HOA1&2 - Module 3 - PREHISTORCI ARCHITECTURE OF KERALA.pptx
(6 7)-1-d-ss-conduction-part2
1. 1D STEADY STATE HEAT
CONDUCTION (2)CONDUCTION (2)
Prabal TalukdarPrabal Talukdar
Associate Professor
Department of Mechanical EngineeringDepartment of Mechanical Engineering
IIT Delhi
E-mail: prabal@mech.iitd.ac.inp
PTalukdar/Mech-IITD
2. Thermal Contact ResistanceThermal Contact Resistance
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Temperature distribution and heat flow lines along two solid plates
pressed against each other for the case of perfect and imperfect contact
3. Consider heat transfer through two metal rods of cross-sectional area A that
are pressed against each other Heat transfer through the interface of these twoare pressed against each 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
Most experimentally determinedMost experimentally determined
values of the thermal contact resistance fall
between 0.000005 and 0.0005 m2∙°C/W (the
corresponding range of thermal contact
gapcontact QQQ
•••
+=
•
where A is the apparent interface area (which is the same as the cross-
sectional area of the rods) and ΔT is the effective temperature
conductance is 2000 to 200,000 W/m2∙°C).erfaceintc TAhQ Δ=
sectional 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
It is related to thermal contact resistance by
erfaceint
c
T
AQ
h
Δ
=
•
(W/m2 oC)
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It is related to thermal contact resistance by
AQ
T
h
1
R erfaceint
c
c •
Δ
== (m2 oC/W)
4. Importance of considerationImportance of consideration
Th th l t t i tThe thermal contact resistance range:
between 0.000005 and 0.0005 m2·°C/W
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5. Two parallel layersTwo parallel layers
⎞⎛ 11TTTT
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−=
−
+
−
=+=
•••
21
21
2
21
1
21
21
11
)(
RR
TT
R
TT
R
TT
QQQ
L L
totalR
TT
Q 21 −
=
•
'
11
1
1
Ak
L
R = '
22
2
2
Ak
L
R =
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21
111
RRR total
+=
21
21
RR
RR
R total
+
=where
6. Combined series-parallelCombined series parallel
TT
Q ∞
• −
= 1
totalR
Q =
RR
convconvtotal RR
RR
RR
RRRR ++
+
=++= 3
21
21
312
'
11
1
1
Ak
L
R = '
22
2
2
Ak
L
R = '
33
3
3
Ak
L
R =
1
3
1
hA
R conv =
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7. Series and parallel composite wall and its thermal circuit
RA
RD
T∞1
A
R∞1
RB
RC RE
RF
R∞2
T∞2
T T TT
2C1 R
111
1
R
11
1
RR ∞∞ +
⎟
⎞
⎜
⎛
++
⎟
⎞
⎜
⎛
+=∑
T1 T3 T4T2
FEDBA R
1
R
1
R
1
R
1
R
1
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
++⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+
TUAQ Δ=
•
(W)
where U is the overall heat transfer coefficient
PTalukdar/Mech-IITD totalR
UA
1
=
8. Complex multi-dimensional problems as 1-D problems
1 Any plane wall normal to the x-axis is isothermal1. Any plane wall normal to the x-axis is isothermal
2. Any plane parallel to x-axis is adiabatic
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9. Heat conduction in cylinderHeat conduction in cylinder
dT
kAQ cylcond −=
•
L2Adr
Q cyl,cond rL2A π=
kdTdr
Q
22 Tcyl,condr
•
∫=∫
Substituting A = 2πrL and performing the integrations give
kdTdr
A
2
1
2
1 TTrr ==
∫−=∫
)rrln(
TT
Lk2Q
12
21
cyl,cond
−
π=
•
cyl,condQ
•
=constant at steady state
cyl
21
cyl,cond
R
TT
Q
−
=
•
(W)
PTalukdar/Mech-IITD Lk2
)rrln(
R 12
cyl
π
=
ln(outer radius/inner radius)
2π(length)(thermal conductivity)
=
10. Heat conduction in sphereHeat conduction in sphere
For sphere sph
21
sphere,cond
R
TT
Q
−
=&
krr4
rr
R
21
12
sph
π
−
= =
outer radius - inner radius
4π(outer radius)(inner radius)(thermal conductivity)
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11. Resistance NetworkResistance Network
cylindrical
2,convcond1,convtotal RRRR ++=
( )
( )12 1rrln1
++=
spherical
( ) 2211 h)Lr2(Lk2hLr2 π
+
π
+
π
2,convsph1,convtotal RRRR ++=
( )
12 1rr1
+
−
+
The thermal resistance network for a cylindrical (or
spherical) shell subjected to convection from
( ) 2
2
221
12
1
2
1 h)r4(krr4hr4 π
+
π
+
π
=
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spherical) shell subjected to convection from
both the inner and the outer sides.
