8.1.1 Grain

Grain is a source of carbohydrates—the raw material for first-generation bioethanol—as well as protein and fat. Wheat and corn grain are made of starch, a natural polymer that consists of glucose molecules, as seen in Fig. 8.1.

The pretreatment of the grain consists of breaking bonds to obtain sugar monomers of glucose. The process consists of liquefaction, and takes place at 90°C and a pH from 6 to 6.5 for 30 min. The starch is broken into maltose. In a second step, the maltose is further hydrolyzed at 65°C and pH 5.5 for 30 min (ie, saccharification) to obtain glucose. These are the reactions:

$2{({\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{O}}_5)}_n+n{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{\alpha }-\text{Amylase}}{\to }n{\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}$

${\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}{+\text{H}}_2\mathrm{O}\stackrel{\mathrm{Glucoamylase}}{\to }{\mathrm{2C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

8.1.2 Lignocellulosic Biomass

Although grain is an interesting source of sugars, its use for biofuel production poses an ethical dilemma because it competes with the food supply chain. Thus, second-generation biofuels are based on the use of biomass that does not compete with the food chain, either in terms of final use of the product or acreage. Lignocellulosic raw materials consist of cellulose, hemicellulose, and lignin. Lignin provides structure to the plant, hemicelluose is responsible for internal structure, and cellulose is the internal polymer. Lignin is a polymer of aromatic monomers, while hemicellulose and cellulose are made of sugar monomers. Fig. 8.2 shows the structure of a plant and the chemical composition of the main components of the plant wall. A number of uses for lignocellulosic raw materials are highlighted: biooil, syngas, sugar, and paper production.

8.1.2.1 Biooil production

This is a thermochemical path used to partially break the biomass into medium-size chemicals. It operates at medium temperatures, and the composition of the products depends on the rate of the process. The faster the process, the higher the liquid content (see Fig. 8.3). Gas and char are also produced. Thus, fast pyrolysis is recommended for chemicals and fuels production. The main drawback of the process is the wide range of products generated and their high corrosivity. An upgrade to the process is therefore required for proper further use of the biooil. Hydrocracking and catalytic cracking are two methods for upgrading (see Chapter 5: Syngas).

8.1.2.3 Sugar production

Sugar production from lignocellulosic biomass uses moderate pretreatment processes (below 200°C) to break plant structures down only as far as their monomers. The pretreatments can be classified into physicochemical, chemical, and enzymatic. The aim is to expose the sugar-containing polymers, namely cellulose and hemicellulose; this makes them more accessible for hydrolysis to sugars. Fig. 8.4 shows the desired effect of pretreatment: breakage of the physical structure into pieces.

8.1.2.3.1 Physicochemical processes

These processes are based on the use of CO₂, steam, or NH₃ under pressure; when the pressure is released, the biomass structure is broken. Among the different processes available, ammonia fiber explosion (AFEX) is one of the few that has reached the industrial development stage. The AFEX process puts biomass into contact with an NH₃ solution (20 atm, 90–180°C) so that when expanded, the biomass structure breaks into pieces. The method does not produce byproducts, and shows high yield. For years the disadvantage was the energy required to recover the NH₃—until recent developments by Prof. Dale’s group at Michigan State University. Fig. 8.5 shows the flowsheet (Martín and Grossmann, 2012a). After the reactor pretreatment, the pressure is released and part of the NH₃ is recovered. The slurry is distilled to recover the rest of the NH₃. Traces of NH₃ remain in the slurry, but they can be used as nutrients in further fermentation stages. Based on the experimental data from Garlock et al. (2012), a correlation that depends on the NH₃-to-biomass ratio, the operating temperature, the water-to-biomass ratio, and the residence time is presented; this can be used to compute the yield. The range of the operating variables is shown in Table 8.1.

$\begin{array}{l}X=0.01\cdot (-88.7919+26.5272\cdot AR-13.6733\cdot \mathrm{water}\_\mathrm{added}+1.6561\cdot T\hfill \\ +3.6793\cdot t-4.4631\cdot A{R}^2-0.0057\cdot T^2+0.0279\cdot t^2-0.4064\cdot AR\cdot t\hfill \\ +0.1239\cdot \mathrm{water}\_\mathrm{added}\cdot T-0.0132\cdot T\cdot t;\hfill \end{array}$ (8.1)

(8.1)

Table 8.1

Typical range of Operating Variables for AFEX Pretreatment

	Lower Bound	Upper Bound
Temperature (°C)	90	180
Ammonia added (g/g)	0.5	2
Water added (g/g)	0.5	2
Residence time (min)	5	30

8.1.2.3.2 Chemical pretreatment

Chemical pretreatment entails the use of alkali or acid solutions, peroxides, or even ozone to break down plant structure. The National Energy Renewable Lab (NREL) has developed a dilute sulfuric acid pretreatment (Aden and Foust, 2009) in which a solution of 0.5–2% H₂SO₄ is put into contact with biomass at 140–180°C and 12 atm for up to 1.5 h. In the expansion, part of the water is recovered and recycled, while the slurry is separated so that the liquid phase is neutralized with CaO, producing gypsum. After filtering the gypsum, both streams are mixed again for further processing. Fig. 8.6 shows the flowsheet for the process.

Based on experimental data, Martín and Grossmann (2014) developed models for the yield to cellulose and hemicellulose as a function of the operation temperature, the acid concentration, the residence time, and the enzyme added in the hydrolysis part, see eqs. (8.2) and (8.3). Table 8.2 shows the range for the operating variables.

$\begin{array}{l}{X}_{\mathrm{cellulose}}=-0.00055171+0.00355819\cdot T+0.00067402\cdot \text{acid}\_\text{conc}\hfill \\ +0.00100531\cdot t-\mathrm{enzyme}\_\text{load}\cdot 0.0394809-0.0186704\cdot T\cdot \text{acid}\_\text{conc}\hfill \\ +0.00043556\cdot T\cdot t+0.0002265\cdot T\cdot \mathrm{enzyme}\_\mathrm{load}\hfill \\ -0.0013224\cdot \text{acid}\_\text{conc}\cdot t-0.00083728\cdot t\cdot \mathrm{enzyme}\_\mathrm{load}\hfill \\ +0.044353\cdot \text{acid}\_\text{conc}\cdot \mathrm{enzyme}\_\mathrm{load}+0.000014412\cdot T^2;\hfill \end{array}$ (8.2)

(8.2)

$\begin{array}{l}{X}_{\mathrm{hemicellulose}}=-0.00015791-0.00056353\cdot T+0.000694361\cdot \text{acid}\_\text{conc}\hfill \\ -0.00014507\cdot t-\mathrm{enzyme}\_\mathrm{load}\cdot 0.01059248-0.02142606\cdot T\cdot \text{acid}\_\text{conc}\hfill \\ +0.000694055\cdot T\cdot t+0.00013559\cdot T\cdot \mathrm{enzyme}\_\mathrm{load}\hfill \\ -0.00145712\cdot \text{acid}\_\text{conc}\cdot t+0.04769633\cdot \text{acid}\_\text{conc}\cdot \mathrm{enzyme}\_\mathrm{load}\hfill \\ -0.00138362\cdot \mathrm{time}\cdot \mathrm{enzyme}\_\mathrm{load}+0.0000059419\cdot T^2\hfill \end{array}$ (8.3)

(8.3)

Table 8.2

Typical range of Operating Variables for Dilute Acid Pretreatment

	Lower Bound	Upper Bound
Temperature (°C)	140	180
Acid concentration (g/g)	0.005	0.02
Residence time (min)	1	80
Enzyme load (g/g)	0.0048	0.0966

8.1.2.3.3 Enzymatic pretreatment

These are slow processes and are not yet competitive.

Once the physical structure of the biomass is broken and the polymers, cellulose, and hemicellulose exposed, the next stage is enzymatic hydrolysis, which produces sugars. The hydrolysis is typically an endothermic set of reactions that takes place at 50°C:

$\begin{array}{cc}{{(\mathrm{C}}_6{\mathrm{H}}_{10}{\mathrm{O}}_5)}_n+n{\mathrm{H}}_2\mathrm{O}\to n{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6& \mathrm{\Delta }H=22.1n\text{kJ}/\text{mol}\end{array}$

$\begin{array}{cc}{{(\mathrm{C}}_5{\mathrm{H}}_8{\mathrm{O}}_4)}_m+m{\mathrm{H}}_2\mathrm{O}\to m{\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_5& \mathrm{\Delta }H=79.0m\text{kJ}/\text{mol}\end{array}$

Alternatively, these polymers can be used for the production of paper.

