Page 177 - Chemical engineering design

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CHEMICAL ENGINEERING
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At 40 C the vapour pressure of water over 45 per cent HNO 3 is 29 mmHg (Perry’s
Chemical Engineers Handbook, 5th edn, pp. 3 65). Take the total pressure as 8 atm. The
mol fraction of water in the outlet gas stream will be given by the ratio of the vapour
pressure to the total pressure:
29 3
mol fraction water D D 4.77 ð 10
760 ð 8
As a first trial, assume that all the water in the inlet stream is condensed, then:
water condensed D 150 kmol D 2700 kg

NO 2 combines with this water to produce a 45 per cent solution:

Reaction 5. 3NO 2 C H 2 O ! 2HNO 3 C NO
For convenience, take as a subsidiary basis for this calculation 100 kmol of HNO 3 (100
per cent basis) in the condensate.
From reaction 5, the mols of water required to form 100 kmol HNO 3 will be:

50 kmol D 900 kg
mass of 100 kmol HNO 3 D 100 ð 63 D 6300 kg
6300 ð 55
water to dilute this to 45 per cent D D 7700 kg
45
So, total water to form dilute acid D 900 C 7700 D 8600 kg.
Changing back to the original basis of 100 kmol NH 3 feed:
Water condensed per 100 kmol NH 3 feed
HNO 3 formed D 100 ð
Total water to form 45 per cent acid, per 100 kmol HNO 3
2700
D 100 ð D 31.4kmol
8600
3
NO 2 consumed (from reaction 5) D 31.4 ð D 47.1kmol
2
NO formed D 31.4 ð 1 D 15.7kmol
2
H 2 O reacted D 15.7kmol
Condensed water not reacted with NO 2 D 150 15.7 D 134.3kmol.

The quantity of unoxidised NO in the gases leaving the cooler-condenser will depend
on the residence time and the concentration of NO and NO 2 in the inlet stream. For
simplicity in this preliminary balance the quantity of NO in the outlet gas will be taken
as equal to the quantity formed from the absorption of NO 2 in the condensate to form
nitric acid:
NO in outlet gas D 15.7kmol

The unreacted oxygen in the outlet stream can be calculated by making a balance over
the unit on the nitric oxides, and on oxygen.