Mass balances, nitrifying plants
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VNH4,02 = 4.6 g 02/g NH!-N
k112A,NH4 = 0.7 (g NH!-N) 112 m- 112 d- 1
A mass balance for the first plant stage can be expressed as follows:
01 · SNH4, 1 - A2· · r A,NH4 = 03 · SNH4,3
On the assumption that the nitrification is oxygen limited (to be controlled after calculating), we know from Expression (6.6) that:
rA,NH4 = (k1,M,02/vNH4,o2 ) · So2,2'~.>, which, if substituted, gives:
01 · SNH4, 1 - A2.(k11.1A.02 /vNH4,02) · So2.2.,.. = 03 · SNH4,3
If we substitute the known values, it is found that:
500 · 100- 20,000(4/4.6) · 511.! = 500 · SNH4,3
SNH4,3 = 22 g NH!-N/m 3
(Control of oxygen limitation:
? SNH4,3 > 0.3 . 802,2
0.3 · So2,2 = 0.3 · 5 = 1 .5
SNH4,3 = 22
that is,
SNH4,3 > 0.3 · So2.2
Oxygen limitation Is ascertained.)
As to the other stage of the filter, ammonium is supposed to be limiting. The mass balance will then be (symbols from Fig 6.1 b):
Os · SNH4,3 - A4. · k1!2A,NH4 · SNH4.4 ' !.> = Os · SNH4,s
The ideal mix can be expressed by the equation SNH4.4 = SNH4,5
By substitution it is found that:
500. 22-20,000. 0.7. SNH4,5 ..... = 500. SNH4,5
from this is found that
SNH4,5 - 0.6 g/m 3
(Controlling whether ammonium is limiting or not:
So2.4 · 0.3 = 2.1
SNH4,4 = 0.6
SNH4,4 < 0.3 · So2.4
that is, ammonium Is limiting).
The outlet from the plant will then be approx. 0.6 g NH!-N/m 3