2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
2.3 Thermal resistance and
the electrical analogy
Electrical analogy
‰
Heat conduction for flat plate
Q=
ΔT
ΔT
=
L / kA Rt
L
Thermal resistance of
flat plate conduction
kA
Heat convection
Rt =
‰
2.3 Thermal resistance and the electrical
analogy
Q=
Tbody − T∞
1 / Ah
Rt =
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2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
1
Ah
‰
‰
heat transfer through the contact plane between two solid surfaces
‰
Surface finish
Material
Pressure
Substance in the interstitial spaces
temperature
The corresponding thermal resistance
1
Rt =
Ah
c
Heat flux
Q=
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2.3 Thermal resistance and
the electrical analogy
College of Energy and Power Engineering JHH
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
1 ∂ ⎛ ∂T ⎞ 1 ∂ T ∂ T q 1 ∂T
+ 2 + =
⎜r
⎟+ 2
2
r ∂r ⎝ ∂r ⎠ r ∂
∂
z
k α
∂t
θ
N
N
ƒ Eqs.
1 ∂ ⎛ ∂T ⎞
⎜r
⎟=0
r ∂r ⎝ ∂r ⎠
ƒ B.C.
⎧T (r = ri ) = Ti
⎨
⎩T (r = ro ) = To
=0
2.3 Thermal resistance and
the electrical analogy
ƒ Heat flux (for a cylinder of length l )
2
=0
Q = −(2π rl )k
=0
Ti − To
∂T
=
∂r ln(ro / ri ) /(2π lk )
ƒ Thermal resistance for cylinder conduction
Rt =
ln(ro / ri )
2π lk
T − Ti
ln(r / ri )
=
To − Ti ln(ro / ri )
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4
Resistances for cylinders and for convection
Radial heat conduction in a tube
ƒ Solution
ΔT
ΔT
=
Rtotal L / kA + 1 / hc A + L / kA
3
Resistances for cylinders and for convection
2
2.3 Thermal resistance and
the electrical analogy
Contact surface can be treated as a
surface with an interfacial
conductance hc W/m2K
ƒ
ƒ
ƒ
ƒ
ƒ
‰
2
Contact resistance
No two solid surface will ever form perfect thermal
contact
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
Thermal resistance of
convection
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
Contact resistance
‰
ΔT
Rt
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1
2.3 Thermal resistance and
the electrical analogy
=
0
5
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2
4
0 1 2
6
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
2.3 Thermal resistance and
the electrical analogy
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
Resistances for cylinders and for convection
‰
Resistances for cylinders and for convection
ƒ Bi>>1
Heat transfer through a
convective boundary condition
ƒ Eqs.
ƒ B.C.
problem of first kind b.c.’s
¾ Bi>20 means we can neglect convection resistance with about 5%
error
1 ∂ ⎛ ∂T ⎞
⎟=0
⎜r
r ∂r ⎝ ∂r ⎠
⎧T (r = ri ) = Ti
⎪
∂T
⎨
⎪h (T − T∞ ) r=ro = −k ∂r
⎩
ƒ Bi<<1
problem of lumped-capacity
¾ Bi<0.1 signals constancy of temperature inside the cylinder with
about ±3%
∂T
∂r
ΔT
=
1 / h2πro l + ln(ro / ri ) /(2πlk )
Q = −(2πrl )k
ƒ Heat flux
r =ro
ƒ Solution
T − Ti
ln(r / ri )
=
T∞ − Ti 1/ Bi + ln(ro / ri )
Bi =
=
hro
k
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2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
2.3 Thermal resistance and
the electrical analogy
ΔT
Rtconv + Rtcond
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7
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
Critical radius of insulation
‰
2.3 Thermal resistance and
the electrical analogy
8
2.3 Thermal resistance and
the electrical analogy
Critical radius of insulation
Example: to insulate a 0.0025m O.R. copper stream
line with 85 magnesia to prevent the steam from
condensing too rapidly. (kmage=0.074W/m·K,
h=20W/m2K)
‰
Find the radius for minimum thermal resistance
dRtotal
d
=
dro
dro
⎛ 1
ln(ro / ri ) ⎞
1
1
⎟
⎜
⎜ 2πr h + 2πk ⎟ = − 2πr 2 h + 2πkr = 0
o
o
o
⎠
⎝
ƒ Critical radius
Copper
pipe
rcrit =
k
h
= 0.0037m
h , T∞
hrcrit
=1
k
ro < rcrit
or Bi =
steam
magnesia
When
Rtotal = Rtconv + Rtcond
ln(ro / ri )
1
=
+
2πk
2πro h
ro } | Rtconv ~
ro } | Rtcond }
College of Energy and Power Engineering JHH
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
insulation } | heat loss }
Until ro / ri = 2.32 or ro = 0.0058m the insulation dose not work
2.3 Thermal resistance and
the electrical analogy
Resistances for spherical shell
‰
1 ∂ ⎛ 2 ∂T ⎞
1 ⎛
∂T ⎞
1
∂ T q 1 ∂T
+ =
⎜r
⎟+
⎜ sin θ
⎟+
∂θ ⎠ r 2 sin 2 θ ∂φ 2 N
∂t
r 2 ∂r ⎝ ∂r ⎠ r 2 sin θ ⎝
k α
N
ƒ B.C.
