While using the eulerian-eulerian two-phase approach, the mathematical equations describing the heat exchange between the downward moving solid phase and the upward moving hot gas phase are based on the equations of energy conservation written for each phase. In this approach, different phases are treated as interpenetrating continuous fluids and the concept of phase volume fraction is need to be introduced.

Model assumptions

· The solid particles conduction effects are neglected as Biot number: B i = h d p k s < 0.1

· The physico-chemical properties of solid and gas phase are constants;

· The solid particles are homogeneous spheres of diameter d_p;

· The gas phase flows upward at constant speed and the solid phase moves counter-currently to it at a constant speed too as depicted Figure 2;

· The heat exchange takes place only in an axial direction (1D problem).

Considering that α is the volume fraction of one phase, ρ the bulk density (kg/m³), C_p the specific heat (J/kg˚C), T the temperature (˚C), v the velocity (m/s) and k the heat conductivity coefficient (W/m˚C). The rate of the convection heat transfer between gas and solid particles is ϕ s g = h A s ( T g − T s ) , in which h is the convection heat transfer coefficient (W/m²˚C), A_s, the specific surface area (m²/m³) given by 6 1 − ε d p , ε , the fractional voidage and d_p, the particles diameter.

Energy conservation equations

For the case where the heat exchange by radiation and by friction is neglected, the energy conservation equations for the control volume dxdydz (Figure 3) and in the time dt can be written as follows:

E i n − E o u t = d E (1)

For solid phase, we have:

E i n = h A s ( T g − T s ) d x d y d z d t − k s ∂ T s ∂ x | x ( d y d z ) s o l i d d t (2)

E o u t = − k s ∂ T s ∂ x | x + d x ( d y d z ) s o l i d d t (3)

By substituting Equation (2) and Equation (3) into Equation (1), we have:

d E = h A s ( T g − T s ) d x d y d z d t + k s ∂ 2 T s ∂ x 2 ( d x d y d z ) s o l i d d t (4)

Dividing by dt on both sides, we obtain:

d E d t = h A s ( T g − T s ) d x d y d z + k s ∂ 2 T s ∂ x 2 ( d x d y d z ) s o l i d (5)

On the other hand, the total amount of heat in ( d x d y d z ) s o l i d is:

d E = ρ s ( d x d y d z ) s c p s d T s (6)

Hence the time rate of change of dE is:

d E d t = ρ s ( d x d y d z ) s c p s d T s d t (7)

In the Eulerian point of view:

d T s d t = ∂ T s ∂ t + ∂ T s ∂ x d x d t (8)

Consequently, by inserting Equation (7) and Equation (8) into Equation (5), we obtain:

ρ s ( d x d y d z ) s c p s ( ∂ T s ∂ t + ∂ T s ∂ x d x d t ) = h A s ( T g − T s ) d x d y d z + k s ∂ 2 T s ∂ x 2 ( d x d y d z ) s o l i d (9)

Since ( d x d y d z ) s = α s d x d y d z and d x d t = v s , we obtain the partial differential equation below:

α s ρ s c p s ( ∂ T s ∂ t + v s ∂ T s ∂ x ) = h A s ( T g − T s ) + α s k s ∂ 2 T s ∂ x 2 (10)

Considering the gas phase and applying the same procedure, we find the equation below:

α g ρ g c p g ( ∂ T g ∂ t + v g ∂ T g ∂ x ) = − h A s ( T g − T s ) + α g k g ∂ 2 T g ∂ x 2 (11)

Fractional voidage

The volume fraction of a phase, α , is defined as the ratio of the volume occupied by this phase to the total volume of the bed:

α s = V s V b e d (12)

Since V b e d = V s + V v o i d , so the volume fraction of the solid phase can be calculated as follows:

α s = 1 − V v o i d V b e d (13)

In the Equation (13), the ratio V v o i d V b e d is called the fractional voidage or void fraction ( ε ). Knowing that the gas phase occupes the volume V v o i d , volume fractions of the solid and gas phases can be written in terms of the fractional voidage as depicted in the Equation (14).

