MASTER’S THESIS
M.Sc PROGRAMME IN MECHANICAL ENGINEERING
Luleå University of Technology
Department of Applied Physics and Mechanical Engineering Division of Energy Engineering
2006:141 CIV • ISSN: 1402 - 1617 • ISRN: LTU - EX - - 06/141 - - SE
MARC CORTINA
Flue Gas Condenser
for Biomass Boilers
Acknowledgements
Firstly I would like to express my gratitude to the international offices of Luleå University of Technology and Enginyeria Tècnica Superior Industrial de Barcelona.
They have made it possible to me to spend this great period in Sweden, and this project.
To Sweden in general, this country that has always helped me, and especially to this University in Luleå, thanks to it I have had at my disposal all the information to make the thesis.
Secondly, I am grateful to my supervisor Roger Hermansson, who suggested this project to me and has helped me whenever it has been necessary. And to Joakim Lundgren, who has also been interested in listening to me and giving his point of view and help in some difficult steps.
I have also received some help from Catalunya, my country. From Bonals and Velo, two teachers from the University of Barcelona who suggested a basic book to handle the condensation topic when it seemed impossible to find the information.
Also my friends Josep Cots, Josep Cortina and Dani appeared to improve the final result with the excel, word and power point programs. And Andi, the student who has spent more time with me in this Erasmus, also to discuss about heat transfer, iterations, and others.
And finally I would also express my gratitude to my family for their unconditional support in my studies and my whole life in general, whatever I do.
Luleå, March 2006
Marc Cortina i Grau
INDEX
1 INTRODUCTION ... 1
1.1 BACKGROUND... 1
1.2 GENERAL DESCRIPTION... 1
1.3 METHOD... 2
2 THEORY... 3
2.1 GAS... 3
2.1.1 Gas composition ... 3
2.1.2 Fuel Flow ... 4
2.1.3 Enthalpy of the Gases ... 5
2.2 WATER... 6
2.3 HEAT EXCHANGER... 7
2.3.1 FLUE GAS CONDENSER (WITHOUT SPRAYING WATER) ... 7
2.3.2 FLUE GAS CONDENSER (SprayING water) ... 21
3 RESULTS AND ANALYSIS... 23
3.1 INTRODUCTION... 23
3.2 GAS COMPOSITION... 23
3.2.1 Fuel mass flow, total gas mass flow, and dry gas mass flow depending on the moisture content ... 23
3.2.2 Gas composition depending on the moisture content... 24
3.2.3 Amount of steam in the stack gases... 24
3.2.4 Dew point... 25
3.3 HEAT RECOVERED... 27
3.4 WATER AND GAS FLOWS AND HEAT TRANSFER COEFFICIENTS... 28
3.4.1 Total water flow... 28
3.4.2 Water flow inside tubes, gas flow outside, and heat transfer coefficients... 28
3.5 AREA NEEDED... 33
3.5.1 FLUE GAS CONDENSER (WITHOUT SPRAYING WATER) ... 34
3.5.2 FLUE GAS CONDENSER (SprayING water) ... 39
3.6 PROTOTYPE STUDY... 41
3.6.1 CONDENSER WITHOUT SPRAYING WATER ... 42
3.6.2 CONDENSER SprayING Water ... 44
3.7 COMPARISON AND FINAL PROTOTYPE... 46
4 CONCLUSIONS AND FURTHER IMPROVEMENTS... 49
4.1 CONCLUSIONS... 49
4.2 IMPROVEMENTS... 49
4.2.1 improvements in the boiler... 49
4.2.2 Improvements in the software ... 50
5 REFERENCES ... 51
6 APPENDIX A ... 52
7 APPENIX B (EXCEL EXPLAMATION) ... 55
TABLE OF SYMBOLS
Symbol Meaning Units
m&w Mass flow rate of water Kg / s
m&wt Mass flow rate of water in each tube Kg / s
m&g Mass flow rate of gas Kg / s
dg
m& Mass flow rate of dry gas Kg / s
λ Air Factor ---
Hi Heating values J/kg fuel
h1,h2 Specific enthalpy at entrance and exit from the exchanger J/kg
Q& Rate of heat recovered W
F, MC Moisture content Kg water / kg fuel
Pout Power output W
η Boiler Efficiency ---
Cpg Specific heat of the gas J/kgK
Cpdg Specific heat of the dry gas J/kgK
Cw Specific heat of the water J/kgK
hwe Latent heat of Vaporization kJ/kg
