Flue Gas Condenser for Biomass Boilers
2006:141 CIV MASTER’S THESIS
Flue Gas Condenser for Biomass Boilers
MARC CORTINA
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
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III
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
IV
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
V
7 APPENIX B (EXCEL EXPLAMATION) ...... 55
VI
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 m & d g 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 ------
Cp g Specific heat of the gas J/kgK
Cp dg Specific heat of the dry gas J/kgK
Cw Specific heat of the water J/kgK
hwe Latent heat of Vaporization kJ/kg
Tr 1,Tr 2 Temperature entrance and Temperature exit of the radiator ºC Temperature entrance and Temperature exit of the heat T1,T 2 exchanger ºC Temperature entrance and Temperature exit of the heat t1,t 2 exchanger ºC T Temperature K t Temperature ºC
pa Atmospheric pressure Pa Re Reynolds Number ------ρ Density Kg / m 3
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 ------2 St Cross section area inside tub m Ф Viscosity Correction ------k Thermal conductivity W / m·K
Nt Number of tubes ------
Ry Rows of tubes in the y direction ------
VII
Rz Rows of tubes in the z direction ------
Sy Separation between rows y direction m
Sz Separation between rows z direction m 2 Ait Area inside tube m
2 Aot Area outside tube m
2 Ai Total area inside tubes m 2 Ao Area outside total tubes m
L, L x 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, Cp gas ·m gas W
Cmax Maximum heat capacity rate, Cp water ·m water W 2 Ui Overall inside heat transfer coefficient W/m ·K 2 Uo Overall outside heat transfer coefficient W/m ·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 p g and p’ g Pa
p’ g pt - p c 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 ------
Marc Cortina Flue Gas Condenser page 1
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 CO 2 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.
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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.
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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 + O 2 → CO 2
H + 0,25O 2 → 0,5H 2O (1)
S + O 2 → SO 2
N + O 2 → NO 2 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 + u oO2 + 3,76 u oN2 → aCO 2 +bH 2O + cSO 2 + dN 2 (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:
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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 = 3,76u o + 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.
= ⋅ m& i Ni⋅ M i (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:
→ λ ⋅ m& N2 m& N2 (in the stoichiometric) (6) → λ − ⋅ m& O2 ( )1 m& O2 (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.
P P m = out = out (8) & fuel η ⋅ η ⋅ − ⋅ Hi 19( 22. 21 7, F)
Marc Cortina Flue Gas Condenser page 5
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.
= ⋅ − Q& m& dg (h1 h2 ) (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
= [ ⋅ + ⋅ ( + ⋅ )] h1 Cp dg t x hwe Cp w t (10)
Where the values for cp dg are taken from Apendix A. And the enthalpy of the gases when they leave the condenser can be calculated as
= [ ⋅ + ⋅ ( + ⋅ )+ − ⋅ ⋅ ] h2 Cp dg t xs hwe Cp w t (x xs ) Cw t (11 ) if there is condensation, and otherwise
= [ ⋅ + ⋅ ( + ⋅ )] h2 Cp dg t x hwe Cp w t (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]:
Marc Cortina Flue Gas Condenser page 6
,0 622 ⋅ p x = ws (13) s − pa pws Where [6]:
exp( 77 ,345 + ,0 0057 ⋅T − 7235 /T ) = g g pws 2,8 (14) Tg
Where T g 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.
= = ∆ = − = ⋅ ⋅ − Pout Q H H1 H 2 m& w Cp w (Tr 1 Tr 2 ) (15)
So as P out , Tr 1 and Tr 2 are not constant, the water flow and Tr 2 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:
Marc Cortina Flue Gas Condenser page 7
P m = out (16) & w ⋅ − Cp w (Tr 1 Tr 2 )
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 N 2, CO 2, 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
Marc Cortina Flue Gas Condenser page 8
-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:
Sz
Sz
Sy Sy
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:
Marc Cortina Flue Gas Condenser page 9
π ⋅ D 2 S = i (17) t 4
= π ⋅ ⋅ Ait Di L (18)
m = & w (19) m& wt Nt
m u = & wt (20) w ρ ⋅ St
-Outside tubes:
= π ⋅ ⋅ Aot Do L (21)
m& g u = (22) ∞g ρ ⋅ ⋅ Ly Lx
2 2/1 S y S y 2 u = u ⋅ if (S − D) < + S − D (23) g max ∞ − y z o (S y Do ) 2
And otherwise (as will be explained later.)
S ⋅ y u∞ 2 = ug max 2/1 (24) S 2 y + 2 − S z Do 2
-General
= ⋅ Nt Ry Rz (25)
Marc Cortina Flue Gas Condenser page 10
= ⋅ Ly Ry S y (26)
= ⋅ Lz Rz Sz (27)
= ⋅ Ai Ait Nt (28)
= ⋅ Ao Aot Nt (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
ρ ⋅u ⋅ D Re = w i (30) µ
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:
+ t1 t2 T = + 273 15. (31) b 2
And very often it will be important to take into account the viscosity correction,
Marc Cortina Flue Gas Condenser page 11
µ φ = (32) µ w 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
= ⋅ ( ⋅ ⋅ ) /1 3 ⋅φ 14.0 Nu 86.1 Re Pr Di (33) within the following ranges:
0,48 < Pr < 16,7 .0 0044 < φ < 75.9
/1 3 Re ⋅ Pr ⋅ D 14.0 i ⋅φ > 2 L
-Transition: (2300 < Re < 10000)
D /1 3 = ⋅ ()2 / 3 − ⋅ /1 3⋅φ 14.0 ⋅ + i Nu .0 116 Re 125 (Pr ) Di (34) L
-Turbulent: (Re > 10000)
Nu = .0 023 ⋅ Re 8.0 ⋅ Pr n (35)
With n=0,3 for cooling, or 0,4 for heating
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The heat transfer coefficient is finally obtained in equation (37) worked out from equation (36)
⋅ h Di Nu = (36) k
And
⋅ = Nu k hw (37) Di
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:
⋅ /1 4 h Do m 36.0 Pr Nu = = C ⋅C ⋅ Re Pr ⋅ 2 max (38) k Pr w
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 (S y-D), but this may not be the case for close spacing in the parallel direction, as when S z is small. The flow enters the tube bank through the area 2 2 1/2 (S y-D) and then splits into the two areas [(S y/2) +S z ] -D. If the sum of these two areas is less than S y-D, then they will represent the minimum flow area. This is why there are
Marc Cortina Flue Gas Condenser page 13 two possible equations to calculate the maximum speed, (23) and (24) which have to be used depending on the geometry.
The maximum Reynolds (Re max ) is calculated as the average one but using the maximum speed instead of the average
ρ ⋅u ⋅ D Re = g max o (39) max µ
So finally again:
⋅ = Nu k hg (40) Do
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, Re max
4 1 < Re max < 2x10 .
The parameters C and m must be taken depending on the conditions from table 2, [4]
Table 2: Parameters Re max , C and m for various Alligned and staggered Tube arrangements.
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Moreover, as it was said, depending on the number of rows this coefficient will need the next correction:
Table 3: Parameter C 2 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:
Marc Cortina Flue Gas Condenser page 15
1 U = (41) i 1 A ⋅ ln ()D / D A 1 + i o i + i ⋅ ⋅π ⋅ ⋅ hw 2 k L Ao ho
= 1 Uo ⋅ () (42) Ao 1 + Ao n Do / Di + 1 ⋅π ⋅ ⋅ Ai hw 2 k L ho
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* = C min /C max (43)
Where
⋅ C = m ⋅Cp or mw Cw depending on which is lower (44) min & g g &
= ⋅ or m ⋅Cp depending on which is higher (45) Cmax m& w Cw & g g
Marc Cortina Flue Gas Condenser page 16
2) Exchanger heat transfer effectiveness
Q E = (46) Qmax
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
U ⋅ A NTU = o (47) Cmin
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:
= ⋅ ( − ) = ⋅ − Q Ch T1 T2 Cc (t2 t1) (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
= ⋅ ( − ) Qmax Cmin T1 t1 (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
= ⋅ ⋅ ( − ) Q E Cmin T1 t1 (50)
Because
E = f (C*, NTU, flow arrangement) (51)
Marc Cortina Flue Gas Condenser page 17 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] :
−1 E = 1− exp ⋅ ()1− exp ()C *⋅NTU (52) C *
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 C min is much lower than C max . 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:
= ⋅ 3 − ⋅ 2 + ⋅ + Tdew 5643 6. x 2880 x 586 85. x 17 .375 (53)
If the temperature of the gases in the exit is higher than T dew , 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
Marc Cortina Flue Gas Condenser page 18
= ⋅ ( − ) Q Cmin T1 Tdew (54)
= ⋅ ( − ) Qmax Cmin T1 t1 (55)
So the effectiveness will be:
Q E = (56) Qmax
And inverting the formula for cross flow,
−1 E = 1− exp ⋅ ()1− exp ()C *⋅NTU ⇒ C * (57) −1 ⇒ NTU = ⋅ ln []1+ C *⋅ln( 1− E) C * So NTU is calculated and then finally
NTU ⋅C A = min (58) o U
b) Area known, and the aim is to calculate the final temperature of the gas:
It is possible to calculate the effectiveness with formula (52), and also
U ⋅ A NTU = o (59) Cmin
the actual heat transferred
= ⋅ = ⋅ ⋅ ( − ) Q Qmax E E Cmin T1 t1 (60)
And finally the temperature with the balance of the gas
= − Q T2 T1 (61) Cmin
Marc Cortina Flue Gas Condenser page 19
With this calculation, the dry part of the heat exchanger is solved. If there is no condensation, the heat recovered and the final temperature of the gas are already known, and if afterwards there is condensation, it is known that it will begin at Tdew and the dry area that has been used to reach it. The next step is to calculate the heat exchanged when the condensation appears.
2.3.1.2 Wet part of the heat exchanger
Once the gas has reached the dew point, the steam that it contains begins condensing. So there are two different kinds of heat transferred from the gas to the water; the sensible heat due to the temperature difference, and the latent heat due to the latent heat of vaporization of the steam when it changes the phase. It means that the usual heat transfer equations used in the dry part are not useful any more, and it is necessary to find a new method. There are really few books that talk about it, the calculations that follow are taken from a classic mass and heat transfer book [1]). When a mixture of a vapour and a noncondensable gas is fed to a condenser and the temperature of the tubes is below the dew point, a film of condensate forms on the tubes. If an equilibrium is presumed to exist on the surface of the condensate film, the partial pressure of the vapour at the tube wall corresponds to the cold condensate pc and the partial pressure of the vapour in the gas film lies between that at the condensate film pc and that in the gas body pv. In order for the vapour in the gas body to continue condensing into the condensate film, it must be driven across the gas film by the difference between the partial pressure of the vapour in the gas body and in the condensate. This passage of one component through another is called diffusion or mass transfer, and has a lot to do with the heat transfer.
