4-Log and CT
What is 4-Log?
• 4-log is a treatment technique that ensures the
removal/inactivation of microorganisms, including
viruses, and is generally measured on a logarithmic
scale (i.e., in terms of orders of magnitude). Many
water systems must provide 4-log treatment due to
microbial contamination of the source water.
What do I need to know?
• To understand 4-log, we
need to understand the
mathematical relationships
between the physical and
environmental components
of a water system (i.e. peak
flow rates, tank and pipe
volumes, water
temperature and pH,
disinfectant residuals, and
baffling factors.) From these
parameters, we can
calculate CT.
What is CT?
•
CT is a measure of disinfection effectiveness. It is the product of the residual
disinfectant concentration (C) in mg/l and the corresponding disinfection contact
time (T) in minutes.
CT (minutes-mg/l) = C x T, where:
• C = residual disinfectant concentration (ultimately measured at the first customer
during peak flow), in mg/l
• T = Time during peak flow, measured in minutes, from the point of disinfection to a
point where the residual is measured before or at the first customer.
•
Results of CT calculations can be compared with published data to determine
whether 4-log virus removal can be achieved.
A little history lesson…
• Approximately 300 B.C. Euclid is thought
to have come up with the concept of ∏ (pi)
• Archimedes is credited with giving an
a
accurate approximation of pi
•In 2600 B.C., the ancient Egyptians
used an approximation of pi in
their monuments
• Indian texts from the 8th Century
B.C. indirectly reference pi
• The Bible indirectly references pi
in 1 Kings 7:23
What is pi?
 No, not that kind of pie…
This pi 
• In its simplest terms, pi is defined as the
ratio of a circle’s circumference to its
diameter.
• Numerically, it’s approximately equal to 3.14
Why is pi important?
• In relation to water systems,
pipes and most tanks are
circular
• In order to determine the
volumes of pipes and tanks,
pi is an integral part of the
formula
Vcylinder = ∏r2L
(where r= ½ the diameter of the pipe
and L= the length of the pipe)
• To simplify the equation and because we need
initial volumes in cubic feet, we must use algebra
to convert the pipe radius to diameter:
V = ∏r2L  ∏(d)2 x L
4
and from inches to feet, so:
V = ∏(d”/12”)2 x L
4
• As a quick example, what is the diameter in
feet of a 2” pipe?
2 inches
12 inches
foot

2 inches x
1 foot = 0.167 foot
12 inches
What is dimensional analysis?
•
•
•
In physics and all science, dimensional analysis is a tool to find or check
relations among physical quantities by using their dimensions. The
dimension of a physical quantity is the combination of the basic physical
dimensions (usually mass, length, time, electric charge, and temperature)
which describe it; for example, speed has the dimension length per unit
time, and may be measured in meters per second, miles per hour, or other
units.
Dimensional analysis is routinely used to check the plausibility of derived
equations and computations.
How many seconds are in a year?
60 sec x 60 min x 24 hr x 365 days = 31,536,000 sec
yr
1 min
1 hr
1 day
1 yr
What are baffling factors and why
must we use them?
• Since poor circulation (i.e. short circuiting) in a storage
or contact tank will reduce the contact time, T as
calculated above is typically multiplied by a conversion
or baffling factor to account for poor circulation. The
diagrams on the following page show typical types of
storage tanks, inlet/outlet pipe configurations, and
corresponding recommended baffling factors (BF).
• Note that bladder tanks and atmospheric or
hydropneumatic tanks with a single inlet and outlet are
not given contact time credit since during a pump-on
cycle, some of the water is bypassing the tank.
What is the BF for Lake Jesup?
Example # 1
100’ of 2” pipe
x 20’ of 4” pipe
well
Cl2 injection
1,000 gal H tank
• Here’s what we know:
–
–
–
–
Well pump flow rate = 25 gpm
pH = 7.9
Temperature = 24⁰C
Cl2 at 1st customer = 1.2 mg/L
x
Sample tap
before or at 1st
customer
Segment 1 (20’ of 4” pipe)
T = V = ∏(d/12”)2 x L x 7.48 gal
Q
4
ft3
Q
(Where T = detention time in minutes, V = volume in gallons, d = diameter in
inches, L = length of segment in feet, and Q = the peak flow rate in gpm)
Find the numerator (the volume) first
V = ∏(4”/12”)2 x 20’ x 7.48 gal
4
ft3
Calculate V (volume in gallons)
V = ∏(4”/12”)2 x 20’ x 7.48 gal
4
ft3
V = (0.785)(0.33333’)2 x 20’ x 7.48 gal
ft3
V = (0.785)(0.11111 ft2) x 20 ft x 7.48 gal
ft3
V = 0.087 ft2 x 20 ft x 7.48 gal
ft3
V = 13.0152 gal
Calculate T (contact time in minutes)
• Remember: T = V
Q
We calculated the volume (V) to be 13.0152 gal
We were given the peak flow rate (Q) of 25 gpm
T = 13.0152 gal
25 gal
min
 13.0152 gal x 1 min = 0.52 min
25 gal
Segment 2 (tank)
• Because it is an H-tank, only 10-20% of the total
volume of 1,000 gallons is usable at cut-in pressure
(lowest water level)
• Additionally, because the tank is not baffled, we
must use a baffling factor of 0.1
T = V x BF
Q
T = 200 gal x 0.1  200 gal x 1 min x 0.1 = 0.8 min
25 gal
25 gal
min
Segment 3 (100’ of 2” pipe)
• Remember T = V
Q
• As in Segment 1, the first step is to calculate volume:
V = ∏(2”/12”)2 x 100’ x 7.48 gal
4
ft3
V = (0.785)(0.166666’)2 x 100’ x 7.48 gal
ft3
V = (0.785)(0.027778 ft2) x 100 ft x 7.48 gal
ft3
V = 0.021805 ft2 x 100 ft x 7.48 gal
ft3
V = 16.3105 gal
Calculate T (contact time in minutes)
• Remember: T = V
Q
We calculated the volume (V) to be 16.31 gal
We were given the peak flow rate (Q) of 25 gpm
T = 16.31 gal  16.31 gal x 1 min = 0.65 min
25 gal
25 gal
min
What is the total T?