Tr1,Tr2 Temperature entrance and Temperature exit of the radiator ºC T1,T2
Temperature entrance and Temperature exit of the heat
exchanger ºC
t1,t2
Temperature entrance and Temperature exit of the heat
exchanger ºC
T Temperature K
t Temperature ºC
pa Atmospheric pressure Pa
Re Reynolds Number ---
ρ Density Kg / m3
uw Average Speed Water m / s
u∞g Average Speed Gas m / s
umax g Maximum Speed Gas m / s
Di Diameter inside m
Do Diameter outside m
µ Dynamic Viscosity Kg / m·s
µw Dynamic Viscosity at the wall Kg / m·s
Nu Nusselt Number ---
Pr Prandt Number ---
St Cross section area inside tub m2
Ф Viscosity Correction ---
k Thermal conductivity W / m·K
Nt Number of tubes ---
Ry Rows of tubes in the y direction ---
Rz Rows of tubes in the z direction ---
Sy Separation between rows y direction m
Sz Separation between rows z direction m
Ait Area inside tube m2
Aot Area outside tube m2
Ai Total area inside tubes m2
Ao Area outside total tubes m2
L, Lx Length of the heat exchanger m
Ly Witdth of the heat exchanger m
Lz Height of heat exchanger m
E Exchanger heat transfer effectiveness
Q&max Maximum heat transfer rate between the two flows W
Cmin Minumum heat capacity rate, Cpgas·mgas W
Cmax Maximum heat capacity rate, Cpwater·mwater W Ui Overall inside heat transfer coefficient W/m2·K Uo Overall outside heat transfer coefficient W/m2·K
Tdew Dew point K
pws Saturation pressure of water Pa
pg Partial pressure of the gas Pa
pc Pressure of condensate film Pa
pv Partial pressure of the steam Pa
pt Pressure total Pa
pgf Log mean of pg and p’g Pa
p’g pt - pc Pa
madw Mass flow of water added (Sprayed) Kg/s
xad Water added (Sprayed) per kg dry air Kg w / kg dry air m’w Total mass rate water flow (After spray) Kg/s
x’ Total water flow (After spray) per kg dry air Kg w / kg dry air
m’g Total gas flow rate (After spray) Kg/s
tb Bulk Temperature ºC
Tb Bulk Temperature K
hw Heat transfer coefficient of the water W/m2·K
hg Heat transfer coefficient of the gas W/m2·K
taspray Temperature of the gas after spray water ºC
tbspray Temperature of the gas before spray water ºC
Q&lat Transfer rate of latent heat W
Q&sens Transfer rate of sensible heat W
Vg Specific volume of the gas mol/kg
Vsteam Specific volume of the steam mol/kg
Mg Molecular weight of the gas ---
Msteam Molecular weight of the steam ---
1 INTRODUCTION
1.1 BACKGROUND
In Sweden district heating is widely used in cities and more dense populated areas.
However there are communities where houses are heated by oil or electricity and the population is dense enough to motivate the building of a district heating netwok. These combustibles are not renewable and collaborate with the greenhouse effect. In addiction, in general each building has its own boiler for heating, what makes the boiler very inefficient since the consumtion of heat in a house is very irregular, and consequently very often the optimum working point for the boiler is far from the real one.
It means that at least two points can be improved. On one hand, it is possible to use biomass instead of the traditional combustibles. It has in general a lower heating value, but it is much more ecological since it is a renewable energy source because it is CO2
neutral. On the other hand, it is possible to create a general heat district, using a boiler powerful enough to heat a certain amount of buildings located in the same area. As each building consumes the energy in different times, it is proved that the general demand of heat is much more regular than for the individual housings. It makes it possible to design accurately all the installation and reach an efficient general working of the heat exchanger.