The rate of diffusion can be calculated with the next formula:
Cp ⋅ µ 3/2 ⋅ g hg = k KG 3/2 (62) µ ⋅ ⋅ ⋅ Cp g pgf M m ρ ⋅ D
Where
p − p ' p = g ´g gf ln( p − p )' g g (63)
Marc Cortina Flue Gas Condenser page 20
and
p '= p − p g t c (64)
and the diffusivity can be calculated [2]
/3 2 T 1 1 D = .0 04357 ⋅ g ⋅ + (65) ⋅ ()/1 3 + /1 3 2 M M pt Vg Vsteam g steam
KG and U are similar in function, K G being to vapour and condensate pressure difference what U is to temperature difference in heat transfer. As we can see, the rates of diffusion and heat transfer do not occur independently. It is also very important to notice that the overall coefficient of heat transfer varies greatly during the condensation of the vapour because this vapour is being removed from the gas body and the percentage of inert gas increases. To establish an equation to solve the thermal problem, it is only necessary to sum up all the resistances in series at an average cross section in each increment of Q. The quantity of heat that leaves the gas film must be equal to the quantity picked up by the cooling water. And the total heat flow across the gas film is the sum of the latent heat carried by vapour diffusion into the condensate film plus the sensible heat removed from the gas because of the temperature difference between the gas and the condense film. The heat load per square meter of surface expressed in terms of the shell-side is:
Q = h ⋅ (T − T )+ K ⋅ M ⋅ h ⋅ ()()p − p = h ⋅ T − T (66) A g g c G v we v c w c w
Where pc is the partial pressure of the vapour at the tub wall, so it will be the partial pressure of the vapour at Tc, the pressure of saturation. Both pv and pc must be calculated with formula (14) at Tg and Tc respectively. The steps used to calculate the final temperature of the gas, or the area needed are outlined below:
1) Split the area in different parts for temperature ranges because as it was said before, some values change and it is not possible to calculate the heat exchanged in the whole condensing area. In the calculations, it has been decided to fix temperature steps, and calculate how much area was needed to cool down the gas from each high temperature until the colder one.
Marc Cortina Flue Gas Condenser page 21
2) Calculate the partial pressures of the gas and the vapour at the established temperatures, and then find the humidity of saturation. With the humidity of saturation and the temperature difference, it is possible to calculate the heat that the gas will release in that area:
C hoose C & m (67)
Where the subindex initial and end are referred to the properties in the higher and lower temperature (in ºC) of the range of the step, respectively.
3) Guess a value of T c, and then calculate p c, and check if the heat flow from the film to the gas and the flow from the water to the film are the same with equation (66). While they are different, change T c until the two values coincide. For each T c, the average temperature of the water must be updated using equation (68) and t 1. Then we will have the heat flux per unit of area, so it will be possible to calculate the area needed with equation (69), where Q is calculated with equation (67).
Q ⋅ A A o T = T + (68) end initial ⋅ Cp w m& w
Q A = (69) Q A
For the next intervals repeat this calculations until the final temperature of the gas is reached.
2.3.2 FLUE GAS CONDENSER (SPRAYING WATER)
The alternative to the indirect condensation is to spray water in the gases before it enters the heat exchanger. The aim of this method is to avoid having dry area in the heat exchanger. A pump should spray just the water needed to bring the vapour of the noncondensable gas to the dew point. The advantage of this method is that more particles would be collected since there would be a bigger wet area. Moreover, the heat exchanged with condensation seems to be more efficient, so the exchanger might be smaller. Besides, the negative point is that there is some waste of energy, which is this water that will not be useful any more because instead of being pre-heated and used to heat the buildings, it will be collected with the rest of condensed water. It would make
Marc Cortina Flue Gas Condenser page 22 sense to think about cleaning this collected water and use it again in the circuit, since it will be quite hot. Anyway, this is not the topic that of project. To decide how much water should be spayed in the gases, there are two basic points. The first one is that when more moisture is spayed in the gas, its dew point is lower. But the temperature of the gas also decreases because the water that is being sprayed will be the water from the condensate, which is expected to be at 45 or 50ºC. So there is a specific amount of steam that brings the mixture to the dew point. The new temperature of the gas mixture is calculated with the next enthalpy balance: the heat absorbed by the water being changed into steam at a certain temperature, is the same that the gas mixture looses because of its temperature decrease.
m ⋅Cp ⋅ (t − t ) = [(h + Cp ⋅t )− C ⋅t ]⋅ x'⋅m (70) & gas gas bspray aspray gas we w aspray w 1 & dg
And once it is known the new temperature, it is necessary to sum this new water flow to the gas flow that leaves the boiler:
m’ g = m g + m adw (71)
m’ w = m w + m adw (72)
x’ = x + x ad (73)
And the next step is to calculate the heat released as it has been explained in the wet part of the heat exchanger of indirect condensation. If the final temperature is known, it will be necessary to do steps until the final temperature, and if the known data is the area, the procedure will be the same, but checking in every step that the total area used does not exceed the area of the heat exchanger.
Marc Cortina Flue Gas Condenser page 23
3 RESULTS AND ANALYSIS
3.1 INTRODUCTION
All the theory that has been explained until now is what has been used to make all the calculations. The program used has been excel. Below there are the results that have been found making use of the theory and the excel tools.
3.2 GAS COMPOSITION
Depending on the gas composition, the heat taken from these gases can be significantly different. The main parameters that must be taken into account are the moisture content of the fuel and the air factor, and the results that are interesting to check are the amount of steam in the stack gases, the dew point of the mixture and the enthalpy they will have. If nothing else is specified, all the results obtained are for a power output of 150kW, a boiler efficiency of 85% and an air factor of 1,6. Below are the tables that show different results for different moisture contents in the fuel and afterwards some comments about them.
3.2.1 FUEL MASS FLOW, TOTAL GAS MASS FLOW, AND DRY GAS MASS FLOW DEPENDING ON THE MOISTURE CONTENT
Table 4: mass flows depending on the moisture content
MC fuel flow (kg/s) gas flow (kg/s) dry gas flow (kg/s) 35% 0,013 0,116 0,106 40% 0,014 0,119 0,108 45% 0,016 0,122 0,110 50% 0,018 0,126 0,112 55% 0,020 0,132 0,115 60% 0,024 0,139 0,119
Marc Cortina Flue Gas Condenser page 24
3.2.2 GAS COMPOSITION DEPENDING ON THE MOISTURE CONTENT
Table 5:gas flow composition (kg/kg of fuel)
MC H2O CO 2 N2 SO 2 O2 35% 0,76 1,42 6,13 0,01 0,70 40% 0,78 1,31 5,65 0,01 0,64 45% 0,80 1,20 5,18 0,01 0,59 50% 0,82 1,09 4,70 0,01 0,54 55% 0,83 0,98 4,23 0,01 0,48 60% 0,85 0,87 3,75 0,01 0,43
Table 6: gas flow composition (kg/s)
MC H2O CO 2 N2 SO 2 O2 35% 9,8 18,2 78,7 0,2 9,0 40% 11,0 18,5 79,9 0,2 9,1 45% 12,5 18,8 81,3 0,2 9,3 50% 14,4 19,2 83,1 0,2 9,5 55% 16,8 19,8 85,4 0,2 9,7 60% 20,1 20,5 88,5 0,2 10,1
3.2.3 AMOUNT OF STEAM IN THE STACK GASES
Table 7:humidity ratio of the stack gases
MC H2O (g/s) Dry gas (g/s) x (kg w/ kg dry air) 35% 9,8 106,1 0,092 40% 11,0 107,7 0,102 45% 12,5 109,6 0,114 50% 14,4 112,0 0,129 55% 16,8 115,1 0,146 60% 20,1 119,2 0,168 *flow units in g/s
Marc Cortina Flue Gas Condenser page 25
Amount of water in the dry gas
140,0
120,0
100,0
mass flow 80,0 (kg/s) H2O 60,0 Dry gas 40,0
20,0
0,0 35% 40% 45% 50% 55% 60% MC (%)
Figure 2: steam in the stack gases as a funtion of to the dry gas flow
We can see that if the fuel has high moisture content, its heating value decreases, and this is the reason why more fuel is needed to achieve the same power output (table 7). Anyway, the dry gas flow and the total gas flow remain almost constant, because although there is more fuel, less air is needed since a big part of it is moisture and does not burn. It is seen in table 7, that shows how the steam flow increases with the moisture content, while the dry gas flow decreases per kilogram of fuel. It is very important to notice in figure 2 how the humidity ratio increases a lot when the moisture content is high. The dew point depends on this ratio, which means that it will have a big influence.
3.2.4 DEW POINT
To determine the dew point is very important because it is the temperature when the first drops in the gas appear. When it happens, then E-NTU method is not useful any more and it is necessary to calculate the heat exchanged with the condensation balance, as it has been said in the theory. Also it must be noticed that when the gas reaches this temperature, the steam that it contains begins releasing the heat of the change of phase, the latent heat of vaporization. Next there are two graphics. In figure 3 it is possible to see the humidity ratio of the mixture depending on the moisture content, and in figure 4, the dew point depending also on the moisture content of the fuel:
Marc Cortina Flue Gas Condenser page 26
Humidity ratio
0,18 0,17 0,16 0,15 0,14 0,13 0,12
x (kg x w/kg(kg dgas) 0,11 0,1 0,09 0,08 35% 40% 45% 50% 55% 60% MC (%)
Figure 3: Humidity ratio of the stack gases as a funtion of to the moisture content
Figure 3 shows that the humidity ratio increases almost lineally with the moisture content in this range of values. Figure 4 shows what happens with the dew point:
Dew point
63 61
59 57
55
53 T dew (ºC) 51 49
47
45 35% 40% 45% 50% 55% 60% MC (%)
Figure 4: Dew Point of the stack gases as a funtion of to the moisture content
Obviously, the dew point also changes significantly with different moisture contents. In figure 4 we see that while with 60% of moisture condensation begins at more than 61ºC, if the fuel has 35% of moisture content, we don’t have condensation until 51ºC. It will have a huge influence in the amount of heat that can be recovered, because with low
Marc Cortina Flue Gas Condenser page 27 moisture contents, it may be difficult to get condensation if the temperature of the water is not quite cold.
