Question

Solve for a), b) and c)

The residual hydrodesulfurization (HDS) process is operated at the capacity of 5,000 barrels per day to produce low sulfur fuel for a power plant. The operating conditions are: Type: Fixed bed plug flow reactor Reactor temperature -400°C Reactor pressure - 2100 psi LHSV (Liquid Hourly Space Velocity) - 0.5/hr-flowrate, bbl/hr)/(reactor volume, bbl) Hydrogen gas rate - 23.7 standard liters/hr The sulfur contents of the fuel oil product must be reduced to 1.0 wt%. The sulfur content of the residual oil feedstock is 3.8 wt% to the HDS a) (20 Points) What is the reactor size in gallons? (1 barrel - 42 gallons) What is the residence time of oil in the plug flow reactor? The hydrodesulfurization reaction is a second order reaction (at constant hydrogen pressure) b) (40 Points) - kc where C-sulfur concentration of the oil, weight % kreaction rate based on Arrhenius" law, k = koek Derive the relationship of sulfur content in the oil and temperature, as 6 - Les can be where Co - sulfur content of the feedstock, weight % G-sulfur content of the desulfuried product, weight % ko-frequency factor E the activation energy of the reaction - 27000 cal/g.mol R -gas constant - 1.987 cal/(mob)(K) e) (40 Points) After ten days operation at 400°C, the sulfur content of the desulfurized product is 1.2 wt% due to the catalyst deactivation. At what temperature should the reactor be raised in order to produce the product fuel oil to meet the specification of 1.0 wt% sulfur content? The activation energy of the desulfurization HDS reaction is 27000 cal/g.mol.

Answer 1

LHSV (liquid hourly space velocity) = 0 5/hr =( flow rate, bbl/hr)/(reactor volume, bbl)

Initial concentration Cao = 3.8 wt%

Final concentration after HDS , Ca = 1 wt%

a) Capacity of plant, vo = 5000 barrel/day

1 barrel = 42 gallon  

5000 barrel /day = 5000*42 gallon/24 hr = 8750 gallon/hr

Flow rate vo = 8750 gallon/hr

Space time = 1/ space velocity = 1/0.5 = 2 hr

Space time tou = Volume of the reactor / flow rate  

Volume of the reactor V = space time *flow rate = 2hr *8750 gallon/hr = 17500 gallon

volume of the reactor V = 1750 gallon

Residence time tou = 2 hr

b)  (-rHDS) = -dC/dt = kC²

-dC/dt = kC²

dC/C² = - kdt

-[1/Cf - 1/Ci] = - kt

Where ,

t= 1/LHSV

k = ko*exp(-Ea/RT)

1/Cf - 1/Ci = 1/(LHSV) * ko*exp(-Ea/RT)

Ci - sulfur content of feedstock(wt%)

Cf - sulfur content of desulfurized product(wt%)

Ea is the activation energy = 27000cal/g.mol

ko is the preexpontial factor.

R is the gas constant = 1.987 cal/gmol.k

C)

Case1: at temperature T1 = 400 C = 673 K

Cf = 1.2 wt%

Ci = 3.8 wt%

1/1.2- 1/3.8 =( ko/LHSV)*exp{(-27000 cal/gmol)*(1.987cal/gmol.k)*673}

0.57 = ko/(LHSV)*exp(-20.19)

ko= 3.3463E8 *LHSV

ko = 1.673E8 hr-1

Case2: for final concentration Cf = 1wt%

At temperature T2 = T k

1/1 - 1/3.8 = (ko/LHSV)*exp(-27000/1.987*T)

0.7368 = 3.3463E8 * exp(-13588.32/T)

2.201E-9 = exp(-13588.32/T)

Taking logarithmic both sides,

-19.93 = - 13588.32/T

T = 681.66 k

Temperature of the reaction for final concentration 1wt%

T = 408.66 C