Moreover, it will make sense to install a flue gas condenser, because a big amount of exhaust gases will leave the burner with a high temperature, what means energy that can be used to preheat the water. Another main advantage would be the capability of the condenser to capture particles in the flue gas.
The aim of this research is to calculate, if this condenser would be profitable, which amount of heat could be recovered, and finally to propose a certain design for the condenser that may be used to achieve this energy saving.
1.2 GENERAL DESCRIPTION
The boiler burns wood and heats water that will be distributed to the houses for heating purposes and to produce hot tap water. Usually, in most of the boilers, the cold water that leaves the houses comes directly back to the boiler, where it is heated again closing the cycle. But in this boiler, when the cold water has left the houses, before it arrives again to the boiler, the hot and wet exhaust gases will be used to preheat it, saving some energy and trying to collect some dangerous particles that otherwise would go to the atmosphere.
To get this transfer of heat from the stack gases to the water, a cross flow heat exchanger is used. The water goes inside thin tubes, and the gas flows across the bank of tubes.
In general, boilers can loose 20% of the combustion energy with the stack gases, and thanks to this flue gas condenser, or heat exchanger, more than 50% of the energy of this stack gases can be recovered depending on the working conditions.
The condensed water is collected in the bottom of the heat exchanger with all the particles that it has collected because of the contact with the gas.
Depending on some parameters, quite different results will be obtained. Of course the power output of the boiler, the air factor and the efficiency will have the main influence in the result, but also the moisture content of the wood will be very important to determine the amount of water that will be condensed. So the objective of the thesis has been to develop a software that can be used to determine the amount of heat that could be recovered and the heat exchanger area that is necessary. It should be easy to vary different varables and estimate their influence on the final result..
1.3 METHOD
The first step is to determinate the composition of the exhaust gases depending on the composition of the combustible and the air factor. They are supposed to be at 200 ºC.
Depending which power output is needed, the water mass flow used to heat the radiators is fixed, and then it is possible to figure out the geometry of the heat exchanger, which will determine the air and water fluxes and the heat transfer coefficients.
Then, if a typical condenser is used, the overall heat transfer coefficient of the heat exchanger must be calculated, for the dry part, and when the condensation begins, a film of water appears on the tubes, and then the temperature of the gas decreases slowly and the most important part of the heat is released by the latent heat of vaporization, depending mainly on the difference of temperatures between the condensation film and the gas, and the diffusion coefficient.
Otherwise, instead of using a normal heat exchanger, it is possible to spray water in the gas, until the dew point is reached. This way, there is condensation in all the heat exchanger, and more water may be collected.
As far as there is condensation, it will be very important to cool down the gases as cold as possible, what means that it will be very important to take the water from the radiators also as cool as possible. Besides that, also due to the condensation, corrosion will appear, because the gas carries a lot of particles that with the water could damage the exchanger. So it will be also an important point to study which materials are appropriat to resist these conditions. But anyway, it will not be studied in this project, which will be concentrated in the heat transfer topic.
2 THEORY
I this chapter the theory used to get the final results is presented step by step, beginning with the gas composition and properties and finally the steps to calculate the theoretical area needed to get a fixed temperature of the gas at the exit of the heat exchanger, or the temperature reached for a fixed heat exchanger geometry.
2.1 GAS
2.1.1 GAS COMPOSITION
The gas composition determines the enthalpy of the stack gases, since each component has different specific heat, molar mass, etc.
To calculate the exhaust gas composition, it is necessary to know the composition of the fuel that is burnt. The real composition of the wood will vary depending on many factors, but it may be quite easy to get a good approximation. in the composition of the fuel used for the calculations is in table 1 [3].
