Download Heat Transfer by Conduction

Heat Transfer and Its Applications
Introduction
1
Heat Exchange
Insulation
2
Nature of heat flow
Heat flows from the object at the
higher temperature to that at the
lower temperature.
How it happens?
3
Three mechanisms :
Conduction (热传导)
Convection (对流传热)
Radiation (热辐射)
4
Heat Transfer by
Conduction
5
Conduction
If a temperature gradient exists in a
continuous substance, heat can flow
unaccompanied by any observable
motion of matter.
Qualitative(定性的)
Quantitative(定量的)
6
Heat transfer by conduction
Assumption:
considering heat flow in homogeneous
( 均一的 ) isotropic( 孤立的 ) solids because
in these there is no convection and the
effect of radiation is negligible.
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Math. Model by Conduction
According to Fourier’s law（傅里叶导热定
律）, the heat flux is proportional to the
temperature gradient and opposite to
x
it in sign.
For one-dimensional heat flow
q
dq
dT
 k
dA
dx
(4.2-1)
Where:
q: heat transfer rate, W=J/s
A: heat transfer area, m2
dq/dA: heat transfer flux, W/m2
T: temperature, K or oC
x: distance, m
dT/dx: temperature gradient, K/m
k: thermal conductivity, W/(m∙K)
8
Thermal conductivity k
The proportionality constant k is a physical property
of the substance. The unit of k are W/(m∙℃) or
W/(m∙K).
For small ranges of temperature, k is constant
For larger temperature ranges, k is approximated by
k  a  bT
(4.2-5)
k vary over a wide range. They are highest for metals and
lowest for finely powdered materials from which air has
been evacuated (排空).
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conductor
k of metal cover a wide range of values, from about
17W/m∙ºC for stainless steel (不锈钢) and 45W/m∙ºC
for mild steel (低碳钢) , to 380W/m∙ºC for copper
and 415W/m∙ºC for silver
For glass and most nonporous materials, the
thermal conductivities are much lower, from
about 0.35 to 3.5.
For most liquid k is lower than that
for solids, with typical values of about
0.17. k decreases by 3 ~ 4 %t for a 10
ºC rise in temperature, except water.
insulator
Gases have the smallest thermal
conductivities, with values as low
as 0.007. For air at 0ºC, k is 0.024
W/m·ºC.
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metals：
thermal conduction results from the motion of free
electrons.
In
conductor
solids ：
poor conductor of electricity, produced by
momentum transport
In
In
most liquids:
thermal conduction results from
momentum transfer between adjacent
vibrating(振动) molecules or atoms.
In
gases:
conduction occurs by the random
motion of molecules.
insulator
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Steady-state conduction
Limiting Condition
Conduction under the condition of constant temperature
distribution is called steady-state conduction. In the steady
state, T is a function of position only, and the rate of heat
flow at any point is a constant.
Heat flow into the tank
Heat flow from the tank
12
Since in steady state there can be neither
accumulation nor depletion of heat within the
slab, q is constant along the path of heat flow.
If x is distance from the hot side.
T
For steady one-dimensional
flow.Eq.(4.2-1) may be written
q
dT   dx
kA
q
(4.2-7)
x
13
Since the only variables in this Eq. are x and T,
direct integration gives
T1  T2
q
T
k
k
A
x2  x1
B
(4.2-8)
Where x2-x1=B= thickness of layer of insulation
T1-T2=ΔT=temperature drop across layer
When k varies linearly with temperature, Eq.(4.2-5), still can be
used rigorously by taking an average value k.
14
Equation (4.2-8) can be written in the form
q T

A R
(4.2-9)
Where R=B/k is thermal resistance between points 1 and 2.
m2∙℃/W
15
Compound resistance in series
16
Compound resistance in series
Consider a flat wall constructed of a series of layers, as
shown in Fig.4.2
Assume that the layers are in excellent
thermal contact, so that no temperature
difference exists across the interfaces
between the layers.
Then, if ΔT is the total temperature
drop across the entire wall
ΔT= ΔTA+ ΔTB+ΔTC
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According to Eq.(4.2-8)
BA
TA  q A
kA A
TA  TB  TC 
BC
BB
TC  qC
TB  qB
kC A
kB A
q A BA qB BB qC BC


 T
Ak A
Ak B
AkC
q A  qB  qC  q
T TA TB TC



R
RA
RB
RC
(4.2-12)
q
T

A BA k A  BB k B  BC kC

T
T

RA  RB  RC
R
(4.2-11)
Where RA, RB and RC=resistance of individual layers
R=overall resistance
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Example 4.2
A flat furnace wall is constructed of 120-mm layer of sil-ocel brick, with a thermal conductivity 0.08 W/(m·℃)，
backed by a 150-mm of common brick, of conductivity 0.8
W/(m·℃), the temperature of inner face of the wall is
1400℃, and that of the outer face is 200℃
BA=0.12m BB=0.15m
To=200℃
Ti=1400℃
T2
kB=0. 8 W/(m∙℃)
kA=0.08 W/(m∙℃)
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Example 4.2
(a) What is the heat loss through the wall, in W per square
meter
T  T0  T3  1400  200  1200℃
BA=0.12m BB=0.15m
T3=200℃
T0=1400℃
T2


