BIOMASS PYROLYSIS
IN DENSE FLUIDIZED BED
Validation case

Validation case


March 14, 2008

Contents

1 Introduction

The pyrolysis process depends on many parameters such as the size of the particles used, their density or their velocity and as a consequence it needs to be correctly modelized. In order to validate the results from Neptune CFD, a simulation on a reference case is undertaken (Lathouwers & Bellan, 2000). The present study is aimed at implementing the hydrodynamics, the chemical reactions, the mass transfers and the energy equation in the NeptuneCFD code developped by EDF, in a simplified geometry.

2 Theory

2.1 Hydrodynamics

In order to take into account polydisperse interactions, a three phase model is used. The phases are gas, sand and biomass and are respectively numbered 1, 2 and 3. The components of the gas and biomass phases are taken into account via scalar transport equations (cf. Section 2.2).

2.2 Chemical kinetics

According to Lathouwers et al. [1], a four steps kinetics can be used to modelize pyrolysis reactions:K

K 1 : virgin(s) → active(s)

K2 : active(s) → tar(g)

K 3 : active(s) → Xchar(s)+ (1- X)gasbio(g)

K 4 : tar(g) → gastar(g)

In the present study, the virgin biomass is considered to be only composed of cellulose (whereas it is also composed of hemicellulose and lignin in Lathouwers’ study). The biomass phase is composed of virgin and active biomass and char. The sand phase is considered inert and only plays the role of heat provider. The gas phase is composed of Nitrogen (inert), tar and gas. The last species (gas) is considered to come either from active biomass or from tar. Hence for facilitating the source terms implementation (cf. Section 2.3), they will be considered as different (respectively named gasbio and gastar) and will finally be added to be compared to Lathouwers’ results. Thus, we define 6 scalars (Xvirgin, Xactive, Xchar, Xgasbio, Xtar, Xgastar) representing the mass fraction of the previous components :

            mk
Xk = mPhaseTransportingk

The constants needed for calculating the reaction rate is provided in table 1.





reaction Ai [s-1] Ei [J.mol-1]






K1 2, 8.1019 242, 4.103



K2 3, 28.1014 196, 5.103



K3 1, 3.1010 150, 5.103



K4 4, 28.106 108.103



Table 1: Chemical kinetics constant

2.3 Mass transfer

2.3.1 Mass transfer between phases

The four reactions listed in Section 2.2 are linked with four mass transfer between species. However, only the reaction K3 and K2 involve mass transfer between phases. The mass transfer involved in these reactions are respectively noted ΓK2 and ΓK3 (these values are defined positive).

      α ρX
Γ K2 =-3τ3(Tact)ive
         2

      α3ρ3Xactive
Γ K3 =  τ3(T )

With

--1--=  A exp (- -Ei)
τi(T)    i     RT

And (Ai, Ei) are constants linked with the chemical kinetics (and are given in table 1).

Hence, in the conservative form, the transport equation for the biomass phase is:

D-(α3ρ3) = - (1- Xc)Γ K3 - Γ K2
Dt
(1)

Where DDt is the particular derivative which contains the local time derivative, the convective and diffusive terms and where αi, ρi are respectively the volume fraction and the density of the ith phase.

The biomass is disapearing into gas, so the source term must be negative and only the part (1 -XcK2, linked with the apparition of gasactive, implies phase exchange.

The source term for the gas phase is the opposite of the biomass source term : (1 -XcK3 + ΓK2. This term (positive) will be noted Γmt, as Mass Transfer.

2.3.2 Mass transfer between species

Let us consider the transport equation for the virgin specie. In the conservative form, this equation is written (with the same convention used in 2.3.1) :

-D-
Dt (α3ρ3Xvirgin) = - Γ K1

However, in Neptune CFD, the source term has to be provided for the non-conservative form of the equation. The latter is obtained by splitting the local derivative term and relating it to the equation of the transporting phase:

D                   DXvirgin        D
Dt-(α 3ρ3Xvirgin) = α3ρ3-Dt---+ XvirginDt-(α3ρ3)

The term -D
Dt(α3ρ3) is replaced by its expression in 1:

X    -D-(α ρ) = X     *(-Γ  )
 virginDt   3 3    virgin     mt

Finally, passing the term above at the other side of the conservative equation, we obtained the equation solved by Neptune CFD:

    DXvirgin
α 3ρ3   Dt   = - Γ K1 - Xvirgin *(- Γ mt)

The transport equation for the other species are obtained with the same manipulation:

α3ρ3DXactive=  +Γ K - (1- Xc )Γ K + Γ K - X  * (-Γ mt)
      Dt         1           2     3   active

    DXchar
α3ρ3--Dt-- = XcΓ K3 - Xchar*(- Γ mt)