Table 1:Composition of the combustible used Weight % Matter As received:
C 60,0 26,7
H2 6,5 2,9
O2 32,0 14,2
N2 0,5 0,2
S 1,0 0,4
Ash 0,5 0,5
MC 55,0 55,0
When it is supposed complete oxidation of these elements with the oxygen of the air introduced; in a hypothetical stoichiometric combustion, the following reactions should occur:
C + O2 → CO2
H + 0,25O2 → 0,5H2O (1)
S + O2 → SO2
N + O2 → NO2
Other compounds than the products of complete oxidation of the elements of the fuel may appear, but in practice, theses few compounds dominate the reactions products, so it is reasonable to ignore the others since they do not have an important effect on the enthalpy of the exhaust gases, which is the aim of the calculation.
So it is necessary to do a mass balance for elements.
CxHyOzNrSt + uoO2 + 3,76 uoN2 → aCO2 +bH2O + cSO2 + dN2 (2)
The mass of each element has to be the same before and after the reaction. The number of moles of oxygen needed can be calculated from:
uo = a + 0,5b + c – 0,5z (3)
and a, b and c can be determined with the next mass balances using equation (2):
Carbon: a = x
Hydrogen: 2b = y (4)
Sulphur: c = t
Nitrogen: d = 3,76uo + 0,5r
Solving these equations it is possible to know the number of moles of each component per mol of combustible, so multiplying this number per the molar mass of each, it is possible to get the mass flow of each component, in kg of each component per kg of combustible.
i i
i N M
m& = ⋅⋅ (5)
Hereafter, in the combustion, it is used more air than the stoichiometric, because otherwise it is very difficult to mix all the fuel with the oxygen. So it will be necessary to take into account that the real oxygen and nitrogen flows after the combustion will not be the ones calculated for the stoichiometric reaction, but they will be:
2
2 mN
N
m& →λ⋅ & (in the stoichiometric) (6)
2
2 ( 1) mO
O
m& → λ− ⋅ & (in the products of the stoichiometric)
So once the mass of each component per kg of fuel is known, to find the mass flow of each component in the stack gases, it will be necessary to calculate the mass flow of the fuel burnt in the boiler.
2.1.2 FUEL FLOW
Each kind of fuel has a certain effective heating value. For biomass this value will strongly depend on the moisture content. For wood fuels the following formula gives a reasonable estimate of the heating value
Hi = 19,22 – 21,7 F (7)
To calculate the fuel flow needed, the power output of the boiler, the efficiency, the moisture content and the heating value of the fuel must be known. Once all these data are fixed, the required fuel flow can be calculated.
) 7 , 21 22 . 19
( F
P H
m P out
i out
fuel = ⋅ − ⋅
= ⋅
η
& η (8)
2.1.3 ENTHALPY OF THE GASES
The exhaust gases in conventional boilers are thrown to the atmosphere at approximately 200ºC, and they are the main part of the losses that the whole boiler installation has. A considerable part of the enthalpy of this gas can be used to preheat the water using a heat exchanger.
So, to estimate which part of the enthalpy of the hot gases can be transferred to the water, it is necessary to calculate first the energy they contain at the temperature in the beginning, and then the energy they will contain in the end, when they leave the heat exchanger. The heat that the stack gases release, is absorbed by the water, and it is the difference between the enthalpy of the gas in the entrance, and the enthalpy in the exit of the condenser.
) (h1 h2 m
Q& = &dg ⋅ − (9)
It is assumed that it will be possible to cool the gases down enough to cross the dew point, were condensation will appear. So the loss of enthalpy of the gases will be due to the change of temperature and also the amount of steam condensed.
The enthalpy of the gases in the beginning, were there is no water, can be calculated like the sum of the enthalpy of each component
( )
[
Cp t x h Cp t]
h1 = dg⋅ + ⋅ we+ w⋅ (10)
Where the values for cpdg are taken from Apendix A.
And the enthalpy of the gases when they leave the condenser can be calculated as
( )
[
Cp t x h Cp t x x C t]
h2 = dg⋅ + s⋅ we+ w⋅ +( − s)⋅ w⋅ (11)
if there is condensation, and otherwise
( )
[
Cp t x h Cp t]
h2 = dg⋅ + ⋅ we+ w⋅ (12)
Where t is the temperature in celsius of the gas in equations (10), (11) and (12).
So it will be necessary to know the humidity of saturation of the gas at the final temperature, what is the amount of steam that the gas can carry at a certain temperature.