8.1.3 Seeds

First-generation biodiesel processes use high oil content seeds as raw materials. The oil is extracted using crushers. The process consists of a number of stages. The seeds are heated and milled, and then the flakes are slightly heated to improve the oil extraction. The cake obtained can be further processed using solvents or by means of cold pressing using a screw. Sometimes both methods are combined to increase the yield. The oil must later be purified. Flash distillation, filtration, or sedimentation can be used for the oil refining step. The biodiesel yields of different seeds are shown in Table 8.3. Fig. 8.7 shows the process.

8.1.4 Algae

Algae are another alternative for the production of biofuels. Due to their high yield to biomass—typically an order of magnitude or two higher than other seeds per area—algae promise higher production. Algae grow by capturing CO₂ (although other carbonous sources can be used), and use sunlight as an energy source. Algae store lipids (up to 70%), starch (up to 50%), protein, and other components. The basic stoichiometry of the algae growing process is as follows:

$106{\mathrm{CO}}_2+16{\mathrm{NO}}_3^{-}+{\mathrm{HPO}}_4^{2-}+122{\mathrm{H}}_2\mathrm{O}+{\mathrm{18H}}^{+}\stackrel{\mathrm{Solar\; radiation}}{\to }{{(\text{CH}}_2\mathrm{O})}_{106}{{(\text{NH}}_3)}_{16}{\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{138O}}_2$

Algae are grown in ponds or photobioreactors. Ponds are the cheaper option, and only civil engineering work is required to create an operational system. Water flows through 20 cm-deep channels into which CO₂ and nutrients are injected. However, ponds present several drawbacks: they are easily contaminated and process control is difficult. Photobioreactors, on the other hand, offer a controlled environment. However, solar exposure is difficult and the investment cost is high. Fig. 8.8 shows examples of both systems.

The algae growth rate is a function of the solar energy:

$\mathrm{Growth}=\frac{I_0\cdot {\eta }_{\max }}{\mathrm{Algae}\mathrm{heating}\mathrm{value}}$ (8.4)

(8.4)

where I₀ (kWh/m²/d) is the solar intensity, η_max is the utilization of the light by the algae (0.045; Walker, 2009), and the algae heating value is 21 kJ/g (Park et al., 2011). Growth rate values of 50 g/m² d are optimistic but feasible. Carbon dioxide consumption is related to algae growth as shown in the following equation:

${\mathrm{CO}}_2{(\mathrm{m}}^3/\text{day})=0.6565\times {\text{Growth}(\mathrm{g}/\mathrm{m}}^2\text{day})+5.0784$ (8.5)

(8.5)

In addition to CO₂, NH₃ (or nitrates) up to 0.8% and phosphates up to 0.6% (both with respect to dry biomass) are needed. Evaporation accounts for water loss of 6.2 m³/day (Sazdanoff, 2006).

Once the algae are produced, the limiting stage for oil production in terms of cost is the harvesting. This step typically consists of flotation followed by centrifugation or filtration and biomass drying to reach moistures below 10%. Alternatively, there is a capillarity-based process that has been proposed by Univenture; it is supposed to yield a biomass cake with 5% moisture and also consume less energy. The dry biomass is processed and oil extracted, as in the seed-based case. The remaining biomass cake consists mainly of starch and protein. The starch can be used to produce ethanol, similar to first-generation techniques for bioethanol production from corn or wheat (Martín and Grossmann, 2013b).

8.1.5 Natural Rubber

Mechanism and kinetics

Radical polymerization is based on double bond opening and free radical chain growth, as per the reaction below:

It requires the use of initiators (I), Ziegler–Natta catalysts. The proposed reaction mechanism consists of three main steps (initiation, propagation, and termination) and is believed to proceed as follows (Odian, 2004):

Initialization:	$\mathrm{I}\stackrel{k_{\mathrm{d}}}{\to }\mathrm{2R}$
Propagation:	$\mathrm{R}+\mathrm{M}\stackrel{k_{\mathrm{i}}}{\to }{\mathrm{P}}_1$
Propagation:	${\mathrm{P}}_n+\mathrm{M}\stackrel{k_{\mathrm{p}}}{\to }{\mathrm{P}}_{n+1}$
Termination by combination	${\mathrm{P}}_n+{\mathrm{P}}_m\stackrel{k_{\mathrm{tc}}}{\to }{\mathrm{D}}_{n+m}$
Termination by disproportion	${\mathrm{P}}_n+{\mathrm{P}}_m\stackrel{k_{\mathrm{td}}}{\to }{\mathrm{D}}_n+{\mathrm{D}}_m$
Transfer to monomer	${\mathrm{P}}_n+\mathrm{M}\stackrel{k_{\mathrm{f}}}{\to }{\mathrm{P}}_1+{\mathrm{D}}_n$

The kinetics proceed as follows:

$\frac{d\left[\mathrm{I}\right]}{dt}=-f\cdot k_{\mathrm{d}}\cdot \left[\mathrm{I}\right]$ (8.6)

(8.6)

where f is the initiator efficiency. Assuming a steady-state for the rates and adding together all of the monomer kinetics, we have:

$\frac{d\left[\mathrm{M}\right]}{dt}=-k_{\mathrm{p}}{(f\cdot k_{\mathrm{d}}[\mathrm{I}]/k_{\mathrm{t}})}^{0.5}\cdot \left[\mathrm{M}\right]$ (8.7)

(8.7)

$\frac{d\mathrm{Q}}{dt}=-\mathrm{\Delta }{H}_{\mathrm{r}}\frac{d\left[\mathrm{M}\right]}{dt}$ (8.8)

(8.8)

The polymerization lasts 15 h at 50°C in a stirred tank. The conversion yield is 75%. The molecular weight of the product is computed using statistical moments. A complete model for a polymerization reactor can be found in Martín (2014). After the reactor step, a stripper is used to recover hydrocarbons. Finally, the rubber crumb slurry is dewatered and dried in an extruder. The dry crumbs are ready for shipment.

8.2.1 Sugars

This section covers the production of a number of chemicals, mainly for the biofuels industry. We also cover the production of antibiotics such as penicillin via fermentation.

8.2.1.1 Bioethanol

Bioethanol can be produced via sugar fermentation. Saccharomyces cerevisiae is capable of anaerobically fermenting glucose at 32–38°C to produce a solution of up to 15% by weight of ethanol. Overpressure is typically used to prevent the entrance of air into the system. Fermentation of pentoses is more complex. Zymomonas mobilis has been identified as appropriate for fermenting pentoses and hexoses simultaneously. In addition to ethanol, byproducts such as glycerol, succinic acid, acetic acid, and lactic acid are also produced. Table 8.4 shows the various reactions and the typical conversions in second-generation bioethanol processes (Aden and Foust, 2009). See Fig. 2.6 for a block diagram of the process.

Table 8.4

Reactions and Conversions in Second-Generation Bioethanol Production

Reaction	Conversion
Glucose→2 Ethanol+2CO2	Glucose 0.92
Glucose+1.2NH3→6 Z. mobilis+2.4H2O+0.3O2	Glucose 0.04
Glucose+2H2O→Glycerol+O2	Glucose 0.002
Glucose+2CO2→2 Succinic acid+O2	Glucose 0.008
Glucose→3 Acetic acid	Glucose 0.022
Glucose→2 Lactic acid	Glucose 0.013
3 Xylose → 5 Ethanol+5CO2	Xylose 0.8
Xylose+NH3→5 Z. mobilis+2H2O+0.25O2	Xylose 0.03
3 Xylose+5H2O→5 Glycerol+2.5O2	Xylose 0.02
3 Xylose+5 CO2→5 Succinic acid+2.5O2	Xylose 0.03
2 Xylose→5 Acetic acid	Xylose 0.01
3 Xylose→5 Lactic acid	Xylose 0.01

The reactions that yield ethanol are exothermic:

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\stackrel{\mathrm{Yeast}}{\to }{\mathrm{2C}}_2{\mathrm{H}}_5\mathrm{OH}+{\mathrm{2CO}}_2\mathrm{\Delta }H=-84.394\text{kJ}/\text{mol}$

$3{\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_5\stackrel{\mathrm{Yeast}}{\to }5{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+{\mathrm{5CO}}_2\mathrm{\Delta }H=-74.986{\text{kJ}/\text{mol}}_{xylose}$

The reaction time is about 24 h at 0.12 MPa, which helps prevent the entrance of air into the system. The maximum concentration of ethanol in the water is 6–8%.

Example 8.1

Model the fermentor for second-generation bioethanol production based on Krishnan et al. (1999).