α s = 1 − ε et α g = ε (14)

Considering all the assumptions mentioned above and using the fractional voidage instead of the volume fractions, the energy conservation equations become:

{ ( 1 − ε ) ρ s C p s ∂ T s ∂ t = ( 1 − ε ) k s ∂ 2 T s ∂ x 2 + h A s ( T g − T s ) + ( 1 − ε ) ρ s C p s v s ∂ T s ∂ x ε ρ g C p g ∂ T g ∂ t = ε k g ∂ 2 T g ∂ x 2 − h A s ( T g − T s ) − ε ρ g C p g v g ∂ T g ∂ x (15)

For ease of notation of the above system of partial differential equations, let us write:

( 1 − ε ) ρ s C p s = a s ;     ( 1 − ε ) k s = b s ;

h A s = c s ;     ( 1 − ε ) ρ s C p s v s = d s ;

ε ρ g C p g = a g ;     ε k g = b g ;

− h A s = c g ;     − ε ρ g C p g v g = d g

{ a s ∂ T s ∂ t = b s ∂ 2 T s ∂ x 2 + c s ( T g − T s ) + d s ∂ T s ∂ x a g ∂ T g ∂ t = b g ∂ 2 T g ∂ x 2 + c g ( T g − T s ) + d g ∂ T g ∂ x (16)

For the system of partial differential equations obtained above (Equation (16)) to be well-posed, we must include some auxiliary conditions to complete the statement of this mathematical model. The number of those auxiliary conditions depends on the highest order derivative in each independent variable (spatial and temporal). Since we have a system of two mixed hyperbolic (first order in x)-parabolic (second order in x) type partial differential equations containing the first order time derivative, we need to introduce two initial conditions and four boundary conditions as shown in Table 1.

Method

The method of lines will be applied to compute the one-dimensional temperature profile within the torrefier. In this method, all spatial derivatives are approximated using finite differences, finite elements or finite volume method, and the semi-discrete (discrete in space but continuous in time) system of ordinary differential equations obtained is solved using the time integration method [5].

In this work, the first spatial derivative is approximated with the centered finite differences schema and the second spatial derivative with the three points centered finite differences schema, all with uniform grid spacing Δx.

{ ∂ T ∂ x ) i = T i + 1 − T i − 1 2 Δ x ∂ 2 T ∂ x 2 ) i = T i + 1 − 2 T i + T i − 1 Δ x 2 (17)

As we define the uniform grid of N nodes, then we obtain a system of 2N linear coupled ordinary differential equations as depicted in the Equation (18) and where T is the vector containing nodal values of temperatures.

∂ T ∂ t = ( A ) 2 N × 2 N T + b (18)

T = [ T 1 T 2 ⋯ T 2 N ] T = [ T s 1 ⋯ T s N T g 1 ⋯ T g N ] T

Then, the entries of the 2 N × 2 N matrix A and the 2 N × 1 vector b for all internal grid points ( i = 2 , ⋯ , N − 1 ; j = N + 2 , ⋯ , 2 N − 1 ) are:

{ A ( i , i + 1 ) = ( b s a s Δ x 2 + d s 2 a s Δ x ) A ( i , i ) = − ( 2 b s a s Δ x 2 + c s a s ) A ( i , i − 1 ) = ( b s a s Δ x 2 − d s 2 a s Δ x ) A ( i , i + N ) = c s a s b i = 0 A ( j , j + 1 ) = ( b g a g Δ x 2 + d g 2 a g Δ x ) A ( j , j ) = ( − 2 b g a g Δ x 2 + c g a g ) A ( j , j − 1 ) = ( b g a g Δ x 2 − d g 2 a g Δ x ) A ( j , j − N ) = − c g a g A ( j , j − N ) = − c g a g (19)

While using boundary conditions defined above we see that the remaining entries (for external grid points) must be written as follows:

{ A ( 1 , 1 ) = − ( 2 b s a s Δ x 2 + c s a s ) A ( 1 , 2 ) = 2 b s a s Δ x 2 A ( 1 , N + 1 ) = c s a s b 1 = 0 A ( N + 1 , N + 1 ) = ( − 2 b g a g Δ x 2 + c g a g ) A ( N + 1 , 1 ) = − c g a g A ( N + 1 , N + 2 ) = ( b g a g Δ x 2 + d g 2 a g Δ x ) b N + 1 = ( b g a g Δ x 2 − d g 2 a g Δ x ) T e n t r e e '

{ A ( N , N ) = − ( 2 b s a s Δ x 2 + c s a s ) A ( N , N − 1 ) = 2 b s a s Δ x 2 A ( N , 2 N ) = c s a s b N = 0 A ( 2 N , 2 N − 1 ) = 2 b g a g Δ x 2 A ( 2 N , 2 N ) = ( − 2 b g a g Δ x 2 + c g a g ) A ( 2 N , N ) = − c g a g b 2 N = 0 (20)

Model parameters

The physico-chemical properties of both phases are reported in Table 2 and Table 3.