To calculate this, assuming that it can be calculated like if it was air, the next formula will be used [6]:
ws a
ws
s p p
x p
−
= 0,622⋅
(13)
Where [6]:
2 , 8
) / 7235 0057
, 0 345 , 77 exp(
g
g g
ws T
T
p + ⋅T −
= (14)
Where Tg is the temperature of the mixture in Kelvin.
The amount of steam that condensates is the difference between the moisture that the gas brings after the combustion, and this humidity of saturation at the exit of the heat exchanger.
Of course, the final temperature of the gas will be very important to take advantage of the energy, because if the moisture content is high, there is a big amount of energy that comes from the condensation of the steam. This means that the temperature of the water when it leaves the radiators should be as low as possible. This will be a very important parameter to take into account to design all the heat installation.
2.2 WATER
Once the fuel flow, the gas flow, and the energy that this gas brings as a function of the temperature is known, the next step is to calculate the properties of the water that will cool down this air.
The two main values that are needed are the temperature when it arrives into the heat exchanger, and also the mass flow. They will determine how much we can bring down the temperature of the stack gases because the heat that the water will absorb is considered to be the same as the heat that the gas will release.
Actually, the temperature of the water depends on the working conditions of all the installation, so it will change depending on the circumstances. For example when it is very cold outside, with difficulty low temperatures of the water can be reached since this water cannot leave the buildings at low temperature, because then it will not be warm enough inside the buildings. Otherwise, when it is not so cold, the temperature of the water may be much lower, and as it has been said, the heat recovered can be much higher.
The mass flow of water will be directly related to the power output of the boiler. The energy that the boiler gives to the buildings is the difference between the enthalpy of the water when it arrives and when it leaves the housings.
)
( 1 2
2
1 H m Cp Tr Tr
H H Q
Pout = =∆ = − = &w⋅ w⋅ − (15)
So as Pout, Tr1 and Tr2 are not constant, the water flow and Tr2 will vary. But with equation 15, assuming a general situation and that all the power output is used for heating the water, the mass flow will be:
) (Tr1 Tr2 Cp
m P
w out
w = ⋅ −
& (16)
In this point of the theory, all the flows and the fluid properties have been explained.
The point that comes now is what to do with them to get heat transfer from the hot flow to the cold one.
2.3 HEAT EXCHANGER
Nowadays many different kinds of heat exchangers exist. They are used in many different industries and with very different fluids, geometries, dimensions, etc. But always with the same finality, to transfer the maximum amount of heat from one fluid to another using the minimum surface.
In some of them there is no condensation, so all the surface is dry and in some others condensation exist in almost all the heat exchanger and of almost all the gas. These are called condensers, and their aim is very often just to condensate a gas.
It is not so usual to have a condenser which aim is to cool down a big amount of noncondensable gas mixed with a condensable gas that crosses its dew point. Usually this kind of exchangers cool down the gas without coming into condensation, since the corrosion caused by the condensed steam mixed with other particles can damage the exchanger. In this research, the gas is mainly formed by N2, CO2, and other noncondensable gases, but depending on the moisture content of the combustible, it’s possible to have a considerable amount of steam that will condense when it reaches the dew point. So more energy will be extracted, but a difficult problem of materials will be caused.
It is quite a peculiar heat exchanger. Regarding to cool down the gases, three different ways of extracting the heat from the gas and transferring it to the water exist. The first one is the indirect condensation withous spraying water. It consists on a normal heat exchanger, with the difference that there will be a wet part. So the materials must be resistant to the corrosion that the wet liquid will provoke. The second option is the same idea and geometry, but spraying part of the water in the hot gases, until the dew point is reached. With this method condensation exists in all the heat exchanger, so much more water is collected and probably much more particles that will be attached to the wet tubes. And the last method is quite a different philosophy [5] that has not been studied in this research, but is consist in a direct contact unit. All the water is spayed in contact with the flue gases, causing condensation and extracting most of the heat.
