(6 7)-1-d-ss-conduction-part2
TRANSCRIPT
1D STEADY STATE HEAT CONDUCTION (2)CONDUCTION (2)
Prabal TalukdarPrabal TalukdarAssociate Professor
Department of Mechanical EngineeringDepartment of Mechanical EngineeringIIT Delhi
E-mail: [email protected]
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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
Consider heat transfer through two metal rods of cross-sectional area A thatare pressed against each other Heat transfer through the interface of these twoare pressed against each other. Heat transfer through the interface of these tworods is the sum of the heat transfers through the solid contact spots and thegaps in the noncontact areas and can be expressed as
Most experimentally determinedMost experimentally determinedvalues 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 byerfaceint
c TAQh
Δ=
•
(W/m2 oC)
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It is related to thermal contact resistance by
AQ
Th1R erfaceint
cc •
Δ== (m2 oC/W)
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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Two parallel layersTwo parallel layers
⎞⎛ 11TTTT⎟⎟⎠
⎞⎜⎜⎝
⎛+−=
−+
−=+=
•••
2121
2
21
1
2121
11)(RR
TTR
TTR
TTQQQ
L L
totalRTTQ 21 −
=•
'11
11 Ak
LR = '22
22 Ak
LR =
PTalukdar/Mech-IITD21
111RRR total
+=21
21
RRRRR total +
=where
Combined series-parallelCombined series parallelTTQ ∞
• −= 1
totalRQ =
RRconvconvtotal RR
RRRRRRRR +++
=++= 321
21312
'11
11 Ak
LR = '22
22 Ak
LR = '33
33 Ak
LR =
1
3
1hA
R conv =
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Series and parallel composite wall and its thermal circuit
RA
RD
T∞1
AR∞1
RB
RC RE
RF
R∞2
T∞2
T T TT2C1 R
1111R
111RR ∞∞ +
⎟⎞
⎜⎛
++⎟⎞
⎜⎛
+=∑T1 T3 T4T2
FEDBA R1
R1
R1
R1
R1
⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+
TUAQ Δ=•
(W)
where U is the overall heat transfer coefficient
PTalukdar/Mech-IITD totalRUA 1
=
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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Heat conduction in cylinderHeat conduction in cylinderdTkAQ cylcond −=
•
L2AdrQ cyl,cond rL2A π=
kdTdrQ
22 Tcyl,condr
•
∫=∫
Substituting A = 2πrL and performing the integrations give
kdTdrA
2
1
2
1 TTrr == ∫−=∫
)rrln(TTLk2Q
12
21cyl,cond
−π=
•cyl,condQ
•
=constant at steady state
cyl
21cyl,cond R
TTQ −=
•(W)
PTalukdar/Mech-IITD Lk2)rrln(R 12
cyl π= ln(outer radius/inner radius)
2π(length)(thermal conductivity)=
Heat conduction in sphereHeat conduction in sphere
For sphere sph
21sphere,cond R
TTQ −=&
krr4rrR21
12sph π
−= = outer radius - inner radius
4π(outer radius)(inner radius)(thermal conductivity)
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Resistance NetworkResistance Networkcylindrical
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
( ) 22
221
12
12
1 h)r4(krr4hr4 π+
π+
π=
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spherical) shell subjected to convection fromboth the inner and the outer sides.
