Consider the end part of a typical process furnace shown in ceramic tube. The inner muffle extends beyond the furnace to a certain length L. The part of the tube inside the furnace is at a higher temperaturetf while the part which extends beyond is exposed to the atmospheric temperature to.
Heat will be conducted out of the furnace by the extension and transferred to the surrounding by convection and radia-tion. Thus, a temperature gradient will be set up in the exten-sion as shown.
If the extension (l) is sufficiently long, the tip (at x = l) will be at atmospheric temperature and there will be no transfer of est and so the heat transfer at base will be maximum. For an extension of limited length, heat will be lost from the tip.
There are many such instances involving extended sur-faces that we will come across. These extensions, when incor-porated intentionally, are called “fins.” They are used to dissipate the heat generated in a limited volume such as inter-nal combustion engines and gear boxes.
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These results are plotted in Figure 3.28(B). This shows
Figure 3.29. The furnace has an inner muffle in the form of a
heat at the end. At the base (x = 0) the temperature is the
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There are two problems associated with extensions that are of our interest.
1. What should be the minimum length of the extension to assure a desired temperature at the tip?
2. How much heat will be lost through such an extension (end losses)?
The differential equation for temperature distribution in the extension is based on heat balance. Heat conducted by the extension cross section will be dissipated to the atmosphere by transfer (convection + radiation) through the surface.
Assume that the heat transfer coefficient h is constant and the conductivity of the extension be l (also constant).
(3.111) Figure 3.29 Cooling (heat loss) from extended surfaces.
d dx m
2 2
2 0
n − =n
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Steady State Heat Transfer 163
where n is the temperature differance tx − to and
(3.112) where r is the perimeter (m) of the extension and A is the cross sectional area (m2). The dimensions of m are
(3.113) A simplified solution to Equation (3.111) above is
(3.114) Here, nx = tx − to and nf = tf − to
The heat lost to the surroundings (i.e., lost from the fur-nace as end loss) is
(3.115) If l is very large cosh ml → ∞ and tanh ml → 1 nx=l = 0
If l is very large so that there is no heat loss at the tip, we can also use the expression
tx = tfe−mx°C and
EXAMPLE 3.24
A ceramic tubular muffle has outer diameter 100 mm and inner diameter 80 mm. A length 0.1 m of the muffle extends out of the furnace which is at 1200°C. The surrounding tem-perature is 30°C
Thermal conductivity of muffle l = 1.0 w/m°C Heat transfer coefficient (average) h = 200 w/m2°C Determine the temperatures along the extension and the heat lost.
m h
= + Ar l
m=
° ×
° × =
w
m C m w
m C m
2 2 m
2 1
nx nf m x
= cosh ( (m− )) cosh
l l
Q=nf hrlAtanh (ml)
Q= tf hrlAw
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164 Industrial Heating
Solution
Cross sectional area of tube A
Perimeter (outer) of the tube
As the muffle has very low conductivity the heat transfer will be mainly near the base. Consider distances of 0.01 m from the base at
x = 0 t = 1200°C
x = 0.01 t = 1200 × e−118 × 0.01 = 369°C x = 0.02 t = 1200 × e−118 × 0.02 = 113°C x = 0.03 t = 1200 × e−118 × 0.03 = 34°C
Thus, heat will be lost through the first 3 cm length of the muffle.
Heat lost will be
EXAMPLE 3.25
Calculate the temperatures along an extension tube made of Fe-Cr-Al alloy. The furnace temperature is 1000°C. The dimensions and other conditions are similar to those in Example 3.24.
Thermal conductivity of alloy tube (l) = 15 w/m°C
= p4
(
1002−802)
×10−6 =2 83 10. × −3 m2DK340X_C03B.fm Page 164 Tuesday, April 26, 2005 2:36 PM
Steady State Heat Transfer 165
Solution
Cross sectional area A = 2.83 × 10−3 m2 Perimeter = 0.314 m l = 15 w/m°C
h = 200 w/m2°C
As the conductivity is much higher and the tube is of finite length, we will use the expression
here m = 38.5 m−1, l = 0.3 m (30 cm) let x = 0.05 m
Similar calculations give the following results
x t°C
0.005 ~ 1000 0.01 ~ 816 0.05 ~ 175
0.1 ~ 30
Showing that at x ~ 10 to 12 cm, the temperature reaches the surrounding temperature.
The heat dissipated is
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166 Industrial Heating
Comparing these results it can be seen that the hot length for metallic muffle is about twice that of ceramic muf-fle. The heat dissipation is also about 2.5 times more.
EXAMPLE 3.26
A furnace enclosure has a surface temperature varying between 200 and 50°C. The atmospheric average temperature is 30°C.
If the enclosure has an emissivity 0.8, determine the heat radiated by the enclosure.
What will be the change in radiation if the surface is given aluminium paint having an emissivity 0.3?
Solution
The thermal radiation flux from the wall to the atmosphere will be
The wall emissivity (unpainted)
2
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Steady State Heat Transfer 167
For the aluminium painted wall = 0.3 q200°C = 0.708 kW/m2
q150°C = 0.401 kW/m2 q100°C = 0.185 kW/m2 q50°C = 0.041 kW/m2
the radiation loss decreases rapidly with the decrease in the wall temperature. Also, the loss from a wall painted with low emissivity paint is much lower (35%) than the bare wall.
Figure 3.30 [Example (3.26)]
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These results are plotted in Figure 3.30 which shows that
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The radiative heat transfer coefficient will be obtained by dividing the calculated flux values by the temperature difference.
If the furnace temperature TF and the wall thickness, etc.
are the same, a bare wall will conduct heat faster than a painted wall. This fact is useful in the design of continuous and batch-type furnaces.
e.g., W/m C unpainted
W/m C painted
C
2
C
2
hr
hr
100
100
1 07
100 30 15 0 401
100 30 5 73
°
°
= − = °
= − = °
.
. .
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