bly lies in establishing the maximum e¬ ect of neglected microscales of roughness in a solution of the contact problem for bodies with multiscale or fractal roughness.
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An analytic expression/or the steady-state interfacial temperature field in a sliding circular Hertzian contact is derived, taking into account the ellipsoidal distribution of the frictional power and the difference between the bulk temperatures of the two bodies, for the case where one surface is stationary and the other rapidly moving with respect to the contact. Other cases may be treated in a similar manner. It is shown that the temperature at any point on the interface can be approximated, to an accuracy improving with velocity, by half the harmonic mean of the two surface temperatures attained at that point if each body were to receive all the frictional power. The resulting maximum flash temperature is 33–38% higher than that given by Blok's widely used formula.
Article
Cross-property connection for a rough interface - that relates its normal incremental compliance to its resistance - allows one to transfer knowledge available in one of the two fields to the other one. This yields results that do not seem to have been discussed earlier. We consider (1) clusters of microcontacts, and implications of available results for resistances of clusters to their elastic compliances, and (2) implications of results for rigid indenters of various shapes pressed against elastic half-space for resistances of these shapes. We also extend the cross-property connection to the shear compliance of contact clusters.
Die technische Physik der elektrischen Kontakte
• R Holm
R. Holm, Die technische Physik der elektrischen Kontakte, Band 4, Edwards Bros. (1944)
• K L Johnson
K.L. Johnson, Contact Mechanics. (Cambridge University Press, Oxford, 1987)