Problem of the Day
2014.Oct.02
Discussion
There is a big difference when concentration is given in units of mg/L versus percent (%). mg/L is
equivalent to a part per million parts (ppM), whereas percent is equivalent to a part per hundred parts
(ppH). Because mg/L is much more dilute, the pounds and pounds-per-day calculations demonstrated
previously will be used. As demonstrated in this Problem of the Day, percent is handled very differently by
WWTT. Once the labeling WWTT advocates is mastered, the “math” basically takes care of itself simply
by canceling units.
In most problems when concentration is given in % it is expressed as so many pounds per 100 pounds.
For example, if the concentration of TS (total solids) is given as 5% TS, this is written exactly the way it
appears substituting “lb” for “%” and putting it over 100 lb: 5 lb TS/100 lb. Here the reader has to stop and
answer the question “per hundred pounds of what?” In this case the answer is “sludge” (abbreviated
“sldg”), so it is completed as 5 lbTS/100 lb sldg. TS are always “part” of sludge. Similarly, if the VS
concentration is given as 74%, it is expressed as 74 lb VS/100 lb of what? In this case, VS are always
“part” of TS: 74 lb VS/100 lb TS.
Many people advocate the use of the conversion factor, 1%/10,000 mg/L, to convert percent to mg/L.
WWTT does not recommend the use of this conversion factor as it only applies when the solution in
question has a density equal to the density of water, 8.34 lb/gal. Even if the density of the solution, or
slurry, is given and used in the calculation, the 1%/10,000 mg/L conversion factor cannot be used
because it is based on the density of water. Instead, WWTT advocates the use of the labeling discussed
in the previous paragraph.
Problem
When doing digester gas produces, always keep in mind how valuable the methane in the gas is.
Problem of the Day: For every pound of VS destroyed, 14.6 cubic feet of anaerobic digester gas are
produced. The volatile content is 74.3% in the influent sludge and 54.3% in the digested sludge. The
TS concentration in the influent sludge is 4.6%. If 225,000 gallons of sludge per day are fed to the
digesters, how many cubic feet of gas are produced each day?
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Solution
If I was doing this problem in front of a class, I would draw a picture of the sludge pump and pipeline
feeding the digester, the digester, the gas piping out the top of the digester, and the digested sludge
coming out the bottom of the digester. Each would be labeled as follows.
Influent Pump and Piping
• 225,000 gal sldg/d
•
4.6%TS = 4.6 lb TS/100 lb sldg (as discussed above)
•
74.3%VS = 74.3 lb VS/100 lb TS (note: these are VS applied, or VSapplied)
•
74.3%VS = 0.743 (for VSR calculation)
•
Sludge density = 8.34 lb sldg/gal sldg (assumed)
The problem statement does not give the density of the sludge. Generally speaking, if the TS
concentration is less than approximately 7%, the density of water can be confidently assumed. If the
reader forgets that density is required, it is helpful to note that both gallons of sludge and pounds of
sludge appear in the list of information given (250,000 gallons of sludge/day and 4.6 pounds of TS/100
pounds of sludge, respectively). The physical constant that allows gallons to be converted to pounds
and pounds converted to gallons is the density. Note, however, it is very important to label the density of
the sludge as shown. You would never want to mistake primary sludge, for example, for water, so label it
accordingly!
Gas Piping
• 14.6 ft3 gas/lb VSdestroyed
Digested Sludge Piping
• 54.3% VS = 0.543 (for VSR calculation)
In all digester-gas type problems, the first step will always be to calculate the VSR.
(0.743 − 0.543)
VSR (%) =
100
0.743 − (0.743 x 0.543)
= 58.9%.
As discussed in the 2014.Sept.28 Problem of the Day, the VSR is expressed in the following manner.
58.9 lb VSdestroyed
58.9% VSR
=
100 lb VSapplied
Labeling the VSR in this manner will basically do the problem for you, as demonstrated below. Note,
“destroyed” and “applied” are abbreviated “dest” and “app” below, respectively.
With the VSR known, the first step in solving the problem, as always, is to identify the units wanted in the
answer—in this case, ft3 of gas/d—which is entered between heavy vertical lines followed by an equals
sign and the blank track. WWTT is very specific in labeling. In this particular example, the reader must
understand that it’s not just “ft3/d,” but “ft3 gas/d.” This is very, very important and cannot be overemphasized.
18030 Brookhurst Street, PMB 573 · Fountain Valley, California 92708 · 866-773-WWTT ·
www.wastewatertechnologytrainers.com
Problem of the Day: For every pound of VS destroyed, 14.6 cubic feet of anaerobic digester gas are
produced. The volatile content is 74.3% in the influent sludge and 54.3% in the digested sludge. The
TS concentration in the influent sludge is 4.6%. If 225,000 gallons of sludge per day are fed to the
digesters, how many cubic feet of gas are produced each day?
ft3 gas
=
d
Notice in the list above, there is only one piece of information given that has “ft3 gas” as a unit. This is part
of the units that the question is asking for in the problem statement. Most of the time, WWTT starts the
railroad track out with the units needed in the answer in the first position in the numerator of the railroad
track as shown.
