Nitric Acid Production - Mass and Energy Balances

The purpose of this process is to produce nitric acid by anhydrous ammonia. Drawing of the flow chart for the process and mass and energy balances are required. Constitutively nitric acid production is that the nitrogen monoxide in the gases formed by the oxidation of ammonia is converted into nitric acid by absorbing it with water. Three different processes are proposed for nitric acid production at the sources. These are,

1. Oxidation and absorption at atmospheric pressure
2. Oxidation and absorption at high pressure (approximately 8 atm)
3. Oxidation at atmospheric pressure and absorption at high pressure
   In this work we will use second option.

Theoretical Knowledge

Nitric acid, which is very common in use, is a clear yellowish liquid. It contains nitrogen, hydrogen and oxygen atoms. The formula is HNO3. Nitric acid can be produced in three ways. These are, reacting the Chilean saltpeter with sulfuric acid, catalytic burning of ammonia and the burning of the air components. Catalytic burning (Ostnvald prodecure) is the most commonly used way of production nowadays. Nitric acid technical (spesific) properties are,

Chemical Name	Nitric Acid
Formula	HNO3
Molecular Weight	63.02 g/mol
Color / Form	Yellowish / Liquid
Density	1,339 ( % 55 ) - 1,1150 ( % 20 ) - 1,3667 ( % 60 ) g/cm³
Freezing Point	17 ° C ( % 20 ) - 22,4 ° C ( % 60 )
Boiling Point	103,4 °C ( % 20 ) - 120,4 °C ( % 60 )

Areas of Usage - Nitric acid is used in fertilizer production, metal industry, paint industry, explosive materials industry, pH balancing, electropolishing etc.

Mass Balances for The Production of Nitric Acid

Let’s suppose that,

• Annual operating time is 8000 hours
• Process yield is %94 over ammonia
• Reactor yield is %96
• The nitric acid concentration to be produced is 58% by weight
• There is 0.2% NO by volume in the exhaust gases
• The risk of explosion may be high if the ammonia being transported is more than 12%. The concentration of ammonia in the inlet should be above 11%

As a basis for mass balances, ammonia entering the reactor is taken as 100 kmol.

Oxidation Unit

The yield of the reaction no 1 that by %96,

NH3(g) + 5/4 O2(g) -» NO(g) + 3/2 H2O(g)

Formed NO	100*(96/100) = 96 kmol
Required Oxygen	96x(5/4) = 120 kmol
Formed water	96x(3/2) = 144 kmol

4% of the ammonia is consumed in the input for nitrogen by reaction No. 2 formation.

NH3(g) + 3/4 O2(g) --» 1/2 N2(g) + 3/2 H2O(g)

Amount of nitrogen produced	4/2 = 2 kmol
Amount of oxygen required	2x3/2 = 3 kmol
Amount of water produced	3x2 = 6 kmol
Total amount of water produced	144 + 6 = 150 kmol
Stoichiometric amount of oxygen	120 + 3 = 123 kmol

NO oxidises to NO2 by courtesty of the excess air in the input

NO(g) + 12 O2(g) --» NO2(g)

If the input ammonia concentration is taken as 11% by volume,

Amount of air in the inlet	100x(100/11) = 909 kmol
Amount of oxygen	909x(79/100) = 718 kmol
Amount of oxygen that does not react	191 - 123 = 68 kmol
Amount of nitrogen in the product stream	718 + 2 = 720 kmol

The amounts of the currents in the oxidation unit:

Mass Balance for The Condenser

The oxidation is completed until it exits in the condenser and the whole of NO is converted into NO2. The water in the gas stream entering the condenser also condenses to form a dilute nitric acid solution containing 40-50% HNO3 by weight. Let’s suppose that,

• 45% of the stream exiting the condenser is nitric acid
• Let's basically assume that there is 100 kmol HNO3 in the liquid stream (condensate) leaving the condenser

