1D STEADY STATE HEAT
CONDUCTION (2)
Prabal Talukdar
Associate Professor
Department of Mechanical Engineering
IIT Delhi
E-mail: prabal@mech.iitd.ac.in
p
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Thermal Contact Resistance
Temperature distribution and heat flow lines along two solid plates
pressed against each other for the case of perfect and imperfect contact
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Consider heat transfer through two metal rods of cross-sectional area A that
are pressed against each other
other. Heat transfer through the interface of these two
rods is the sum of the heat transfers through the solid contact spots and the
gaps in the noncontact areas and can be expressed as
•
•
Q = Q
Most experimentally determined
Most
experimentally determined
values of the thermal contact resistance fall between 0.000005 and 0.0005 m2∙°C/W (the corresponding range of thermal contact conductance is 2000 to 200,000 W/m2∙°C).
•
contact
+ Q
gap
•
Q = h c A Δ T int
erface
where A is the apparent interface area (which is the same as the crosssectional area of the rods) and ΔTinterface is the effective temperature
difference at the interface. The quantity hc, which corresponds to the
convection heat transfer coefficient, is called the thermal contact
conductance and is expressed as
•
h
c
Q
=
Δ T
(W/m2 oC)
A
int
erface
It is related to thermal contact resistance by
R
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c
=
1
h c
=
Δ T
int
•
Q
erface
A
(m2 oC/W)
Importance of consideration
Th th
The
thermall contact
t t resistance
i t
range:
between 0.000005 and 0.0005 m2·°C/W
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Two parallel layers
•
•
Q = Q
•
1
+ Q
2
=
T1 − T
R1
2
+
T1 − T
R 2
2
= (T1 − T
R1 =
•
Q =
T1 − T
R total
⎛ 1
1
) ⎜⎜
+
R 2
⎝ R1
L1
k 1 A 1'
R
2
=
⎞
⎟⎟
⎠
k
L
2
2
A
'
2
2
where
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2
1
R
total
=
1
1
+
R1
R 2
R
total
=
R1R 2
R1 + R
2
Combined series-parallel
series parallel
•
Q =
T1 − T
R total
∞
R1 R 2
+ R 3 + R conv
R1 + R 2
R total = R 12 + R 3 + R conv =
R1 =
R
conv
L1
k 1 A 1'
=
1
hA
R
3
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2
=
L
k
2
2
A
'
2
R
3
=
L
k
3
3
A 3'
Series and parallel composite wall and its thermal circuit
RD
RA
R∞1
RC
T∞1
∑ R = R ∞1 +
R∞2
RE
T∞2
RF
RB
1
⎛ 1
1 ⎞
⎜⎜
⎟⎟
+
⎝ RA RB ⎠
+ RC +
1
⎛ 1
1
1 ⎞
⎜⎜
⎟⎟
+
+
⎝ RD RE RF ⎠
+ R ∞2
•
Q = UA Δ T
(W)
where U is the overall heat transfer coefficient
UA =
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1
R total
T1
T2
T3 T4
Complex multi-dimensional problems as 1-D problems
1 Any plane wall normal to the x-axis is isothermal
1.
2. Any plane parallel to x-axis is adiabatic
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Heat conduction in cylinder
•
Q cond ,cyl = − kA
dT
dr
A = 2 π rL
L
•
r
∫ r2= r
1
Q cond ,cyl
A
T
dr = − ∫ T2=T kdT
1
Substituting A = 2πrL and performing the integrations give
•
Q cond ,cyl
T − T2
= 2 πLk 1
ln( r2 r1 )
•
Q cond
, cyl
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=
T1 − T 2
R cyl
R cyl =
•
Q cond ,cyl
=constant at steady state
(W)
ln( r2 r1 )
2 πLk
=
ln(outer radius/inner radius)
2π(length)(thermal conductivity)
Heat conduction in sphere
T1 − T2
&
Q
=
cond ,sphere
R sph
For sphere
r −r
Rsph = 2 1
4πr1r2k
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=
outer radius - inner radius
4π(outer radius)(inner radius)(thermal conductivity)
Resistance Network
cylindrical
R total = R conv ,1 + R cond + R conv , 2
=
ln (r 2 r1 )
1
1
+
+
(2 π r1 L )h 1
2 π Lk
( 2 π r2 L ) h 2
spherical
R total = R conv
=
1
(4 π r )h
1
The thermal resistance network for a cylindrical (or
spherical) shell subjected to convection from
both the inner and the outer sides.
