Flues and Chimneys
朱信
Hsin Chu
Professor
Dept. of Environmental Engineering
National Cheng Kung University
1
1. Functions of the Flue System


2
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.


3
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.


4
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.


5
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.


6
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.


7
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.

8
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.
2. Chimney Heat Transfer
2.1 Heat Transfer Mechanism
 The rate of heat transfer affects the
temperatures in the system, which in turn
determine the safety margin in respect of
possible corrosion problems, and also the
pressure difference due to buoyancy forces,
known as the chimney draught.
9

10
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.


11
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).
2.2 U-value of a Chimney

12
Consider the cross-section sketch of a
chimney wall shown in Fig. 11.1 (next slide).
There are three sequential stages in the
steady-state heat transfer process:
(1) Convective heat transfer from the hot flue
gas to the inside surface of the chimney.

(2) Heat transfer between the inside and
outside 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
combined mode of heat transfer
(conduction/convection together with
radiation) will be operating.
14

(3) Convective heat transfer from the outside
surface of the chimney to the ambient air.
Under still air conditions this will be by
natural convection, but in general the
action of wind will induce forced
convection from the chimney.
15

16
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


17
In equation (1) the heat transfer across the
fabric has been expressed in terms of a
fabric heat transfer coefficient, hf, and a
cross-sectional area, Af, over which this
operates.
The contributory terms to (hfAf) depend on
the construction of the chimney, as will be
shown later.

The above equations can be summarized giving:
 1
1
1
Q


hA h A h A
f
f
o o
 i i

18

   t g  to 

(2)
The rate of heat transfer is conventionally represented
in the form:
Q = UoAo (tg-to)
(3)
where Uo represents the overall heat transfer
coefficient for the exchanger and Ao represents the
area which is associated with it.


Any of the three area Ai, Af or Ao could be used for
this 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):
Uo 
19
1
Ao
hi Ai

Ao
h f Af
 h1o
(4)

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  2 ro L
Ai  2 ri L
Af 
2 (r0  r1 ) L
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:
U
1
1
hi
+ (ro k- ri )  h1o
(5)
where k is the thermal conductivity of the gas.
21


22
Above expression is generally sufficiently
accurate for most practical purposes.
For convenience, we can replace (ro- ri) with
x, the thickness of the fabric layer.

The thermal resistance of the fabric layer is then:
x
k

23
( K / m2 / W )
For a composite construction of n layers, the total thermal
resistance is found by adding up the individual layer
resistances:
xi

( K / m2 / W )
ki


24
A common configuration for chimneys of less
than 15 m height is to have an outer skin of
aluminum 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
affected by the emissivity of the surfaces and
the width of the gap.


In general, the thermal resistance of an air gap
increases with its width, but the value remains
substantially constant at separations greater
than 20 mm.
The following values can be taken:
6 mm
20 mm+
Low emissivity surface
0.18
0.35
the units being K/m2/W.
25
High emissivity
0.1
0.18

26
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.
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
where ν is the kinematic viscosity of the flue gas.
28


29
For a circular cross-section the characteristic
dimension d is given by the diameter of the
chimney.
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
range of flue gas velocity (v) to be between 6
m/s and 20 m/s.


30
The kinematic viscosity of a gas varies with
temperature, and, although values for individual
gases are readily available, it is not possible to
calculate accurately the viscosity of a mixture from
the values and volume fractions of its constituents.
The estimated kinematic viscosity of a flue gas
resulting from the combustion of a natural gas with
20% excess air is compared with the values for air in
Table 11.1 (next slide).
Table 11.1 Kinematic viscosities for typical flue gas and air
Temperature (℃)
50
100
150
200
250
300
350
31
Kinematic viscosity (m2/s×106)
Flue gas
Air
17
22
27
33
40
46
54
18
23
27
32
36
41
46

32
At a temperature of t ℃ the kinematic
viscosity 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.

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 
 210,900
6
32 10
which is well into the turbulent regime.
33

The emissivity of the flue gas is low, hence the predominant
mode 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
where the Nusselt number, Nu, is given by:
Nu 
hi d i
k
and the prandtl number, Pr, by
Pr 
34
cp 
k
(6)

35
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.



36
Once again, a value for k for the gas mixture could
be estimated from a knowledge of the individual gas
properties and the mixture composition.
Here it is also an acceptable approximation to use
values 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.


37
Example 1:
Estimate the internal coefficient of heat transfer in a
500 mm diameter chimney. The flue gas velocity is
10 m/s and the gas temperature 250℃.
Solution:
Start by working out the gas kinematic viscosity and
thermal conductivity:
ν=(0.1335+0.925×10-3×250) ×10-4=3.64×10-5 m2/s

k=0.02442+0.6992×10-4×250=0.0419 W/m/K
10  0.5
5
Re 


1.37

10

3.64 105
vdi

38
The Nusselt number, Nu, is
Nu=0.02×(1.37×105)0.8=257

The coefficient of heat transfer:
Nuk 257  0.0419
hi 

 21.6 W / m2 / K
di
0.5

39
An approximate value of hi can be obtained more quickly
from Fig. 11.2 (next slide), where values of hi × di (internal
film coefficient × chimney diameter) are plotted as a
function of v × di (flue gas velocity × chimney diameter) for
flue gas temperatures ranging from 100 to 300℃.