α1ρ1DXgasactive-= (1- Xc)Γ K3 - XgasactiveΓ mt
       Dt

    DXtar
α1ρ1 Dt  =  Γ K2 - XtarΓ mt

    DXgas
α1ρ1-----tar=  Γ K4 - XgastarΓ mt
      Dt

2.4 Energy transfer

Energy Pyrolysis is an endothermic phenomenon. Hence, it is necessary to solve the energy equation in order to get physical results. The convective exchanges are already modelized in Neptune CFD. The enthalpy differences Δh2, Δh3, Δh4 in the reaction K2, K3, K4 are respectively 255 kJ/kg, 20 kJ/kg and 42 kJ/kg. These enthalpy must be carefully considered at two locations: in the enthalpy exchanges between phases (similar to mass transfer) and in the temperature calculation. Indeed, by default the temperature of the kth phase is deduced from its enthalpy using the following relation:

hk = cp,k(Tk- T0k)+ h0k

With

cp,k

heat capacity of the kth phase [J.kg-1.K-1]

Tk0

reference temperature of the kth phase [K]

hk0

enthalpy of formation at Tk0 [J.kg-1]

By default, the enthalpy of formation is supposed to be null and Tk = T0 + Hk-
Cp,k. As we need to modelize an endothermic reaction, we cannot use this convention but we have to calculate enthalpies of formation thanks to the enthalpy variations Δhi. K2 is endothermic whereas K3 and K4 are exothermic. The system that needs to be verified is:

h0  = h0   + Δh2
 tar   active

X h0   +(1 - X )h0    = h0   - Δh
  cchar        c  gasactive   active     3

h 0gastar = h0tar- Δh4

Let us impose that all solid species have a null enthalpy of formation: hvirgin0 = hactive0 = hchar0 = 0 J.kg-1. Then, the enthalpies of formation of the gaseous species are:

h0tar = Δh2

          - Δh
h0gasactive =-----3--
         (1- Xc )

 0
hgastar = Δh2- Δh 4

As there is no restriction on the enthalpy of formation of the nitrogen (inert component of the gaseous phase), we impose : hN20 = 0 J.kg-1

If we take “m kg” of the gas phase, the enthalpy of this system is:

Hgas = Hgasactive + Htar +Hgastar + HN 2

mhgas = mgasactivehgasactive + mtarhtar+ mgastarhgastar + mN2HN 2

hgas = Xgasactivehgasactive + Xtarhtar+ Xgastarhgastar + XN 2hN 2

If we decompose the enthalpy into sensitive enthalpy and enthalpy of formation, we obtain:

                        0         0
hgas = (∑ Xk)cp,gas(Tgas- Tgas)+ ∑ Xkhk

As Xk = 1, the temperature of the gas phase is calculated from its enthalpy using this relation:

                      0
Tgas = T0gas+ hgas--∑-Xkhk
               cp,gas


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Figure 1: Enthalpy transfer


In the same time that mass is moving between phases, energy is exchanged too. This exchange is illustrated on figure 1. Let us consider a single biomass particle in a cell, only composed of active at t. At t + dt, the mass dm initially in the active specie has transformed into tar. The variation of the enthalpy of the biomass particle between t and t + dt is noted dH and is equal to the enthalpy of the formed tar at the temperature of the particle (htar(Tbio) = cp,gas(Tbio -Tgas0) + htar0):

dH = dm htar(Tbio)

Thus, the energy needed by this endothermic reaction is supposed to be taken from the biomass particle and not from the surrounding gas phase. Dividing the above equation by dt, we get:

dH-   dm-
 dt =  dt htar(Tbio)

Hence, after adding the mecanism of diffusion-convection (integrated into the particular derivative) and the reaction K2, we obtain the equation of the enthalpy of the biomass phase in the conservative form:

D--
Dt (α3ρ3hbio) = htar(Tbio)* (-Γ K2)+ hgasactive(Tbio)* (-(1- Xc)Γ K3)

The source term of enthalpy has to be provided in the non conservative form in Neptune CFD. By doing the same splitting than in Section 2.3.2, we obtain:

   -D-
α3ρ3Dt (hbio) = -htar(Tbio)Γ K2 - hgasactive(Tbio)*(1- Xc)Γ K3 - hbio*(- Γ mt)

The equation for the gas phase is similar:

α1ρ1 D-(hgas) = +htar(Tbio)Γ K2 + hgasactive(Tbio)* (1 - Xc)Γ K3 - hgasΓ mt
    Dt

2.5 Efficiency variables

One of the main objective of Lathouwers et al. [1] is to optimize the tar production. To verify the consistency of our results, two performance variables are calculted : the yield and the Differential Reactor Efficiency (DRE). These variables are calculated for all species. For example, the definition of the tar yield is :

     Mtar-+Ωtar
ηtar =   Mfeed

Where Mtar, Ωtar, Mfeed are respectively the instaneous mass of tar in the reactor, the mass of tar that has left the reactor through the output and the mass of biomass that has been injected through the feed.