3.3 HEAT RECOVERED
Next there is a first approximation of the theoretical heat that can be recovered in the condenser of a small boiler, of 150kW of power output, working in normal conditions, which in this case means that it has been assumed that the gas will arrive to the heat exchanger at 200ºC and will leave at 54ºC. Whereas T 1 will be constant of 200ºC in all the calculations, the temperature of the water returns to the boiler will be one of the most important parameters that will be changed in the calculations, having a very big influence on the heat that will be possible to recover.
Theoretical Heat Recovered
45 40
35 30
25
20 Q Q (kW) 15 10
5
0 35% 40% 45% 50% 55% 60% MC (%)
Figure 5: Heat recovered depending on the moisture content
Figure 5 shows how, with these conditions, the heat recovered with a moisture content of 60% would be two times bigger than the heat recovered if it was 40%. Of course there is a huge difference. It means that it is already possible to deduce that the amount of moisture that the fuel carries will be determinant to decide if the condenser will be useful. Also in figure 5 it is possible to see how the slope increases a lot between 40% and 45%. This is because between this ranges of temperatures condensation appears, and then, most of the heat recovered is not due to the temperature difference of the gases, it is due to the steam that is condensing. When condensation begins, really small changes of temperature in the gas release a lot of heat. Although x increases almost lineary with the moisture content, the heat recovered growth is not linear because the value of the gas mass flow gows up when the moisture content is higher.
Marc Cortina Flue Gas Condenser page 28
3.4 WATER AND GAS FLOWS AND HEAT TRANSFER COEFFICIENTS
Up till now only theoretical values has been shown. Without thinking about the heat exchanger, just assuming difference of temperatures. From now on, it will be necessary to calculate the real values of the water and the gas flow, determined for the geometry of the heat exchanger. Also temperatures for the water and the gas will be assumed to be able to calculate the heat transfer coefficients with the fluids properties.
3.4.1 TOTAL WATER FLOW
To fix the total water flow, it is necessary to forget for a while the heat exchanger, and be focused in the buildings, because the available water flow will depend on which temperature of the water is needed in the entrance and exit of the radiators of the housings. It has been assumed that in general, the water will arrive at the houses at 80ºC and will leave them at 50ºC. As it has been said, it will depend a lot on many circumstances, but to do the calculations and get the results it is necessary to establish some hypothesis. All the results that come next are based on this hypothesis, but it is possible to change them if some specific cases need to be studied. So if the water temperature difference is 30ºC and the power output is 150kW, the following result is obtained using equation (16)
Table 8: Water flor in the circuit ∆T water P output T av Cp water Water flow 30 ºC 150 kW 70 ºC 4,21 kJ/kg 1,2 kg/s
3.4.2 WATER FLOW INSIDE TUBES, GAS FLOW OUTSIDE, AND HEAT TRANSFER COEFFICIENTS
In general if the water flows with a high speed inside the tubes, the heat transfer coefficient is also higher. However, there are some values that shouldn’t be exceeded, because then a lot of energy is lost with friction with the tube walls. The parameters that establish the water heat transfer coefficient are the inside diameter of the tubes and the total number of tubes in the heat exchanger, because the whole water flow must be split in all these tubes. The outside heat transfer coefficient depends on some more variables. In this case, since there are many tubes, regarding to table 3 the coefficient for the tube bank can be considered the same as the coefficient for one single tube, so C 2=1. The basic parameters that define it are: the outside diameter of the tubes and the separation between them, which defines the maximum gas speed (formula 23 or 24) It will be favourable to have a high maximum gas speed, but once again, it is not possible to decide the geometry only with the aim of having a good heat transfer coefficient, it is also important not to have big pressure losses, because then a lot of energy is lost. It is
Marc Cortina Flue Gas Condenser page 29 important to find a reasonable value for the heat transfer coefficient avoiding important pressure lost. To check how all the parameters affect to the area of the heat exchanger, the flow, the heat transfer coefficients, and the overall heat transfer coefficient, different tables have been elaborated with the aim of getting information to be able to decide about which geometry should be chosen for the heat exchanger.
a) Fixing the number of tubes to 1200 and the tube length to 50cm, the following results are obtained:
- Total area inside the tubes
- Total area outside the tubes
- Total cross section area inside the tub
- Water speed inside the tube
- Heat transfer coefficient of the water
Depending on the inside and outside diameter of the tubes.
Table 9:Water side characteristics
Di Do Ai Ao St uw hw 5 6 8,5 10,2 236 0,051 346 6 7 10,2 11,9 339 0,035 288 7 8 11,9 13,6 462 0,026 247 8 9 13,6 15,3 603 0,020 216 10 11 17,0 18,7 942 0,013 173 12 14 20,4 23,8 1357 0,009 144 14 16 23,8 27,1 1847 0,007 124 17 19 28,8 32,2 2724 0,004 102 20 22 33,9 37,3 3770 0,003 86 23 25 39,0 42,4 4986 0,002 75 26 28 44,1 47,5 6371 0,002 67 30 32 50,9 54,3 8482 0,001 58
2 *Units: D i, D o in mm; A it , A ot , u w, h w in SI units; S t in cm
The most significant result is the big influence of the diameter of the tubes in the water flow, when the number of tubes is fixed. The total seccion of the tubes increases so the water speed decreases. Consequently the heat transfer coefficient also decreases, as is shown in figure 6. It is important to mention that this graphic is just theoric to check
Marc Cortina Flue Gas Condenser page 30 what happens with the water heat transfer coefficient. For the highest values of the diameter, the number of tubes would be reduced.
Water Heat Transfer coeficient
400 350 300 250 200 150 h (W/m2) 100 50 0 5 6 7 8 10 12 14 17 20 23 26 30 D (m)
Figure 6: Water heat transfer coefficient depending on the inside tube diameter
High speed of the water has the advantage of a very high heat transfer coefficient. It will be important to take into account this point to calculate the overall heat transfer coefficient. Otherwise, it is also important to notice that the total inside and outside area of the heat exchanger is linearly related to the tub diameter, while the cross section area increases exponentially. So the number of tubes and their diameter are the two variables that control the area of the heat exchanger that will reach the aim of a certain heat transfer rate.
b) Fixing the number of tubes to 1200, 40 vertical rows and 30 horizontal with a tube length of 50cm, a vertical separation equal to the horizontal, and a moisture content of 40%, the following results are obtained:
- Height and width of the heat exchanger
- Maximum velocity of the gas
- Free velocity of the gas
- Heat transfer coefficient of the gas
Marc Cortina Flue Gas Condenser page 31
depending on the horizontal separation between tubes, for the following outside diameters.
Diameter outside tubes = 10 mm
Table 10: gas heat transfer coefficient Sy Sx Ly Lz umax uav hg 15 15 0,45 0,60 1,74 0,58 54,8 17 17 0,51 0,68 1,24 0,51 46,3 19 19 0,57 0,76 0,97 0,46 40,9 21 21 0,63 0,84 0,79 0,41 37,0 23 23 0,69 0,92 0,67 0,38 34,0 25 25 0,75 1,00 0,58 0,35 31,7 27 27 0,81 1,08 0,51 0,32 29,7 31 31 0,93 1,24 0,41 0,28 26,8
Diameter outside tubes = 15 mm
Table 11: gas heat transfer coefficient Sy Sx Ly Lz umax uav hg 20 20 0,60 0,80 1,74 0,43 44,8 23 23 0,69 0,92 1,09 0,38 35,4 26 26 0,78 1,04 0,79 0,33 30,2 29 29 0,87 1,16 0,62 0,30 26,8 32 32 0,96 1,28 0,51 0,27 24,3 35 35 1,05 1,40 0,43 0,25 22,4
Diameter outside tubes = 20 mm
Table 12: gas heat transfer coefficient Sy Sx Ly Lz umax uav hg 25 25 0,75 1,00 1,74 0,35 38,8 28 28 0,84 1,12 1,09 0,31 30,7 31 31 0,93 1,24 0,79 0,28 26,1 34 34 1,02 1,36 0,62 0,26 23,2 37 37 1,11 1,48 0,51 0,23 21,0 40 40 1,20 1,60 0,43 0,22 19,4
Marc Cortina Flue Gas Condenser page 32
Diameter outside tubes = 30 mm
Table 13: gas heat transfer coefficient Sy Sx Ly Lz umax uav hg 35 35 1,05 1,40 1,74 0,25 31,7 38 38 1,14 1,52 1,09 0,23 25,0 41 41 1,23 1,64 0,79 0,21 21,3 44 44 1,32 1,76 0,62 0,20 18,9 47 47 1,41 1,88 0,51 0,18 17,2 50 50 1,50 2,00 0,43 0,17 15,8
Both the outside diameter of the tubes and the horizontal separation between them have a direct influence on the heat transfer coefficient of the gas (see from table 10 until 13), which is much lower than the one for the water (table 9). The first is that the tubes must be as thin as possible because both are on the side to get high coefficients for both, the gas and the water. The limitation to use very thin tubes is the mechanical properties, because depending on how long they are, they must resist the strength of the gas without bending. Besides that, to get a high heat transfer it is also important to keep the tubes as close as possible, but in this case with the limitation of the pressure loss, since the maximum speed of the gas is much higher than the free speed before flowing across the tubes. In this project the pressure loss is not studied, but it will be necessary to take reasonable values, which make that the maximum speed does not exceed three times the speed of the free flow. While thin tubes are positive to get a high efficiency and also make it possible to have a small heat exchanger, big diameters and separation between tubes make it less efficient and also bigger. With all the information that has been given until now, it is already possible to state that the aim will be to build a heat exchanger with the tubes as thin as possible and also as close as possible while it does not have negative effects in other aspects. To confirm this statement, some values for the overall heat transfer coefficient have been calculated for different values of the tube diameter and separation between them, with the same hypothesis as in the previous calculations
Marc Cortina Flue Gas Condenser page 33
Table 14: General characteristics and coefficients depending on the geometry and disposition of the tubes
Di Do Sy Sz Ly Lz hw hg Uo Ao 8 11 17 17 0,51 0,68 213,4 47,7 36,5 20,7 8 11 25 25 0,75 1,00 213,4 31,2 26,0 20,7 11 14 21 21 0,63 0,84 155,2 39,2 29,6 26,4 11 14 30 30 0,63 0,84 156,1 25,8 21,3 26,4 14 17 26 26 0,78 1,04 122,0 31,3 23,9 32,0 14 17 35 35 1,05 1,40 122,0 22,2 18,2 32,0 17 20 30 30 0,90 1,20 100,4 27,4 20,7 37,7 17 20 40 40 1,20 1,60 100,4 19,4 15,8 37,7 20 24 36 36 1,08 1,44 85,4 22,8 17,3 45,2 20 24 50 50 1,50 2,00 85,4 15,5 12,7 45,2
These results are very clear, and confirm what has already been said, that it makes a big difference to the overall heat transfer coefficient which dimensions of tubes are used. To make the next calculations, that are required to find the heat recovered, and the area needed, it is necessary to decide one specific geometry.
