11-Flues and Chimneys
11-Flues and Chimneys
11-Flues and Chimneys
朱信
Hsin Chu
Professor
Dept. of Environmental Engineering
National Cheng Kung University
1
1. Functions of the Flue System
The function of the flue in combustion equipment can
be summarized quite simply: it is for the safe and
effective disposal of the products of combustion.
This focuses on two considerations, namely bringing
the products of combustion to the outlet of the flue at
the required conditions (such as temperature and
velocity), and ensuring that the location of this outlet
is such that the environmental impact of the
discharge is controlled.
2
There are two ways in which the dispersal of
the combustion products can be effected.
If the fuel has a very low sulfur content (such
as natural gas) then it is often possible to
dilute the products of combustion with
ambient air and to discharge the diluted
mixture at a low level.
3
This may be desirable economically as it
avoids the need to construct a high-level
discharge, or it might be aesthetically
appealing if the presence of a chimney, either
outside the building or appearing at the
rooftop, is thought to be undesirable.
The majority of flue systems discharge the
combustion products at a high level via a flue
or chimney.
4
Contemporary chimneys are generally of
circular cross-section and are of metal
fabrication.
An important consideration in chimney
design not discussed here is the wind loads
imposed on the structure, which can be
carried by the fabric of the chimney itself or
by external bracing such as fins or wires.
5
In addition to the steady-state loads on a
chimney, vibrations can be induced by the
regular shedding of vortices from the
cylindrical surface (the same mechanism
which causes telephone wires to hum).
The spiral “strakes” which are often attached
to the upper part of a chimney are there to
reduce this effect.
6
The force for moving the flue gas within the system
can come from the buoyancy of the hot gas within
the flue, from external fan power or from a
combination of both.
In addition to maintaining the correct flow rate of gas
through the flue it is essential to maintain the
temperature of the gas within the flue system above
the water vapor dew point, or the acid dew point if a
high sulfur content fuel is being burned.
7
In terms of mechanisms, we are interested in
the heat transfer and fluid flow performance
of the flue, recognizing that both these
considerations must be integrated in the
engineering design of the flue system.
8
2. Chimney Heat Transfer
9
The most common situation is where the
temperature, composition and flow rate of the
gas entering the flue are known, and the
objective is to find the temperature of the gas
as it is discharged from the flue.
10
The approach adopted here is to consider
the flue in classical heat exchanger terms.
This approach involves two stages:
evaluating an overall thermal conductance
(U-value) followed by an analysis of the
performance of the flue, taking into account
the flows of the system fluids (flue gas and
ambient air).
11
2.2 U-value of a Chimney
12
(2) Heat transfer between the inside and o
utside surfaces of the chimney.
If the chimney is of solid construction, the
mechanism for this will be conduction.
If there is an air gap present, then a c
ombined mode of heat transfer (conduction/
convection together with radiation) will be o
perating.
14
(3) Convective heat transfer from the outside
surface of the chimney to the ambient air.
Under still air conditions this will be by n
atural convection, but in general the action
of wind will induce forced convection from th
e chimney.
15
These processes can be represented thus:
Q = hiAi(tg-tsi)
Q = hfAf(tsi-tso) (1)
Q = hoAo(tso-to)
where
Q is the heat flux (W)
h is the heat transfer coefficient (W/m2/K)
A is the area (m2)
t is the temperature (℃)
Subscripts:
f: fabric
g: gas
i: inside
o: outside
s: skin
16
In equation (1) the heat transfer across the fa
bric has been expressed in terms of a fabric
heat transfer coefficient, hf, and a cross-secti
onal area, Af, over which this operates.
The contributory terms to (hfAf) depend on th
e construction of the chimney, as will be sho
wn later.
17
The above equations can be summarized giving:
�1 1 1 �
� ( g o)
Q� + + = t -t (2)
�h A h A h A �
�i i f f o o �
18
Any of the three area Ai, Af or Ao could be used for th
is purpose, but here the outside surface area of the
chimney (the largest of the three) has been used.
An expression for the overall U-value of the chimney
is obtained by dividing equation (2) by equation (3):
1
Uo = Ao
hi Ai + h fAAo f + h1o (4)
19
If the chimney is of circular cross-section and does
not have an appreciable taper, the areas in the above
expression are given by:
Ao = 2p ro L
Ai = 2p ri L
2p (r0 - r1 ) L
Af =
ln(ro / ri )
where L is the length (height) of the chimney.
20
If the thickness of the insulation/fabric is small
compared with the radius, then equation (4)
simplifies to that of one-dimensional heat transfer
with all the areas being equal.