The definition of the tar DRE is :

ϑtar = Mtar-+-Ωtar
      Mproducts

With

Mproducts = Mgasactive+tar + Ωgasactive+tar + Mtar + Ωtar+ Mchar+ Ωchar

The time evolution of these parameters are calculated in section 4.

3 Simulation

3.1 Geometry

The geometry used in the present work is similar to the one of Lathouwers et al. [1]. The geometric dimensions of the reactor is presented on figure 2. The mesh size is 40*148 cells.


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Figure 2: Geometry


3.2 Initial conditions

The reactor is initially filled with sand with a volume fraction of 0.6 under the height of 0.326 m. The six mass fractions are taken to zero. Thus no biomass is present in the reactor at the initial state.

3.3 Boundary conditions

The fluidization velocity is 0.72 m.s-1 and only nitrogen is injected there (α1 = 1), at T = 800K. The injection velocity through the feed point is 1 m.s-1 with a volume fraction of biomass of 1.92 * 10-4 and a volume fraction of nitrogen of 0.999808. The biomass is only composed of virgin there (Xvirgin = 1) and the biomass temperature is 400K whereas the nitrogen temperature is 800K. The initial temperature is 800K for all phases.

4 Results

4.1 Isotherm case


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Figure 3: Volume fraction of sand and biomass


The volume fraction of sand is illustrated on figure 3 ref hydrodynamics.png at t = 1.9 s. It can be noticed that the polydisperse interactions closely link the biomass and sand particles locations, the former are well mixed in the sand bed. We can distinguish three zones. One solid zone with bubbles at the bottom of the reactor separated from the fluid zone by a “splash” zone.

In this case, chemical kinetics are considered as adiabatic reactions and the biomass is injected at the temperature of 800 K. Thus the fluidized bed doesn’t need to bring the heat to biomass for reactions. Hence, the sand is playing no role in this case. This is confirmed by looking at the figure 4 which show the partial density of biomass (virgin and active), tar, gas and char. One can see that the biomass begins to react as soon as it enters into the reactor. That is why our results in this case are different of Lathouwers’ ones. In his study, the char production takes place at the bottom of the reactor because the biomass is heated there and in the same time it is mixted with sand (figure 5). However, there are some similarities : first, the maximum of biomass density is near the injection. There is no storage, it is quicky consumed. Then, one can observe on Lathouwers results, that the second concentration peak of biomass is linked to the char production peak, and also those of gas and tar. So we can think that it is the place where reactions are. In our case of study, most of the reactions take place at the injection point.


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Figure 4: Partial density of biomass, gas and char at t=2.5 s



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Figure 5: Lathouwers’ results : partial density of biomass, gas and char at t=2.5 s


4.2 Fully resolved case


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(a) Virgin biomass
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(b) Active biomass
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(c) Char

Figure 6: Mass fraction of solid products



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(a) Gas active
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(b) Tar
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(c) Gas tar

Figure 7: Mass fraction of gaseous products


The mass fractions of the species involved in the four step kinetics are plotted on figures 6 and 7 at t = 5 seconds and at t = 279 seconds. These contours are not easy to interpret as they are not linked with the presence of their transporting phase. To understand those fields, one can imagine that the Xk of the solid species are the mass percentages of the kthspecie in each biomass particules. Thus, if there is no particle in a cell, these mass fractions don’t have a lot of sense. That is why it is more appropriate to look at partial densities : αphaseρphaseXk.


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(a) virgin biomass
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(b) active biomass
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(c) char

Figure 8: Partial densities of solid products



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Figure 9: Partial densities of gaseous products


In order to compare our results with Lathouwers’ one, the instantaneous fields of partial densities at t = 5 seconds and at t = 279 seconds are illustrated on figure 8 and figure 9. The fields at 5 seconds are comparable to Lathouwers’ fields at 3 seconds (cf. figure 5). The fields at t=279 seconds are considered to be representative of a permanent state. This assumption is well confirmed by the yields and DREs plots showed on figure 11. There are non negligible differences with Lathouwers’ results regarding the partial density fields. Indeed, in his results, the partial density of biomass (virgin and active) is maximum at the injection point (2 kg.m-3) and a peek of biomass (0.5 kg.m-3) in the lower right corner is producing a lot of char, gas and tar. In our case, the maximum of partial density of biomass is also at the injection point but is higher (=5kg.m-3). There is also a secondary peek of biomass at the opposite side of the injection point that is higher (=2kg.m-3) than Lathouwers’ one. Regarding the products, the partial densities are lower than those of Lathouwers. One could think that this difference is due to lower temperature fields, however, the temperature are plotted on figure 10 and it can be seen that the temperature field is similar to the one from Lathouwers.