Solution

The kinetic model for ethanol production via fermentation is based on the Michaelis–Menten mechanism:

$\mathrm{E}+\mathrm{S}\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}\mathrm{ES}\stackrel{k_2}{\to }\mathrm{E}+\mathrm{P}$

In the case of ethanol production, there is both substrate and product inhibition:

$\mathrm{EI}+\mathrm{S}\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}\mathrm{E}+\mathrm{S}+\mathrm{I}\underset{k_{-1}}{\overset{k_1}{\rightleftarrows }}\mathrm{ES}+\mathrm{I}\stackrel{k_2}{\to }\mathrm{E}+\mathrm{P}+\mathrm{I}$

Assuming a steady-state for [E], [EI], and finally [ES], the modified Monod kinetics are given as follows:

$\mu =\frac{{\mu }_m\cdot \mathrm{S}}{K_s+\mathrm{S}+{\mathrm{S}}^2/K_i}$

Next, the product inhibition is included:

$\frac{\mu }{{\mu }_0}=\left(1-{\left(\frac{\mathrm{P}}{{\mathrm{P}}_m}\right)}^{\beta }\right)$

The kinetics models for the different species involved are given in the equations that follow, where G represents glucose and X represents xylose. The parameters for the fermentation are given in Table 8E1.1 (Krishnan et al., 1999).

$\begin{array}{l}\text{Cells}:\hfill \\ {\mu }_g=\frac{{\mu }_{m,g}\cdot \mathrm{S}}{K_{s,g}+\mathrm{S}+{\mathrm{S}}^2/K_{i,g}}\left(1-{\left({\displaystyle \frac{\mathrm{P}}{{\mathrm{P}}_m}}\right)}^{{\beta }_g}\right)\hfill \\ {\mu }_x=\frac{{\mu }_{m,x}\cdot \mathrm{S}}{K_{s,x}+\mathrm{S}+{\mathrm{S}}^2/K_{i,x}}\left(1-{\left({\displaystyle \frac{\mathrm{P}}{{\mathrm{P}}_m}}\right)}^{{\beta }_g}\right)\hfill \\ \frac{1}{\mathrm{X}}\frac{d\mathrm{X}}{dt}=\frac{\mathrm{G}}{G+X}{\mu }_g+\frac{\mathrm{X}}{G+X}{\mu }_x\hfill \end{array}$

$\begin{array}{ccc}\begin{array}{l}\text{Product}:\hfill \\ {\nu }_{\mathrm{E},g}=\frac{v_{m,g}\cdot \mathrm{S}}{K_{s,g}+\mathrm{S}+{\mathrm{S}}^2/K_{i,g}}\left(1-{\left({\displaystyle \frac{\mathrm{P}}{{\mathrm{P}}_m}}\right)}^{{\gamma }_g}\right)\hfill \\ {\nu }_{\mathrm{E},x}=\frac{v_{m,x}\cdot \mathrm{S}}{K_{s,x}+\mathrm{S}+{\mathrm{S}}^2/K_{i,x}}\left(1-{\left({\displaystyle \frac{\mathrm{P}}{{\mathrm{P}}_m}}\right)}^{{\gamma }_x}\right)\hfill \\ \frac{1}{\mathrm{X}}\frac{d\mathrm{P}}{dt}=({\nu }_{\mathrm{E},x}+{\nu }_{\mathrm{E},g})\hfill \end{array}& \mathrm{with}& \begin{array}{c}{\mu }_{m,g}=0.152\cdot {\mathrm{X}}^{-0.461}\\ {\mu }_{m,x}=0.075\cdot {\mathrm{X}}^{-0.438}\\ v_{m,g}=1.887\cdot {\mathrm{X}}^{-0.434}\\ v_{m,x}=0.16\cdot {\mathrm{X}}^{-0.233}\end{array}\end{array}$

$\begin{array}{l}\text{Sustrate}:\hfill \\ -\frac{d\mathrm{S}}{dt}=\frac{1}{Y_{\mathrm{X}/\mathrm{S}}}\frac{d\mathrm{X}}{dt}+m\mathrm{X}=\frac{1}{Y_{\mathrm{P}/\mathrm{S}}}\frac{d\mathrm{P}}{dt}\hfill \\ -\frac{d\mathrm{S}}{dt}=\frac{1}{Y_{\mathrm{P}/\mathrm{S}}}\frac{d\mathrm{P}}{dt}\hfill \\ -\frac{d\mathrm{xylo}}{dt}=\frac{1}{Y_{\mathrm{P}/\mathrm{S}}}({\nu }_{\mathrm{E},x}\mathrm{X})\hfill \\ -\frac{d\mathrm{glu}}{dt}=\frac{1}{Y_{\mathrm{P}/\mathrm{S}}}({\nu }_{\mathrm{E},g}\mathrm{X})\hfill \end{array}$

Table 8E1.1

Kinetic Parameters for Fermentation

Parameter	Glucose Fermentation	Xylose Fermentation
μm (h−1)	0.662	0.190
νm (h−1)	2.005	0.250
Ks (g/L)	0.565	3.400
Ks′ (g/L)	1.342	3.400
Ki (g/L)	283.700	18.100
Ki′ (g/L)	4890.000	81.300
Pm (g/L)	95.4 for P≤95.4 g/L
	129.9 for 95.4≤P≤129 g/L	59.040
Pm′ (g/L)	103 for P≤103 g/L
	136.4 for 103≤P≤136.4 g/L	60.200
β	1.29 for P≤95.4 g/L
	0.25 for 95.4≤P≤129 g/L	1.036
γ	1.42 for P≤95.4 g/L	0.608
m (h−1)	0.097	0.067
YP/S (g/g)	0.470	0.400
YX/S (g/g)	0.115	0.162

The model for solving the equations can be written in any software. The following is example code for MATLAB:

[a,b]=ode15s('Fermen',[0 36],[200,50,0,1]);

plot(a,b(:,1),'k',a,b(:,2),'k--',a,b(:,3),'ko',a,b(:,4),'k.')

xlabel('t(h)')

ylabel('g/L')

legend('Glucose','Xylose','Ethanol','Cells')

function Reactor=Fermen(t,x)

Glucose=x(1);

xylose=x(2);

ethanol=x(3);

cells=x(4);

v_m_g=2.005;

v_m_x=0.250;

K_s_g=0.565;

K_i_g=283.7;

K_s_x=3.4;

K_i_x=18.1;

K_sp_g=1.341;

K_ip_g=4890;

K_sp_x=3.4;

K_ip_x=81.3;

m_g=0.097;

m_x=0.067;

Y_P_S_g=0.470;

Y_P_S_x=0.4;

P_m_x=59.04;

P_mp_x=60.2;

Beta_x=1.036;

gamma_x=0.608;

if ethanol <95.4;

P_m_g=95.4;

Beta_g=1.29;

gamma_g=1.42;

else

P_m_g=129;

Beta_g=0.25;

gamma_g=0;

end

if ethanol <103;

P_mp_g=103;

else

P_mp_g=136;

end

if cells < 5;

mu_m_g=0.152*(cells)^(-0.461);

v_m_g=1.887*(cells)^(-0.434);

mu_m_x=0.075*(cells)^(-0.438);

v_m_x=0.16*(cells)^(-0.233);

else

mu_m_g =0.662;

v_m_g= 2.005;

mu_m_x=0.190;

v_m_x= 0.25;

end

mu_g=mu_m_g*Glucose*(1-(ethanol/P_m_g)^Beta_g)/(K_s_g+Glucose+Glucose^2/K_i_g);

mu_x=mu_m_x*xylose*(1-(ethanol/P_m_x)^Beta_x)/(K_s_x+xylose+xylose^2/K_i_x);

v_g=v_m_g*Glucose*(1-(ethanol/P_mp_g)^gamma_g)/(K_sp_g+Glucose+Glucose^2/K_ip_g);

v_x=v_m_x*xylose*(1-(ethanol/P_mp_x)^gamma_x)/(K_sp_x+Glucose+Glucose^2/K_ip_x);

Reactor(1,1)=-(1/Y_P_S_g)*v_g*cells;

Reactor(2,1)=-(1/Y_P_S_x)*v_x*cells;

Reactor(3,1)=(v_g+v_x)*cells;

Reactor(4,1)=cells*((Glucose)*mu_g/(Glucose+xylose)+(xylose)*mu_x/(Glucose+xylose));

The profile of the species over time can be seen in (Fig. 8E1.1)

8.2.1.3 Ibutene

Sugars can also be a source of valuable chemicals. Ibutene, a C4 chemical, can be produced from glucose fermentation. The advantage of this process is that the product is in the gas phase and thus no energy-intensive dehydration is required. Ibutene must be separated from the CO₂ (Martín and Grossmann, 2014).