Problem of the Day: For every pound of VS destroyed, 14.6 cubic feet of anaerobic digester gas are
produced. The volatile content is 74.3% in the influent sludge and 54.3% in the digested sludge. The
TS concentration in the influent sludge is 4.6%. If 225,000 gallons of sludge per day are fed to the
digesters, how many cubic feet of gas are produced each day?
ft3 gas
14.6 ft3 gas
=
d
lb VSdest
Once the railroad track is started and the information given in the problem precisely labeled, doing the
problem is simply a matter of canceling out the units that are not needed.
How are we going to cancel out “lb VSdest?”
This unit does not appear in the information given in the problem statement as listed above. But wait!
Notice how we have expressed the VSR calculated above: 58.9 lb VSdest/100 lb VSapp. This, then, is
entered into the railroad track next and units canceled.
Problem of the Day: For every pound of VS destroyed, 14.6 cubic feet of anaerobic digester gas are
produced. The volatile content is 74.3% in the influent sludge and 54.3% in the digested sludge. The
TS concentration in the influent sludge is 4.6%. If 225,000 gallons of sludge per day are fed to the
digesters, how many cubic feet of gas are produced each day?
ft3 gas
14.6 ft3 gas
58.9 lb VSdest
lb VSdest
100 lb VSapp
=
d
Next, lb VSapp (for lb VSapplied) are entered for cancelation. This unit shows up in only one other place.
Problem of the Day: For every pound of VS destroyed, 14.6 cubic feet of anaerobic digester gas are
produced. The volatile content is 74.3% in the influent sludge and 54.3% in the digested sludge. The
TS concentration in the influent sludge is 4.6%. If 225,000 gallons of sludge per day are fed to the
digesters, how many cubic feet of gas are produced each day?
ft3 gas
14.6 ft3 gas
58.9 lb VSdest
74.3 lb VSapp
lb VSdest
100 lb VSapp
100 lb TS
=
d
Next, lb TS are entered for cancelation. This unit shows up in only one other place.
18030 Brookhurst Street, PMB 573 · Fountain Valley, California 92708 · 866-773-WWTT ·
www.wastewatertechnologytrainers.com
Problem of the Day: For every pound of VS destroyed, 14.6 cubic feet of anaerobic digester gas are
produced. The volatile content is 74.3% in the influent sludge and 54.3% in the digested sludge. The
TS concentration in the influent sludge is 4.6%. If 225,000 gallons of sludge per day are fed to the
digesters, how many cubic feet of gas are produced each day?
ft3 gas
14.6 ft3 gas
58.9 lb VSdest
74.3 lb VSapp
4.6 lb TS
lb VSdest
100 lb VSapp
100 lb TS
100 lb sldg
=
d
Next, lb sldg are entered for cancelation. This unit shows up in only one other place.
Problem of the Day: For every pound of VS destroyed, 14.6 cubic feet of anaerobic digester gas are
produced. The volatile content is 74.3% in the influent sludge and 54.3% in the digested sludge. The
TS concentration in the influent sludge is 4.6%. If 225,000 gallons of sludge per day are fed to the
digesters, how many cubic feet of gas are produced each day?
ft3 gas
14.6 ft3 gas
58.9 lb VSdest
74.3 lb VSapp
4.6 lb TS
8.34 lb sldg
lb VSdest
100 lb VSapp
100 lb TS
100 lb sldg
gal sldg
=
d
Next, gal sludge are entered for cancelation. This unit shows up in only one other place.
Problem of the Day: For every pound of VS destroyed, 14.6 cubic feet of anaerobic digester gas are
produced. The volatile content is 74.3% in the influent sludge and 54.3% in the digested sludge. The
TS concentration in the influent sludge is 4.6%. If 225,000 gallons of sludge per day are fed to the
digesters, how many cubic feet of gas are produced each day?
ft3 gas
14.6 ft3 gas
58.9 lb VSdest
74.3 lb VSapp
4.6 lb TS
8.34 lb sldg
225,000 gal sldg
lb VSdest
100 lb VSapp
100 lb TS
100 lb sldg
gal sldg
d
=
d
All the units have canceled except those needed in the answer, ft3 gas/d, so the math is done. The
arithmetic completes the problem:
14.6 x 58.9 x 74.3 x 4.6 x 8.34 x 225,000 ÷ 100 ÷ 100 ÷ 100 = 551,523 ft3 gas/d.
Way cool!
18030 Brookhurst Street, PMB 573 · Fountain Valley, California 92708 · 866-773-WWTT ·
www.wastewatertechnologytrainers.com