2.3.1 FLUE GAS CONDENSER (WITHOUT SPRAYING WATER)
It is the most typical idea of heat exchangers. One of the fluids flows inside tubs, in this case the water, and the other one flows across the tubes. Depending on the direction of the fluxes, they can be classified in [7]:
-Parallel flow: the hot and the cold fluids flow in the same direction.
-Conterflow: the hot and the cold fluids flow in opposite directions
-Cross flow: the cold and hot fluids flow in perpendicular directions.
In all of them it is possible to find all kinds of geometries. Different numbers of passes of the cold fluid, of the hot one, etc. In this case, a cross flow heat exchanger has been chosen, where the water flows inside the tubes and the gas upwards or downwards across the tubes bank. The main reason is that it is simple; compact, easy to clean, and not very expensive; and also efficient, since the water has a much bigger heat capacity than the gas. But it will be explained later. First it is necessary to talk about the geometry, which will determine firstly the heat transfer coefficients for the water and the gas and secondly the overall heat transfer coefficient of the exchanger.
So the main important geometrical data of the heat exchanger is the diameter of the tubes, thelength, and the separation between them. Depending on the dimensions, and on the shape it will have, there will be more vertical or horizontal rows of tubes. The heat transfer characteristics will also change if the geometry of the tube banks is staggered or in-line. In figure 1 both configurations are shown:
Figure 1: Tubes arrangement
These geometrical values will define the water speed inside the tubes and the average and maximum gas speed outside the tubes; as well as the total outside and inside area of the heat exchanger. Naturally these values will be essential to determine the heat transfer characteristics of the exchanger, so next there are all the formulas to calculate them.
-Inside the tub:
Sy
Sz
Sy
Sz
4
2 i t
S = π⋅D
(17)
L D
Ait =π⋅ i⋅ (18)
t w
wt N
m m&
& = (19)
t wt
w S
u m
= ⋅ ρ
&
(20)
-Outside tubes:
L D
Aot =π⋅ o⋅ (21)
x y
g
g L L
u m
⋅
= ⋅
∞ ρ
&
(22)
)
max (
o y
y
g S D
u S
u = ∞ ⋅ − if z o
y
y S S D
D
S −
+
<
−
2 / 1 2 2
) 2
( (23)
And otherwise (as will be explained later.)
o z
y
y
g
D S S
u S
u
−
+
⋅
=
∞
2 / 1 2 max 2
2
2 (24)
-General
z y
t R R
N = ⋅ (25)
y y
y R S
L = ⋅ (26)
z z
z R S
L = ⋅ (27)
t it
i A N
A = ⋅ (28)
t ot
o A N
A = ⋅ (29)
2.3.1.1 Dry part of the heat exchanger
While there is no condensation, the heat flux and the variation of the temperature is calculated as follows:
2.3.1.1.1 Heat transfer coefficient of the water
The method to find the heat transfer coefficient of a fluid is always the same. The geometrical parameters determine the fluid properties, and then the flow type. Once it is known, the Nusselt number must be calculated and then the heat transfer coefficient is established.
For the water flowing inside a tube, the Reynolds number is calculated with the formula below
µ ρ⋅uw⋅Di
=
Re (30)
It will determine what kind of flow is there in the tubes, and depending on this, different formulas will be used to find the Nusselt number.
The water properties (in appendix A.2) are taken at the bulk temperature which is in Kelvins:
15 . 2 273
2
1+ +
=t t
Tb (31)
And very often it will be important to take into account the viscosity correction,
µw
φ = µ (32)
that is used to correct the Nusselt number because of the wall effect.
Next the different flows are defined, and how to calculate the Nusselt number with each one:
-Laminar: (Re < 2300)
Many different formulas can be found, but one of the most general and commonly used is equation (33), Seider-Tate (1936), [2], which gives a good result for a big range of properties
(Re Pr )1/3 0.14
86 .
1 ⋅ ⋅ ⋅ ⋅φ
= Di
Nu (33)
within the following ranges:
0,48 < Pr < 16,7 75 . 9 0044
.