3NO2 + H2O --» 2HNO3 + NO

Amount of water required to form 100 kmol HNO3	50 kmol = 900 kg
Mass of 100 kmol HNO3	100x63 = 630 kg
Amount of water required to dilute 630 kg acid as 45%	(6300x55) / 45 = 7700 kg
Total amount of water required to obtain dilute acid	900 + 7700 = 8600 kg
Formed HNO3	(Amount of water preduced per 100 kmoles of NH3 in the inlet) / ( Amount of water required to make 100 kmol of HNO3 to 45% solution) = 100x(2700/8600) = 31,4 kmol
Amount of NO2 consumed in the No.5 reaction	31,4x(3/2) = 47,1 kmol
Amount of NO reacted	31,4x(1+2) = 15,7 kmol
Amount of water reacted	15,7 kmol

The amount of water that is condensing but not reacting with NO2 = 150 -15,7 = 134,3 kmol

The amount of NO in the gas stream from the condenser can be equal to the amount of NO2 that forms HNO3 by absorbing it in the condensate. Therefore, the amount of NO in the gas stream from the condenser = 15.7 kmol

Mass Balance for Nitrogen Oxides

The total amount of NO + NO2 entering the unit = No.4 = 96 kmol

31.4 kmol , leaves the unit as HNO3. For this reason, the total amount of NO + NO2 in the gas stream = 96 -31,4 = 64,6 kmol

Assuming this is 15.7 kmol of NO, the amount of NO2 in the outgoing gas stream = 64,6-15,7 = 48,9 kmol

Mass Balance for Oxygen

Let’s suppose that,

The amount of oxygen not reacted is x moles

The amount of oxygen resulting from the unit = The amount of oxygen in stream 6 + The amount of oxygen in stream 4

= [NO/2 + NO2 + x ] + [3/2 *HNO3 +H2O/2] = (171 + x) kmol
= [15,7/2 + 48,9 + x ] + [3/2 *31,4 +134,3/2] = (171 + x) kmol

Amount of oxygen entering the unit = Amount of oxygen in stream 5

= [NO/2 + O2 + H2O] = [96/2 + 68 + 150/2] = 191 kmol

191= 171 + x x=20kmol (it is the amount leaving the unit without entering the reaction)

To calculate how much of the water vapor in the condenser input stream is concentrated and how much remains in the vapor phase, as a preliminary experiment, suppose that the entire water vapor is concentrated.

Amount of water in the condenser leaving stream	Water mole fraction x Total flow rate
Total flow rate of this stream (except water vapor)	804,6 kmol
The mole fraction in this stream of water	4,77x10^3
Amount of water vapor	4,77x10^3x804,6 = 3,8 kmol
The condensed water vapor content	134,3 - 3,8 = 130,5 kmol
Amount of water in stream 6	3,8 kmol = 68,4 kg
Amount of water in stream 7	134,3 - 3,8 = 130,5 kmol = 2349 kg

Total flow rate of 6 and 7 = 23588,4 + 4327,2 = 27915,6 kg

Total flow rate of incoming stream = 27915 kg

Absorption Unit

Let’s suppose that,

NO2 in the gas stream entering the absorption unit is absorbed by water and a %60 by weight nitric acid solution is formed. The amount of oxygen in the incoming stream should be the amount that will increase NO to NO2. The amount of oxygen in the waste gases leaving the absorption unit should be kept at around %3

By taking advantage of the combination of 6 currents, the amount of NO in the current entering the unit = 15.7 kmol of oxygen = 20 kmol

The amount of oxygen needed to oxidize NO2 in the input stream = 15,7 / 2 = 7,85 kmol

The amount of free oxygen in the incoming stream = 20 -7,85 = 12,15 kmol

Amount of oxygen required = (48,9 + 15,7)x(1/4) = 16,15 kmol

16,15-12,15 = 4 mol oxygen should be sent

If y mol of secondary air stream is sent to the absorption unit, 0.21 kmol of oxygen will be in this stream. This oxygen will be consumed for 4 kmol NO oxidation and the amount of oxygen in the waste gases will be 0,21y - 4 kmol.

Nitrogen will not undergo any change in the absorption unit. Therefore, amount of nitrogen in exhaust gases; equal to the sum of the amount of nitrogen in the stream coming from the condenser to the absorption unit plus the amount of nitrogen in the secondary air.