3.5 AREA NEEDED
With the purpose to decide all the characteristics of the heat exchanger, the program will be used to calculate the area needed to cool down the gases of different moisture contents for varied water temperatures. Depending on these results, all the parameters will finally be decided and then the behaviour of one specific heat exchanger for different conditions will be checked, to calculate the amunt of heat that can be recovered. It must be decided about some variables regarding all the results presented before. The length and the width of the heat exchanger will be fixed, and the height is the variable that will be free. Also the diameter and the separation between the tube rows must be established now. Comparing with [9] it seems reasonable to take the following values and in table 15 they are shown togeter with the ones that they imply:
Table 15: Decided Geometry
2 Di (mm) Do (mm) Lx (m) Ry Sy (mm) Ly (m) Uo (W/m K) 8 11 0.5 30 17 0.51 36,5
It is important to remember that all the calculations are done assuming an air factor of 1.6, a power output of 150kW and an efficiency of 85%, a gas temperature of 200ºC and Sy equal to S z. Now it is possible to calculate the total area needed, the wet and dry area, for different moisture contents and different initial temperatures of the cold water leaving the houses. It will give the height and the number of vertical rows of the heat exchanger, since the
Marc Cortina Flue Gas Condenser page 34 other two horizontal dimension variables are already fixed. It will be assumed that the gases will not be cooled down more than 5 degrees above the initial water temperature, because when the temperatures are so close, the area needed increases too much in relation to the heat that is recovered and it would not be worthwhile any more.
3.5.1 FLUE GAS CONDENSER (WITHOUT SPRAYING WATER)
First, the heat recovered and the area needed will be calculated if the classical heat exchanger was used. In this case without spraying water and transferring all the heat with convection and conduction through the tubes. Two situations have been studied:
a) Final temperature of the gases of 55ºC and water available at 50ºC and moisture contents from 40% to 55%.
Table with the results:
Table 16: Heat recovered and area needed depending on the moisure content Final temperature of the gases 55ºC and cooling water available at 50ºC
MC 55% 50% 45% 40% Q total 30,3 24,5 19,9 19,1 Q dry part 21,2 20,4 19,8 19,1 Q wet part 9,1 4,1 0,1 0,0 Q sensible 21,9 20,7 19,8 19,1 Q laten 8,5 3,8 0,1 0,0 A tot 19,2 15,6 12,7 12,4 A dry 11,3 11,8 12,7 12,4 A wet 7,9 3,8 0,1 0,0
The flollowing Graphics are obtained:
Marc Cortina Flue Gas Condenser page 35
Heat Recovered (Tgas=55)
35,0
30,0
25,0
20,0 Q (kW) Q dry 15,0 Q w et 10,0
5,0
0,0 55% 50% 45% 40% MC %)
Figure 7: Heat Recovered in the dry and wet part. Final temperature of the gases 55ºC and cooling water available at 50ºC
A needed (Tgas=55)
20,0 18,0 16,0 14,0 12,0 A (m2) 10,0 A w et 8,0 A dry 6,0 4,0 2,0 0,0 55% 50% 45% 40% MC %)
Figure 8: Dry and wet area needed. Final temperature of the gases 55ºC and cooling water available at 50ºC
Marc Cortina Flue Gas Condenser page 36
Q recovered and A needed (Tgas=55)
35,0
30,0
25,0
Q (kW), 20,0 A (m2) 15,0 Q A 10,0
5,0
0,0 55% 50% 45% 40% MC %)
Figure 9: Total heat recovered and area needed.Final temperature of the gases 55ºC and cooling water available at 50ºC
When the moisture content is high, more heat is recovered because of the condensation, but also more area in needed. Of course if the moisture content is low, this problem does not appear, since with water at 50ºC it does not make sense, or it is even impossible to cross the dew point. So the first conclusion is that if the aim of the heat exchanger is to extract also the heat from the condensation of the steam, it will be very important to have at one’s disposal combustible with high moisture content.
b) Final temperature of the gases of 50ºC and water available at 45ºC and moisture contents from 40% up to 55%.
Table with the results:
Table 17: Results depending on the moisure content. Final temperature of the gases 50ºC and cooling water available at 45ºC
MC 55% 50% 45% 40% Q total 38,4 32,4 27,6 23,8 Q dry part 21,2 20,4 19,8 19,4 Q wet part 17,2 11,9 7,8 4,4 Q sensible 22,6 21,4 20,5 19,8 Q laten 15,8 11,0 7,1 4,0 A tot 22,0 18,9 16,5 14,6 A dry 9,8 10,0 10,3 10,8 A wet 12,2 8,9 6,1 3,7
Marc Cortina Flue Gas Condenser page 37
The flollowing Graphics are obtained:
Heat Recovered (Tgas=50)
40,0 35,0 30,0 25,0
Q (kW) 20,0 Q w et 15,0 Q dry 10,0 5,0
0,0 55% 50% 45% 40% MC %)
Figure 10: Heat Recovered in the dry and wet par. Final temperature of the gases 50ºC and cooling water available at 45ºC
A needed (Tgas=50)
25,0
20,0
15,0 A (m2) A w et 10,0 A dry
5,0
0,0 55% 50% 45% 40% MC %)
Figure 11: Dry and wet area needed. Final temperature of the gases 50ºC and cooling water available at 45ºC
Marc Cortina Flue Gas Condenser page 38
Q recovered and A needed (Tgas=50)
40,0 35,0 30,0
25,0 Q (kW), 20,0 A (m2) Q 15,0 A 10,0
5,0 0,0 55% 50% 45% 40% MC %)
Figure 12: Total heat recovered and area needed. Final temperature of the gases 50ºC and cooling water available at 45ºC
If the temperature of the gases is 50ºC instead of 55ºC, and the moisture content is high, almost 10kW more can be recovered with almost the same area. Furthermore, if the moisture content is of 55%, it is possible to recover more than 1.5 times the energy that would be recovered with a moisture content of 40%, with less than 8 more square meters of outside surface area of the tubes. So according to these results, it is now clear that there are two main factors that will determine the amount of heat possible to recover, the temperature of the water and the moisture content of the fuel. It is also important to notice that the heat exchanger becomes more efficient if the water is colder. In the next table there is the ratio of the heat exchanged per square meter in the dry and wet part for the two cases:
Cooling down to 55ºC:
Table 18: Heat recovered per square meter
MC 55% 50% 45% 40% Dry Part 1,9 1,7 1,6 1,5 Wet Part 1,2 1,1 1,0 ---- Total 1,6 1,6 1,6 1,5 *units: kW/m 2
Marc Cortina Flue Gas Condenser page 39
Cooling down to 50ºC:
Table 19: Heat recovered per square meter
MC 55% 50% 45% 40% Dry Part 2,2 2,0 1,9 1,8 Wet Part 1,4 1,3 1,3 1,2 Total 1,7 1,7 1,7 1,6 *units: kW/m 2
It is also important to stress (see table 18 and 19) that if the moisture content is high, the efficiency is slightly higher. Cooling down to 55ºC, if the moisture content is 55%, the heat recovered per square meter in the dry area is around 1.9, whereas the heat recovered per square meter in the wet part is around 1.2. It means that the dry part of the heat exchanger is more efficient, but a considerable amount of energy will be left if we do not get condensation by cooling down below the dew point. In the case of cooling down to 50ºC, with a moisture content of 55%, it results in 2.2 and 1.4 for the dry and wet part respectively. This result introduces the possibility to consider to spray water, because with a higher amount of steam in the mixture, the heat exchanged per square meter increases. Morover, condensation would then take place in the whole heat exchanger with the advantage that more particles may be collected. This is what is shown next.
3.5.2 FLUE GAS CONDENSER (SPRAYING WATER)
After the water is sprayed, the gas enters in to the heat exchanger saturated. It means that if the amount of water added is the correct one, there will not be any dry area. Below there is the area needed for the same situations as the previous heat exchanger, cooling the mixture down to 55ºC and 50ºC.
a) Final temperature of the gases of 55ºC and water available at 50ºC and moisture contents from 40% up to 55%.
Table with the results:
Table 20: Heat recovered, Area used and Heat recovered per square meter. Final temperature of the gases 55ºC and cooling water available at 50ºC
MC 55% 50% 45% 40% Q total 29,7 21,8 17,3 14,0 A total 18,5 12,7 9,6 7,5 Q/A 1,6 1,7 1,8 1,9
Marc Cortina Flue Gas Condenser page 40
*units: kW/m 2
The flollowing Graphics are obtained:
Q recovered and A needed (Tgas=55)
30,0
25,0
20,0 Q (kW), 15,0 A (m2) Q 10,0 A
5,0
0,0 55% 50% 45% 40% MC %)
Figure 13: Energy recovered and area needed. Final temperature of the gases 55ºC and cooling water available at 50ºC
b) Final temperature of the gases of 50ºC and water available at 45ºC and moisture contents from 40% up to 55%.
Table with the results:
Table 21: Heat recovered, Area used and Heat recovered per square meter. Final temperature of the gases 50ºC and cooling water available at 45ºC
MC 55% 50% 45% 40% Q total 37,6 29,6 24,9 21,5 A total 19,6 14,8 12,1 10,3 Q/A 1,9 2,0 2,0 2,1 *units: kW/m 2
The flollowing Graphics are obtained:
Marc Cortina Flue Gas Condenser page 41
Q recovered and A needed (Tgas=50)
40,0 35,0 30,0 25,0 Q (kW), 20,0 A (m2) Q 15,0 A 10,0 5,0 0,0 55% 50% 45% 40% MC %)
Figure 14: Energy recovered and area needed. Final temperature of the gases 50ºC and cooling water available at 45ºC
The results show how the exchanger with sprayed water is more efficient if the water temperature is low and the moisture content too. Spaying water seems to be good, but it is important to make sure that the new dew point will be higher enough than the water temperature. Otherwise a lot of heat is lost heating the water and vaporizing it. It is also important to say that if the aim is to take the maximum amount of heat, spraying water will not be the good choice, since there is always some energy lost. To check which option is the best, the same heat exchanger must be used with the same conditions and then compare the results.