$\begin{array}{cc}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\stackrel{\mathrm{Yeast}}{\to }{\mathrm{C}}_4{\mathrm{H}}_8+{\mathrm{2CO}}_2+{\mathrm{2H}}_2\mathrm{O}+2\mathrm{ATP}& \mathrm{\Delta }H=-41.9\text{kJ}/\text{mol}\end{array}$

8.2.1.4 Furans

Dehydration of C6 sugars yields hydroxymethyl furfural, an intermediate in the production of dimethyl furfural (DMF). C5 sugars can produce furfural as presented in the chemical reactions below:

8.2.1.5 Butanol

Sugars can also be fermented to produce butanol. The typical reaction is the so-called ABE fermentation in which, apart from butanol, both ethanol and acetone are produced. The maximum concentration of butanol is obtained after 48 h of fermentation (Papoutsakis, 1984). The reaction looks like this:

$\begin{array}{l}100\mathrm{Glucose}\to 2.944{\mathrm{C}}_4{\mathrm{H}}_{7.2}{\mathrm{N}}_{0.8}{\mathrm{O}}_2+12.5\mathrm{Acetic}\mathrm{acid}+4.3\mathrm{Butiric}\mathrm{acid}+112.27{\mathrm{H}}_2+211{\mathrm{CO}}_2\hfill \\ +56\mathrm{Butanol}+22.4\mathrm{Acetone}+9.3\mathrm{Ethanol}+6.3\mathrm{Acetoine}+38.16{\mathrm{H}}_2\mathrm{O}\hfill \end{array}$

8.2.1.6 Penicillin

Penicillin was discovered in 1928 when Alexander Fleming, a biologist, realized that one of his experimental setups was contaminated but did not grow any bacteria. He tracked down the culprit—penicillin—and thus the first natural antibiotic was found. The first purification of penicillin is credited to Howard Florey and Ernst Chain by 1939. There are currently around 50 drugs that are classified as penicillins.

Sugars such as glucose are used as substrates to produce penicillin via aerobic fermentation. Oxygen is fed to a stirred tank. The resulting reaction is exothermic and requires constant cooling. The pH, dissolved oxygen, and temperature of the reaction are tightly controlled. It is a fed-batch operation where the substrate is added in small increments. This change in volume must be included in the analysis. Due to the viscosity of the medium and mass transfer limitations, bubble column reactors are also used. The process for the production of the penicillin is modeled on Monod kinetics, as presented in Eq. (8.9). Further details of the model can be found in Birol et al. (2002). The reactor operates at 20–24°C and pH 6.5. After 40 h, penicillin is secreted, and growth stops after about 7 days.

$\begin{array}{l}\text{Biomass}:\begin{array}{l}\frac{d\mathrm{X}}{dt}=\mu \mathrm{X}-\frac{\mathrm{X}}{V}\frac{dV}{dt}\hfill \\ \mu =\left[{\mu }_x\frac{\mathrm{S}}{K_x\mathrm{X}+\mathrm{S}}\frac{C_{\mathrm{L}}}{K_{\mathrm{OX}}\mathrm{X}+C_{\mathrm{L}}}\right]\hfill \end{array}\hfill \\ \text{Penicillin}:\begin{array}{l}\frac{d\mathrm{P}}{dt}={\mu }_{\mathrm{pp}}\mathrm{X}-K\mathrm{P}-\frac{\mathrm{X}}{V}\frac{dV}{dt}\hfill \\ {\mu }_{\mathrm{pp}}=\left[{\mu }_{\mathrm{p}}\frac{\mathrm{S}}{K_{\mathrm{p}}+\mathrm{S}+{\mathrm{S}}^2/K_{\mathrm{I}}}\frac{C_{\mathrm{L}}^p}{K_{\mathrm{OP}}\mathrm{X}+C_{\mathrm{L}}^p}\right]\hfill \end{array}\hfill \\ \text{Substrate}:\frac{d\mathrm{S}}{dt}=-\frac{\mu }{Y_{X/S}}\mathrm{X}-\frac{{\mu }_{\mathrm{pp}}}{Y_{P/S}}\mathrm{X}-m_x\mathrm{X}+\frac{F}{V}-\frac{\mathrm{S}}{V}\frac{dV}{dt}\hfill \\ \text{Dissolved}\text{oxygen}:\frac{dC}{dt}=-\frac{\mu }{Y_{X/O}}X-\frac{{\mu }_{\mathrm{pp}}}{Y_{P/O}}\mathrm{X}-m_{\mathrm{O}}\mathrm{X}+k_{\mathrm{L}}a(C*-C)-\frac{C}{V}\frac{dV}{dt}\hfill \\ \text{Global}\text{mass}\text{balance}:\frac{dV}{dt}=\sum _{\mathrm{in}}F-F_{\mathrm{loss}}\hfill \end{array}$ (8.9)

(8.9)

The reaction generates energy, and only if this energy is properly removed can the reaction be considered isothermal. The mass transfer between the gas and liquid phases depends on the hydrodynamics and properties of the system. The term k_La refers to the liquid film resistance, k_L, and the contact area, a, between the phases. The two-film theory considers the resistance in both the stagnant liquid film and the gas–liquid film at interphase:

$N_{\mathrm{A}}=k_{\mathrm{L}}(c_{\mathrm{LB}}-c_{\mathrm{L}i})=k_{\mathrm{G}}(c_{\mathrm{G}i}-c_{\mathrm{GB}})=K(c_{\mathrm{LB}}-c_{\mathrm{GB}})$ (8.10)

(8.10)

It has been experimentally proven that in gas–liquid processes the main resistance is usually the resistance of the liquid phase (k_L).

Although k_L and a can be determined separately, the area corresponds to that provided by the bubble dispersion in the tank, and can be estimated using either Eqs. (8.11–8.12) as per Gogate et al. (2000):

${\epsilon }_{\mathrm{G}}=0.21{\left[\frac{P_g}{V}(1-{\epsilon }_{\mathrm{G}})\right]}^{0.27}{u}_{\mathrm{G}}^{0.65}$ (8.11)

(8.11)

$a=\frac{6·{\epsilon }_{\mathrm{G}}}{d_{eq}}$ (8.12)

(8.12)

or Eq. (8.13) from Calderbank (1958):

$a=1.44\left[\frac{{\left(P_g/V\right)}^{0.4}\cdot {\rho }^{0.2}}{{\sigma }^{0.6}}\right]{\left(\frac{u_{\mathrm{G}}}{U_{\mathrm{B}}}\right)}^{0.5}$ (8.13)

(8.13)

k_L is a function of the diffusivity of the gas to the liquid and the gas-liquid contact time (Higbie, 1935). The complexity of evaluating these two variables (k_L and a) separately has led to the tradition of using the empirical prediction of k_La:

$k_{\mathrm{L}}a=k\cdot {\left(\frac{P_g}{V}\right)}^{\alpha }\cdot u_{\mathrm{G}}^{\beta }$ (8.14)

(8.14)

where u_G is the superficial gas velocity in the cross sectional area of the tank and P_g is the aeration power as a function of the gas flow rate, Q_c, the impeller diameter, T, and the liquid volume, V, the stirrer angular velocity, N, and the power in absence of aeration, P, as given by Eq. (8.15):

$\frac{P_g}{P}=0.1\cdot {\left(\frac{N^2{T}^4}{g\cdot T_i\cdot V^{2/3}}\right)}^{-1/5}\cdot {\left(\frac{Q_c}{N\cdot V}\right)}^{-1/4}$ (8.15)

(8.15)

Biological media are characterized by high viscosities that affect the diffusion of gas into the liquid phase. Thus, Eq. (8.14) is corrected by the viscosity as follows:

$k_{\mathrm{L}}a=k\cdot {\left(\frac{P_g}{V}\right)}^{\alpha }\cdot u_G^{\beta }\cdot {\left(\frac{\mu }{{\mu }_{\mathrm{water}}}\right)}^{\delta }$ (8.16)

(8.16)

Furthermore, in the presence of solids, as in the case of the Penicillium, Eq. (8.14) can also be corrected by the fraction of solids, X, as follows:

$k_{\mathrm{L}}a=k\cdot {\left(\frac{P_g}{V}\right)}^{\alpha }\cdot u_{\mathrm{G}}^{\beta }\cdot X^{\gamma }$ (8.17)

(8.17)

with k=33.59, α=−0.0463, β=0.94, and γ=−1.012 (Kielbus-Rapala and Karcz, 2011).

8.2.2 Syngas

Syngas is a versatile raw material (see Chapter 5: Syngas). No further discussion regarding its production, composition adjustment, or cleanup is offered here, only its use within the biofuels industry (Martín and Grossmann, 2013a).

Syngas fermentation is one way to produce ethanol. Syngas with an H₂-to-CO ratio equal to 1 and can be fermented anaerobically at 32–38°C. The reaction conversion is around 70%, but ethanol inhibits the process and thus its concentration in water cannot typically exceed 5%.