13. Radial heat conduction through cylindrical systems
t
T
Cg
z
T
.k
z
T
r.k
r
1
r
T
r.k
rr
1
2
∂
∂
ρ=+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
φ∂
∂
φ∂
∂
+⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
&
0
dr
dT
r
dr
d
=⎟
⎠
⎞
⎜
⎝
⎛
Integrating the above equation twice T=C ln r + C
⎠⎝
Subject to the boundary conditions, T=T1 at r = r1 and T=T2 at r = r2
Integrating the above equation twice, T=C1ln r + C2
( )
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
=
1
2
1221
1
2
12
r
r
ln
rlnTrlnT
rln
r
r
ln
TT
T
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⎠⎝⎠⎝ 11 rr
15. Critical Radius of InsulationCritical Radius of Insulation
1. Steady state conditions
2. One-dimensional heat flow only in
the radial direction
r2
Insulation
T h
3. Negligible thermal resistance due to
cylinder wall
4 Negligible radiation exchange
r1 Thin wall
T∞ , h
T4. Negligible radiation exchange
between outer surface of insulation
and surroundings
Ts
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16. Critical Radius of InsulationCritical Radius of Insulation
• Practically, it turns out that adding insulation in cylindrical and spherical
exposed walls can initially cause the thermal resistance to decrease, therebyp y , 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:
r Insulation
T1 T s T∞Q&
r2
r1
Insulation
Thin wall
T∞ , h hr2
1
2π
( )
k2
rrln 12
π
1 Thin wall
Ts
Rt’
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17. An insulated cylindrical pipe exposed
to convection from the outer surface
and the thermal resistance networkand the thermal resistance network
associated with it.
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18. • To find rcr, set the overall thermal resistance dRt’/dr = 0 and
solve for r:
( )
rh2
1
k2
rrln
R i
t
π
+
π
=′ ri = inner radius
0
hr2
1
kr2
1
dr
Rd
2
t =
π
−
π
=
′
k k2
• For insulation thickness less that r the heat loss increases
h
k
rr cr ==
h
k2
rcr =Similarly for a sphere
• For insulation thickness less that rcr 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):
– cylinder mm3m003.0
·K)W/(m10
W/(m·K)03.0
h
k
r
2cr ====
– sphere
)(
mm6
h
k2
rcr ==PTalukdar/Mech-IITD
19. Values of r1, h and k are constant
To see the condition maximizes or
minimizes the total resistance
11Rd2
At r =k/h
hr
1
kr2
1
dr
Rd
3
2
2
2
2
2
total
2
π
+
π
−=
Total thermal resistance per unit length
At r2=k/h
r
ln 2
⎟⎟
⎞
⎜⎜
⎛ ( )
0
hk2
1
k2
1
k
1
hk
1
dr
Rd
2322
2
total
2
>
π
=⎟
⎠
⎞
⎜
⎝
⎛
−
π
=
Heat transfer per unit length
Always positive, total resistance
at k/h is minimumhr2
1
k2
r
ln
R
2
1
total
π
+
π
⎟⎟
⎠
⎜⎜
⎝=
iTTQ −
= ∞
•
k
r = (m)p g
Optimum thickness is associated with r2
,
totalRL
0
2
=
dr
dRtotal
h
r cylinder,cr = (m)
2dr
0
hr2
1
kr2
1
2
22
=
π
−
π
h
k
r2 =
20. ( )
( ) cm1m01.0
CmW5
CmW05.0
h
k
r
o2
o
min
insulationmax,
max,cr ==≈=
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 of insulation material increases heat transfer-Addition of insulation material increases heat transfer
Critical radius of insulation for spherical shell: h
k2
r sphere,cr =
23. 0
2
+
•
qTd
02
=+
kdx
( ) 2
CxCx
q
xT ++=
•
Boundary conditions:
( ) 21
2
CxCx
k
xT ++−=
y
( ) 1,sTLT =−
L
TT
C ss
2
1,2,
1
−
=
( ) 2,sTLT =
L2
22
1,2,2
2
ss TT
L
k
q
C
+
+=
•
•
PTalukdar/Mech-IITD 22
1
2
)( 1,2,1,2,
2
22
ssss TT
L
xTT
L
x
k
qL
xT
+
+
−
+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=
•
24. sss TTT ≡= 2,1,
xqL
⎟
⎞
⎜
⎛
•
22
1)( sT
L
x
k
qL
xT +⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−= 2
1
2
)(
L
•
2
If the surface temperature of the heat generating body is
Put x = 0so T
k
qL
TT +=≡
2
)0(
2
If the surface temperature of the heat generating body is
unknown and the surrounding fluid temperature is T∞
Using energy balance
Find temperature gradient
from the above Eq. at x = L)(| ∞= −=− TTh
d
dT
k sLx •
g gy
We can obtain the surface temperature
q
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dx
h
Lq
TTs
•
∞ +=
25. 0
1
=+⎟
⎠
⎞
⎜
⎝
⎛
•
k
q
dr
dT
r
dr
d
r ⎠⎝ kdrdrr
1
2
2
C
k
rq
d
dT
r +−=
•
1
2kdr
21
2
ln
4
)( CrC
k
rq
rT ++−=
•
Boundary conditions:
4k
dT TT )(
s
o
T
r
k
rq
rT +⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−=
•
2
22
1
4
)(
0| 0 ==r
dr
dT so TrT =)(
rq
TC o
2
•
ork ⎟
⎠
⎜
⎝4
( ) ))(2(2
∞
•
−Π=Π TTLrhLrq soo
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01 =C
k
q
TC o
s
4
2 +=
h
rq
TT o
s
2
•
∞ +=
