Page 178 - Chemical engineering design

FLOW-SHEETING
Balance on oxides
Total NO C NO 2 entering D NO in stream 4 D 96 kmol 155
Of this, 31.4 kmol leaves as nitric acid, so (NO C NO 2 ) left in the gas stream
D 96 31.4 D 64.6kmol.
Of this, 15.7 kmol is assumed to be NO, so NO 2 in exit gas D 64.6 15.7
D 48.9 kmol.

Balance on oxygen
Let unreacted O 2 be x kmol. Then oxygen out of the unit will be given by:
NO 3 H 2 O

C NO 2 C x C HNO 3 C
2 gas 2 2 acid
stream 6 stream 7

15.7 3 134.3
D C 48.9 C x C ð 31.4 C D 171 C x kmol
2 2 2

NO
Oxygen into the unit D C O 2 C H 2 O
2 stream 5
96 150
D C 68 C D 191 kmol
2 2
Equating O 2 in and out:

unreacted O 2 ,x, D 191 171 D 20.0kmol
As a first trial, all the water vapour was assumed to condense; this assumption will
now be checked.
The quantity of water in the gas stream will be given by:

mol fraction ð total flow.

The total flow of gas (neglecting water) = 804.6 kmol, and the mol fraction of water
3
was estimated to be 4.77 ð 10 .
3
So, water vapour D 4.77 ð 10 ð 804.6 D 3.8kmol
And, mols of water condensed D 134.3 3.8 D 130.5kmol.
The calculations could be repeated using this adjusted value for the quantity of water
condensed, to get a better approximation, but the change in the acid, nitric oxides, oxygen
and water flows will be small. So, the only change that will be made to the original
estimates will be to reduce the quantity of condensed water by that estimated to be in the
gas stream:
Water in stream 6 3.8kmol D 68.4kg
So, water in stream (7) D 134.3 3.8 D 130.5kmol D 2349 kg.