${\text{3CO}+\text{3H}}_2\to {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+{\mathrm{CO}}_2$

Syngas can also be used to produce ethanol and other fuels via catalytic synthesis.

Ethanol production requires an H₂-to-CO ratio of 1 for a global conversion of 60%. The process, known as mixed alcohol synthesis, is based on the hydrogenation of CO over a catalyst, similar to Fischer–Tropsch (FT) processes. The reaction is carried out at 68 bar and 300°C to obtain a range of alcohols of small molecular weight, including methanol, ethanol, propanol, and also butanol and pentanol in smaller amounts. The reactions with their conversions are as follows:

$\mathrm{CO}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2+{\mathrm{CO}}_2;X_{{\mathrm{CO}}_2}=0.219$

$\mathrm{CO}+2{\mathrm{H}}_2\to {\mathrm{CH}}_3\mathrm{OH};X_{\mathrm{MeOH}}=0.034$

$\mathrm{CO}+3{\mathrm{H}}_2\to {\mathrm{CH}}_4+{\mathrm{H}}_2\mathrm{O};X_{{\mathrm{CH}}_4}=0.003$

$2\mathrm{CO}+4{\mathrm{H}}_2\to {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+{\mathrm{H}}_2\mathrm{O};X_{\mathrm{EtOH}}=0.282$

$2\mathrm{CO}+5{\mathrm{H}}_2\to {\mathrm{C}}_2{\mathrm{H}}_6+{\mathrm{2H}}_2\mathrm{O};X_{{\mathrm{C}}_2{\mathrm{H}}_6}=0.003$

$3\mathrm{CO}+6{\mathrm{H}}_2\to {\mathrm{C}}_3{\mathrm{H}}_7\mathrm{OH}+2{\mathrm{H}}_2\mathrm{O};X_{\mathrm{PropOH}}=0.046$

$4\mathrm{CO}+8{\mathrm{H}}_2\to {\mathrm{C}}_4{\mathrm{H}}_9\mathrm{OH}+3{\mathrm{H}}_2\mathrm{O};X_{\mathrm{ButOH}}=0.006$

$5\mathrm{CO}+10{\mathrm{H}}_2\to {\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{OH}+4{\mathrm{H}}_2\mathrm{O};X_{\mathrm{PentOH}}=0.001$

Methanol, H₂, and FT fuel production from syngas is detailed in Chapter 5, Syngas.

8.2.3 Oil

Oil obtained from various sources can be transformed into biodiesel via transesterification. The reasons for not using oil itself as fuel are its higher viscosity and its differences from current diesel specifications (which do not allow its proper combustion in engines). Apart from transesterfication, hydrotreating emulsions can also be used to break oil into smaller pieces that can be used in diesel engines. This section focuses on transesterification:

$\begin{array}{lll}{\mathrm{CH}}_2-\mathrm{OOC}-{\mathrm{R}}_1\hfill & \mathrm{R}\text{'}-\mathrm{OOC}-{\mathrm{R}}_1\hfill & {\mathrm{CH}}_2-\mathrm{OH}\hfill \\ |\hfill & \hfill & |\hfill \\ {\mathrm{CH}}_2-\mathrm{OOC}-{\mathrm{R}}_2+\text{3R}\text{'}\text{OH}\leftrightarrow \hfill & \mathrm{R}\text{'}-\mathrm{OOC}-{\mathrm{R}}_2+\hfill & \mathrm{CH}-\mathrm{OH}\hfill \\ |\hfill & \hfill & |\hfill \\ {\mathrm{CH}}_2-\mathrm{OOC}-{\mathrm{R}}_3\hfill & \mathrm{R}\text{'}-\mathrm{OOC}-{\mathrm{R}}_3+\hfill & {\mathrm{CH}}_2-\mathrm{OH}\hfill \end{array}$

Although different short-chain alcohols can be used, methanol’s low cost and fast reaction kinetics have made it the alcohol of choice for transesterification. Therefore, biodiesel is also known as fatty acid methyl esters (FAME). The low price of FAME is due to its fossil source, and therefore the word “biodiesel” loses its meaning. Lately, ethanol has been investigated as an alternative to methanol. It is currently produced in biorefineries and its products are called fatty acid ethyl esters (FAEE).

The transesterification reaction is governed by an equilibrium that can be driven to products by controlling the alcohol-to-oil ratio, the operating pressure (supercritical conditions only) and temperature, the catalyst load, and the residence time. Common catalysts are alkalis and acids, but these suffer from a series of disadvantages. Alkali catalysts, in spite of high conversion, are sensitive to the presence of water and free fatty acids (FFA); therefore, a pretreatment with an acid catalyst is necessary. The yield of this pretreatment using a solid acid catalyst can be computed using Eq. (8.18) (Martín and Grossmann, 2012b) as a function of the methanol-to-FFA ratio, the operating temperature, and the residence time:

$\begin{array}{l}{X}_{\mathrm{FFA}}=31.03104+1.486403123\cdot T-6.97793097\cdot RM+19.77691899\cdot t\hfill \\ -0.00018078\cdot T^2-0.16677756\cdot R{M}^2-1.6230585\cdot t^2\hfill \\ +0.02516368\cdot T\cdot RM-0.41625815\cdot T\cdot t+2.37322062\cdot RM\cdot t;\hfill \end{array}$ (8.18)

(8.18)

Once treated, the oil is transformed into FAME using KOH or NaOH as a catalyst. The use of KOH has the advantage that it can later be easily separated from the liquid mixture by precipitation using H₃PO₄ because K₃PO₄ is only slightly soluble in water. Before the neutralization step, the biodiesel must be washed, which increases the water consumption of the facility. A model to predict the yield of the transesterification process as a function of the methanol-to-oil ratio, the temperature of operation, and the catalyst load (Martín and Grossmann, 2012b) is given by Eq. (8.19). Table 8.6 shows the typical range of the operating variables.

$\begin{array}{l}X=74.6301+0.4209\cdot T+15.1582\cdot \mathrm{Cat}+3.1561\cdot RM–0.0019\cdot T^{\mathrm{2}}-0.2022\cdot T\cdot \mathrm{Cat}\hfill \\ –0.01925\cdot T\cdot RM-4.0143\cdot {\mathrm{Cat}}^2–0.3400\cdot \mathrm{Cat}\cdot RM-0.1459\cdot R{M}^2\hfill \end{array}$ (8.19)

(8.19)

Homogeneous acid catalysts such as H₂SO₄ have the advantage of not being sensitive to the presence of FFAs; however, the reaction kinetics are slow. Furthermore, as in the case of KOH or NaOH, a washing step is required before neutralization. Sulfuric acid can be neutralized using CaO. The reaction between CaO and sulfuric acid produces CaSO₄ (gypsum), which precipitates and can be easily separated.

In addition to homogeneous catalysis, heterogeneous methods are also being studied as a way to simplify the biodiesel separation stages. The catalyst can easily be recovered after the reaction. A model to compute the yield of heterogeneous catalyst using methanol as the alcohol can be seen in Eq. (8.20) as a function of the methanol-to-oil ratio, the temperature of operation, and the catalyst load (Martín and Grossmann, 2012b). Table 8.7 shows the range of the operating variables for this case.

$X=-73.6+2.5\cdot T+24.9\cdot \mathrm{Cat}+8.8\cdot RM-0.01\cdot T^2-1.29\cdot {\mathrm{Cat}}^2-0.39\cdot R{M}^2-0.26\cdot T\cdot \mathrm{Cat}$ (8.20)

(8.20)

The use of supported enzymes and a noncatalyzed reaction under supercritical conditions are other alternatives that have been proposed, but the economics are not as good. Models for the yield of heterogeneous catalysts can also be found in Martín and Grossmann, 2012b.

As mentioned above, ethanol can also be used. Severson et al. (2013) developed several models to predict the yield of the reaction as a function of the ethanol-to-oil molar ratio, the operating temperature, the catalyst load, and the residence time. The two most promising processes used either KOH or enzymes as catalysts. The models are presented below, and Tables 8.8 and 8.9 show the range of the operating variables.

