thermodynamic analysis of ethanol from co2 and ch4
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Assignment 1
Martínez Cajigas, Marcelo Ernesto 1; Sánchez Trujillo, Alexander 2 Escalado y Simulación de Bioprocesos, Escuela de Procesos y Energía, Universidad Nacional de Colombia, Sede
Medellín, Colombia. Cra. 80 x Cl. 65. Barrio Robledo.
[email protected], ingeniería biológica; [email protected]; ingeniería biológica
1. The conventional process to obtain ethanol have pros and cons, the pros of thistraditional route can be: the high availability and low cost of raw material and thisraw material is a high pollutant, and the cons be principally the hydrolysis ofcellulose, hemicellulose for obtain monomers, the high cost of the hydrolysis, thelow yield and the generation of secondary subproducts as lactic acid, succinic acid,acetic acid, formic acid, other subproduct of the hydrolysis is Furfural for thehydrolysis of hexoses and Hidroximetylfurfural (HMF) for pentoses this compoundsbe inhibitors.
Other consideration in this process is the free energy of Gibbs ∆ (Ghosh et al.,2013) in a study thermodynamic parameters of hemicelluloses hydrolysis ondifferent concentrations have that the ∆ in the process is positive, around of 140KJ/mol. This value this value says that the reaction needs energy.2. Substrate sources for this process do not require pretreatment in many cases isthe result of other industrial processes (PWT plant wastewater treatment) or theextraction of oil.
The advantages of this process is the reduction of greenhouse gases anddenitrification, from an environmental point of view is favorable.
3.
. . + Signs considerations in the reaction:
Methane and carbonic dioxide are carbon source, Methane is also energy sourceand nitric acid is nitrogen source.
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Balancing the equation:
Atomic Balance
Carbon
=1
Hydrogen
4 =2 2 Nitrogen
=0.2 Oxygen
2 3 =0.4 Electron Balance
= = = . ,= . Nitric acid is considered the nitrogen source with reduction grade 5
= = . Solving the equations, (Tijhuis, Van Loosdrecht, & Heijnen, 1993), the stoichiometriccoefficients are:
Theorical stoichiometry (Nielsen, 2014)
. . . . . . + a. Herbert-Pirt substrate distribution
= =4.5 69000(11298) R=8.314 J/mol.K
T=306K
=0.44 / =∆=0.0067 Assuming heterotrophic microorganism and using the Heijnen correlation, with =1 =8 because the methane is the energy source.
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=200 18 6. [ 3.8⁄ . 3.6 0.4] =1088.094 /
b. Thermodynamic analysis of the product reaction
. . . . . Standard conditions (T= 25°C, P= 1 Bar, 1 mol/L, pH=7).(S. J. Heijnen, 1994)
∆°′= 1.7550.750.25394.3590.4080.7168.40.20 1.7 2 ∆°′= 115.124 / With the effect of temperature in
∆
The formation enthalpy for compounds is:
∆ 306= 50,343 ∆ 306=68.568 ∆ 306= 394.363 ∆ 306= 79.133 ∆ 306= 236.361 The free energy of Gibbs
∆306= 114.767 / c. Using the Herbert-Pirt equation, the growth and production reactions find thespecific rates in terms of and ∆306
∆=114.9012 4 ∗1 2 41.75 4=65.581
||=∆
||=17.366 =17.366 0.00 8 17.366 0.00676.2 12 =0
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=11.061 0.00 4.
a)
Balance
4 2 3 20.40.224 2 2 + (J. J. Heijnen & Van Dijken, 1993)
Nitrogen Balance
The is only produced by the catabolic reaction0.2=
22 4
=2.212 0.00 Nitrogen balance
0.22 =0 =4.624 0.001 Carbon balance
2 =0 =5.755 0.01
Charge balance
=0 Oxygen balance2 3 0.4 =0
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=24.9840.0854 In the overall reaction, apply the first law of thermodynamic
4 2 3 20.40.224 2 2 + =0 =17.366 0.00 =11.061 0.00
=2.212 0.00
=4.624 0.001 =5.755 0.01=24.9840.0854 =∆ We can replace the specific velocities in the overall reaction
b. Graphics
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c.