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 / m2 / K
41
Outside Surface Coefficient


42
The heat transfer processes taking place on the
outside surface of the chimney are rather more
complex than is the case for the inside surface
coefficient.
Radiation heat transfer does have a part to play-the
surface of the chimney will exchange heat by
radiation to the surrounding environment, and the
convective heat loss will be affected by the prevailing
wind speed.


43
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.


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:
Nu  0.26Re0.6 Pr 0.3
44

Assuming a constant value for Prandtl number for
air of 0.74, this expression simplifies to:
Nu  0.24Re0.6

(8)
In this case the Reynolds and Nusselt numbers refer
to the outside diameter, do, of the chimney.
The outside convective heat transfer coefficient is
then given by:
hc ,o 
45
Nuk
do

46
For the case of forced convection the
temperature of the air has only a small effect
on the value of ho, and Fig. 11.3 (next slide)
shows the variation of hc,o with windspeed for
a range of chimney outside diameters.


48
Example 2:
Estimate the outside convective heat transfer
coefficient for a 750 mm diameter chimney exposed
to a wind 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

The Reynolds number is then:
10  0.75
5
Re 

5.43

10
13.8 106

49
Nu =0.024(5.43×105)0.6 (from equation (8))
=662.4


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
0.75
=21.9 W/m 2 /K
hc ,o 
50


51
The heat transfer from the outside surface of the
chimney 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 with
respect to temperature and is affected by the nature
and configuration of the participating surfaces.

52
However, in circumstances such as these a linear
approximation can be made to the radiative heat
transfer component and an outside film coefficient
defined in terms of the convective and radiative
components thus
ho  hc,o   hr ,o
where εis the emissivity of the surface and hr,o is a
linearized radiation surface heat transfer coefficient,
which is dependent on the surface temperature of the
chimney and the temperature of the surroundings.


53
For typical situations a reasonable value for
hr,o is 5 W/m2/K.
The emissivity will depend on the outside
finish of the chimney, and some example
values are given in Table 11.2 (next slide).
Table 11.2 Emissivities of some surfaces
Material
Polished aluminium
Oxidized aluminium
Polished steel
Rolled sheet steel
Galvanized Zinc
Brick
Glazed surface
Non-metallic paint
54
Emissivity
0.10
0.18
0.07
0.66
0.21
0.93
0.90
0.90

55
If we want to take into account the effects of
radiation 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

56
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.
Overall U-value


57
Bringing the inside and outside film coefficients
together with the thermal resistance of the chimney
fabric into the evaluation of the overall U-value of a
chimney is best illustrated by a simple example.
Example 3:
A flue has a height of 15 m, an external diameter of
750 mm and a construction consisting of an internal
steel lining, 3 mm thick, of 50 mm mineral wool
insulation (conductivity 0.04 W/m/K), and an outer
skin of 1.6 mm alloy with a polished outer surface.


58
The internal and external heat transfer coefficients
have 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 / m2 / K
20  0.04  23

59
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.
2.3 Temperature Distribution in a Chimney


60
In order to evaluate the overall heat transfer from the
flue to the surroundings it is necessary to take into
account the flow pattern of the two fluids concerned,
i.e., the flue gases rising in the flue and the air
flowing over the outside of the chimney.
The latter can be considered as being at a constant
temperature to, hence we can construct an energy
balance about a short section of the chimney:
dQ = -U(t - to) dA
(9)
dQ = Wdt
(10)


Where dA represents the surface area of the
small section under consideration and dt is the
small temperature 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
product of the mass flow rate and specific heat
for each of the species present, that is:
W = (mcp )
61
kW/K


62
Evaluation of this expression can be a bit timeconsuming.
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.



Equation (9) is a rate equation and equation (10) is
an energy balance on the gas in the control volume;
note that in these equations, heat lost from the gas is
treated 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

W

0
dA  
t2
t1
dt
(t  t0 )
Giving:



A
(t  t )
UA
 ln 2 0
W
(t1  t0 )
Hence:
t2  t0  (t1  t0 )e(UA/W )
64
(11)


65
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.


66
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
is to keep the temperature of the gas high by
limiting the temperature drop between the
gas and the inner lining of the chimney.

67
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.

Thermal resistance is the reciprocal of conductance
we can write
t g  ts
t g  to


Uo
hi
Hence
U
ti  t g  (t g  to )
hi
68
(12)



69
Maintaining internal surfaces temperature well above
the gas dew point is an important design
consideration.
Estimation of the surface temperatures in a flue is
illustrated 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
outside air temperature of -3℃.