Using ethanol (KOH) we have the following:

$\begin{array}{l}X=22.94293+113.88\cdot \mathrm{Cat}+2.828881\cdot RE-1.02734\cdot T-1.44522\cdot \mathrm{Cat}\cdot RE\hfill \\ +0.250723\cdot \mathrm{Cat}\cdot T+0.023375\cdot RE\cdot T-41.4402\cdot {\mathrm{Cat}}^2-0.07568\cdot R{E}^2+0.006226\cdot T^2;\hfill \end{array}$ (8.21)

(8.21)

Using ethanol (enzymatic) we have the following:

$\begin{array}{l}X=3.624996-1.64904\cdot T+17.91299\cdot t-7.60104\cdot RE+10.59497\cdot \mathrm{Cat}-0.49902\cdot \mathrm{water}\_\mathrm{add}\hfill \\ +0.014332\cdot T^2-0.65091\cdot t^2-0.33241\cdot R{E}^2-0.31632\cdot {\mathrm{Cat}}^2+0.00692\cdot \mathrm{water}\_{\mathrm{add}}^2\hfill \\ -0.0407\cdot T\cdot t+0.17485\cdot T\cdot \mathrm{RE}-0.0138\cdot T\cdot \mathrm{Cat}-0.0156\cdot T\cdot \mathrm{water}\_\mathrm{add}-0.0601\cdot t\cdot RE\hfill \\ -0.4629\cdot t\cdot \mathrm{Cat}+0.11014\cdot t\cdot \mathrm{water}\_\mathrm{add}+0.43481\cdot RE\cdot \mathrm{Cat}+0.21369\cdot RE\cdot \mathrm{water}\_\mathrm{add}\hfill \\ -0.09614\cdot \mathrm{Cat}\cdot \mathrm{water}\_\mathrm{add};\hfill \end{array}$ (8.22)

(8.22)

Kinetics: The use of surface of response analysis for evaluating the yield of the transesterification equilibrium is an efficient way to evaluate a process that depends on a number of input variables such as temperature, concentration of species, catalyst load, and time. However, a more detailed analysis based on the reaction mechanism provides better insight. The process to produce biodiesel is a set of three reactions in equilibrium. Let TG be triglycerides, DG diglycerides, MG monoglycerides, ME methyl ester, and GL glycerol:

$\begin{array}{l}\mathrm{TG}+\mathrm{MeOH}\stackrel{k_1}{\to }\mathrm{DG}+\mathrm{ME}\hfill \\ \mathrm{DG}+\mathrm{ME}\stackrel{k_2}{\to }\mathrm{TG}+\mathrm{MeOH}\hfill \\ \mathrm{DG}+\mathrm{MeOH}\stackrel{k_3}{\to }\mathrm{MG}+\mathrm{ME}\hfill \\ \mathrm{MG}+\mathrm{ME}\stackrel{k_4}{\to }\mathrm{DG}+\mathrm{MeOH}\hfill \\ \mathrm{MG}+\mathrm{MeOH}\stackrel{k_5}{\to }\mathrm{GL}+\mathrm{ME}\hfill \\ \mathrm{GL}+\mathrm{ME}\stackrel{k_6}{\to }\mathrm{MG}+\mathrm{MeOH}\hfill \end{array}$ (8.23)

(8.23)

The reactor design based on the kinetics is as follows:

$\begin{array}{l}\frac{d\mathrm{TG}}{dt}=-k_1\left[\text{TG}\right]\left[\text{MeOH}\right]+k_2\left[\text{DG}\right]\left[\text{ME}\right]\hfill \\ \frac{d\mathrm{DG}}{dt}=k_1\left[\text{TG}\right]\left[\text{MeOH}\right]-k_2\left[\text{DG}\right]\left[\text{ME}\right]-k_3\left[\text{DG}\right]\left[\text{MeOH}\right]+k_4\left[\text{MG}\right]\left[\text{MeOH}\right]\hfill \\ \frac{d\mathrm{MG}}{dt}=k_3\left[\text{DG}\right]\left[\text{MeOH}\right]-k_4\left[\text{MG}\right]\left[\text{MeOH}\right]-k_5\left[\text{MG}\right]\left[\text{MeOH}\right]+k_6\left[\text{ME}\right]\left[\text{GL}\right]\hfill \\ \frac{d\mathrm{ME}}{dt}=k_1\left[\text{TG}\right]\left[\text{MeOH}\right]-k_2\left[\text{DG}\right]\left[\text{ME}\right]+k_3\left[\text{DG}\right]\left[\text{MeOH}\right]-k_4\left[\text{MG}\right]\left[\text{MeOH}\right]\hfill \\ +k_5\left[\text{MG}\right]\left[\text{MeOH}\right]-k_6\left[\text{ME}\right]\left[\text{GL}\right]\hfill \\ \frac{d\mathrm{GL}}{dt}=+k_5\left[\text{MG}\right]\left[\text{MeOH}\right]-k_6\left[\text{ME}\right]\left[\text{GL}\right]\hfill \\ \frac{d\mathrm{MeOH}}{dt}=-k_1\left[\text{TG}\right]\left[\text{MeOH}\right]+k_2\left[\text{DG}\right]\left[\text{ME}\right]-k_3\left[\text{DG}\right]\left[\text{MeOH}\right]+k_4\left[\text{MG}\right]\left[\text{MeOH}\right]\hfill \\ -k_5\left[\text{MG}\right]\left[\text{MeOH}\right]+k_6\left[\text{ME}\right]\left[\text{GL}\right]\hfill \\ k_i=k_{ic}{C}_{\mathrm{cat}}+k_{\mathrm{in}}\hfill \\ i=1,\mathrm{..}.,6\hfill \end{array}$ (8.24)

(8.24)

The rate constants depend on the oil, the catalyst, and the alcohol. Table 8.10 (Issariyakul and Dalai, 2012) shows some KOH 1% values from the literature; for this example equilibrium is achieved from 10 min onwards.

8.2.4 Biogas

Biogas is a mixture of mainly methane (50–70%) and CO₂ (20–40%), and is produced from the anaerobic digestion (AD) of biomass. The various sources discussed above can be used as raw materials for its production. Biogas is an interesting product that provides further value to wastes, and also showcases a treatment technology (AD) for waste disposal. Apart from the use of biomass, biogas is typically produced from sludge and animal manure (either in slurry form or as a solid). Not only is biogas produced, but also digestate, an interesting fertilizer. The value of this fertilizer can be computed using the NPK index, representing the amount of nitrogen, phosphorous, and potassium.

8.2.4.2 Process analysis

8.2.4.2.1 Kinetics

The simplest process analysis is to consider a first-order rate model for the substrate utilization:

$\frac{d\mathrm{S}}{dt}=-k\mathrm{S}$ (8.25)

(8.25)

Contois in 1959 proposed a modified form of the Monod model for the first stages, and a Haldane function for the methanogenesis step (Mairet et al., 2012). pH and the gas–liquid mass transfer step must also be controlled.

$\begin{array}{l}\text{Hydrolysis}-\text{acidogenesis}:\hfill \\ {\mu }_i({\mathrm{S}}_i,{\mathrm{X}}_i)=\frac{{\mu }_i\cdot {\mathrm{S}}_i}{K_{si}{\mathrm{X}}_i+{\mathrm{S}}_i}\hfill \\ \text{Methanogenesis}:\hfill \\ {\mu }_3=\frac{{\mu }_3\cdot {\mathrm{S}}_3}{K_{s3}+\mathrm{S}{}_3+{\mathrm{S}}_3^2/K_{i,3}}\frac{K_{I{\mathrm{NH}}_3}}{K_{I{\mathrm{NH}}_3}+{\mathrm{NH}}_3}\hfill \end{array}$ (8.26)

(8.26)

Complete models can be found elsewhere. The anaerobic digestion model (ADM-1; Batstone et al., 2002) is the basis for most analyses.

8.2.4.2.2 Gas composition

The actual composition of the gas depends not only on the operating conditions, but also on the composition of the feedstock. Apart from first principle analysis based on mass and energy balances, the biochemical reaction used for a basic analysis can be considered as follows:

${\mathrm{C}}_n{\mathrm{H}}_a{\mathrm{O}}_b+\left(n-\frac{a}{4}-\frac{b}{2}\right){\mathrm{H}}_2\mathrm{O}\leftrightarrow \left(\frac{n}{2}-\frac{a}{8}+\frac{b}{4}\right){\mathrm{CO}}_2+\left(\frac{n}{2}+\frac{a}{8}-\frac{b}{4}\right){\mathrm{CH}}_4$ (8.27)

(8.27)

If nitrogen from amino acids and protein is also considered, the equation becomes:

${\mathrm{C}}_n{\mathrm{H}}_a{\mathrm{O}}_b{\mathrm{N}}_c+\left(n-\frac{a}{4}-\frac{b}{2}+\frac{3c}{4}\right){\mathrm{H}}_2\mathrm{O}\leftrightarrow \left(\frac{n}{2}-\frac{a}{8}+\frac{b}{4}-\frac{3c}{8}\right){\mathrm{CO}}_2+\left(\frac{n}{2}+\frac{a}{8}-\frac{b}{4}+\frac{3c}{8}\right){\mathrm{CH}}_4+c{\mathrm{NH}}_3$ (8.28)

(8.28)

There are some statistical studies that provide the yield of gas as a function of the time, C-to-N ratio, and loading concentration (kg VS/m³), all of which are highly dependent on the biomass used as feedstock.