=3∆ . 69000(11298) ∆=65.658 / =8
=1088.094 /
=0.44 /
= . − 5.For =0.02 ℎ−
. . . . . . . .. 2 4 Molar flux = 10000mol/h =0.5 =0.5 10000 ℎ∗0.2 =2000 /ℎ
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10000 ℎ∗0.5795 =5795 /ℎ10000 ℎ∗1.5684 =15684 /ℎ 10000 ℎ∗0.4177 =4177 /ℎ
10000 ℎ∗2.5921 =25921 /ℎ 10000 ℎ∗0.0886 . . =886 . ./ℎ Gas s tream
The gas stream input is composed for methane and carbon dioxide in the same
proportion
and for the stoichiometry is needed,
16603 this amount is
=0.5= =31368 /ℎInput of CO2 minus the CO2
2ℎ = ∗0.5 5795 /ℎ =9889ℎ 10000ℎ 2000ℎ=21889 /ℎ =0.45
=0.09 =0.46 Liquid st ream
= 4177 =886 /ℎ 2/ℎ = 26807 =886 26807 ℎ=0.033=25921 26807 ℎ=0.967
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̅ =0.967 (18 ) 0.033 (23.2 )̅ =18.17 ℎ/
=487.08 ℎ/ℎ
886 /ℎ487.08 ℎ/ℎ=1.82 / The Henry constants for gases is higher than 11000 bar, it´s very longer andsolubility is low, for instance ethene more soluble among the gases in this process
Henry constants
CO2 1700 bar
N2 89000 bar 300K
O2 45000 bar
CH4 40200 bar 298K
C2H4 11700 bar
6. According to the fraction, the proprieties of the broth approximates to the water.
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7.
=21889 /ℎ =0.45 =0.09
=0.46
=4177 =26807 =0.033=0.967
= 5.4454 =31368 /ℎ =0.5 =0.5
8. If we don´t remove the N2 and CO2 of the recycle this is accumulated in thereactor, increasing the pressure.
a. It is removed with an absorber with water as solvent. The Henry constant isby ethene 11000 bar, compared with other gases is the most soluble.
b. The heat produced by the reaction is removed with cooling water, avoidingevaporation.
9.
With =0.02
ℎ− fix the stoichiometry . . . . . . . .. 2 4
Also we use the value of the stoichiometry for the production of carbon dioxide, (Stockar, Urs von, John F. Carpenter, Eva y. Chi, A. Espah Borujeni, M.t. Gude et al., n.d.)
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36143 ∗0.5 5795 =12ℎ = =
=10000 12276.5
= 36143 =/ =2000/ =12626.5/ =10000/
10.
a.
= ∗ =0 ℎ
=1.2 0.05 (A) = ∗ ( ) =18071,5=3∗∗
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= = =36143 /ℎ/ℎ
( )
Using gas ideal equation
/ℎ = ∗ ∗ / =4.49∗10 ~33° / ℎ
= .∗ =16000 = 0.1000 = 360 /ℎ = 306
= 44900 = 1.309/ℎ = 3∗ 361 = 12 0.5∗ = 2.0652 = ./ ∗0 = 3.6369
= 7.5109 = 2.1519 b. Recycle Fr [100 to 20000]
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c. 2=30 2=9.24 2 4=0.7 =1.2 0.05∗
= ∗ 2 = ∗ 2 = ∗ 2 4 = 39.0978
= 12.0421
= 0.9123 /
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b. = 4.1780 1 = ∗1000∗
1 = 176.3406
12. For a Lab bioreactor you have a control culture with microorganism interestwithout concentrations of ethene inhibitory and culture with concentrations ofinhibition. Later is measured biomass concentrations to different values of etheneproduced. It is observes where it stops the growth and performance in curve.If show inhibition is necessarily diminished the pressure for stimulated the pass agas phase.
13. The water evaporation act with sorbent and drag gas molecules in caseinhibition is the grand advantage. No necessarily with methane or Carbon dioxidethat are substrate and their low solubility in the medium will be a problem.
14. If u is small will be susceptible because the dilution rate is more probability, thatcause washout in the bioreactor, for perturbation at system. It is necessary a goodcontrol system that avoid future problems.
References
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Heijnen, S. J. (1994). Thermodynamics of microbial growth and its implications for process
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Ghosh S.,Haldar S.,Ganguly A., Chatterjee P. (2013). Investigations on the kinetics and
thermodynamics of dilute acid hydrolysis of Parthenium hysterophorus L. substrate
Chemical Engineering Journal. 229. 111-117
Nielsen, J. (2014). Bioreaction Engineering Principles . Psychological Science (Vol. 25).
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