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2
+
1
,1
+ R sph + R conv
,2
r 2 − r1
1
+
4 π r1 r 2 k
( 4 π r2 2 ) h
2
Multilayered cylinder
R total = R conv ,1 + R cyl ,1 + R cyl , 2 + R cyl , 3 + R conv , 2
=
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ln (r 2 r1 ) ln (r 3 r 2 ) ln (r 4 r 3 )
1
1
+
+
+
+
h 1A 1
2 π Lk 1
2 π Lk 2
2 π Lk 3
h 2A
4
Radial heat conduction through cylindrical systems
1 ∂ ⎛ ∂T ⎞ 1 ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞
∂T
⎜⎜ k.r
⎟⎟ + ⎜ k. ⎟ + g& = ρC
⎜ k.r
⎟+ 2
∂r ⎠ r ∂φ ⎝ ∂φ ⎠ ∂z ⎝ ∂z ⎠
∂t
r ∂r ⎝
d ⎛ dT ⎞
⎜r
⎟=0
dr ⎝ dr ⎠
Integrating the above equation twice
twice, T=C1ln r + C2
Subject to the boundary conditions, T=T1 at r = r1 and T=T2 at r = r2
T =
T 2 − T1
T ln r 2 − T 2 ln r1
ln (r ) + 1
⎛r ⎞
⎛r ⎞
ln ⎜⎜ 2 ⎟⎟
ln ⎜⎜ 2 ⎟⎟
⎝ r1 ⎠
⎝ r1 ⎠
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C1
ddT
Q = − kA r
|r = r1 = − k.2πr1L.
dr
r1
Q = − k.2πr1L.(T2 − T1 ).
=
2πkL(T1 − T2 )
⎛ r2 ⎞
ln⎜⎜ ⎟⎟
⎝ r1 ⎠
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1
⎛r
r1 ln⎜⎜ 2
⎝ r1
⎞
⎟⎟
⎠
Critical Radius of Insulation
1. Steady state conditions
2. One-dimensional heat flow only in
the radial direction
3. Negligible thermal resistance due to
cylinder wall
4 Negligible radiation exchange
4.
between outer surface of insulation
and surroundings
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Insulation
r2
T∞ , h
r1
Thin wall
Ts
Critical Radius of Insulation
•
•
Practically, it turns out that adding insulation in cylindrical and spherical
exposed
p
walls can initiallyy cause the thermal resistance to decrease,, thereby
y
increasing the heat transfer rate because the outside area for convection
heat transfer is getting larger. At some critical thickness, rcr, the thermal
resistance increases again and consequently the heat transfer is reduced.
To find an expression for rcr, consider the thermal circuit below for an
insulated cylindrical wall with thermal conductivity k:
& T1
Q
Insulation
r2
T∞ , h
r1
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Thin wall
Ts
Ts
ln(r2 r1 )
2πk
Rt ’
T∞
1
2πr2h
An insulated cylindrical pipe exposed
to convection from the outer surface
and the thermal resistance network
associated with it.