The expression for the U-value is then:
1
U = 1 (ro - ri ) 1
ih + k o
+h (5)
21
Above expression is generally sufficiently acc
urate for most practical purposes.
For convenience, we can replace (ro- ri) with
x, the thickness of the fabric layer.
22
The thermal resistance of the fabric layer is then:
x
( K / m2 / W )
k
For a composite construction of n layers, the total thermal
resistance is found by adding up the individual layer
resistances:
xi
S ( K / m2 / W )
ki
23
A common configuration for chimneys of less
than 15 m height is to have an outer skin of a
luminum alloy, with a low-emissivity surface,
combined with a steel inner lining with an air
gap of around 6 mm between the two metals.
The heat transfer across an air gap is affecte
d by the emissivity of the surfaces and the wi
dth of the gap.
24
In general, the thermal resistance of an air gap i
ncreases with its width, but the value remains su
bstantially constant at separations greater than 2
0 mm.
The following values can be taken:
Low emissivity surface High emissivity
6 mm 0.18 0.1
20 mm+ 0.35 0.18
25
To evaluate equation (5), values are needed
for the inside and outside surface coefficients
of heat transfer, together with the thermal
resistance of the chimney fabric and a means
of estimating its effective heat transfer area.
26
Inside Surface Coefficient
The flow of gas in the chimney may well be driven by
buoyancy forces, but these are generated by the
difference in density between the hot flue gas and
the ambient air.
Under these circumstances the contribution to the
buoyancy force provided by the difference in
temperature between the flue gas and the inside
surface of the chimney will be very small, and heat
transfer from the gas to the chimney inside surface
will take place by forced convection.
27
The flue gas flow regime (whether the flow is laminar
, transitional or turbulent) is indicated by the value of
Reynolds number:
Re = vdi/ν
28
For a circular cross-section the characteristic
dimension d is given by the diameter of the c
himney.
To get some perspective on this situation we
can take a range of diameters between 0.25
m and 2 m, and consider the applicable rang
e of flue gas velocity (v) to be between 6 m/s
and 20 m/s.
29
The kinematic viscosity of a gas varies with temperat
ure, and, although values for individual gases are rea
dily available, it is not possible to calculate accuratel
y the viscosity of a mixture from the values and volu
me fractions of its constituents.
The estimated kinematic viscosity of a flue gas result
ing from the combustion of a natural gas with 20% ex
cess air is compared with the values for air in Table 1
1.1 (next slide).
30
Table 11.1 Kinematic viscosities for typical flue gas and air
Kinematic viscosity (m2/s×106)
Temperature (℃)
Flue gas Air
50 17 18
100 22 23
150 27 27
200 33 32
250 40 36
300 46 41
350 54 46
31
At a temperature of t ℃ the kinematic viscosit
y of air is approximated by:
ν= (0.1335 + 0.925×10-3 t)10-4 m2/s (7)
giving a value at 200℃ of 0.000032 m2/s.
32
The Reynolds number for a flue gas velocity of 9 m/s
in a chimney of 750 mm diameter is thus:
9 �0.75
Re = -6
= 210,900
32 �10
which is well into the turbulent regime.
33
The emissivity of the flue gas is low, hence the predominant mo
de of heat transfer is forced convection. A relationship for forced
convection from a turbulent gas flow inside a cylindrical tube is :
Nu = 0.023 Re0.8 Pr0.4 (6)
where the Nusselt number, Nu, is given by:
hi di
Nu =
k
and the prandtl number, Pr, by
cp m
Pr =
k
34
Most gases have a value of Pr of about 0.74,
and this value is substantially independent of
temperature, hence equation (6) simplifies to;
Nu = 0.02 Re0.8
and the coefficient of heat transfer is
obtained knowing the diameter of the tube, d,
and the thermal conductivity of the gas, k.
35
Once again, a value for k for the gas mixture could b
e estimated from a knowledge of the individual gas p
roperties and the mixture composition.
Here it is also an acceptable approximation to use va
lues for air.
The thermal conductivity of air can be obtained from:
kair = 0.02442 + 0.6992×10-4 t W/m/K
where t is the temperature in ℃. At 200℃ this gives
a value of 0.0384 W/m/K.
36
Example 1:
Estimate the internal coefficient of heat transfer in a
500 mm diameter chimney. The flue gas velocity is 1
0 m/s and the gas temperature 250℃.