The convective heat transfer coefficient between spherical particles of sugarcane bagasse and gas is calculated using the following Ranz-Marschall correlation [7] because it suits for packed bed of uniform spheres [8]:

h = k g d p ( 2 + 1.8 × R e 0.5 × P r 1 / 3 ) (21)

In which R e p = ρ g × v g × d p μ g is the Reynolds number and P r = μ g × C p g k g is the Prandtl number.

In addition, since in a moving bed we have to avoid fluidization of the bed, the

relative velocity between solid particles and gas (v_r) must remain less than minimum fluidization velocity (v_mf). That means the upper extrem value of the relative velocity is v_mf.

While taking into account the above descriptions, the relative velocity can be written as follow:

v r = v g − v s = v m f (22)

For sugarcane bagasse of mass flow rate of m ˙ b , continuously flowing downward the reactor of diameter D, the axial speed of the bed (solid particles) can be calculated as depicted in the Equation (23):

v s = 4 × m ˙ b π × D 2 × ρ s (23)

Hence, by combining the two above Equation (22) and Equation (23), the gas phase velocity can be written as:

v g = v m f + 4 × m ˙ b π × D 2 × ρ s (24)

Considering the mathematical model developed in this work, the calculations will be made for a pilot scale cylindrical moving bed reactor, rated to process 5 kg/h of sugarcane bagasse and having a diameter of 30 cm.

Table 4 shows the minimum fluidization velocity of sugarcane bagasse as function of particle size.

Calculations

Before to solve the 2 N × 2 N initial value problem Equation (18) obtained from the spatial discretization of the system of partial differential equations Equation (16), we first need to analyze its stiffness for the choice of an effective numerical method. To do that, the eigenvalues of the matrix A was calculated.

We can notice that from around 250 lines, the stiffness ratio (the ratio of the magnitudes of the real parts of the largest and smallest eigenvalues) depends heavily on the number of lines. Thus, as we increase the number of lines, the accuracy of the numerical solution increases [9], but the problem become stiffer meaning that explicit methods work extremely slowly [10] due to the need of a very small Δt to avoid numerical instability (Figure 4).

Table 5 shows a comparison of performance of implicit and explicit Runge-Kutta algorithms as function of numbers of lines.

Thus, we used the built-in Matlab solver ode15s based on implicit Runge-Kutta algorithm instead of ode45 based on explicit Runge-Kutta algorithm with a residence time of 10 minutes.

The temperature profiles along the axial direction are given in Figure 5 and Figure 6 for particles that are respectively 1 mm and 2 mm in diameter. One can see that the height of the reactor column must be at least 30 cm for that are 1 mm in diameter and 108 cm for particles that are 2 mm in diameter.

Furthermore, it is noticed that the effect of particles diameter on the minimum column height is not negligible. These temperature profiles indicate that the torrefaction of larger biomass particles require a reactor with a larger minimum column height. This is due to the fact that the heat transfer through the bed depends on biomass particles size (diameter). Indeed, large biomass particles have a resistance to the thermal propagation superior to that of small ones [11] [12].

Figure 7 depicts a qualitative diagram for biomass temperature as it passes through the torrefier column. The wet biomass enters the torrefier from the top at ambient temperature (point O), and as it gradually moves down, its temperature rises from ambient temperature. At point A, biomass reaches the drying temperature of 100˚C. Between A and B, the temperature remains constant while biomass loses its moisture. Then, between B and C, the temperature of biomass continues to rise to the desired temperature of torrefaction at which biomass begins to lose a fraction of its weigh in the form of volatiles. Finally, this temperature is maintained.

Comparing Figure 5, Figure 6, and Figure 7, the shape of results presents good agreement with the one’s obtained qualitatively because of the similarity of trends.

Therefore, it can be concluded that the method of lines, proposed in this work, is useful for the estimation of the minimum column height of an integrated fixed or moving bed biomass torrefaction reactor.