Ammonia Emmissions 4

[bif]

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.

 

[sshep]

Perry has a table for partial pressure of Ammonia, water, and total vapor pressure over an aqueous solution. This data will give you the concentration you need. From your description air makes up the difference in partial pressure relative to atmospheric. By the ideal gas law you should be able to calculate just about anything else you might need as well. I think it will be very easy for you.

best of luck, sshep

 

[mbeychok]

bif:

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)

 

[virk]

At 20°C (ambient conditions) vapor pressure above a 5%-NH3/H2O-solution is about 68 mbar. Concentration of the NH3/H2O-vapor is about 66%. So 4,5% of the gas volume above is ammonia gas. Perhaps 3% of the mass above is ammonia gas.

Perhaps this (is correct) and helps.

virk

 

[25362]

The calculation would be as follows:

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%

 

[quark]

virk and 25362,

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,

 

[25362]

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 ?

 

[quark]

No. I am ok with algebra but I couldn't get a hint of 2818. Tried using Avagadro's law to get volume fractions but unsuccessful.

 

[quark]

I thought somebody would reply me or redflag my post. Still couldn't make how 500L becomes 2818 mass units.

Any help from the chemical guys?

 

[sshep]

quark,
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

 

[mbeychok]

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)

 

[quark]

Sshep and MBeychok,

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,