Solution:
Start by working out the gas kinematic viscosity and t
hermal conductivity:
ν=(0.1335+0.925×10-3×250) ×10-4=3.64×10-5 m2/s
37
k=0.02442+0.6992×10-4×250=0.0419 W/m/K
vdi 10 �0.5
Re = = = 1.37 �10 5
n 3.64 �10-5
The Nusselt number, Nu, is
Nu=0.02×(1.37×105)0.8=257
38
The coefficient of heat transfer:
Nuk 257 �0.0419
hi = = = 21.6 W / m 2 / K
di 0.5
An approximate value of hi can be obtained more quickly f
rom Fig. 11.2 (next slide), where values of hi × di (internal fi
lm coefficient × chimney diameter) are plotted as a functio
n of v × di (flue gas velocity × chimney diameter) for flue g
as temperatures ranging from 100 to 300℃.
39
For the above example v × di=10×0.5=5 m2/s, giving
a value of hi × di from the graph at 250℃ of 10.8 W/m
/K, resulting in an estimated value of:
hi = 10.8 �0.5 = 21.6 W / m 2 / K
41
Outside Surface Coefficient
42
In calm conditions, the convective heat
transfer from the outside surface of the
chimney will be by natural convection.
As the wind velocity increases, forced
convection will become the dominant
mechanism.
43
A chimney exposed to the wind approximates to a
cylinder with its axis at right-angles to the direction of
the flow.
The relevant heat transfer relationship, valid for
Reynolds numbers between 103 and 105 is:
44
Assuming a constant value for Prandtl number for ai
r of 0.74, this expression simplifies to:
45
For the case of forced convection the temper
ature of the air has only a small effect on the
value of ho, and Fig. 11.3 (next slide) shows t
he variation of hc,o with windspeed for a range
of chimney outside diameters.
46
Example 2:
Estimate the outside convective heat transfer coeffici
ent for a 750 mm diameter chimney exposed to a win
d of 10 m/s.
Assume the air temperature to be 5℃.
Solution:
At 5℃ the kinematic viscosity of air from equation (7)
is:
ν = (0.1335+0.925×10-3×5)10-4
= 0.0000138 m2/s
48
The Reynolds number is then:
10 �0.75
Re = = 5.43 �10 5
13.8 �10-6
49
The thermal conductivity of air at 5℃ is:
k = 0.02442+(0.6992×10-4×5)
= 0.0248 W/m/K
giving an outside convective film coefficient of
662.4 �0.0248
hc ,o =
0.75
=21.9 W/m 2 /K
50
The heat transfer from the outside surface of the chi
mney to the environment also contains a contribution
from a radiative exchange to the surroundings.
This is not an easy mechanism to quantify accurately
, as radiative heat transfer is a non-linear process wit
h respect to temperature and is affected by the natur
e and configuration of the participating surfaces.
51
However, in circumstances such as these a linear ap
proximation can be made to the radiative heat transfe
r component and an outside film coefficient defined in
terms of the convective and radiative components thu
s
ho = hc ,o + e hr ,o
where εis the emissivity of the surface and hr,o is a lin
earized radiation surface heat transfer coefficient, whi
ch is dependent on the surface temperature of the chi
mney and the temperature of the surroundings.
52
For typical situations a reasonable value for h
r,o is 5 W/m /K.
2
53
Table 11.2 Emissivities of some surfaces
Material Emissivity
Polished aluminium 0.10
Oxidized aluminium 0.18
Polished steel 0.07
Rolled sheet steel 0.66
Galvanized Zinc 0.21
Brick 0.93
Glazed surface 0.90
Non-metallic paint 0.90
54
If we want to take into account the effects of r
adiation in Example 2, for the case of a dull
metal finish (ε=0.2), we have:
ho = 21.96 + 0.2 × 5
= 22.96 W/m2/K
55
If the chimney is insulated, the contribution of
variation in ho to the overall heat transfer
process is quite small, hence the
approximations introduced above are
justifiable.
56
Overall U-value
57
The internal and external heat transfer coefficients ha
ve been estimated as 20 and 23 W/m2/K respectively.
Calculate the overall U-value.
Solution:
For equation (6), neglecting the thermal resistance of
the two metal layers:
1
U o = 1 0.05 1 = 0.746 W / m 2 / K
20 + 0.04 + 23
58
The relative magnitudes of the three terms in
the denominator of the above expression
show that by far the most significant term in
the case of an insulated flue is the thermal
resistance to conduction across the insulated
layer, and thus in these circumstances a high
level of accuracy in the evolution of the film
coefficients is unwarranted.
