3 replies to this topic

[keby7lkv]

Hi,

Currently I am designing a scrubber for final year design project. The CATOX(catalytic oxidation reactor) would oxidize the organic contained and the outlet streams passing through expander and recover the energy. The outlet condition of CATOX is 9.5bar and 350C. After reducing the pressure into 1.3bar, the stream is cooled to 70C and send for vent gas treatment.

The flowrate into the scrubber is very large(8000kmol/h), and at 1.3bar, 70C. The scrubber is to eliminate the HBr contained, which is around 80ppm and required to be treated into 1.5ppm vol/vol basis. The main component of the inlet is nitrogen. With that large amount of flowrate, I have calculated the L/G for the system, normally the L/G would be larger for absorption, means that definitely the L required for absorption would be more than 8000kmol/h. With such a large amount of flowrate, is that possible to design it? I have calculated the diameter, with using larger size of packing, the diameter is 4-5m. I am afraid of the super huge size absorber is required for the vent gas treatment.

Is there any other method to get rid of nitrogen before putting the air into the system? or pre-treatment is required before scrubbing?

I have attached the details about inlet stream of scrubber.

thanks.

Attached Files

•  data for inlet stream of scrubber.xls   114KB   60 downloads

[kkala]

Gas flowrate is about 176000 m3/h. In a fertilizer plant (1980) gases from drier were of that magnitude and were splitted into two separate lines, each one with treatment and fan. This splitting remained when gas treatment was modernized, placing a scrubber and a spray column at each line.
So a way out in your case could be to install two absorbers in parallel; in a symmetrical configuration, to split gas flow as evenly as possible.
Water of 8000 kmol/h makes 144 ton/h, or a volumetric flow of 144 m3/h. Rather insignificant, compared to gas flow rate.
A way (that could be investigated whether it is feasible) to get rid of nitrogen is to introduce oxygen instead of air. This would increase temperature after burning, which may be harmful on materials and catalyst. And O2 should be prepared in the factory.
As pretreatment I can only think of further cooling of gases in E-131. They are saturated in H2O. If cooled to 50 oC (instead of 70 oC), there shall be some condensation of H2O, about 0.3116/0.9884 - 0.1234/1.1767 ~ 0.2 kmol H2O / kmol N2 (0.3116 and 0.1234 being H2O sat press in bara at 70 and 50 oC, total gas press=1.3 bara). Condensation is assumed to transfer HBr into liquid phase, as observed in practice for HF absorption in wet scrubbers (assumption that this is also valid for HBr). This absorption can be significant, but I suppose hard to simulate.

Edited by kkala, 31 March 2011 - 01:12 PM.

[keby7lkv]

kkala:

The 8,000 kmol/h is the gas stream; I have calculated the k of HBr at that condition and I get around 5-6. With considering the L/G, indireactly it means the L required is 40,000 kmol/h. I am wondering, if i use caustic soda solution as solvent, the HBr will have reaction with the caustic soda. Can I assume that to be a constant driving force between the two phases? With assuming HBr in water is reacted with caustic soda and no accumulation of HBr in water stream?

[kkala]

Even if water flow is 40000 kmol/h=720000 kg/h~720 m3/h, the volumetric flow is not significant compared to volumetric gas flow. So two identical cleaning systems, each of 88000 m3/h gas & 360 m3/h water, can be a way out.
Gas film resistance is the controlling factor in HBr absorption (if assumed same as of HF). So HBr transfer rate from gas phase to water could be roughly simplified to K*(PG-PE)*A, where
PG=partial HBr pressure in gas, from 80E-6*1.3=104E-6 Bar to 1.5E-6*1.3=1.95E-6 Bar
PE=equilibrium vapor pressure of HBr over its aqueous solution at operating temperature (70 oC).
A=contact surface of gas / water
K=(mass) transfer coefficient of HBr.
Using soda or lime solution instead of water would reduce HBr concentration in water to practically zero, so PE=0. This would promote mass transfer only in the case when PE is high enough to get close to 1.95E-6 Bar.
For instance, if PE=1E-7 Bar, almost nothing would be gained by soda or lime. If PE=1E-6 Bar, soda or lime would help. If PE=1E-5 Bar, soda or lime is necessary. It is noted that PE depends on the HBr concentration in water, which is a function of the pure water flow in.
So look into HBr vapor pressure over its aqueous solution to see benefit (if any) by using alkaline solution. Since output is only 1.5 ppm HBr (v/v), alkaline solution may be the indicated case.
Even so, driving force would not be constant. This is PG-PE. It would be PE=0, but PG would vary as above along the gas path.
P.S.
I worked (1975-81) in fertilizer industry, where gaseous HF / SiF4 emissions had to be abated. Fluoride solubility in water is high, e.g. 5% HF corresponds to 0.1 mm Hg of HF at 25 oC, that is to 131 ppm HF in gaseous phase. Gas film resistance is the controlling factor in HF absorption. Water was used as an absorber, even at the phosphoric acid plant. Soda would reduce pure water consumption, but its cost made its choice uneconomical. Nevertheless F emission standard was about 15 mg / Nm3 of gas at that time, that is about 17.7 ppm (v/v), not 1.5 ppm.