Ammonia Emmissions
Ammonia Emmissions
(OP)
I have a small storage tank (~1m³ vol) used to hold 5% aq Ammonia.
The tank is at atmospheric pressure and under ambient conditions. Can someone remind me how I can calculate the concentration of NH3 vented off, when the tank is re-filled? A typical re-fill would be 500 litres. The tank is vented via an open pipe to atmosphere.
The tank is at atmospheric pressure and under ambient conditions. Can someone remind me how I can calculate the concentration of NH3 vented off, when the tank is re-filled? A typical re-fill would be 500 litres. The tank is vented via an open pipe to atmosphere.





RE: Ammonia Emmissions
best of luck, sshep
RE: Ammonia Emmissions
The United States Environmental Protection Agency (EPA) has a software program called "Tanks" which is free and can be downloaded at:
www.epa.gov/ttn/chief/software/tanks/index.html
It calculates the vapor emissions (i.e., losses) from storage tanks.
Milton Beychok
(Contact me at www.air-dispersion.com)
RE: Ammonia Emmissions
Perhaps this (is correct
virk
RE: Ammonia Emmissions
The vapor pressure of the solution: ~ 7% of the atmosphere, ie 7% mol (= volume). Of this, ammonia 5% and water 2%.
On a mass basis:
ammonia 5 × 17 = 85 = 3 %
water 2 × 18 = 36 = 1.3%
air 93 × 29 = 2697 = 95.7%
RE: Ammonia Emmissions
Excuse my ignorance. You both didn't mention partial pressure of ammonia but, yet, got the figures. Is there any approximate method for this or you guys deliberately omitted it for simplification? Secondly, I didn't get 85 = 3% and so on. Perhaps, my brain is rusting.
Regards,
RE: Ammonia Emmissions
quark
I indeed did mention the VP when I said that 5% (out of 7%) of an atmosphere is the vapor pressure of ammonia at the given conditions, while 2% are for water's V.P.; the remaining 93% are taken by air.
When converting moles to mass, 85 out of a total of 2818 (=85+36+2697) is 3%.
Is it now clear ?
RE: Ammonia Emmissions
RE: Ammonia Emmissions
Any help from the chemical guys?
RE: Ammonia Emmissions
I think 25362's mixed vapor basis was 100 moles which he calculates to weigh 2818 mass units. This was an intermediate result to convert our partial pressures into mass fractions and was not directly related to the 500 liters of displacement.
Because you are a highly regarded mechanical engineer in this forum, we won't whip you too hard on these type conversions. -sshep
RE: Ammonia Emmissions
Quark:
Perhaps if we go into more detail for 25362's calculations, it might be helpful. First, some assumptions:
(1) Bif's aqua ammonia is a solution of 5 weight percent of ammonia in water. (He didn't explicitly state that it was weight percent.)
(2) The storage temperature of the aqua ammonia is 20 degree C.
Then, some givens:
(1) Atmospheric pressure is 14.696 psia
(2) mole % = volume % = partial pressure % for gases
(3) Molecular weights: NH3 = 17, H20 = 18, air = 29
Evidently, 25362 found these vapor pressure data for a 5 weight % solution of ammonia in water at 20 deg C:
(1) vapor pressure of NH3 above the solution = 0.735 psia
(2) vapor pressure of H2O above the solution = 0.295 psia
(3) total vapor pressure of the solution = 1.030 psia
Thus, the composition of the vapor above the solution:
NH3 = 100(0.735/14.696) = 5 partial press. % = 5 mol %
H20 = 100(0.295/14.690) = 2 partial press. % = 2 mol %
Air = 100 - 5 - 2 = 93 mol %
Converting the vapor composition from mol % to wt. %:
NH3 = (5 mols)(17 lb/lb-mol) = 85 lbs = 3.0 %
H2O = (2 mols)(18 lb/lb-mol) = 36 lbs = 1.3 %
Air = (93 mols)(29 lb/lb-mol) = 2697 lbs = 95.7 %
Note: 100 total mols with a total mass of 2818 lbs means that the vapor above the solution has a molecular weight of 28.18 lb/lb-mol.
Also note: The vapor pressures that I found in Perry's Chemical Engineers' Handbook are slightly different from those found by 25362, but not enough to be significant.
I hope this helps,
Milton Beychok
(Contact me at www.air-dispersion.com)
RE: Ammonia Emmissions
Thanks for the clarification. Infact, I did go through Perry's and was able to get the solution upto the calculation of percentage mols of each gas but I foolishly overlooked the 100mols thing.
Thanks for your patience as well.
Regards,