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Figure 10: Contours of the temperature of the gas (left), sand (middle), and biomass (right) at t = 5s


As it is difficult to measure differences quantitatively on 2D colored fields, the yields are plotted on figure 11 (a) and (b). The figure 11 (b) shows the five first seconds of the simulation. This is why we have chosen to compare our results at 5 seconds to the results of Lathouwers at 3 seconds : at 3 seconds the yields in our study are too low to be comparable with Lathouwers results. At 5 seconds, the reactions are high enough to have yields at reasonable and comparable values and the DREs are stationary.


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Figure 11: Yields



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(a)
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(b)

Figure 12: DREs


The DREs are plotted on figure 12. The DRE unstationary phase is illustrated on figure 12 (b). Before t = 1 s, the DRE is senseless as the mass of products is almost zero. After this phase where the biomass particles are heating up, a similar behaviour as Lathouwers can be observed : at first the DRE of tar is higher than the others, endothermic mass transfer is occuring with the reaction K2 and the temperature of the biomass phase quickly decrease. Then, as the energy activation is lower for the reaction K3, char and gasactive relative production increase. Then, the equilibrium is reached as the temperature stays quasi-stationary. The table 2 lists the yields and DREs obtained in our study and in Lathouwers case at the end of the unstationary phase for DREs. The temperature is said to be quasi-stationary because in the meantime of the reaction rates timescales (τ1(700K) = 0.04 s), the temperature is not decreasing. However, the temperature of the sand bed and the gas phase is slowly decreasing, as the energy provided at the fluidization entry by the gas is not sufficient to balance the energy needed for biomass pyrolysis. This illustrates the necessity to control the heat injected in the reactor in order to produce the wanted products. This also shows the advantage of a Dual Fluidized Bed at the industrial level (cf. Application to the industrial case).





Yields forThe present study ( t=5s)Lathouwers’ study ( t= 3s )






Char 0.01 0.14



Gas 0.03 0.14



Tar 0.23 0.49






DREs forThe present study ( t=5s)Lathouwers’ study ( t= 3s )






Char 0.04 0.14



Gas 0.11 0.17



Tar 0.85 0.69



Table 2: Comparison of yields and DREs

In order to check that the timestep is not too large in comparison with the reaction rates timescales, the figure 13 plots the ratio dt
τi for the reactions K1 to K4. It can be noticed that the ratio stays at low values (maximum: 0.06), which confirms the quasi-stationary state and the sufficiently small timestep.


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Figure 13: dt
τ for reactions K1 to K4 at t =279s



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(a) Char
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(b) Gas
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(c) Tar

Figure 14: Influence of the fluidization temperature on yields


Finally to check the influence of the temperature of the fluidization gas and the sand bed, the efficiency parameters are calculated using several operating conditions. Three temperatures are chosen : 700 K, 800K and 950K. The yields are presented on figure 14. The influence on yields is obvious : the more the temperature increase the more the yields increase rapidly.


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(a) Char
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(b) Gas
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(c) Tar

Figure 15: Influence of the fluidization temperature on DREs


The DREs are plotted on figure 15. Regarding the DREs, the temperature is not a key parameter. The increase of temperature only tends to lower the unstationary phase duration. The increase of temperature doesn’t change the relative production : at quasi-stationary state, 80 % of the mass of products is tar, 15 % is gas and 5 % is char.

The difference between our study and Lathouwers one must come from the fact that we took a virgin biomass only composed of cellulose, whereas Lathouwers also take into account the hemicellulose and the lignin components that introduces completely different kinetic constants. A future development could consist in adding these components and their kinetic constants in the Neptune CFD model.

5 Conclusion

A simplified model of pyrolysis containing the most important phenomena has been succesfully implemented in the NeptuneCFD code. Although the composition of biomass is simplified the results are still consistent with the ones from Lathouwers [1]. Future developments will consist in taking into account the physical property differences between the components. Indeed, char has not the same density than active biomass and a change in the biomass phase density must be considered. At last but not least, the mass transfer should be accompanied with a decreasing of the biomass particles diameter in order to have a better modelization of pyrolysis and its influence on hydrodynamics.

References

[1]   D. Lathouwers and J. Bellan. Modeling of dense gas solid reactive mixtures applied to biomass pyrolysis in a fluidized bed. International Journal of Multiphase Flow, 27:2155–2187, 2001.