Multilayered cylinderMultilayered cylinder
2,conv3,cyl2,cyl1,cyl1,convtotal RRRRRR ++++=
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( ) ( ) ( )423
34
2
23
1
12
11 Ah1
Lk2rrln
Lk2rrln
Lk2rrln
Ah1
+π
+π
+π
+=
Radial heat conduction through cylindrical systems
tTCg
zT.k
zTr.k
r1
rTr.k
rr1
2 ∂∂ρ=+⎟
⎠⎞
⎜⎝⎛
∂∂
∂∂+⎟⎟
⎠
⎞⎜⎜⎝
⎛φ∂∂
φ∂∂+⎟
⎠⎞
⎜⎝⎛
∂∂
∂∂
&
0drdTr
drd
=⎟⎠⎞
⎜⎝⎛
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
rrln
rlnTrlnTrln
rrln
TTT
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⎠⎝⎠⎝ 11 rr
Cd
1
11rrr r
C.Lr2.k|drdTkAQ
1π−=−= =
⎟⎟⎠
⎞⎜⎜⎝
⎛−π−=
1
21
121
rrlnr
1).TT.(Lr2.kQ
⎠⎝ 1
⎟⎟⎠
⎞⎜⎜⎝
⎛−π
=2
21rln
)TT(kL2
⎟⎠
⎜⎝ 1r
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Critical Radius of InsulationCritical Radius of Insulation
1. Steady state conditions
2. One-dimensional heat flow only in the radial direction
r2 Insulation
T h3. Negligible thermal resistance due to
cylinder wall
4 Negligible radiation exchange
r1 Thin wallT∞ , h
T4. Negligible radiation exchange between outer surface of insulation and surroundings
Ts
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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, thereby p y , yincreasing 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 InsulationT1 T s T∞Q&
r2
r1
Insulation
Thin wallT∞ , h hr2
12π
( )k2rrln 12
π1 Thin wall
TsRt’
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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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• To find rcr, set the overall thermal resistance dRt’/dr = 0 and solve for r:
( )rh2
1k2rrlnR i
t π+
π=′ ri = inner radius
0hr2
1kr2
1drRd
2t =
π−
π=
′
k k2
• For insulation thickness less that r the heat loss increases
hkrr cr ==
hk2rcr =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/(m10W/(m·K)03.0
hkr 2cr ====
– sphere
)(
mm6hk2rcr ==
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Values of r1, h and k are constant
To see the condition maximizes or minimizes the total resistance
11Rd2
At r =k/h
hr1
kr21
drRd
32
22
22
total2
π+
π−=
Total thermal resistance per unit length
At r2=k/h
rln 2⎟⎟⎞
⎜⎜⎛ ( )
0hk2
1k21
k1
hk1
drRd
23222
total2
>π
=⎟⎠⎞
⎜⎝⎛ −
π=
Heat transfer per unit length
Always positive, total resistance at k/h is minimumhr2
1k2r
lnR
2
1total π
+π
⎟⎟⎠
⎜⎜⎝=
iTTQ −= ∞
•
kr = (m)p g
Optimum thickness is associated with r2,
totalRL
02
=dr
dRtotal
hr cylinder,cr = (m)
2dr0
hr21
kr21
222
=π
−π h
kr2 =
( )( ) cm1m01.0
CmW5CmW05.0
hk
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: hk2r sphere,cr =
Summary• Table 3 3
SummaryTable 3.3
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1D Conduction with Heat Generation
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02
+
•
qTd 02 =+kdx
( ) 2 CxCxqxT ++=
•
Boundary conditions:
( ) 212CxCx
kxT ++−=
y
( ) 1,sTLT =−L
TTC ss
21,2,
1
−=
( ) 2,sTLT =
L2
221,2,2
2ss TT
Lk
qC+
+=
•
•
PTalukdar/Mech-IITD 221
2)( 1,2,1,2,
2
22ssss TT
LxTT
Lx
kqLxT
++
−+⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
•
sss TTT ≡= 2,1,
xqL ⎟⎞
⎜⎛
•22
1)( sTLx
kqLxT +⎟⎟
⎠
⎞⎜⎜⎝
⎛−= 21
2)(
L•
2
If the surface temperature of the heat generating body is
Put x = 0so Tk
qLTT +=≡2
)0(2
If the surface temperature of the heat generating body is unknown and the surrounding fluid temperature is T∞
Using energy balanceFind temperature gradientfrom the above Eq. at x = L)(| ∞= −=− TTh
ddTk sLx •g gy
We can obtain the surface temperature
q
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dx
hLqTTs
•
∞ +=
01=+⎟
⎠⎞
⎜⎝⎛
•
kq
drdTr
drd
r ⎠⎝ kdrdrr
1
2
2C
krq
ddTr +−=
•
12kdr
21
2
ln4
)( CrCkrqrT ++−=
•
Boundary conditions:
4k
dT TT )( so TrkrqrT +⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
•
2
22
14
)(0| 0 ==rdr
dT so TrT =)(
rqTC o2
•
ork ⎟⎠
⎜⎝4
( ) ))(2(2∞
•
−Π=Π TTLrhLrq soo
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01 =C kqTC o
s 42 +=
hrqTT o
s 2
•
∞ +=