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• To find rcr, set the overall thermal resistance dRt’/dr = 0 and solve for r:
R′t =
ln(r ri )
1
+
2πk
2πrh
ri = inner radius
dR′t
1
1
=
−
=0
2
dr
2πkr 2πr h
k
r = rcr =
h
Similarly for a sphere
2k
rcr =
h
• For
For insulation thickness less that r
insulation thickness less that rcr the heat loss increases the heat loss increases
with increasing r and for insulation thickness greater that rcr the heat loss decreases with increasing r
• If k = 0.03 W/(m∙K) and h = 10 W/(m2∙K):
k 0.03 W/(m·K)
= 0.003m = 3mm
– cylinder rcr = =
2
h
10 W/(m
( ·K))
2k
= 6mm
– sphere rcr =
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h
Values of r1, h and k are constant
To see the condition maximizes or
minimizes the total resistance
d 2 R total
dr2 2
=−
1
2πkr22
+
1
πr23 h
At r2=k/h
Total thermal resistance per unit length
d 2 R total
⎛r ⎞
ln⎜⎜ 2 ⎟⎟
r
1
= ⎝ 1⎠+
2πk
2πr2 h
dr2 2
R total
Heat transfer p
per unit length
g
1
⎛1 1 ⎞
− ⎟=
>0
⎜
2 k 2k
3 2
⎠ 2πk h
π(k h ) ⎝
1
Always positive, total resistance
at k/h is minimum
•
Q T∞ − Ti
=
L
Rtotal
Optimum thickness is associated with r2, dRtotal
1
1
−
=0
2πkr2 2πr22 h
=
dr2
r2 =
k
h
rcr ,cylinder =
=0
k
h
(m)
rcr , max =
k max, insulation
h min
≈
(
0 .05 W m o C
(
5 W m 2o C
)
) = 0.01m = 1cm
We can insulate hot water pipes and steam lines without
worrying the critical radius of insulation
Insulation of electric wires:
-Radius of electric wires may be smaller than the critical radius
-Addition
Addition of insulation material increases heat transfer
Critical radius of insulation for spherical shell:
rcr ,sphere =
2k
h
Summary
• Table 3.3
Table 3 3
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1D Conduction with Heat
Generation
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•
2
d T
q
+
= 0
2
dx
k
•
T (x ) = −
q 2
x + C1x + C 2
2k
Boundaryy conditions:
T (− L ) = T s ,1
T (L ) = T s , 2
•
qL 2
T (x) =
2k
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C1 =
T s , 2 − T s ,1
•
C
2
2L
T s , 2 + T s ,1
q
2
=
L +
2k
2
T s , 2 + T s ,1
⎛
x 2 ⎞ T s , 2 − T s ,1 x
⎜⎜ 1 − 2 ⎟⎟ +
+
2
2
L ⎠
L
⎝
Ts ,1 = Ts , 2 ≡ Ts
•
qL 2
T (x) =
2k
⎛
x2 ⎞
⎜⎜ 1 − 2 ⎟⎟ + T s
L ⎠
⎝
•
qL
L2
T (0) ≡ To =
+ Ts
2k
Put x = 0
If the surface temperature of the heat generating body is
unknown and the surrounding fluid temperature is T∞
Find temperature gradient
dT
| x = L = h(Ts − T∞ ) • from the above Eq.
Using
g energy
gy balance − k
q at x = L
d
dx
We can obtain the surface temperature
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qL
Ts = T∞ +
h
•
1 d ⎛ dT ⎞ q
⎟+ = 0
⎜r
r dr ⎝ dr ⎠ k
•
dT
q r2
r
=−
+ C1
d
dr
2k
•
q r2
T (r ) = −
+ C 1 ln r + C 2
4k
•
Boundary conditions:
dT
|r = 0 = 0
dr
T ( ro ) = T s
•
•
C1 = 0
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q ro
T (r ) =
4k
C 2 = Ts +
q ro
4k
2
(
2
⎛
r2 ⎞
⎜⎜ 1 − 2 ⎟⎟ + T s
ro ⎠
⎝
)
q Π ro2 L = h ( 2 Π ro L )( T s − T ∞ )
•
q ro
Ts = T∞ +
2h