The amount of nitrogen in the waste gas = 720 + 0,79y kmol

If we assume that there is only oxygen and nitrogen in the flue gas stream, and that the amount of other gases is negligible, the percentage of oxygen in this stream is %3. y = 141,6 kmol

The amount of oxygen in the waste gas stream = (141,6x0,21) - 4 = 25,7 kmol

The amount of nitrogen in the waste gas stream = (141,6x0,79)+ 720 = 831,8 kmol

If we assume that the amount of NO in the waste gas stream is 0.2%, the amount of NO in the waste gas stream = total flow rate x 0,002 = (831,8 + 25,7)x0,002 = 1,7 kmol

The amount of oxygen in the waste gas stream = 25,7 + 1,7x(1/4 + 1/2) = 27,0 kmol

Absorbed nitrogen oxides = 48,9 +15,7) - 1,7 = 62,9 kmol = 3962,7 kg

Using the 6-way reaction method, the necessary stoichiometric amount of water = (62,9/4)x2 = 31,5 kmol

If the concentration of acid stream from the absorption unit is 60% by weight, amount of water required for dilution = = (3962,7 / 0,6)x0,4 = 2641,8 kg = 146,8 kmol

The amount of acid formed is based on the reaction of 6,
31,5 kmol H2O x 4kmolHNO3/2kmolH2O = 63 kmol HNO3

Energy Balances of Nitric Acid Production

Ammonia Vaporizer
Ammonia will be collected under pressure as liquid. The saturation temperature at 8 atm is 20 °C. Let's assume that the feed is done at 15 ° C, at the evaporator ambient temperature.

Specific heat at 8 bar	4,5 kJ/kgK
Covert heat at 8 bar	1186 kJ/kg
Vaporizer inlet velocity	731 kg/st
The inlet temperature required to raise the temperature to 20 ° C and evaporate;	731x[4,5x(20-15) + 1186] = 883413,5 kJ/st
With the addition of 10% excess for heat losses	1,1x883413,5 = 971754,9 kJ/st ≈972 MJ/st

Mixing Point

For Air	---------
Feed rate	11272,9 kg/st
Feed temperature	230 °C
Cp Air	1 kJ/kgK

For NH3 Vapor	---------
Feed rate	731 kg/st
Feed temperature	20 °C
Cp NH3 Vapor	2,2 kJ/kgK

The energy balance around the confluence point,

11272,9x1(230 - t3 )+731x2,2x(20 - t3) = 0
t3 = 204 °C

Energy Balance in The Absorption Tower

Heat source in the absorption column will be the same as the condenser and the same calculation method will be used.

Tower	---------
Heat in the secondary air	1754,8x1(40-25) = 0,018 GJ/st
Heat in tail gases (at 25 ° C)	0 °C
Heat in the feed water (at 25 ° C)	0 °C
Oxidized NO	(202,5-21,9)/30 = 6,02 kmol/st
Produced heat	6,02x57120 = 0,34 GJ/st
Formed HNO3	1704/63 = 27,05 kmol/st
Produced heat 2	27,05x63640 = 1,72 GJ/st
Dilution temperature of %60 at 25 ° C	27,05x14207 = 0,38 GJ/st
Condensed water	29,4 - 26,3 = 3,1 kg/st
Latent heat at 40 ° C	2405 kJ/st
Heat at reference temperature	4,18x(40 - 25) = 63 kJ/kg
Heat released out	3,1x(2405 + 63) = 7,6x10-3 GJ/st
Heat in the ascending acid	0,64x(40 - 25 ) = 0,11 GJ/st

---

References:

• Keleti, C. (1085), Nitric Acid and Fertilizer Nitrates ( Fertilizer Science & Technology Series), CRC Press
• http://www.essentialchemicalindustry.org/chemicals/nitric-acid.html
• R.M. Harrison and H.A. McCartney, (1979) Some Measurements of Ambient Air Pollution Arising from the Manufacture of Nitric Acid and Ammonium Nitrate Fertilizer, Atmos