3.6 PROTOTYPE STUDY
This is the last step to design the heat exchanger. After calculating how much area is needed to reach certain temperatures of the gas with some stipulated values for the temperature of the cold water, now it is time to decide which heat exchanger we want to have. The aim is to find a heat exchanger that works properly under the normal conditions. As it has been seen, depending on the temperature of the water and on the composition of the fuel, different areas are needed to extract a certain amount of heat. As more area is used, more heat is recovered, but also more expensive is the heat exchanger and depending on the conditions, with less area it would be possible to extract almost the same energy. To decide about the area, it will be assumed that most of the time the conditions will be favorable, so it will work with cooling water under 50ºC and high moisture content in
Marc Cortina Flue Gas Condenser page 42 the fuel. It means that the area should be quite big to take advantage of a significant amount of heat. These are the dimensions that have been chosen:
Table 22: Prototyc Characteristics
Di Do Lx Ly Lz Sy Sz Nt Ao 8 11 0,5 0.51 0.49 17 17 1042 18
3.6.1 CONDENSER WITHOUT SPRAYING WATER
It will be studied for two different water temperatures
a) Water available at 50ºC and moisture contents from 40% up to 55%.
Table 23: Results depending on the moisture content.Water available at 50 ºC and area of 18m 2
MC 55% 50% 45% 40% Q total 29,6 26,1 23,3 20,9 Q wet part 8,4 5,7 3,5 1,5 Q dry part 21,2 20,4 19,8 19,4 Q sensible 21,8 20,9 20,1 19,5 Q latent 7,8 5,3 3,2 1,4 T end gases 55,4 54,1 52,9 52,1 A dry 11,3 12,0 13,0 14,6 A wet 6,8 6,2 5,3 3,6
Marc Cortina Flue Gas Condenser page 43
Heat Recovered
30,0
25,0
20,0
Q (kW) 15,0 Q wet Q dry 10,0
5,0
0,0 55% 50% 45% 40% MC %)
Figure 15: Heat recovered in the wet and in the dry part. .Water available at 50 ºC and area of 18m 2
b) Water available at 45ºC and moisture contents from 40% up to 55%.
Table 24 : Results depending on the moisture content.Water available at 45 ºC and area of 18m 2
MC 55% 50% 45% 40% Q total 35,9 31,9 28,6 25,8 Q wet part 14,7 11,4 8,8 6,5 Q dry part 21,2 20,4 19,8 19,4 Q sensible 22,3 21,4 20,6 20,0 Q latent 13,5 10,5 8,0 5,9 T end gases 51,7 50,4 49,2 48,4 A dry 9,6 9,9 10,4 11,0 A wet 8,5 8,1 7,7 7,2
Marc Cortina Flue Gas Condenser page 44
Heat Recovered
40,0
35,0
30,0
25,0
Q (kW) 20,0 Q wet 15,0 Q dry
10,0
5,0
0,0 55% 50% 45% 40% MC (%)
Figure 16: Heat recovered in the wet and in the dry part. Water available at 45 ºC and area of 18m 2
The results are quite reasonable. 23 m 2 seem to be a good value for the area because the final temperature of the gases is not much higher than the temperature of the water when the moisture content is high, and in the opposite case, when the moisture content is low, it does not arrive to be too close. It means that a big amount of the energy can be recovered even for high values of the moisture content, and also that there is not wasted surface area for the cases when the fuel is dryer.
3.6.2 CONDENSER SPRAYING WATER
Below are the results for exactly the same heat exchanger, but spraying water:
Table 25: Prototype Characteristics
Di Do Lx Ly Lz Sy Sz Nt Ao 8 11 0,5 0.51 0.49 17 17 1042 18
The same heat exchanger is also studied for the two different water temperatures as in the previous case
Marc Cortina Flue Gas Condenser page 45
a) Water available at 50ºC and moisture contents from 40% up to 55%.
Table 26: Heat Recovered, t final gases and heat per square meter. Water available at 50ºC and area of 18m 2
MC 55% 50% 45% 40% Q 30,5 24,6 21,1 18,4 T 54,5 53,3 52,6 52,2 Q/A 1,7 1,4 1,2 1,0
Q recovered
35,0
30,0
25,0
20,0
15,0
10,0
5,0
0,0 55% 50% 45% 40%
M C ( %)
Figure 17: Q recovered. Water available at 50ºC and area of 18m 2
b) Water available at 45ºC and moisture contents from 40% up to 55%.
Table 27: Heat Recovered, t final gases and heat per square meter. Water available at 45ºC and area of 18m 2
MC 55% 50% 45% 40% Q 37,6 31,3 27,6 24,6 T 49,9 48,5 47,8 47,4 Q/A 2,1 1,7 1,5 1,4
Marc Cortina Flue Gas Condenser page 46
Q recovered
40,0 35,0 30,0 25,0 20,0 15,0 10,0 5,0 0,0 55% 50% 45% 40% M C ( %)
Figure 18: Q recovered. Water available at 45ºC and area of 18m 2
In this case it seems that 18 m 2 is too large area in some cases. In table 27 it is possible to see that the final temperature of the gas when the moisture content is 40% is only 2.4 degrees higher than the water temperature in the beginning. For high moisture contents the heat exchanger works slightly better with spraying water, but when the fuel is dry, it is not able to reach the same heat recovered as the without spraying water condenser. The reason of this change of behaviour is due to the small difference of temperatures between the condensed gas and the water in tubes. On the one hand, when there is condensation, the heat transfer is better, but on the other hand, if the new dew point is not high enough, the temperatures are too close to reach a good heat transfer, and a big part of the area can be quite useless. It means that the spraying water heat exchanger works better as smaller it is, since the temperature difference between the gas and the water is higher.
3.7 COMPARISON AND FINAL PROTOTYPE
The heat exchanger of 18 m 2 works more or less the same way if water is spayed or not, or maybe a little bit better without spraying water. With high moisture contents the one with sprayed water works slightly better, and the opposite for low moisture contents. The heat exchanger with the spray water works with less temperature difference than the other one, and the result is that if the area is big; many tubes are almost wasted because they exchange almost no heat. It gives the idea that maybe the best solution would be a heat exchanger with sprayed water, but smaller. 18 square meters was too much for the spraying water heat exchanger. It is supposed that for smaller areas it should work better, and now it will be compared the 18 square meters without spraying water heat exchanger, to a 14 square meters heat exchanger working in the same conditions. Below there are the results after spraying water:
Marc Cortina Flue Gas Condenser page 47
a) Water available at 50ºC and moisture contents from 40% up to 55%.
Table 28: Heat Recovered, t final gases and heat per square meter. Spraying water heat exchanger of 14 square meters
MC 55% 50% 45% 40% Q 28,5 23,2 19,9 17,5 T 55,6 54,1 53,4 52,8 Q/A 2,0 1,7 1,4 1,3
Comparison with the indirect exchanger of 18 m 2 working in the same conditions
Comparison Indirect Spray
30
25
20
Q(kW) 15 Indirect 10 Spray
5
0 55% 50% 45% 40% MC (%)
Figure 19: Heat recovered without spraying water and 18 m 2 compared to spraying with 14 m 2
b) Water available at 45ºC and moisture contents from 40% up to 55%.
Table 29: Heat Recovered, t final gases and heat per square meter. Spraying water heat exchanger of 14 square meters
MC 55% 50% 45% 40% Q 35,6 29,8 26,3 23,6 T 51,3 49,7 48,8 48,2 Q/A 2,5 2,1 1,9 1,7
Marc Cortina Flue Gas Condenser page 48
Comparison with the indirect exchanger of 18 m 2 working in the same conditions
Comparison Indirect Spray
40
35
30
25
Q(kW) 20 Indirect 15 Spray
10
5
0 55% 50% 45% 40% MC (%)
Figure 20: Heat recovered without spraying water and 18 m 2 compared to spraying with 14 m 2
It seems to be quite a difficult choice. Now the spraying water exchanger is more efficient, but unless the moisture content is very high and the water temperature low, it is not able to recover the same amount of energy, and as tables 19 and 20 show, it works worse for dry gases or high water temperatures. The question is if saving 4 square meters is reason enough to loose this heat. In fact, it must also be taken into account that the smaller condenser is supposed to collect much more particules, a big benefit. If the price of the heat exchanger increases to much with the surface area, or if it is not possible to have a big heat exchanger, to spray water is the best option. The conclusion is that to extract a big amount of heat in all the situations, which means to always bring the temperature of the gas close to the temperature of the water, a big heat exchanger without spraying water must be used. Otherwise, if it is more important to have a small heat exchanger, the condenser with sprayed water will be more efficient, although in some cases more heat will be left, mainly when the moisture content is low. Also is important to say, that when the moisture content of the fuel is constant, it will be more advisable to use a spraying water exchanger. Otherwise, if the moisture content of the fuel varies, a without spraying water heat exchanger will work better, since the area needed for this kind of heat exchanger does not depend so strongly on the moisture content of the fuel.
Marc Cortina Flue Gas Condenser page 49
4 CONCLUSIONS AND FURTHER IMPROVEMENTS
4.1 CONCLUSIONS
The main conclusion is that the heat exchanger can be very useful and rentable because it is possible to recover quite a big amount of heat while the particules are collected with the wet tubes. It must be said that the good operation of the exchanger will depend a lot on the circumstances when it is being used. As we have seen in the results, it will mainly depend on the moisture content of the fuel and the available temperature of the cooling water that is heated. If the fuel has high moisture content, the heating value is lower so more flow is needed, but it is also assumed to be cheaper, and probably some energy will be saved because it has not been necessary to dry it. It entails that all the installation must be designed to work with these conditions. From the fuel introduced in the boiler until the radiators that are in the houses.
I must also be said that it would be very interesting to check which amount of particulesi can be collected thanks to the condensation. The small particles that the usual boilers emit are very dangerous for the humans, and it would really be interesting, if it was confirmed, that in addiction to recover more heat with the condensation, there is another important problem solved, or at least improved. From my modest points of view, the philosophy of the district heating with an exchanger is a really efficient, ecological and economical way of heating. If it is possible to make it work under constant conditions, I think that it is possible to speak about a very efficient system, environmentally friendly and not too sophisticated, that is easy to manage.
