AN APPROACH TO OPTIMIZE COSTS DURING ULTRA-LOW HYDRODESULFURIZATION OF A BLEND CONSISTING OF DIFFERENT OIL STREAMS

Silva, J. I. S. da; Secchi, A. R.

doi:10.1590/0104-6632.20180354s20170370

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Brazilian Journal of Chemical Engineering

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Process Systems Engineering • Braz. J. Chem. Eng. 35 (4) • Dec 2018 • https://doi.org/10.1590/0104-6632.20180354s20170370 copy

AN APPROACH TO OPTIMIZE COSTS DURING ULTRA-LOW HYDRODESULFURIZATION OF A BLEND CONSISTING OF DIFFERENT OIL STREAMS

Authorship SCIMAGO INSTITUTIONS RANKINGS

ABSTRACT

The presence of contaminants in diesel oil is severely controlled. These components affect the diesel properties, causing impacts on the emissions of pollutants and compromising its commercialization. The Brazilian legislation imposes the use of diesel S10. In this work, a hydrotreatment unit fed with multiple diesel streams was optimized using a phenomenological mathematical model. A three-phase model of a tricked-bed reactor was considered. The optimization of the process was performed with the interior point method considering the hydrodesulfurization reaction of diesel. It was proposed to minimize the operating cost, keeping the amount of sulfur in the final product below the maximum allowed value. The EMSO process simulator was used to model, simulate and optimize the hydrotreatment process. It was observed that the optimization strategy drove the contaminant content into the diesel S10 specification range, as well as reducing the process operating cost by increasing the feed temperature and hydrogen flowrate and adjusting the feed composition blend of the available diesel streams.

Key words:
Hydrotreatment; Diesel S10; Optimization

System Informations	Values
Organic sulfur concentration (ppm)	400 (oil A)
	300 (oil B)
	200 (oil C)
Feed (cm³/s)	0.03 (oil A)
	Mass balance (oil B)
	0.02 (oil C)
	0.1 (TBR inlet)
T_MABP - Average boiling temperature (^oC)	476
Surface gas velocity (cm/s)	0.28
Length of the TBR (cm)	35
Equivalent particle diameter (cm)	2.54×10^-1 (catalyst CoMo/Al2O₃)
Operational pressure (MPa)	5.3
Liquid Hourly Space Velocity, LHSV, (h^-1)	1
Costs (US$/cm3)	1.3×10^-3 (oil A)
	1.4×10^-3 (oil B)
	1.5×10^-3 (oil C)
	2.0×10^-3 (hydrogen)
Costs of heat in the mixer (US$/kJ)	3.0

Details of the optimization problem	Values
Number of equality constraints	4980
Number of inequality constraints	6
Degrees of freedom of the optimization problem	4
Number of discrete points	110

Informations	Original Solution	Optimized Solution
u_G (cm/s)	0.28	0.21
F_A (cm³/s)	0.03	0.018
F_B (cm³/s)	0.05	0.022
F_C (cm³/s)	0.02	0.06
T_o (K)	653	699
OF (US$/h)	356.5	404.09
Sulfur concentration at the exit of the reactor (ppm)	160	9.84

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[1] E-mail: josesilva@peq.coppe.ufrj.br; arge@peq.coppe.ufrj.br

Gas-liquid mass-transfer coefficient	$\frac{K_{i}^{L} a_{L}}{D_{i}^{L}} = 7 {(\frac{G_{L}}{μ_{L}})}^{0.4} {(\frac{μ_{L}}{D_{i}^{L} ρ_{L}})}^{0.5}$
Liquid-solid mass-transfer coefficient	$\frac{K_{i}^{S}}{D_{i}^{L} a_{S}} = 1.8 {(\frac{G_{L}}{a_{S} μ_{L}})}^{0.5} {(\frac{μ_{L}}{D_{i}^{L} ρ_{L}})}^{\frac{1}{3}}$
Liquid-solid heat-transfer coefficient	$h_{L S} = j_{H} d_{p L} u_{L} ρ_{L} {(\frac{c_{p L} μ_{L}}{k_{L}})}^{- \frac{2}{3}}$
Oil density	$∆ ρ_{T} = (T_{L} - 520) [0.0133 + 152.4 {(∆ ρ_{p} + ρ_{0})}^{- 2.45}] - {(T_{L} - 520)}^{2} [8.1 \times 10^{- 6} - 0.0622 \times {10^{- 0.764}}^{(∆ ρ_{p} + ρ_{0})}]$ $∆ ρ_{P} = (\frac{P}{1000}) [0.167 + (16.181 \times 10^{- 0.0425 \times ρ_{0}})] - {(\frac{P}{1000})}^{2} \times 0.01 [0.299 + (263 \times 10^{- 0.0603 \times ρ_{0}})]$ $ρ (T_{L}, P) = ρ_{0} + ∆ ρ_{P} - ∆ ρ_{T}$
Liquid heat capacity	$C_{P}^{L} = 4.1868 (\frac{0.415}{\sqrt{{(ρ (T_{L}, P))}^{15.6}}} + 0.0009 [T_{L} - 288.15])$
Henry coefficient	$𝓗_{i} = \frac{v_{N}}{λ_{i} ρ_{L}}$
Specific surface area	$a_{S} = \frac{6}{d p} (1 - \in)$ ; See Table 1
Diffusivity	$D_{i}^{L} = 8.93 \times 10^{- 8} (\frac{v_{L}^{0.267}}{v_{i}^{0.433}}) (\frac{T_{L}}{μ_{L}})$
Dynamic viscosity of the liquid	$μ_{L} = 3.141 \times 10^{10} {(T_{L} - 460)}^{- 3.444} {[\log_{10} (A P I)]}^{y}$ $y = 10.313 [\log_{10} (T_{L} - 460)] - 36.447$
Molar volume	$v_{i} = 0.285 \times v_{i, c}^{1.048}$ $v_{c}^{m i x} = 7.5214 \times 10^{- 3} (T_{M A B P}^{0.2896}) (d_{15.6}^{- 0.7666})$ ; See Table 1
Solubility of H₂S	$λ_{H_{2} S} = e x p (3.367 - 0.00847 T_{L})$
Solubility of H₂	$λ_{H_{2}} = - 0.559729 - 0.42947 \times 10^{- 3} T_{L} + 3.07539 \times 10^{- 3} (\frac{T_{L}}{ρ_{20}}) + 1.94593 \times 10^{- 6} T_{L}^{2} + \frac{0.835783}{ρ_{20}^{2}}$
Kinetic model of HDS reaction	See Equation 11, with E_a = 131 993 J/mol; k₀ = 4.266×10⁹ cm³/(g s)×(cm³/mol)^0.45 and HDS reaction enthalpy Δ_HDS = -251 000 J/mol $K_{H_{2} S} = k_{0, H_{2} S} e x p (\frac{∆ H_{a d s}}{R T_{S}})$ ; k_0,H₂S = 41 770 cm³/mol; ΔH_ads = 2 761 J/mol