59
2.3 Temperature Distribution in a Chimney
60
Where dA represents the surface area of the sma
ll section under consideration and dt is the small t
emperature drop of the flue gases in this section.
The term W represents the thermal capacity rate
of the flue gas, and is obtained by summing the p
roduct of the mass flow rate and specific heat for
each of the species present, that is:
W = S(mc p ) kW/K
61
Evaluation of this expression can be a bit time-
consuming.
However, an approximate value of 1.476 kJ/m3/K can
be assumed for the volumetric specific heat of the
flue gas which, when multiplied by the volume flow
rate of the gas (m3/s) gives a value for W.
The volume flow rate of flue gas can be estimated
from the gas velocity and the internal cross-sectional
area of the chimney.
62
Equation (9) is a rate equation and equation (10) is a
n energy balance on the gas in the control volume; n
ote that in these equations, heat lost from the gas is t
reated as a negative quantity.
Eliminating dQ from these equations gives:
-U(t-to)dA = Wdt
Rearrange:
U dt
- dA =
W t - to
63
Which can be integrated:
U A dt
t2
-
W �
0
dA = �
t1 (t - t )
0
Giving:
UA (t - t )
- = ln 2 0
W (t1 - t0 )
Hence:
t2 = t0 + (t1 - t0 )e - (UA / W ) (11)
64
This expression gives the flue gas
temperature t2 at the outlet of a section of the
flue of area A, from a starting temperature of
t1.
It is most appropriately applied to the entire
flue but the expression can be solved for any
number of flue section, so giving a
temperature distribution along the flow path.
65
As the thermal capacity rate of the flue gas is
quite high, the ratio UA/W is generally small,
giving a low temperature drop in the flue gas
as within the chimney.
The main function of the chimney insulation i
s to keep the temperature of the gas high by l
imiting the temperature drop between the gas
and the inner lining of the chimney.
66
In the steady state, the ratio of the
temperature drop across the inner boundary
layer to the overall temperature difference
between the gas and the outside air is equal
to the ratio between the thermal resistance of
the inner boundary layer to the resistance of
the entire chimney wall.
67
Thermal resistance is the reciprocal of conductance
we can write
t g - ts Uo
=
t g - to hi
Hence
U
ti = t g - (t g - to )
hi (12)
68
Maintaining internal surfaces temperature well above
the gas dew point is an important design considerati
on.
Estimation of the surface temperatures in a flue is illu
strated in the following example.
Example 4:
Estimate the flue gas temperature on leaving and the
internal surface temperature at the flue exit for a 15
m high flue with a U-value of 3.5 W/m2/K at an outsid
e air temperature of -3℃.
69
The internal diameter of the flue is 1.25 m an
d the external diameter 1.5 m. The flue gas v
elocity is 4 m/s and the gas temperature on e
ntering is 275℃.
Solution:
The first task is to find the bulk flue gas temp
erature at the chimney exit from equating (1
1).
70
The capacity rate of the gas must first be obtained
from the gas velocity and the internal cross-sectional
area:
p
W = 1.476 � �1.252 �4 = 7.24 kW / K
4
The outside surface area, A, of the chimney is:
A = p �1.5 �15 = 70.69 m 2
so UA=0.247 kW/K
71
The temperature of the gas at the flue exit is:
t2 = -3+(275+3)e-(0.247/7.24)
= 265.7℃
The next step is to estimate the internal film
coefficient – the approximate method
outlined in the previous section will be used.
72
The value of (vdi) is 4 × 1.25 = 5, hence (hidi) will be 1
0.7 (from Fig. 11.2). Hence:
hi = 10.7/1.25 = 8.56 W/m2/K
We can now solve equation (12) to get the internal su
rface temperature:
3.5
ts = 265.7 - ( 265.7 + 3)
8.56
= 155.8o C
73
A U-value of 3.5 W/m2/K is a fairly modest lev
el of thermal insulation but would be typical o
f a double skin flue with a low-emissivity air g
ap as insulation.
The example does, however, illustrate the im
portance of the temperature gradient across t
he chimney as opposed to that along the flow
path of the flue gas.
74
3. Pressure Loss
77
In the majority of installations, the
combustion air is supplied to the burner by
an integral fan, which will have sufficient duty
to propel the air through the boiler itself.
In these circumstances all the flue draught is
available for providing the energy required to
move the flue gas through the chimney.
78
Taking the point A in Fig. 11.4, the pressure d
ifference due to the buoyancy of the column
of hot gas in the chimney is simply equal to t
he relative weights of the columns of hot gas
and surrounding air:
Δp = zg (ρg-ρa) (13)
In this expression ρg represents the mean de
nsity of the flue gas.