4.2 IMPROVEMENTS
4.2.1 IMPROVEMENTS IN THE BOILER
The boiler with the heat exchanger has a very high efficiency. Working in good conditions, some manufacturers affirm that the global efficiency can be even higher than 90%. Really few elements in the field of the energy can speak of such good efficiencies. It means that although everything can be improved, if this boiler works properly, really little more energy will be possible to supply. Anyway, there are still some losses of energy. In the chimney, after leaving the condenser, the gas is at its dew point. It means that it will need some energy to be able to go upwards without continue condensing. Not much, but this energy could be saved
Marc Cortina Flue Gas Condenser page 50 if somehow it was possible to extract the steam from the gases, because then there would be neither condensation nor corrosion and the gases could leave through the chimney without any problem. The other point where there are losses is in the water that is condensed and collected in the bottom of the exchanger. Especially if water is sprayed to the stack gases. This water is supposed to be hotter than the water that leaves the condenser but it can not be mixed into the district heating water and the excess temperatures is so low that is not wortahile to use a heat exchanger.
4.2.2 IMPROVEMENTS IN THE SOFTWARE
In the software there are much more aspects to improve than in the boiler installation. This software only lays the foundations to calculate accurately the heat that can be exchanged in the condenser. The most evident improvement, from my point of view, is the way of calculating the heat transfer coefficient of the gas across the tube banks. It is very difficult to really estimate this value, and there are a lot of different coefficients that can be used to approximate it closer to the reality, but they are based on the experience, so they are different in each case. They correct the entrance effect, the flow next to the wall that does not flow between the tubes, etc. Another possible improvement would be to calculate better also the heat resistances from the water to the condense film, since it has been assumed that only the heat transfer coefficient of the water determines the heat flow from the condense film to the water. It may not be a big imprecision, since in [1] it is assumed that the film of condensate is equivalent to the tub, and the resistance of the tub is negligible. Anyway, small differences here have an important effect because Tc is really determinant to know the heat transferred because of the condensation of the steam. Another improvement could be to calculate the properties of the gas not in the average temperature between the entrance and the exit of the heat exchanger, but to take into account that it does not cool down linearly, but it is exponential. Anyway, it would not change the results significantly, because the gas properties do not change so much in the range of temperatures that the condenser is working. Finally, changing the topic a little bit to speak not directly about heat transfer, say that instead of having four different excel programs depending on if water is sprayed or not, and if the temperature or the area is known, maybe it would be possible to have one compact program with the possibility of introducing the type of calculation that we want to make. It would need some informatic work and quite a complex actualization of the values, but of course it is possible.
Marc Cortina Flue Gas Condenser page 51
5 REFERENCES
[1] Donal Q.Kern: Process Heat Transfer, Mc Graw Hill, ISBN 0-07-463217-5
[2] J. P. Holman: Heat Transfer, Mc Graw Hill, ISBN 0-07-844785-2
[3] Björn Kjellström, Jenny Lindberg and Gudrun Keikkala: Combustion And Gasificaton In Theory And Pratctice, Luleå University Of Technology
[4] Bejan, Adrian, Kraus, Allan: Heat Transfer Handbook John Wiley&Sons Available. [online] http://www.knovel.com/knovel2/Toc.jsp?BookID=725
[5] Boiler Burnen, Flue Gas Condenser. Available. [online] http://www.energysolutionscenter.org/BoilerBurner/Eff_Improve/Efficiency/Flue_ Condensers.asp
[6] MOIST AIR (Psychometry ), Luleå University Of Technology
[7] Heat Exchanger Technology Knowledge. Available [online] http://www.engineersedge.com/heat_exchanger/heat_exchanger_menu.shtml
[8] Yunus A. Çengel: Thermodinamics And Heat Transfer Mc Graw Hill ISBN 0-07- 114109-X
[9] Shell And Tube Heat Exchanger. Available [online] http://www.engr.pitt.edu/Chemical/undergrad/lab_manuals/shell_tube.pdf
Marc Cortina Flue Gas Condenser page 52
6 APPENDIX A
A.1 Gas properties at atmospheric pressure
N2 Specific Dynamic Kinematic Thermal Termal Prandtl T (K) Density Heat Viscosity Viscosity Conducivity Diffusivity Number (ρ) (Cp) (µ) () (λ) (α) (Pr) 300 1,142 1040,8 1,784E-05 1,563E-05 0,0262 2,204E-05 0,71 400 0,854 1045,9 2,198E-05 2,574E-05 0,0334 3,734E-05 0,69 500 0,682 1055,5 2,570E-05 3,766E-05 0,0398 5,530E-05 0,68
CO2 Specific Dynamic Kinematic Thermal Termal Prandtl T (K) Density Heat Viscosity Viscosity Conducivity Diffusivity Number (ρ) (Cp) (µ) () (λ) (α) (Pr) 300 1,797 871,0 1,499E-05 8,321E-06 0,0166 1,059E-05 0,77 400 1,342 942,0 1,932E-05 1,439E-05 0,0246 1,946E-05 0,74 500 1,073 1013,0 2,326E-05 2,167E-05 0,0335 3,084E-05 0,70
H2O Specific Dynamic Kinematic Thermal Termal Prandtl T (K) Density Heat Viscosity Viscosity Conducivity Diffusivity Number (ρ) (Cp) (µ) () (λ) (α) (Pr) 300 0,586 2041,0 9,100E-06 2,160E-05 0,0181 2,036E-05 1,03 400 0,554 2000,0 1,344E-05 2,420E-05 0,0264 2,330E-05 1,00 500 0,441 1977,0 1,704E-05 3,860E-05 0,0357 3,870E-05 0,96
O2 Specific Dynamic Kinematic Thermal Termal Prandtl T (K) Density Heat Viscosity Viscosity Conducivity Diffusivity Number (ρ) (Cp) (µ) () (λ) (α) (Pr) 300 1,301 920,3 2,063E-05 1,586E-05 0,0268 2,235E-05 0,71 400 0,976 942,0 2,554E-05 2,618E-05 0,0346 3,768E-05 0,66 500 0,780 972,2 2,991E-05 3,834E-05 0,0417 5,502E-05 0,70 According to [2] and [8].
Marc Cortina Flue Gas Condenser page 53
A.2 Water properties
= + ⋅ − ⋅ 2 + ⋅ 3 Cp w 2820 11 82. T .0 03502 T .0 00003599 T
λ = − .0 3835 + .0 00525 ⋅T − .0 000006265 ⋅T 2 ρ = 741 .966 + .1 9613 ⋅T − .0 00371211 ⋅T 2
1830 2 −13 73. + + .0 0197 ⋅T − .0 0000147 ⋅T µ = 10 T µ ⋅Cp Pr = λ *All the temperature must be in Kelvin
A.3 Other Important Properties
A.3.1 Molar Mass
H2O CO2 N2 SO2 O2 0,0180 0,0440 0,0280 0,0641 0,0320
These are the molar masses that have been used in kg per mole. [3].
A.3.2 Water Latent Heat of Vaporization
The value of the latent heat of vaporization has been assumed as 2502 kJ/kg [6]
A.3.3 Molecular Weight
The value of the molecular weight of the gas has been assumed as: 29.9 The value of the molecular weight of the steam has been assumed as: 18.8
Both according to [2]
Marc Cortina Flue Gas Condenser page 54
A. 3.4 Atmospheric pressure
The value of the atmospheric pressure has been assumed as: 101325 Pa .
A.3.5 Thermal Conductivity
The value for the thermal conductivity of the tubes has been assumed as 150 W/(m·ºC)
Marc Cortina Flue Gas Condenser page 55
7 APPENDIX B (EXCEL EXPLANATION)
Next there is a brief explanation of the excel program that has been used to achieve the results. Each excel sheet has its own explanation, since each one corresponds to one topic.
B.1 SHEETS Gas comp
This first page calculates the composition of the stack gases. The fuel composition is specified, as well as the air factor, the moisture content and the relative humidity of the air used for the combustion. With this information, using the theory form [4] as it has been explained in the theory, the moles of each component per mole of fuel are calculated.
CalcWorkshhet CALCULATE THE FLOW GASES AND THE COMPOSITION OF GASES VARYING MOISTURE CONTENT AND AIR FACTOR
Combustible Content Humidity HR: 0,6000 Aar 0,5000 FI: 0,0074 MC 40 Air Factor As Matter recieved: LAMBA: 1,60 C 60,0000 35,7000 H2 6,5000 3,8675 O2 32,0000 19,0400 N2 0,5000 0,2975 S 1,0000 0,5950 Ash 0,5000 0,5000 MC 40,0000 40,0000 100,0000 Calculation Sheet
Analysis Analysis O2 Species Molar Mass (g/Kg) (mol/kg) (mol/Kg) H2O CO2 N2 SO2 O2 C 12,0100 357,0000 29,7252 29,7252 29,7252 H2 2,0200 38,6750 19,1460 9,5730 19,1460 O2 32,0000 190,4000 5,9500 -5,9500 N2 28,0100 2,9750 0,1062 0,1062 S 32,0600 5,9500 0,1856 0,1856 0,1856 Ash 5,0000 #¡DIV/0! 0,0000 MC 18,0200 400,0000 22,1976 0,0000 22,1976
Sum O2: 33,5338
H2O CO2 N2 SO2 O2 Nitrogen in Air: nN2=3,76*O2 126,0872 126,0872 Theoretical dry air: nad 159,6211 Moisture content in air 1,1812 1,1812 Total theoretical air: nao 160,8023 Theoretical amount of gas: g0 198,6291 42,5248 29,7252 126,1934 0,1856 Theoretical amount of dry gas 156,1043 Amount of Oxygen in total gas 20,1203 20,1203 Nitrogen 75,6523 75,6523 Moisture 0,7087 0,7087 Total Amount Of Gas 295,1104 43,2335 29,7252 201,8458 0,1856 20,1203
Marc Cortina Flue Gas Condenser page 56
BASE: Kg of fuel mol/kg fuel 0,0227 Nm3/Kg fuel Theoretical dry Air 159,6211 3,6234 Theoretical Air 160,8023 3,6502 Total dry Air 255,3937 5,7974 Total Air 257,2836 5,8403 Theretical amount of gas 198,6291 4,5089 Theoretical Dry Gas 156,1043 3,5436 Total Gas 295,1104 6,6990 Total Dry Gas 251,8769 5,7176 Total H2O 43,2335 0,9814
Amount of water. BASE: Kg of dry gas mol H2O/mol dry gas Total H2O 0,1716
Mc
The table in this page is the amount of water per kg of fuel in the stack gases. It is just a result, it is unimportant for the further calculations.
Flow of fuel
Depending on the moisture content, the heating value of the fuel is calculated and there are tables where the fuel flow needed is calculated for different efficiencies and power outputs when the moisture content changes from 20% to 40%.