• Total T = Segment 1 + Segment 2 + Segment 3
• Total T  0.52 min + 0.8 min + 0.65 min = 1.97 min
Now let’s calculate the theoretical CT
CT = C x T
CT = 1.2 mg x 1.97 min
L
CT = 2.36 mg-min
L
Does this system meet 4-log CT
requirements?
• From the instructions for MOR Form 62-555.900(3)
• The required CT to meet 4-log at 24⁰C is 2.2 mg-min/L
• 2.36 > 2.2, so yes this system meets 4-log at the given
parameters
• Would this system still meet CT requirements at 18⁰C?
CT values at 18°C
• 2.36 < 3.4, so no this system does not meet 4-log at 18°C
• What can this system do to meet 4-log CT requirements
at worst case scenario?
– Hire an engineer to modify the treatment plant by adding
storage, treatment options which provide a log credit, or tank
baffling.
– Operational changes such as increasing disinfectant residuals,
relocating injection point, or increasing minimum tank water
level
*Beware of simultaneous compliance issues (DBP production due to
increased residuals or detention times)
Adding storage tanks to increase CT
WTP modification to increase CT
What would our disinfectant residual
have to be to meet 4-log at 18⁰C?
CT = C x T  C = CT  C = 3.4 mg-min
T
L
1.97 min
= 1.73 mg/L
 C = 3.4 mg-min x
1
L
1.97 min
Treatment processes which provide log
credit
• Currently in the Central District, there are two
types of treatments providing a 2-log credit
– Conventional filtration
• Must maintain combined filter effluent turbidity below
1 NTU at all times.
– Reverse osmosis
• Must maintain a permeate effluent salt passage below
5% at all times.
Example #2
Wells
Cl2 injection
Clarifier
x
NH3
x
10’x10’x10’
Clearwell
10’ of 10”
Transfer
pumps
GST
100’ of 10”
VFD
HSP
House
Plant office
•
•
•
•
Wells – 3 @ 50 gpm each
Transfer pumps – 2 @ 100 gpm
each, shut off at 5’
HSP – 2 @ 100 gpm each
GST – diameter 20’
- max water depth 18’
- min water depth 10’
•
Here’s what we know:
– pH = 10.2
– Temperature = 18⁰C
– Free chlorine residual = 1.0
mg/L
– Combined residual = 0.8 mg/L
Segment 1 - Clearwell
T = V x BF
Q
T = 500 ft3 x 7.48 gal
ft3
200 gal
min
x 0.3 BF
500 ft3 x 7.48 gal x 1 min x 0.3 BF
ft3 200 gal
T = 18.7 min x 0.3 BF
T = 5.61 min
•
CTfree = C x T
1.0 mg/L x 5.61 min
CTfree = 5.61 mg-min/L
Segment 2 – Pipe
T = V = ∏(10”/12”)2 x 110’ x 7.48 gal
4
ft3
Q
200 gal
min
T = 2.24 min
• CTChloramine = 0.8 mg/L x 2.24 min
CTChloramine = 1.8 mg-min/L
Segment 3 - GST
T = V = ∏(20 ft)2 x 10 ft x 7.48 gal x 0.5 BF
Q
4
ft3
200 gal
min
T = 117.4 min x 0.5 BF
T = 58.7 min
• CTChloramine = 0.8 mg/L x 58.7 min
CTChloramine = 46.96 mg-min/L
Does this system meet 4-log CT
requirements?
• Since part of the CT is from free chlorine and part of it is
from chloramines, a direct comparison is not possible.
• Therefore, we must use ratios of the respective CT.
– If the sum of the ratios is greater than 1.0, then we meet 4-log.
Does this system meet 4-log CT
requirements?
Total CTratio = CTfree Segment 1 + CTchloramine Segment 2 + CTchloramine Segment 3
Required CTFree
Required CTChloramine
Required CTChloramine
Total CTratio = 5.61 + 1.8 + 46.96
25.2 845 845
= 0.22 + 0.002 + 0.055
= 0.278
• 0.278 < 1.0, therefore the system does not meet
4-log at the given parameters
Does this system meet 2-log CT
requirements?
• Since this is a conventional filtration plant, and they
continuously monitor and keep their CFET below 1
NTU, they receive a 2-log credit. Therefore, they only
have to meet 2-log CT through disinfection.
Does this system meet 2-log CT
requirements?
Total CTratio = 5.61 + 1.8 + 46.96
12.6 364 364
= 0.445 + 0.005 + 0.129
= 0.579
• 0.579 < 1.0, therefore the system still does not
meet 4-log at the given parameters
Questions?
• Email: [email protected]