79
As was shown above, the temperature drop as the
gas passes through the chimney is quite small, so
there would be only a small error incurred if the
temperature of the gas at the flue inlet was used to
evaluate the available draught.
The density of the air is primarily influenced by its
temperature, but the density of the flue gas is
dependent on both the temperature and composition
of the gas, which will vary with different fuels and the
air-to-fuel ratio.
80
The equation of state for an ideal gas:
m
pV = RT
M
can be written in terms of the gas density, ρ:
r
p= RT
M
or
pM
r=
RT
81
At standard atmospheric pressure of 1.01325 bar and
noting that the universal gas constant, R, is 8.314 kJ/
kmol/K:
M
r = 12.1873 (kg/m 3 ) (14)
T
This expression can be substituted for the two densit
y terms in equation (13), noting that the average mole
cular weight for air is 28.84 and incorporating g in the
numerical constant:
Dp �Mg
�
= 118.9 Tg - 28.84 (Pa/m)
z � Ta �
82
A further simplification can be made if we consider
the molecular weight of the flue gas, Mg.
This will obviously depend on the chemical
constitution of the fuel and on the air-to-fuel ratio.
Some typical values for Mg range from 27.86
(natural gas burned at 20% excess air) to 30.0
(bituminous coal burned at 20% excess air).
83
Within this range it is reasonable to take an
approximate value equal to the mean molecular
weight of air, i.e. 28.84, leading to the simplified
expression for the chimney draught:
Dp
= 3, 429 �1
- 1 �
(15)
z �
Tg Ta �
84
As an example, if the ambient temperature is
-5℃ (268K) and the mean flue gas
temperature is 250℃ (523K), then the
draught, given by equation (14), is -6.24
Pa/m.
85
3.2 Flow Resistance
87
For the case of the friction in the conduit of
the chimney itself K is represented as:
K = f (L/D)
where f is the friction factor which depends
on both the roughness of the surface and the
Reynolds number of the flow.
L: length
D: diameter
88
For a fuller treatment of the relationships for
pressure loss in pipes and ducts, the reader
is referred to any standard text on fluid
mechanics or the chartered Institution of
Building Services Engineers, London
(CIBSE) and American Society of Heating,
Refrigerating and Air Conditioning Engineers
(ASHRAE) handbooks.
89
If an estimate of the flue gas velocity is
available, an initial estimate of the pressure
loss in the flue system can be made by
substituting a value for K of 5.0 in equation
(16).
90
At a later stage, when the design of the flue system is more
established, the pressure drop for the system can be obtain
ed from summing up the individual contributions to the press
ure loss multiplied by their respective velocity pressures:
Dp = ( K1 + SK f + K D ) vp1 + K 2vp2
Here K1 is the loss coefficient at the inlet to the flue duct. Th
e second term represents the summation of all the individual
losses from fittings such as bends, dampers or connectors.
91
For values of these, the reader is referred ag
ain to the handbooks; approximate example
values are 0.8 for a smooth round 900 bend a
nd 1.25 for a mitred 900 bend.
For the duct loss KD an estimate can be mad
e from:
KD = L/(30 di)
92
With regard to the exit loss K2 is equal to 1.0 and is
associated with the velocity pressure in the chimney
duct vp1 if the chimney discharges as a free jet into
the atmosphere, but there will be an additional
contributing factor (based on the velocity pressure
vp2 in the smaller outlet diameter) if a tapering
conical cap is used to increase the efflux velocity of
the jet.
A typical value of K2 for a gradual contraction is 0.20.
93
Example 5:
For the chimney described in Example 4, find the chi
mney draught and the pressure drop in the flue if a c
ap is used to accelerate the efflux velocity to
8 m/s.
Solution:
In Example 4, the gas temperature on entering was 2
75℃ and the gas temperature on leaving was 265.7
℃.
The mean temperature is thus 270.3℃ (543 K).
94
The outside air temperature is -3℃ (270K), hence
the draught, from equation (15) is:
Dp
= 3, 429 ( 543
1
- 270
1
) Pa/m
z
Dp = -95.8 Pa
96
If the flow is accelerated to 8 m/s at the outlet
, the velocity pressure is:
vp2 = 5.2 × 82/42 = 20.8 Pa
Assuming a K-factor of 0.2 for the reducer, a
nd 1.0 for the discharge of the free jet, the tot
al pressure drop is given by:
Δp = (0.4×5.2)+(1.2×20.8)=27 Pa #
97