X sat
With formula (13), the saturation humidity ratio that the gas has for a specific temperature is calculated. Furthermore, with the graphic that defines this ratio of saturation for each temperature, and using the excel tool that finds the formula to approximate a curve from a graphic, it is possible to calculate the dew temperature for each humidity ratio, which will be very useful in further pages to know when will condensation appear, if it does.
Heat Exchanged
This is one of the most important sheets in the program. Defining the main parameters this sheet calculates all the mass flows and the heat recovered if the gas is cooled down to the temperatures that have been specified, but distinguishing between the heat exchanged from the dry part, the wet part, the sensible heat, the latent heat, the heat recovered from the dry gases and the heat recovered from the steam.
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heat exchanged EXAMPLE OF THE HEAT EXCHANGED
Hipotesis
MC 40% Value that we can decide "at least now" Efficiency 85% Consequences of our deign Power Output (kW) 150 Constants or propeties Aar 0,50% Final Results Heat. Val. (MJ/kg) 12,49 Hipotesis LAMBA: 1,60 t1 (celsius) 200 t2 (celsius) 54 t dew (celsius) 53,07 x sat. (kg/kg) (t2) 0,1077
Values
H2O CO2 N2 SO2 O2 Amount of Gas (mol/kg fuel) 43,23 29,73 201,85 0,19 20,12 Molar mass (kg/mol) 0,02 0,04 0,03 0,06 0,03 Amount of Gas (kg/kg fuel) 0,78 1,31 5,65 0,01 0,64 Amount of Gas (mol/s) 0,61 0,42 2,85 0,00 0,28 Amount of Gas (kg/s) 0,01 0,02 0,08 0,00 0,01 Volume of Gas (m3/s) [1] 0,02 0,01 0,06 0,00 0,01
M dry gas (kg/mol) 0,0302 Fuel Flow (Kg/s) 0,0141 x (mol w/mol d gas) 0,1716 x(kg water/kg dry gas) 0,1023 Dry gas flow (kg/fuel) 7,6176 Total gas flow (kg/s) 0,1187 Dry gas flow (kg/s) 0,1077 Dry gas flow (mol/s) 3,5595
Exist condensation? 1 "if x>xs-->1, otherwise 0"
Constants (200 celsius)
Cp CO2 (kJ/kmolK) 43,84 Cp N2 (kJ/kmolK) 29,52 Cp SO2(kJ/kmolK) 45,8 Cp O2 (kJ/kmolK) 30,88
Cp H2O (gas) (kJ/kgK) 1,84 Cp H2O (liquid) (kJ/kgK) 4,19 Sensible heat recovered H H2O (kJ/kg) 2502 H initial (1) (kW) 26,36 Enthalpy of the dry gas H end (2) (kW) 7,17 H1-H2 sensib (KW) 19,18 H = fuel flow*t* Σi(gi*Cpi) Condensation heat recovered H initial (1) (kW) 22,30 H end (2) (kW) 6,02 H initial (1) (kW) 27,55 H end (2) (kW) 28,88 Enthalpy of the steam and water H1-H2 cond (KW) -1,34
H moisture = x*(Cpwg*t1+hwe)*(dry gas flow) H moisture and water = [(xs*condensation+x*(1-condesation))*(Cpwg*t2+hwe)+((x-xs)*condesation)*(Cpwl*t2)]*(dry gas flow)
H moisture (1) (kw) 31,60 H moisture and water (2) (kw) 30,04 "if function is included"
Totall Power recovered (kw) = H1-H2 17,84
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Geometry and initial conditions
Here the exchanger topic is tackled. Following the theory the water flow is calculated and all the geometry is defined. As a result there is the water flow in each tube. All these values will obviously be used afterwards to tackle the heat transfer coefficients.
Geometry & Initial Cond
HEAT EXCHANGER GEOMETRY AND INITIAL CONDITIONS Tubes mm and mm2 SI units Value that we can decide "at Din 8 0,008 least now" Thikness Tube 2 0,002 Consequences of our deign Dout 11 0,011 Constants or propeties Stube 50 0,0001 Final Results L (mm) 500,0 0,5 Hipotesis Ain tube 12566400 0,013 Aout tube 17278800 0,017 Rows y 30 30 Rows z 58 58 Ntubes 1736 1736 Stotal tubes 87273 0,09 Ain total tubes 21818181818 22 Aout total tubes 30000000000 30 Lx (m) 500,0 0,50 Ly (m) 510 0,51 Lz (m) 984 0,98
Geometry (Staggered)
Sy 17 0,017 Sd 24,0 0,0240 Sz 17 0,017 Sx-D 6,0 0,006 Sz-D 6 0,006
Boolean expressions
Diagonal max speed? 0
Calculation Total Water flow (at Tav) SI units ∆T water (K) 30 30 P output (kw) 150 150000 T water av (celsius) 70 343,15 Cp water (kJ/kg K) 4,21 4206,59 Water flow (kg/s) 1,2 1,2
T water av in tube (celsius) 52,0 325,2 Density Water (T=52cels) 987,2 Water flow (m3/s) 0,00120
Water Flow (in each tube)
Water flow in tube (kg/s) 0,001
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Water flow in tube (m3/s) 6,93E-07 Water speed (m/s) 0,0138
Gas temperatures
T initial 200 T end 54
Water temperatures
T initial 45 T end 49
h water
The properties of the water are calculated with the formulas from appendix A.2 for the average temperature. With them and using the theory the Reynolds number defines the kind of flow and depending on it an if function decides the right formula to calculate the Nusselt number and finally the heat transfer coefficient of the water flowing in the specified conditions inside the tubes.
h water HEAT TRANSFER COEFFICENT OF THE WATER Water propeties
specific heat (cp) 2820+11,82*T-0,03502*T^2+0,00003599*T^3 conductivity ( λ) -0,3835+0,00525*T-0,000006265*T^2 density ( ρ) 741,966+1,9613*T-0,00371211*T^2 Viscosity 10^(-13,73+1830/T+0,0197*T-0,0000147*T^2) Pr µ*cp/ λ
Inital state
T (celsius) 45 Value that we can decide "at least now" T (Kelvin) 318,15 Consequences of our deign Constants or propeties cp (J/kg) 4194,81 Final Results conductivity (W/m °C) 0,65 Hipotesis density (kg/m3) 990,22 Viscosity 0,00063 Viscosity (wall) 0,00058 Pr 4,07
Final state
T (celsius) 49,00 T (Kelvin) 322,15
cp (J/kg) 4196,67
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conductivity (W/m °C) 0,66 density (kg/m3) 988,55 Viscosity 0,00059 Viscosity (wall) 0,00054 Pr 3,77
Average state
Tav (celsius) 47,00 Tav (Kelvin) 320,15
cp (J/kg) 4195,74 conductivity (W/m °C) 0,66 density (kg/m3) 989,40 Viscosity 0,00061 Viscosity (wall) 0,00056 Pr 3,92
Calculations
Reynols 178,60 Laminar? 1 (Re<2300)=>1 Trans 0 (Re>2300 and Re<10000)=>1 Nusselt LAM 2,32 Turbul 0 (Re>10000)=>1 Heat tr Coef (w/m2) if LAM 189,73 Eq 6.10? 1 (Re*Pr*d/L>10)=>1 Nusselt TRA -18,36 Heat tr Coef (w/m2) if TRA -1503,21
Nusselt TUR 3 Heat tr Coef (w/m2) if TUR 205,87
Heat tr Coef (w/m2) 189,73 h gas
In a similar way as the water, the heat transfer coefficient is calculated. But now, instead of using formulas to calculate the gas properties, they are taken from appendix A.1 and calculated for an average state, which is supposed to be the average temperature between the entrance and the exit of the heat exchanger. In the end, the parameters C and m are chosen with a function depending on the Reynolds number and the Nusselt number can be calculated to fins finally the heat transfer coefficient of the gas flowing across the tube bank.
H gas HEAT TRANSFER COEFFICENT OF THE GAS Gas propeties
N2 T (K) Density Cp µ Viscosity Lambda Alfa Pr 300 1,142 1040,8 1,784E-05 1,563E-05 0,0262 2,204E-05 0,71
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400 0,854 1045,9 2,198E-05 2,574E-05 0,0334 3,734E-05 0,69 500 0,682 1055,5 2,570E-05 3,766E-05 0,0398 5,530E-05 0,68
CO2 T (K) Density Cp µ Viscosity Lambda Alfa Pr 300 1,797 871,0 1,499E-05 8,321E-06 0,0166 1,059E-05 0,77 400 1,342 942,0 1,932E-05 1,439E-05 0,0246 1,946E-05 0,74 500 1,073 1013,0 2,326E-05 2,167E-05 0,0335 3,084E-05 0,70
H2O T (K) Density Cp µ Viscosity Lambda Alfa Pr 300 0,586 2041,0 9,100E-06 2,160E-05 0,0181 2,036E-05 1,03 400 0,554 2000,0 1,344E-05 2,420E-05 0,0264 2,330E-05 1,00 500 0,441 1977,0 1,704E-05 3,860E-05 0,0357 3,870E-05 0,96
O2 T (K) Density Cp µ Viscosity Lambda Alfa Pr 300 1,301 920,3 2,063E-05 1,586E-05 0,0268 2,235E-05 0,71 400 0,976 942,0 2,554E-05 2,618E-05 0,0346 3,768E-05 0,66 500 0,780 972,2 2,991E-05 3,834E-05 0,0417 5,502E-05 0,70
Average T (K) Density Cp µ Viscosity Lambda Alfa Pr 300 1,205 1098,0 1,680E-05 1,506E-05 0,0240 2,012E-05 0,75 400 0,912 1110,4 2,104E-05 2,386E-05 0,0314 3,327E-05 0,72 500 0,728 1128,1 2,484E-05 3,530E-05 0,0386 4,992E-05 0,71
Inital state
T (celsius) 200 T (Kelvin) 473,15
Density Cp µ Viscosity Lambda Alfa Pr T initial (473.15) 0,778 1123,3 2,382E-05 3,223E-05 0,0367 4,545E-05 0,72
Final state
T (celsius) 54,00 T (Kelvin) 327,15
Density Cp µ Viscosity Lambda Alfa Pr T final (328,15) 1,125 1101,3 1,795E-05 1,745E-05 0,0260 2,369E-05 0,74
Average state
Tav (celsius) 127,00 Tav (Kelvin) 400,15
Density Cp µ Viscosity Lambda Alfa Pr T av (400,65) 0,911 1110,4 2,105E-05 2,388E-05 0,0314 3,330E-05 0,72 T wall (325,15)) 1,084 1041,8 1,867E-05 1,767E-05 0,0276 2,513E-05 0,71
Calculation Total Gas flow (at Tav)
Gas flow (kg/s) 0,119 Gas flow (m3/s) 0,130
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V (m/s) 0,511 V max (m/s) 1,447
Calculations Calcul Q and T
Reynols 243,16 Reynols (max) 688,96 Nusselt av (p483) 16,69 Heat tr Coef (w/m2) 47,71
C 0,71 m 0,5
Energy condense
Probably this is the most complex calculation. As it has been exposed, it is not possible to calculate directly the heat exchanged in the condensing part of the exchanger with the data that is known. It will be necessary to calculate some intermediate values and iterate. To begin, it is necessary to calculate the pressure of the gas at the dew point, and also the humidity ratio of saturation. Then, a lower temperature is chosen, and the same properties are calculated, as well as the mean molecular weight.