$y({\mathrm{m}}^3/\text{kg}\text{VS})=f(\mathrm{retention}\mathrm{time},\mathrm{SC}({\text{kg}\text{VS}/\mathrm{m}}^3),\mathrm{C}/\mathrm{N})$ (8.29)

(8.29)

8.3.1 Ethanol Dehydation

Ethanol production via fermentation yields a diluted water–ethanol mixture, from 5–15% depending on the raw material. The separation is carried out by removing solids first and using a distillation column known as a beer column. Next, rectification and/or molecular sieves are used to produce anhydrous ethanol, which is ready for use as a fuel. Alternatively, extractive distillation has also been suggested.

A beer column allows recovery of most of the ethanol. However, the operation is energy-intensive. Therefore, the use of multieffect columns has been suggested so that the energy consumption can be cut by one-third and the cooling needs approximately by half. The operation of such columns was discussed in Chapter 2, Chemical processes. Fig. 8.9 shows the scheme of the operation.

Example 8.3

A single beer column operates with an L/D ratio of 1. The feed, 6 kg/s of a water–ethanol mixture, has 15% by weight of ethanol. The distillate has 80% purity by weight and the residue 2%. Compute the distillate and the residue, as well as the number of trays required for the column. Determine the energy removed and provided.

Solution

We assume that the flows are constant along the column. Thus the rectifying and stripping operating lines are as follows:

$y_n=\frac{L}{V}{x}_{n-1}+\frac{D}{V}{x}_D$

$y_m=\frac{L\text{'}}{V\text{'}}{x}_{m-1}+\frac{W}{V\text{'}}{x}_W$

Because the data is by weight, we must convert it into molar ratios for each of the streams involved in the column operation. x is the mass fraction, F is the total mass flow rate, and M is the molar weight: $y_i=\frac{x_i\cdot F/M_i}{\sum _i{x}_i\cdot F/M_i}$ (Table 8E3.1).

Table 8E3.1

Molar and Mass Compositions

	x	y
Feed	0.15	0.0645933
Distillate	0.8	0.61016949
Residue	0.02	0.00792254

We formulate the mass balance as follows:

$F=D+R$

$F\cdot x_f=D\cdot x_D+R\cdot x_R$

By solving the system we get D=1 kg/s and R=5 kg/s. Then, we use the McCabe method to compute the number of trays. Due to the nonideal behavior of the mixture, experimental data is used to plot the equilibrium curve. Next, using D, R, and the L/D ratio we plot the operating lines:

$V=D+L$

$L\prime =F+L(\mathrm{assuming\; feed\; as\; saturated\; liquid})$

$V\prime =V$

$L/D=1\to \frac{L}{V}=\frac{D}{(1+(L/D))·D}=0.5$

q line represents the intersection between the stripping and rectifying lines. Thus we have seven trays required for such separation (see Fig. 8E3.1). The energy involved in the condenser, Qw, and the reboiler, Q_H, are computed as follows:

$Q_{\mathrm{W}}=\lambda \cdot V=(0.8{\lambda }_{\mathrm{EtOH}}+0.2{\lambda }_{\mathrm{EtOH}})(L/D)\cdot D=(9266/46\cdot 4.18\cdot 0.8+0.2\cdot 9723/18\cdot 4.18)\cdot 1\cdot 1=1125\text{kJ}/\mathrm{s}$

$Q_{\mathrm{H}}=\lambda \cdot V\prime =(0.02{\lambda }_{\mathrm{EtOH}}+0.98{\lambda }_{\mathrm{EtOH}})(L/D)\cdot D=2230\text{kJ}/\mathrm{s}$

Final dehydration: Fuel-quality ethanol is achieved only by surpassing the azeotrope (96% by weight). Therefore, a number of strategies have been suggested and evaluated, such as extractive distillation, pervaporation, and the use of molecular sieves. Among them, molecular sieves use silica gel or zeolites as adsorbent beds so that they retain the water. The operation of these units is similar to pressure swing adsorption (PSA) systems, except for the fact that there is no pressurization–expansion stage (see chapters Air and Syngas for an evaluation of adsorbent beds). Next, a stream of hot air is used to regenerate the beds. Pervaporation consists of the use of membranes permeable to one of the components so that by evaporation, ethanol is recovered. The use of molecular sieves is the technology of choice in most industrial facilities, and provides an advantage in terms of energy consumption.

Example 8.4

A second-generation bioethanol production facility processes biomass that contains cellulose (assume glucose for calculations), hemicellulose (xylose), and lignin. The conversion from glucose to ethanol is 90%, and that from xylose is 80%. Ninety-six percent (96%) of the ethanol produced is recovered. If the fractions of cellulose and hemicellulose in the feedstock are the same, the consumption of energy is 7000 kJ/kg of ethanol produced, and it is possible to obtain up to 19,500 kJ/kg of lignin. Compute the minimum amount of lignin a certain biomass needs to contain for the facility to be energy self-sufficient.

Solution

Assume 100 kg of initial biomass. We formulate the mass and energy balances for the plant composition, the ethanol production, and the energy required and produced as follows:

$\begin{array}{l}{X}_{\mathrm{G}}+X_{\mathrm{X}}+X_{\mathrm{L}}=1\hfill \\ X_{\mathrm{G}}=X_{\mathrm{X}}\hfill \\ \mathrm{EtOH}=BX{}_{\mathrm{G}}\cdot \frac{2·46}{180}\cdot Con{v}_{\mathrm{G}}\cdot Sep+B{X}_{\mathrm{X}}\cdot \frac{5·46}{3·150}\cdot Con{v}_x\cdot Sep\hfill \\ {\mathrm{Energy}}_{\mathrm{consumed}}=7000·\mathrm{EtOH}\hfill \\ {\mathrm{Energy}}_{\mathrm{produced}}=BX{}_{\mathrm{L}}\cdot 19500\hfill \\ {\mathrm{Energy}}_{\mathrm{consumed}}={\mathrm{Energy}}_{\mathrm{produced}}\hfill \end{array}$

Solving this system of equations we get the results shown in Table 8E4.1:

Table 8E4.1

Summary of Results

			Fermentation
Biomass	100	PM	Conv	Ethanol	Recovery
Cellulose	0.42196144	180	0.9	19.4102261	0.96	18.6338171
Hemicellulose	0.42196144	150	0.8	25.8803015	0.96	24.8450894
Lignin	0.15607813
Total	1.000001
Energy (kJ/kg Lig)	19500	304352.345
Consumption (kJ/kg et)	7000	304352.345	0

8.3.2 Hydrocarbon and Alcohol Mixture Separation

FT processes and mixed alcohol synthesis require the use of distillation to separate the products. FT liquid fuels are separated using petrochemical-based technologies (see also chapters: Chemical processes and Syngas). In the case of mixed alcohols, a sequence of distillation columns is required; see Chapter 2, Chemical processes for the heuristics behind the technique, and Fig. 2.1 for a block flowsheet of the process. The separation of butanol mixtures is more complex due to the inmiscibility between water and ethanol. Recently a hybrid extraction–distillation scheme was proposed that uses mesitylene in an extraction column that processes the fermentation broth. This setup is followed by a sequence of three distillation columns operating at 1, 2, and 0.5 bar, respectively, to separate the solvent, mesitylene, butanol, acetone, and ethanol (Kramer et al., 2011). Ibutene purification is the simplest in the sense that it is a gas produced in the fermentation. Therefore, it exits the reactor together with CO₂ and oxygen. PSA and membrane systems can be used for its purification.

8.3.3 Alcohol Recovery: Biodiesel

As presented in the analysis of the equilibrium towards biodiesel, an excess of alcohol is typically used to drive the reaction into biodiesel. However, the decision concerning the excess of alcohol can be made not just at the reactor level (aiming at higher conversions), but also at the process level. The reason is because for a larger excess of alcohols, the conversion increases, as well as the energy required to recover the excess. Therefore, there is a tradeoff to be solved (Martín and Grossmann, 2012b). Fig. 8.10 shows the flowsheet for the production of biodiesel using heterogeneous catalysts.