0 <φ <
Pr 2
Re 0.14
3 / 1
>
⋅
⋅ ⋅ φ
L Di
-Transition: (2300 < Re < 10000)
( )
+
⋅
⋅
⋅
−
⋅
=
3 / 1 14
. 0 3 / 1 3
/
2 125 (Pr )
Re 116 .
0 L
D D
Nu φ i i (34)
-Turbulent: (Re > 10000)
Nu =0.023⋅Re0.8⋅Prn (35)
With n=0,3 for cooling, or 0,4 for heating
The heat transfer coefficient is finally obtained in equation (37) worked out from equation (36)
k D
Nu = h⋅ i (36)
And
i
w D
k
h = Nu⋅ (37)
2.3.1.1.2 Heat transfer coefficient of the gas
Even more important than heat transfer coefficient of the water inside tubes, will be the heat transfer coefficient of the gas to the cylinders in cross flow, because it will have a bigger influence in the overall heat transfer coefficient, since it is more difficult to get a good heat transfer coefficient with the gas than with the water.
The procedure is almost the same as before, but with gas outside tubes it is slightly more complicated. Firs the theoretical heat transfer coefficient for the gas flowing across one single tube must be calculated, and the it must be corrected depending on the number of tubes and their disposition because the heat transfer from a tube depends on its location within the bank.
In the range of lower Reynolds numbers, typically the tubes in the first row show similar heat transfer to those in the inner rows. For higher Reynolds numbers, flow turbulence leads to higher heat transfer from inner tubes than from the firs row. The heat transfer becomes invariant with tube location following the third or fourth row in the mixed-flow regime, occurring above Reynolds maximum.
Once again, there are a lot of different ways of calculating the Nusselt number, but one of the most used is the following one:
4 / 1 36
. 0 max
2 Pr
Pr Pr
Re
⋅
⋅
⋅
⋅ =
=
w
o C C m
k D Nu h
(38)
The Reynolds maximum is calculated with the maximum speed of the gas with equation (30). For a staggered arrangement, in most of the cases it will occur through the minimum frontal area (Sy-D), but this may not be the case for close spacing in the parallel direction, as when Sz is small. The flow enters the tube bank through the area (Sy-D) and then splits into the two areas [(Sy/2)2+Sz2
]1/2-D. If the sum of these two areas is less than Sy-D, then they will represent the minimum flow area. This is why there are
two possible equations to calculate the maximum speed, (23) and (24) which have to be used depending on the geometry.
The maximum Reynolds (Remax) is calculated as the average one but using the maximum speed instead of the average
µ ρ⋅ug ⋅Do
= max
Remax (39)
So finally again:
o
g D
k
h = Nu⋅ (40)
and all the gas properties are taken in the average temperature between the entrance and the exit of the heat exchanger.
This formula is valid in the ranges:
0,7 < Pr < 5000, Remax
1 < Remax < 2x104.
The parameters C and m must be taken depending on the conditions from table 2, [4]
Table 2: Parameters Remax, C and m for various Alligned and staggered Tube arrangements.
Moreover, as it was said, depending on the number of rows this coefficient will need the next correction:
Table 3: Parameter C2 for various tube rows and configurations
2.3.1.1.3 Tub resistance
Between the two fluids, there is the metal tub that will also obstruct a little bit the heat transfer between them. However, the wall will be very thin, and with a very high heat conduction coefficient. Consequently, it will have a really small influence on the overall heat transfer coefficient.
2.3.1.1.4 Overall heat transfer coefficient
Once the resistances are known, the overall heat transfer coefficient is determined and facilitates the heat flux from the hot stream to the cold one. It´s value depends on the heat transfer coefficients, the conductivity of the wall, and the thickness of the tub.
It can be referred to the outside, or inside area, and is calculated as follows:
( )
o o i i o i w i
h A
A L k
D D A h
U 1
2 / ln 1
1
⋅
⋅ +
⋅
⋅ + ⋅
=
π
(41)
( )
o i o o
w i o o
h L k
D D n A h A U A
1 2
/ 1
1
⋅ +
⋅
⋅ + ⋅
=
π
(42)
The high values in the denominator dominate the final value of the expression, this is the reason why the lowest heat transfer coefficient has the biggest influence. This is the reason why when the geometry is decided, what becomes the most important is to reach a high gas heat transfer coefficient.