RA 
BA 0.12

 1.5 m2 ℃/W
k A 0.08
RB 
BB 0.15

 0.1875 m2 ℃/W
0.8
kB


According to Eq.(4.2-9)
kB=0. 8 W/(m∙℃)
kA=0.08 W/(m∙℃)
1200
q
T

 711.1W m 2 

1.5  0.1875
A RA  RB
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Example 4.2
b) To reduce the heat loss to 600 W/m2 by adding a layer
of cork (软木) with k= 0.2 W/(m·℃) on the outside of
common brick, how many meters of cork are required?
BA=0.12m BB=0.15m BC=? m

q
 600 W m 2
A

According to Eq.(4.2-9)
q
T
1200


 600 W m 2
A RA  RB  RC 1.5  0.1875  RC

To=200℃
Ti=1400℃
T2
T3


BC BC
RC 

 0.3125 m2 ℃/W
k C 0.2
kC=0. 2 W/(m∙℃) B  0.0625m  62.5mm
C
kB=0. 8 W/(m∙℃)
kA=0.08 W/(m∙℃)
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
Heat flow through a cylinder
Consider the hollow cylinder represented by Fig.4.3
The inside radius of the cylinder is ri, the outside
radius is ro, and the length of the cylinder is L.
ri, Ti
The temperature of outside surface is To, and the
that of inside surface is Ti. The thermal
conductivity of the material of which the cylinder
is made is k.
L
ro, T0
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It is desired to calculate the rate of heat flow outward for this case.
At radius r from the center, the heat flow rate is
q and the area through which it flows is A.
At steady state the heat flow rate is constant.
Since heat flows only in the r
direction, Eq. (4.2-7) becomes
ri, Ti
r
L
q
q
dT

 k
A 2rL
dr
(4.2-13)
ro, T0
23
Rearranging Eq(4.2-13) and integrating between limits gives

ro
ri
dr 2Lk To
2Lk

dT 

T
i
r
q
q
2Lk
ln ro  ln ri 
q

Ti
To
dT
r
Ti  To 
2Lk Ti  To 
q
ln ro ri 
ri, Ti
L
(4.2-14)
the logarithmic (对数) mean
rL 
ro  ri
ln ro ri 
q
(4.2-15)
k AL Ti  To 
ro  ri
AL 
2Lro  ri 
ln ro ri 
(4.2-16)
ro, T0
(4.2-17)
24
The logarithmic mean is less convenient than the arithmetic (算术)
mean, and the latter can be used without appreciable error for
thin-walled tubes, where ro/ri is nearly 1.
96%
rL 
The ratio of the logarithmic mean to the arithmetic
mean is a function of ro/ri as shown in Fig4.3
ro  ri
ln ro ri 
(4.2-15)
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Formula Derivation
r1
r0 k 2
2 1
k1
r2
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Formula Derivation
r1
r0 k 2
2 1
k1
r2
According to Eq.(4.2-14)
q1 
2Lk1 T1  T0 
ln r0 r1 
T1  T0  T1
q2 
2Lk 2 T2  T1 
ln r1 r2 
T2  T1  T2
Steady-state conduction
q  q1  q2
 ln ri 1 ri 
q1 ln r0 r1  q2 ln r1 r2 
 q 

T  T1  T2 

2

Lk
2Lk1
2Lk 2
i
i 

Heat loss per meter for multi-layer cylinder
q

L
T
ln ri 1 ri 
i 2k
i
27
Exercise
A steel tube with dimension Ф60 ×3mm is backed by a 30-mm
thick cork, with a thermal conductivity of 0.043 W/(m·℃) , and
is followed by a 40-mm layer of some insulating material, of
0.07 W/(m·℃) . The temperature of the outer face of steel tube
is -110℃, and that of the outer face of insulating material is 10℃.
What is the heat loss per meter tube through the wall?
28
Summary
Conduction—If a temperature gradient exists in a continuous
substance, heat can flow unaccompanied by any observable
motion of matter.
For steady one-dimensional conduction
q
T

A BA k A  BB k B  BC kC
(4.2-11)
Additional formula
q

L
T
ln ri 1 ri 
i 2k
i
Uniform units
Draw schematic diagram
29
Problem 4.1
A layer of pulverized cork 152-mm thick is used as a layer of
thermal insulation in a flat wall. The temperature of the cold
side of the cork is 4.4 ℃, and that of the warm side is 82.2 ℃.
The thermal conductivity of the cork at 0 ℃ is 0.036 W/(m ·℃),
and that at 93.3℃ is 0.055 W/(m ·℃). The area of the wall is
2.32m2. What is the rate of heat flow through the wall in watts?
30
Steps & traps
1、draw a schematic diagram
2、write out the known quantity and unify the units
3、analysis the relationship between the known’s and unknown’s
4、be careful with some details such as diameter, temperature
dependent physical properties, average temperature
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Problem 4.3
A tube of 60-mm outer diameter (OD) is insulated with a 50mm layer of silica foam, for which the conductivity is 0.055
W/(m · ℃ ), followed with a 40-mm layer of cork with a
conductivity of 0.05 W/(m ·℃). If the temperature of the outer
surface of the pipe is 150℃ and the temperature of the outer
surface of the cork is 30℃, calculate the heat loss in watts per
meter of pipe.
32
Problem 4.5
The outer diameter of a steel tube is 150mm. The tube wall
is backed by two insulating layers to reduce the heat loss.
The ratio of thermal conductivity of two insulating
materials is k2/k1=2 and both insulating materials have the
same thickness 30mm. If the temperature difference
between pipe wall and outer surface of insulating material
is constant, which insulating material should be backed
inside to enable less heat loss?
33