Nitrogen Sweeping Calculation 2

[laffan]

During calculation to purge vessel to get H2S below 10% ppm ..... we found that we need five cycle to get our value, by nitrogen pumping.
Note:
Vessel volume 11 m3
pumping with nitrogen 7 barg take five time each time around one hours, amount nitrgen used in each pumping 74 Nm3/h
Question:
if we use sweeping method shall we get less time and less amount of Nitrogen... and which is better?

Thanks

 

[25362]

As I understand it, please correct me if I'm wrong, you are interested in changing from pressure purging to sweep-through purging.

The required nitrogen volume of the latter would depend on the degree of mixing, itself affected by the location of the inlet and outlet nozzles, the vessel geometry, and the composition of the original gas.

The theoretical equation is obtained after assuming 100% (perfect) mixing, constant T, P, and no H[sub]2[/sub]S entering with the sweep gas:

Q.t = V ln(C[sub]1[/sub]-C[sub]2[/sub])​

Where C[sub]1[/sub], C[sub]2[/sub] are the initial and final concentrations of H[sub]2[/sub]S, Q is the volumetric flow rate, t is time, and V is the volume of the vessel being purged.

Because perfect mixing is not achieved in practice, a large safety factor may need to be applied; a factor that can be determined by actual trials.

To my grasping the sweep-trough method is especially applied to equipment not rated for pressure or vacuum. The selection of the better process, to reach the same result, is based on economics (time, volume of inert gas, etc.).

The usual method used in hydrotreaters is combining pressure and vacuum purging.

 

[laffan]

thanks for replying
can you correct my calculation
I used the equation where:
Q= 74.5 Nm3/h
Concentration 24000 ppm to 10 ppm
V = 11 m3
the Time = 1.489 hours
this is right

 

[25362]

There is a typo in the formula I quoted it should be

Q.t = V ln(C[sub]1[/sub]/C[sub]2[/sub])​

Thus: Q = (11/1.489) ln(24000/10) = 57.5 m[sup]3[/sup]/h

I've seen multiplying safety factors (to account for imperfect mixing) in the range of 4-5 depending on the vessel's geometry and location of the venting nozzles.
It is advisable to determine such by actual testing.

Hot nitrogen would diffuse quicker than ambient nitrogen bringing down the safety factor.

 

[laffan]

thaks and best regards

 

[engrx]

I am not sure what your system looks like, but I have seen a vessel purge done with natural gas to a fuel or flare system, followed by nitrogen. This allows you to sweep the H2S/hydrocarbon to fuel/flare without blowing H2S all over the place (or putting excess nitrogen to fuel/flare).

I am very impressed that there is a calculation out there for determining purge volumes- kudos! Normally we purge and check (wash-rinse-repeat).

 

[sshep]

Pressuring and depressuring is the most reliable method for purging a vessel, and in most cases the fastest. The best way to shorten the time is to not bring the pressure up so high that your N2 flow is reduced at the end of the cycle. Example: if you are using 7bar N2 and bring the pressure to 7 bar, the flow of N2 into the system can become quite small towards the end if a hose is used. Also, depressure as rapidly as possible to the lowest pressure possible. Note: the latter point is especially effective for freeing a process of liquid water (dry out) as the shock wave can lift free water pockets from the process in a manner that flow through purging cannot come close to duplicating.

Generally the only advantage to an open purge is that it doesn't require any attention, but when speed is required pressuring/depressuring is the way to go.

best wishes,
sshep

 

[ChEMatt]

25362,

Can you site the source of your equation?

Thanks!

Onwards,

Matt

 

[reena1957]

Dear ChEMatt,
Way back (30yrs), we had a course with the name of Transport Phenomena ( and a book by the same name by Bird, Stewart and Lightfoot) which helped us to derive all sorts of differential equations and integrate them to get some smart looking formulae like the one 25362 had given. Only that he remembered it and the rest of us have forgotten.
Though this forum is not meant for this type of derivation material, I beg the indulgence of moderators to allow me to add the following:
My recollection is as follows:
Assume there is a well stirred vessel with volume V M3 and concentration of contaminant c Kg/M3 and there is a constant in/outflow equal to QM3/hr. The outflow will have the same conentration c Kg/M3 as the vessel. Let us assume the inflow(purging medium) has a contaminant concentration of cP Kg/M3. The material balance across the vessel is:
Rate of outflow of contaminant - rate of inflow of contaminant = rate of accumulation of contaminant in vessel
Q x c - Q x cP = - V x dc/dt where t is time and d the differential

dt = - V / Q d [ c - cP ]
----------
[ c - cP ]
Integrating this eqn, between Cinitial and Cfinal, you get
Time = V / Q ln [ cinitial - cP ]
------------------
[ cfinal - cP ]

In case of eqn given by 25362. the purge gas is pure Nitrogen so cP= 0 and both eqns match.
Best wishes

 

[25362]

Crowl and Louvar: Chemical Process Safety: Fundamentals with Applications, Prentice Hall, ISBN 0-13-129701-5.

 

[ifreeman]

I hope this article help.
George R.Kinsley,Jr., "Properly Purge and Inert Storage Vessels", CEP Feb., 2001

You may find this article through google.