2.3.1.1.5 Ε-NTU method
The most common methods to calculate the heat exchanged are the LMTD and the E- NTU methods. In this case, it is better to use the E-NTU method, because in the cross flow fluxes, it is necessary to correct the Logarithmic Mean Temperature Difference by some coefficients that in this range of temperature difference (between the gas and the water), may be very imprecise, since they do not appear in the usual tables of the books that explain this topic.
Besides that, nowadays it is much more common to use a method based on the effectiveness of the heat exchanger in transferring a given amount of heat.
The E-NTU method is based in three values:
1) Capacity rate ratio
C* = Cmin/Cmax (43)
Where
g
g Cp
m
Cmin = & ⋅ or m&w⋅Cw depending on which is lower (44)
w
w C
m
Cmax = & ⋅ or m&g⋅Cpg depending on which is higher (45)
2) Exchanger heat transfer effectiveness
Qmax
E = Q (46)
This is the ratio of the actual heat transferred divided by the maximum heat that could be reached if the exchanger worked in ideal counterflow.
3) Number of transfer units
Cmin
A
NTU =U ⋅ o (47)
The number of transfer units is a measure of the size of the exchanger.
The actual heat transfer is given by the enthalpy balance, with the next equation:
(T1 T2) C (t2 t1)
C
Q = h⋅ − = c⋅ − (48)
And the maximum possible heat transfer for the exchanger, would be the heat transferred to the fluid with lower m&⋅Cp, if the temperature change was the maximum temperature difference in the heat exchanger. It should be the difference between the entrance temperature of the gas and the entrance temperature of the water. So the maximum possible heat transfer is expressed as
( 1 1)
min
max C T t
Q = ⋅ − (49)
Observe that the value of E will range between zero and unity and that for a given E and Qmax , the actual heat transfer can be written as
( 1 1)
min T t
C E
Q = ⋅ ⋅ − (50)
Because
E = f (C*, NTU, flow arrangement) (51)
each exchanger has its own effectiveness relationship depending on their specific heat capacities, the area of the heat exchanger, and the flow arrangement. In this heat exchanger, which is in cross flow with the water unmixed and the gas mixed, the effectiveness is [4] :
( )
( )
⋅ − ⋅
−
−
= C NTU
E C 1 exp *
* exp 1
1 (52)
In this formula it is possible to see why the cross flow arrangement is a good choice.
The reason is that the efficiency will be very high since Cmin is much lower than Cmax. The physical reason of this high effectiveness is that while the water flowing in the tubes cools down the gas, the water temperature increases very slowly, so it can cool all over the tube with a high temperature difference. It could be almost considered that the temperature of the tubes is constant if you compare it to the temperature difference between the gas in the entrance and in the exit.
Refered to the heat exchanger design, two different problems can be raised depending on the available information or aims. On the one hand, if a final temperature is required, the area must be calculated, and on the other hand, if the area is established, the final temperature is what must be found.
In general, in this heat exchanger, the aim is to get condensation, so it will be known the final temperature of the dry gas, which will be the dew point. It means that it will be necessary to calculate the area needed to get condensation. Afterwards the heat in the wet part will be calculated, but do not forget that now it is refered only to the dry part.
Otherwise, if the fuel is very dry, or the heater must work very hard because it is very cold outside, in some cases maybe it is not possible to arrive to the dew point, and then the aim will be to calculate which is the final temperature of the gas, so in this case the area will be known, and the final temperature of the gas unknown.
Next it is explained how to solve both cases:
a) Final temperature of the gas known, and the aim is to calculate the area needed:
To calculate the dew point:
375 . 17 85
. 586 2880
6 .
5643 ⋅ 3− ⋅ 2+ ⋅ +
= x x x
Tdew (53)
If the temperature of the gases in the exit is higher than Tdew, there will not be condensation, and if it is lower, it will be necessary to calculate the area needed until Tdew is reached with this method, and later use a different method for the condensing part.
The final temperature of the gas is used to calculate the actual heat transfer
