VOC calculation - confusion - please help

[chemks2012]

Dear all,

I need your help. I am confused with VOC calculation.
During tanker offloading, the outbreathing rate from a tank is 50Nm3/hr. The breather valve was originally set at +25mbarg which was then reset at +35mbarg. The original design document says the increased set pressure has reduced the VOC emission by 10% w/w. I tried calculating the VOC reduction but it turns out as 1.5% and not 10%. Could anybody help please? Or am I missing anything here. Assume storage temperature of 30degC, molecular weight of solvent stored as 88kg/kmol and vapour pressure of pure solvent is 200mm Hg.

Your help would be highly appreciated.
Regards,
KS

 

[Latexman]

Did you account for the mass of VOC that did not get omitted that attributed to the higher pressure in the tank? Or, did you just look at steady state rates?

Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

 

[EmmanuelTop]

It all depends on the quantity (volume) of solvent you pump in the tank. Once you get into steady state with actual tank pressure = set pressure of the vent valve, you will be venting the same amounts of gas, regardless of the set pressure. This is purely the gas volume displaced by the solvent volume.

The only case without venting is if you can perform the entire loading operation without reaching the vent valve set point. Meaning: if you can pump the entire quantity of solvent before tank pressure reaches 35 mbarg.

In all other cases in between, the actual amount of venting is dependent on the volume of solvent being pumped into the tank and the tank initial pressure. Whether this decrease of venting is 1%, 10% or 100% (theoretically possible if you pump small volumes of solvent and the initial tank pressure is below 35 mbarg), it depends on these two parameters: volume of solvent, and the initial pressure in the tank vapor space.

Dejan IVANOVIC
Process Engineer, MSChE

 

[chemks2012]

Hi Latexman and Dejan

Thanks for your input. I just want to calculate emissions at steady state please assuming the tank is completely empty and pumping rate into the tank is 50m3/hr.
In fact, I just want to see the difference in VOC % emission for the set pressure of 20mbarg [SORRY, original set pressure is 20mbarg and NOT 25mbarg] and 35mbarg.

Looking forward to hear from you.
Regards
KS

 

[EmmanuelTop]

In this case, reduced emissions would correspond to the difference of the mass of vapor occupying fixed (constant) volume of tank vapor space at 35 mbarg and 20 mbarg. The lower is the tank liquid level, the higher is the difference in vapor mass for these two cases.

Dejan IVANOVIC
Process Engineer, MSChE

 

[chemks2012]

Hi Dejan,

Thanks for your reply. Can you please show and/or attach a quick calculation please as I am completely confused.
It would be great if you include any assumptions you make.

Thanks in advance.
Regards,
KS

 

[EmmanuelTop]

Assuming isothermal loading operation (no change in temperature of the tank), this is how the process develops:

t=0: tank pressure is 0 mbarg (or whatever pressure below 20 mbarg), loading commences.
t=t₁: tank pressure is 20 mbarg, liquid level in the tank is H₁, remaining vapor space is V₁. From gas state equation you calculate the number of moles of gas in the tank vapor space, n₁.

Now, calculate how much less volume will this amount of gas occupy at 35 mbarg (V₂<V₁). The volume difference between the two cases (V₁-V₂) is the incremental volume of solvent you can load into the thank, without having the breather valve opened to atmosphere (or storage flare, wherever). This is at the same time the volume of gas you didn't vent, and which you would vent if the set pressure had remained 20 mbarg. So this volume equals the amount of reduced venting.

This is simplified hand calculation method. You can play with some fancy dynamic tools (Aspen for example), and follow the real-time charts of tank pressure, breather valve opening, and actual vent flow. It is quite interesting.



Dejan IVANOVIC
Process Engineer, MSChE

 

[chemks2012]

Hi Dejan

Thanks very much for your help/input.

P1 = 103.325 kPaa (= 20mbarg)
V1 = 10m3 (assumed)
T1 = 303K (= 30degC)
n1 = 0.4101 kmol
m1 = 0.4101*88 = 36.09kg

P2 = 104.825 kPaa (= 35mbarg)
V2 = ?
T2 = 303K (= 30degC)

P1V1 = P2V2

V2 = 9.85m3

V2 - V1 = 10 - 9.85 = 0.1431m3

Density of vapor at NTP is
= (101.325 * 88)/(8.314*293*1) = 3.66kg/m3

Mass of vapor not emitted if set pressure is increased to 35mbarg = 0.1431*3.66 = 0.52kg

% reduction = 0.52*100/36.09 = 1.45%

Hope I haven't missed anything.
Just wondering why we are not accounting for vap our pressure here?

Thanks in advance.
KS

 

[EmmanuelTop]

The calculation seems OK to me. Vapor pressure cancels out because it appears in both equations (initial and final condition).

This % reduction - in terms of absolute mass or mass flow through the vent - can be quite significant for large storage tanks. If you consider, for example, large LNG tanks with 150,000 m3 volume, for similar case and neglecting other factors (heat leak being the key contributor), this 1.5% would result in containing additional 7,500 kg of gas within the tank. This is quite significant figure, especially if you have to flare it otherwise.


Dejan IVANOVIC
Process Engineer, MSChE

 

[EmmanuelTop]

One note: in my example I have used the same density as in your case, not the density of LNG vapors.

Dejan IVANOVIC
Process Engineer, MSChE

 

[chemks2012]

Hi Dejan,

Thanks for your help and input, much appreciated.

Regards,
KS