The distillation column or the flash separation to recover the alcohols must work in such a way that the temperature of operation is below 150°C to avoid glycerol decomposition. This column presents the particular feature that the feed and the bottom stream are phases. Thus, the vapor pressure is that given by both phases as follows:

$P_{\mathrm{T}}=P_{v,\mathrm{polar}}+P_{v,\text{non}-\text{polar}}=\sum _{i=\text{water},\text{glycerol},\text{alcohol}}{x}_{i}P{v}_i+\sum _{j=\text{biodiesel},\text{oil}}{x}_{j}P{v}_j$ (8.30)

(8.30)

Once the alcohol has been recovered, polar and nonpolar phases are separated at around 40–60°C for ethanol and methanol, respectively. In the case of a homogeneous catalyst, water is added in the separation to help remove the catalyst. The polar phase contains glycerol, alcohols and the homogeneous catalyst. The nonpolar phase contains biodiesel and unconverted oil. The polar phase is neutralized, if needed, and later distilled to purify the glycerol. In the case where acid homogeneous catalysts are used, CaO is recommended for neutralization (as presented above). In the case where KOH is used, H₃PO₄ is recommended. The nonpolar phase is also distilled to purify the biodiesel. The distillate should be below 250°C to avoid decomposition. As a result, this column and the alcohol recovery columns are meant to operate under vacuum. Furthermore, the column processes an immiscible mixture, and therefore it has to be taken into account in boiling and dew point computations.

Glycerol, the main byproduct, has become an interesting carbon source for the production of chemicals such as H₂ or syngas via reforming, ethanol, via fermentation, glycerol ethers, together with ibutene, polyesters, etc.

Example 8.5

An algae-based biodiesel facility grows algae with 40% by weight oil (molecular weight 884 kg/kmol), 25% starch (assumed to be glucose), 10% protein, and the rest inert material. The aim is to produce ethanol from the starch and biodiesel from oil (assume a molecular weight of 310 kg/kmol). The conversion of the sugar fermentation is 90%, and the transesterification reaction has a conversion of 95%. Determine the amount of ethanol, biodiesel, and glycerol sold to the market if we use part of the ethanol produced as a transesterifying agent.

Solution

We perform the mass balances based on the stoichiometry of the reactions below:

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\stackrel{\mathrm{Yeast}}{\to }2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+2{\mathrm{CO}}_2$

$\begin{array}{lll}{\mathrm{CH}}_2-\mathrm{OOC}-{\mathrm{R}}_1\hfill & \mathrm{R}\text{'}-\mathrm{OOC}-{\mathrm{R}}_1\hfill & {\mathrm{CH}}_2-\mathrm{OH}\hfill \\ |\hfill & \hfill & |\hfill \\ {\mathrm{CH}}_2-\mathrm{OOC}-{\mathrm{R}}_2+3R\text{'}\text{OH}\leftrightarrow \hfill & \mathrm{R}\text{'}-\mathrm{OOC}-{\mathrm{R}}_2+\hfill & \mathrm{CH}-\mathrm{OH}\hfill \\ |\hfill & \hfill & |\hfill \\ {\mathrm{CH}}_2-\mathrm{OOC}-{\mathrm{R}}_3\hfill & \mathrm{R}\text{'}-\mathrm{OOC}-{\mathrm{R}}_3+\hfill & {\mathrm{CH}}_2-\mathrm{OH}\hfill \end{array}$

Let Conv_G be the conversion of glucose, Conv_L the lipids conversion, B the total initial biomass, and X_i the fraction of component i in the biomass (S starch, L Lipids):

$\begin{array}{l}\mathrm{EtOH}=BX{}_{\mathrm{S}}\cdot \frac{2·46}{180}\cdot Con{v}_G-3\cdot \frac{46}{884}BX{}_{\mathrm{L}}\cdot Con{v}_L\hfill \\ \mathrm{FAEE}=BX{}_{\mathrm{L}}\cdot \frac{3·310}{884}\cdot Con{v}_L\hfill \\ \mathrm{Glycerol}=BX{}_{\mathrm{L}}\cdot \frac{1·92}{884}\cdot Con{v}_L\hfill \end{array}$

Solving the system we get the following results (Table 8E5.1):

Table 8E5.1

Summary of Results

					Prod EtOH		Prod Biodiesel		Products
Biomass	100		PM	Conv	kmol	kg	kmol	kg	kg
Starch	0.40	40.00	180.00	0.22				0.00	0.00
Oil	0.25	25.00	884.00	0.03				0.00	0.00
Protein	0.10							0.00	0.00
Others	0.20							0.00	0.00
FAEE			310.00				0.08	26.30	26.30
EtOH			46.00		0.40	18.40	−0.08	−3.90	14.50
Glycerol			92.00				0.03	2.60	2.60

8.3.4 Penicillin Purification

The process for purification of penicillin obtained via fermentation is more complex than the previous ones mentioned. Biomass must be separated from the culture. Next, penicillin is extracted, and if a solvent is used, a recovery stage must be added to the process. Subsequently, the liquid phase containing the penicillin is recovered by centrifugation. A second extraction is followed by crystallization (either via cooling or drowning) to purify it. Finally, washing and drying allow powder production. It is typically stored as a potassium or sodium salt. See Chapter 4, Water for the operation of crystallizers.

8.4.1 Rankine Cycle

The most widely used thermodynamic cycle is the regenerative Rankine cycle. High-pressure, superheated steam is obtained by burning the biomass. Alternatively, this energy can be obtained from burning any fossil fuel (as in thermal plants) or by using solar or geothermal energy. In the case of solar energy, concentrated solar power is required so that solar energy can reach an intensity capable of producing high-temperature steam. The variability in solar energy availability becomes a challenge for the operation of such a plant. Typically, a heat transfer fluid such as molten salts is used to buffer the absence of solar energy during the nighttime. However, in the long term, a different solution must be used. Hybrid plants, for example, which combine the use of different energy sources such as biomass and solar and allow continuous energy production over time (Vidal and Martín, 2015).

Once the steam is produced, it is fed to a steam turbine. The turbine is divided into three stages: high, medium, and low pressure. In the high-pressure turbine the steam expands into a moderated pressure. The steam is sent back to the boiler to be reheated. Next, the stream is expanded in the medium- and low-pressure turbines. Several extractions at different pressures are obtained from the turbines. These streams are used to reheat the condensed stream to close the cycle. A thermal plant scheme based on a Rankine cycle is presented in Fig. 8.11. The energy generated in the expansions is computed as an enthalpy difference. The temperature–entropy (TS) diagram of the cycle is shown in Fig. 8.11.

$W=\eta ·\sum _i(H_k-H_{k-1}),\forall i\in \mathrm{Expansion}$ (8.31)

(8.31)

The enthalpy is a function of the temperature and the pressure. See Appendix B (Thermodynamic Data) to compute the enthalpy. The isentropic efficiency of the turbine (typically from 0.7–0.9) is computed using Eq. (8.32):

$\eta =\frac{H_{\text{steam},(\text{out})}-H_{\text{steam},(\text{in})}}{H_{\text{steam},(\text{out},\text{isoentropy})}-H_{\text{steam},(\text{in})}}$ (8.32)

(8.32)

where the entropy of the stream is also a function of the pressure and temperature. For a detailed model of the cycle shown in Fig. 8.11, we refer the reader to the supplementary material in Vidal and Martín (2015).

8.4.2 Brayton Cycle

Compressed air is used to burn a fuel (ie, syngas) in a combustion chamber. When burned, the gas heats up to a high temperature that can be computed using Eq. (8.33):

$\begin{array}{l}{\mathrm{C}}_n{\mathrm{H}}_m+(n+m/2)\cdot {\mathrm{O}}_2\to n{\mathrm{CO}}_2+(m/2)\cdot {\mathrm{H}}_2\mathrm{O}\hfill \\ \mathrm{CO}+½{\mathrm{O}}_2\to {\mathrm{CO}}_2\hfill \\ {\mathrm{H}}_2+½{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}\hfill \end{array}$

$0=\sum _{i\in \mathrm{Product}}{n}_i\left(\mathrm{\Delta }{H}_{f,i}^{298}+{\int }_{298}^{T_{\mathrm{out}}}C{p}_{i}dT\right)-\sum _{i\in \mathrm{Reactants}}{n}_i\left(\mathrm{\Delta }{H}_{f,i}^{298}+{\int }_{298}^{T_{\mathrm{in}}}C{p}_{i}dT\right)$ (8.33)

(8.33)

Next, it is fed to a gas turbine. In the gas turbine the superheated gas is expanded isentropically to produce power. The energy involved in the compression of the air and that obtained in the expansion can be computed using Eq. (8.34), where k is around 1.4. The cycle diagrams can be seen in Fig. 8.12.

$W(\mathrm{Comp})=(F)\cdot \frac{8.314\cdot k\cdot (T_{\mathrm{in}}+273.15)}{((Mw)\cdot (k-1))}\frac{1}{{\eta }_{\mathrm{s}}}\left({\left(\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}\right)}^{\frac{k-1}{k}}-1\right)$ (8.34)

(8.34)