Afterwards, T c must be assumed, P c, p’ gas , p gf , D and K G are calculated. Then Q/A of the condensed steam, the gas and the water are calculated. Q/A condensate plus Q/A sensible must be equal to Q/A of the water. If they are not equal, it is necessary to calculate the T c that would get the equilibrium for these conditions. Then, the program calculates the difference between this hypothetical T c and the T c that was assumed, splits this value per 30, and adds it to the T c supposed in the beginning. It is the only way that I have found to make T c converge to a value that finally makes equal the heat transferred from the gas to the condensate film, and the one transferred form the condensate film until the the water. Then the first step is done, and the area needed is calculated dividing the Q that the gas has lost in this step per the Q/A found. Then the same calculations must be repeated changing the temperature interval. T dew must be substituted for the final temperature that was taken in the last calculation, and another actual final temperature must be assumed. These steps must be repeated until the final temperature reaches the temperature to which the gases will be cooled down (in the case it is known) or until the total area used will be the area of the heat exchanger (in case this area is the known information)
It would be very long to do it by hand, because it takes a lot of iterations until T c has converged, so a easy program is designed with the visual basic, that is able to programme a while function that calculates the new T c for each step until it has converged. It also saves the values of all the heats and areas needed in each step and later they are written in the “energy dry part and total” sheet.
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Energy condens WET PART OF THE HEAT EXCHANGER (CALCULATION OF THE AREA)
Temperatures T1 T2 Value that we can decide "at least now" celsius 46,27 46,17 Consequences of our deign K 319,42 319,32 Constants or propeties Final Results Pressions (Pa) Hipotesis P1 P2 Next Value for iteration P total 101325 101325 P vapour 10197 10145 Molar Mass P gas 91128 91180 M vapour (kg/mol) 0,028 Amount of water M inert gas (kg/mol) 0,030 Xs (kg w/kg dry air) 0,0696 0,0692 Xs (mol w/mol d air) 0,0997 0,0991
Step by step Calculation Specific Volume
First step Vvapour (mol/kg) 18,80 Vgas (mol/kg) 29,90 Properties at av conditions K and Pa celsius & atm Gas properties (T end h water) T av 319,4 46,2 P tot av 101325 1 h gas (W/m2) 47,71 P steam av 10171,3 0,100 cp 1101,35 P gas av 91153,7 0,900 µ 1,795E-05 Water cond (mol w/mol d gas) 0,0006 Lambda 0,026 Latent heat 2500000 Values to calculate Kg (at Tend of h water) Density 1,13 SI units celsius & atm Mm (mean) (kg/mol) 0,029
Tc 319,3120654 46,16 Pc 10142 0,1001 Calcul Q and T p'gas 91183,3 0,8999 pgf 0,90 Kd 6,128E-04 Kg 7,37992E-05 Tc 319,31 46,16 46,16
Heat fluxes (W/m2)
Q/A cond steam 153,0 Q/A gas sensible 2,7 Q/A water 155,9
Heats (W)
q cond 105,9 q dry gas 11,2 q steam 1,4 q water -8,2 q total 110,2 T final water 45,7 318,8
Dimensions
A1 needed (m2) 0,7 Rows z dry 1,4 Lz dry (m) 0,023
Total Results
A total wet (m2) 18,7 Q total wet (W) 9032 T end gases (celsius) 46,2
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Energy dry part This is the main sheet of the entire program. Here all the parameters that will define the exchanger are fixed. It follows the steps shown in the theory. If the final temperature is known, it assumes a concrete area to find all the properties, and calculates the area used in the dry part, and then the area calculated in the “Energy condense” is added to finally show the total area needed. This area is different from the area that was assumed to make all the calculations, so all the calculations must be repeated with the new area, this will be more close to the reality. Again the program visual basic is used to iterate until the new area calculated is close enough to the area assumed. Otherwise, if the area is known, all the geometry is fixed. The program then calculates the final temperature of the water, as well as the Q transferred in the dry and wet part and so on.
Energy dry part DRY PART OF THE HEAT EXCHANGER (CALULATION OF THE HEAT EXCHANGED AND FINAL GAS TEMPERATURE)
Parametres to fix Calcul Q and T MC 40% Efficiency 85% Power Output (kW) 150 Value that we can decide "at least now" Aar 0,50% Consequences of our deign LAMBA: 1,6 Constants or propeties t1 gas (celsius) 200 473,15 (K) Final Results t2 gas (celsius) 54 327,15 (K) Hipotesis t1 water (celsius) 45 318,15 (K)
Din (mm) 8 Thikness Tube (mm) 1,5 L (mm) 500 Rows y 30 Rows z 57,9 Aout total tubes 30,0 Value that we fix Sy 17 Sz 17
Properties of the fluids
h in tubes 189,73 h out tubes 47,71 λ tubes 150
Overall Heat-Transfer Coefficient
Uin (W/m2) 48,73 Uout (W/m2) 35,44
E-NTU method dry part (calculate heat and area without condensation) [untill Tdew]
Condenation? 1 in kW and K t dew (celsius) 53,1 326,2 Cmin 131,8 Cmax 4987,1 C* 0,0264 Q (W) 19359,5 E 0,948 Ntu 3,0771 A (m2) 11,4 Rows z dry 22,1 Lz dry (m) 0,38
E-NTU (Without condensation) [Calculate Q and T end gases]
Marc Cortina Flue Gas Condenser page 65
Ntu 8,07 E 1,00 Q (W) 20408 T end Gases (celsius) 45,1
Q recovered (kW)
Q total 28391,9 Q wet part 9032,5 Q dry part 19359,5 Q sensible 20244,3 Q latent 8147,6 T end gases 46,2 A dry 11,4 A wet 18,7
Spray Water
This sheet is only used when some water is sprayed into the gases. It takes the initial humidity ratio, adds a small amount of water, and calculates the new T dew with formula (53) and the new temperature of the gas with formula (70). The program adds water until the new temperature of the gases reaches the dew point. Then this water added is taken into account in the gas flux, and the flows are correspondingly changed. The exception is the water flow, because as it is so much bigger than the water removed to be sprayed to the gas, its effect is negligible.
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B.2 Funtions in Visual Basic
Below is some of the programing with Visual Basic that makes the iterations automatically.
To Find the right values to Calculate Nusselt : (In “H gas” sheet)
Private Sub CommandButton1_Click() ChooseCm End Sub
Public Sub ChooseCm() If Range("b72") > 40 Then Range("b76") = 0.71 Range("b77") = 0.5 Else Range("b76") = 1.04 Range("b77") = 0.4 End If
End Sub
To calculate all the results updating the area until and iterating: (in “Energy condens” sheet)
Public Sub CalculTc() Range("c35") = 60
While Abs(Range("c41") - Range("c35")) > 0.001 Range("c35") = Range("d41") Wend
End Sub
Public Sub CALCQA_Click()
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CalculateQA End Sub
Public Sub CalculateQA() Sheets("H gas").ChooseCm Awet = 0 Qlatent = 0 Qtotwet = 0 If Sheets("heat exchanged").Range("c38") = 1 Then T2 = Sheets("heat exchanged").Range("b15") Atot = Sheets("Energy dry part").Range("b21") Aused = Sheets("Energy dry part").Range("b47") Qsens = Sheets("Energy dry part").Range("b44")
While Atot > Aused T1 = T2 Range("b7") = T1 T2 = T1 - 0.1 Range("c7") = T2 CalculTc Aused = Aused + Range("b60") Qtotwet = Qtotwet + Range("b55") Qsens = Qsens + Range("b52") + Range("b53") Qlatent = Qlatent + Range("b51") + Range("b54") Awet = Awet + Range("b60") Wend Range("b66") = Awet Range("b67") = Qtotwet Range("b68") = T2 Sheets("Energy dry part").Range("b60") = Sheets("Energy dry part").Range("b44") + Qtotwet Sheets("Energy dry part").Range("b61") = Qtotwet Sheets("Energy dry part").Range("b62") = Sheets("Energy dry part").Range("b44")
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Sheets("Energy dry part").Range("b63") = Qsens Sheets("Energy dry part").Range("b64") = Qlatent Sheets("Energy dry part").Range("b65") = Range("b68") Sheets("Energy dry part").Range("b66") = Sheets("Energy dry part").Range("b47") Sheets("Energy dry part").Range("b67") = Range("b66") Else Range("b66") = 0 Range("b67") = 0 Range("b68") = Sheets("Energy dry part").Range("b56") Sheets("Energy dry part").Range("b60") = Sheets("Energy dry part").Range("b55") Sheets("Energy dry part").Range("b61") = Qtotwet Sheets("Energy dry part").Range("b62") = Sheets("Energy dry part").Range("b55") Sheets("Energy dry part").Range("b63") = Sheets("Energy dry part").Range("b55") Sheets("Energy dry part").Range("b64") = Qlatent Sheets("Energy dry part").Range("b65") = Sheets("Energy dry part").Range("b56") Sheets("Energy dry part").Range("b66") = Sheets("Energy dry part").Range("b21") Sheets("Energy dry part").Range("b67") = Range("b66")
End If
End Sub
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B.3 General philosophy
The most important of the excel sheets, is that they begin with the firsts calculations that are needed to find all the final results, and it continues step by step from the firsts pages until the last ones. But anyway, in the sheet “energy dry part and total”, the boiler values must be written (power output, moisture content, efficiency, initial temperature of the gas and final temperature of the gas, or area, depending on which is the known information) and also the geometric ones (diameter, rows, longitude, separation between rows...). Once these values are written, all the other values from the first page until the last one are updated. It means that everything is calculated for the variables that are chosen in the final sheet. It is very important, to have all the values of all the sheets depending and related only on some parameters that are chosen in a main page. It makes the program useful and easy to use for different cases, what is supposed to be very important.