|
CEE
680 |
|
Fall
2011 |
Water
Chemistry
In
Spring when temperatures are on the rise, it is common
for conventional activated sludge wastewater treatment plants to start
nitrifying. When this happens, the pH
drops due to the following overall stoichiometry:
Where
What
is the rate of nitrification in moles of ammonia per minute, if the pH drops
from 7.3 to 6.9 in 4 hours? (hint: conventional activated sludge systems are intensively
aerated or mixed with air)
This is an open system, because the
activated sludge is intensively aerated.
In an open system, total carbonate is not conservative, but alkalinity
is.
Change in alkalinity is determined by the
change in the value of (CB-CA) going from the left side
of the above equation to the right side.
On the left side (CB-CA) = (22-0) or +22. On the right side (CB-CA)
= (0-21) or -21. The net change is
therefore -43 equivalents of alkalinity formed per "mole" of cells
produced. We can also conclude that
there are 22 moles of ammonia oxidized per "mole" of cells
produced. Combining, this gives us, 22
moles of ammonia oxidized per 43 equivalents of alkalinity consumed.
We also can calculate the rate of change
of alkalinity from the rate of change of pH.
To do this we need to refer to the open system logC vs pH diagram for
the carbonates. Once we know the pH, we
can find alkalinity from the following approximation of the ENE: Alk or (CB-CA)
= [HCO3-].
If you use 6.3 as the pK1 value
you get:
The initial conditions:
pH = 7.3,
Alk
= 1.0x10-4 or 10-4.0 equivalents/L =
Final conditions (after 4 hours):
pH = 6.9
Alk
= 3.98x10-5 or 10-4.4 equivalents/L
The difference in alkalinity is:
=
(10-4.0 - 10-4.4) = 6.05 x 10-5 equivalents/L
Therefore the rate of change of alkalinity
is
=
(6.05x10-5)/4 hrs
= 1.51 x 10-5 equivalents/L/hr
And combining this with the above
stoichiometry, we get:
|
nitrification rate |
=
(1.51 x 10-5 equivalents alkalinity consumed/L/hr) x (22
moles of ammonia oxidized per 43 equivalents of alkalinity consumed) |
|
|
= 7.74 x 10-6 moles ammonia
oxidized/L/hr = 1.29 x 10-7 moles ammonia
oxidized/L/min |
If you use 6.35 as the pK1
value you get:
The initial conditions:
pH = 7.3,
Alk
= 8.94x10-5 or 10-4.05 equivalents/L =
Final conditions (after 4 hours):
pH = 6.9
Alk
= 3.55x10-5 or 10-4.45 equivalents/L
The difference in alkalinity is:
=
(10-4.05 - 10-4.45)/ = 5.40 x 10-5
equivalents/L
Therefore the rate of change of alkalinity
is
=
(5.40x10-5)/4 hrs
= 1.35 x 10-5 equivalents/L/hr
And combining this with the above
stoichiometry, we get:
|
nitrification rate |
=
(1.35 x 10-5 equivalents alkalinity consumed/L/hr) x (22
moles of ammonia oxidized per 43 equivalents of alkalinity consumed) |
|
|
= 6.90 x 10-6 moles ammonia
oxidized/L/hr = 1.15 x 10-7 moles ammonia
oxidized/L/min |
You
are treating an alkaline groundwater by alum coagulation followed by base
addition for corrosion control. Your
target pH for the water in the distribution system is 8.5. Analysis of the finished water prior to base
addition shows a pH of 7.2, and an alkalinity of 250 mg/L as CaCO3. How much of either one of the following two
bases would you have to add (in mg/L) to reach your target pH.
|
|
a. Caustic Soda (NaOH) |
|
|
|
b. Soda Ash (Na2CO3) |
|
This
is a closed system. Recall that:
We
also know that:
Alk = 250 mg/L as CaCO3 =
5x10-3 equivalents/L
Since caustic soda does not contain
carbonates, CT is a constant for this case.
@ pH 7.2,
Next, it's a simple matter of calculating
the change in alkalinity that accompanies the change in pH.
@ pH 8.5,
|
|
Alk |
= (0.97832+2*0.015505)*5.6245x10-3 |
|
|
Alk |
= 5.67699x10-3 equivalents/L |
|
|
delta Alkalinity |
=
5.67699x10-3 equivalents/L
- 5x10-3
equivalents/L |
|
|
|
=
6.77x10-4 equivalents |
and
this change in alkalinity must come from the caustic soda added:
Caustic Soda Dose =
6.77x10-4 equ. * 40g/equ.
Caustic Soda Dose = 27.1 mg/L
This problem is similar to part
"a", except that CT is augmented by addition of the soda
ash. It can be directly related to the
soda ash dose (SAD), knowing that one mole of soda ash contains one mole of
carbonate species:
CT-final
= CT-initial + SAD
We can then use this and estimate final
Alkalinity (Alkfinal) as a function of SAD using the two different
equations for alkalinity, then solve both for SAD. (N.B. you can also solve this problem by
recognizing that addition of soda ash does not change the acidity. You then use this acidity to solve for final
carbonate concentration which then gives you the SAD).
|
|
Soda Ash Dose |
= (6.83x10-4
moles/L)*(106 g/mole) |
|
|
|
= 0.0724 g/L |
|
|
|
= 72 mg/L |
Repeat
#2, but this time consider that your water treatment
system has a large uncovered finished-water reservoir. Make the extreme-case assumption that the
water will reach equilibrium with CO2 in the atmosphere,
after base addition, but before entry into the distribution system.
This is an open system problem. To solve, you can examine the open system
diagram and determine the Alkalinity needed to achieve a pH of 8.5. This is where the an
imaginary horizontal (CB-CA) line would intersect the
bicarbonate line at pH 8.5. The
alkalinity level needed is about 10-2.8 equivalents/L or about 1.5 meq/L.
Since this is below the 5 meq/L we're starting with, no additional
alkalinity is needed. In fact, 3.5 meq/L
of acidity would be needed if it was deemed undesirable for the pH to exceed
8.5.