=0
Q = −(4π r 2 )k
=0
1 ∂ ⎛ 2 ∂T ⎞
⎜r
⎟=0
r 2 ∂r ⎝ ∂r ⎠
Ti − To
∂T
=
∂r
1 ⎛1 1⎞
⎜ − ⎟
4π k ⎝ ri ro ⎠
ƒ Thermal resistance for spherical shell
⎧T (r = ri ) = Ti
⎨
⎩T (r = ro ) = To
ƒ Solution
2.3 Thermal resistance and
the electrical analogy
ƒ Heat flux
2
ƒ Eqs.
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
10
Resistances for spherical shell
Radial heat conduction in a spherical shell
=0
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9
Rt =
1 ⎛1 1⎞ 1 δ
⎜ − ⎟=
4π k ⎝ ri ro ⎠ π k di d o
1
where δ = (do − di )
2
T − Ti 1 ri − 1 r
=
To − Ti 1 ri − 1 ro
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11
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12
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
2.3 Thermal resistance and
the electrical analogy
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
Resistance for thermal radiation
‰
Resistance for convection and radiation
Radiant heat flux
‰
Qnet = A1 F1−2εσ (T14 − T24 ) = A1 F1−2εσ (T12 + T22 )(T1 + T2 )(T1 − T2 )
2 +( ΔT )2 / 2
=2Tm
=2Tm
An electrical resistor cooled by convection and
radiation
=ΔT
ΔT
ΔT
=
=
1
Rt
A1 F1−2εσ (2Tm2 + (ΔT ) 2 / 2)2Tm
ƒ When T1 and T2 are close,
2.3 Thermal resistance and
the electrical analogy
1
1
1
=
+
Rtequiv Rtrad Rtconv
(ΔT ) 2 / 2 << 2Tm2
Q=
Qnet = A1 (4 F1−2εσTm3 )ΔT
Tresistor − Tair
Rtequiv
≡hrad
Rtrad
1
=
A1 hrad
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13
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
14
2.4 Overall heat transfer
coefficient, U
Definition
‰
Overall heat transfer coefficient
U=
Q = UAΔT
Q = UAΔT =
2.4 Overall heat transfer coefficient, U
College of Energy and Power Engineering JHH
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
U=
‰
2.4 Overall heat transfer
coefficient, U
Heat transfer through a composite wall.
ΔT T flame − Tbolling water
Q=
=
L
1
1
∑ Rt
+
+
h A kA hb A
Q=
=
Q
1
=
AΔT 1 + L + 1
h k hb
1
Rttotal = Rtconv +
h ≈ 200 W/m 2 K
(
L = 0.001 m k Al = 160 W/m ⋅ K
hb ≈ 5000 W/m 2 K
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16
Example of overall heat transfer coefficient
Heat transfer through the bottom of a tea kettle
U=
1
Rt A
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
Example of overall heat transfer coefficient
‰
ΔT ΔT
=
1
Rt
UA
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15
2.4 Overall heat transfer
coefficient, U
Q
AΔT
17
1
1
+
)
Rt pin Rt sawdust
+ Rtconv
U=
ΔT
∑ Rt
T∞l − T∞r
1
1
+
k
A
⎛
hA
k A
⎜⎜ p p + s s
L
⎝ L
⎞
⎟⎟kA
⎠
+
1
hA
Q
1
=
2
1
AΔT
+
h ⎛ k p Ap ks As ⎞
+
⎜
⎟
LA ⎠
⎝ LA
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18
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
2.4 Overall heat transfer
coefficient, U
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
Typical values of U
Fouling resistance
‰
The fouling of a pipe
U based on Ai =
1
ri ln(ro / rp ) ri ln(rp / ri )
ri
1
+
+
+ Rf +
ro ho
kinsul
k pipe
hi
ƒ Fouling resistance for a unit area of pipe
College of Energy and Power Engineering JHH
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
2.4 Overall heat transfer
coefficient, U
1
1
−
U old U new
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19
2.4 Overall heat transfer
coefficient, U
Rf ≡
2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
Fouling resistance
20
2.5 Summary
Summary
‰
What have been done
‰
What have not considered
ƒ Heat diffusion equation
ƒ How to determine h
ƒ Electric analogy to steady
heat flow
ƒ How to design heat
exchanger
ƒ Overall heat transfer
coefficient
ƒ h and U vary with position
ƒ Some practical problem
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2. Heat conduction concepts, thermal resistance,
and the overall heat transfer coefficient
21
2.5 Summary
Homework
‰
‰
‰
‰
‰
‰
2.8
2.9
2.10
2.20
2.30
2.35
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23
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22