70
The internal diameter of the flue is 1.25 m
and the external diameter 1.5 m. The flue
gas velocity is 4 m/s and the gas temperature
on entering is 275℃.
Solution:
The first task is to find the bulk flue gas
temperature at the chimney exit from
equating (11).

The capacity rate of the gas must first be obtained
from the gas velocity and the internal cross-sectional
area:

W  1.476  1.252  4  7.24 kW / K
4

The outside surface area, A, of the chimney is:
A   1.5 15  70.69 m2
so UA=0.247 kW/K
71


72
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.


73
The value of (vdi) is 4 × 1.25 = 5, hence (hidi) will be
10.7 (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
surface temperature:
3.5
ts  265.7 
 265.7  3
8.56
 155.8o C


74
A U-value of 3.5 W/m2/K is a fairly modest
level of thermal insulation but would be
typical of a double skin flue with a lowemissivity air gap as insulation.
The example does, however, illustrate the
importance of the temperature gradient
across the chimney as opposed to that along
the flow path of the flue gas.
3. Pressure Loss
3.1 Chimney Draught
 The chimney contains a column of hot gas
which has a density considerably lower than
that of the air surrounding it (Fig. 11.4, next
slide).
 This difference in density produces an
apparent “suction” at the base of the flue
which is known as the “chimney draught”.
75


77
In some instances (mainly smaller
combustion equipment) this pressure
difference is sufficient to overcome the
resistance to flow through the boiler and the
flue, and to provide the correct efflux velocity
at the top of the flue.
In cases where this natural draught is
insufficient, fan assistance can be employed.


78
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.


79
Taking the point A in Fig. 11.4, the pressure
difference due to the buoyancy of the column
of hot gas in the chimney is simply equal to
the relative weights of the columns of hot gas
and surrounding air:
Δp = zg (ρg-ρa)
(13)
In this expression ρg represents the mean
density of the flue gas.


80
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.

The equation of state for an ideal gas:
pV 
m
RT
M
can be written in terms of the gas density, ρ:
p
or

81

M
RT
pM
RT

At standard atmospheric pressure of 1.01325 bar and
noting that the universal gas constant, R, is 8.314
kJ/kmol/K:
  12.1873

M
T
(kg/m3 )
This expression can be substituted for the two
density terms in equation (13), noting that the
average molecular weight for air is 28.84 and
incorporating g in the numerical constant:
p
Mg


 118.9 Tg  28.84
T
a


z
82
(14)
(Pa/m)



83
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).

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:
p
 3, 429  T1g  T1a 


z
84
(15)

85
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.
3.2 Flow Resistance


86
In a natural-draught appliance, the buoyancy
force produced in the flue has to move the air
through the boiler and the flue system.
Where the burner incorporates a forced air
supply via a fan, it is most likely that the
designer will have to ensure that an
adequate pressure difference is available to
overcome the resistance of the flue system
itself.

87
The basis for quantifying the pressure (or
energy) losses when a fluid flows through a
system containing resistances due to bends,
fittings and the friction of the pipe itself is:
Δp = K (v2/2g)
(16)
where K is the pressure loss coefficient and
the bracketed term is the velocity pressure of
the fluid.

88
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

89
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.

90
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).

At a later stage, when the design of the flue system is
more established, the pressure drop for the system can be
obtained from summing up the individual contributions to
the pressure loss multiplied by their respective velocity
pressures:
p   K1  K f  K D  vp1  K 2 vp2

91
Here K1 is the loss coefficient at the inlet to the flue duct.
The second term represents the summation of all the
individual losses from fittings such as bends, dampers or
connectors.


92
For values of these, the reader is referred
again to the handbooks; approximate
example values are 0.8 for a smooth round
900 bend and 1.25 for a mitred 900 bend.
For the duct loss KD an estimate can be
made from:
KD = L/(30 di)


93
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.


94
Example 5:
For the chimney described in Example 4, find the
chimney draught and the pressure drop in the flue if
a cap is used to accelerate the efflux velocity to
8 m/s.
Solution:
In Example 4, the gas temperature on entering was
275℃ and the gas temperature on leaving was
265.7℃.
The mean temperature is thus 270.3℃ (543 K).

The outside air temperature is -3℃ (270K), hence
the draught, from equation (15) is:
p
1
1
 3, 429  543
 270

z
p = -95.8 Pa

To evaluate the pressure drop it is first necessary to
estimate the density of the flue gas.
This is done via equation (14):
  12.1873
95
Pa/m
M
29
 12.1873
 0.651 kg / m3
T
543

The velocity pressure in the flue duct is then:
42
vp1  0.651  5.2 Pa
2

Assuming a friction factor of 0.033, the pressure loss
coefficient for the duct, KD, is;
15
K D  0.033
 0.396  0.4
1.25
96


97
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,
and 1.0 for the discharge of the free jet, the
total pressure drop is given by:
Δp = (0.4×5.2)+(1.2×20.8)=27 Pa #