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7.1: How Much For How Long?

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    7145
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    As part of the Safe Drinking Water Act, there are a set of regulations known as the Surface Water Treatment Rule (SWTR.) The first version of the SWTR enacted in 1989 required inactivation percentages for Giardia of 99.9% and for viruses of 99.99%, among other things. In 1998, the Interim Enhanced Surface Water Treatment Rule (IESWTR) was introduced adding the 99% inactivation of Cryptosporidium to the list. In 2002, the Long Term 1 Enhanced Surface Water Treatment Rule was announced with the Long Term 2 Enhanced Surface Water Treatment Rule (LT2ESWTR) following soon after in 2003. This text focuses on the basic elements of CT calculations and the inactivation of Giardia and viruses.

    In order to be in compliance with the SWTR, drinking water treatment plants must meet the following inactivation requirements:

    • Giardia lamblia – 3.0 Log or 99.9% Inactivation
    • Viruses – 4.0 Log or 99.99% Inactivation
    • Cryptosporidium parvum – 2.0 Log or 99% Inactivation

    The table below compares the Log and Percent Inactivation values.

    Table 7.1. Log Percent and Inactivation Values

    Log Inactivation

    Expressed as Log

    Log Value

    Percent Inactivation

    0.5

    100.5

    3.162

    68.38

    1.0

    101.0

    10

    90.00

    2.0

    102.0

    100

    99.00

    3.0

    103.0

    1,000

    99.90

    4.0

    104.0

    10,000

    99.99

    5.0

    105.0

    100,000

    99.999

    6.0

    106.0

    1,000,000

    99.9999

    7.0

    107.0

    10,000,000

    99.99999

    Updated versions of the act were published in 2002 and 2003 as the LT1ESWTR and LT2ESWTR respectively. A simple calculation can be used to determine the percent inactivation or the log inactivation if one or the other is known. The following formula can be used to calculate the percent inactivation if the log inactivation is known.

    Converting from Log to Percentage

    Example

    • 0.5 Log
      • (1 - (1/10log inactivation)) x 100
      • (1 - (1/100.5)) x 100
      • (1 - (1/3.16)) x 100 = 68%
    • 3.0 Log
      • (1 - (1/10log inactivation)) x 100
      • (1 - (1/103)) x 100
      • (1 - (1/1000) x 100 = 99.9%

    The following formula can be used to calculate the log inactivation if the percent inactivation is known.

    Converting from Percentage to Log

    Example

    • 68.38%
      • 100/(100 - Percent Inactivation) = Log Inactivation
      • 100(100 - 68.38) = 3.162
      • 3.162 = 100.5 = 0.5 Log

    Use the Log or exponent function on your calculator to solve 100.5 to get 3.162.

    • 99.9%
      • 100/(100 - Percent Inactivation) = Log Inactivation
      • 100(100 - 99.9) = 1,000
      • 1,000 = 103 = 3 Log

    Although the above formulas can be used, we typically only deal with Giardia (3 Log or 99.9%), viruses (4 Log or 99.99%), and Cryptosporidium (varies) so you only need to remember a few values. Cryptosporidium will not be discussed in this class due to the complexities of the LT2ESWTR regulations.

    Surface water must go through some type of treatment. Disinfection can be used as the sole means of meeting the CT requirements. However, various treatment processes account for some of the inactivation or removal of pathogens. Therefore, the SWTR provides “credits” toward the inactivation of Giardia and viruses. For example, the CT requirement for viruses is 4 Log. If a treatment process received 1 Log credit, then the disinfection requirement for viruses would be 3 Log (4 – 1 = 3.) The table below shows the various credits and resulting disinfection requirements.

    Table 7.2. Treatment Credits and Log Inactivation Requirements

    Treatment

    Log Inactivation Requirements

    Removal Credit Logs

    Required Log Inactivation from Disinfection

    Giardia

    Viruses

    Giardia

    Viruses

    Giardia

    Viruses

    Conventional

    3

    4

    2.5

    2

    0.5

    2

    Direct Filtration, DE, or Slow Sand

    3

    4

    2

    1

    1

    3

    As previously mentioned CT stands for Concentration and Time. Concentrations are expressed in mg/L and Time in min, therefore CT is expressed as (mg/L · min). When solving CT problems, the concentration of the disinfectant is typically provided in the question. However, there may be times when the chemical dosage formula is needed. In order to calculate the time (contact time to be exact), the detention time or “plug flow” formula will be needed.

    Time is defined as the time the disinfectant is in contact with the water to the point where the Concentration is measured. These times are easily calculated through pipelines and reservoirs, but sometimes they can be difficult to calculate through various treatment processes. In this case, Tracer Studies are sometimes conducted. T10 represents the time for 10% of an applied tracer mass to be detected through a treatment process or, the time that 90% of the water and pathogens are exposed to the disinfectant within a given treatment process. Some problems will require the calculation of the contact time while others will provide T10 values.

    Once the actual CT values have been calculated, the final step in the CT calculation process involves CT Tables. The U.S. Environmental Protection Agency (USEPA) as part of the SWTR, created a series of tables that list the type of disinfectant, the pH of the water, the concentration of the disinfectant, the contact time, and the pathogen in question. Using all this information, the required CT (mg/L · min) values can be found. For your reference, the CT Tables are provided at the end of this text. They can be confusing at first, but once you understand what information you need to look for, the CT values can be easily found.

    In the paragraph above, two terms were used; “actual CT” and “required CT.” Actual CT is the actual concentration through the treatment process and the actual time the disinfectant is in contact with the water. For example, if the contact time is 10 minutes and the concentration is 0.2 mg/L then the actual CT is 0.2 mg/L times 10 minutes which equals 2 mg/L · min. The CT Tables give you the required CT, the required concentration time needed to inactivate either Giardia or viruses. The ratio of the actual CT and the required CT is then calculated. If the actual CT is equal to or greater than the required CT then the ratio is equal to or greater than 1.0 and CT is met. If the actual CT is less than the required CT then the ratio would be less than 1.0 and CT would not be met. The following examples show how to calculate CT problems from start to finish.

    Example

    A typical question will provide the pH, the temperature, the pathogen of interest, the type of disinfectant, the dosage or a way to calculate the dosage, and the type of treatment. An example of this data is provided below.

    • pH – 7.5
    • Temperature – 10°C
    • Disinfectant – Free chlorine
    • Dosage – 0.2 mg/L
    • Pathogen – Giardia
    • Treatment – Direct Filtration

    Use this information to find the appropriate CT Table to use. You should discover that Table C-3 is the correct table to use for this data set. For a quick reference, this table is presented on the next page.

    -kPOMin1vILrpzMKjyR9mZieiR48ok2dUMEynTxxu-Y1h0E0c-HZT0VVMLQB7KI6TMZeo5DMjebrhim1ZQqzB-vHySJh8X7MTICxwT3fWdZAgF555ID3r-EEZGpyYkK3w3JATtM
    Figure \(\PageIndex{1}\)

    Finding Required CT

    The title of the table tells you which CT value the table will provide. This table is for Giardia, with free chlorine as the disinfectant, at a temperature of 10°C. Now you need the other information (pH and dosage concentration.) There are seven (7) boxes in the table each with different pH values. Look for the one that says pH = 7.5. On the far left of the table, you can find the varying disinfectant concentrations, starting with less than or equal to 0.4 mg/L going up to 3 mg/L. Since the problem states a dosage of 0.2 mg/L, you’ll need to use the first row of the table. You now need the last bit of information, the treatment process. In this instance, it is Direct Filtration. Remember, Giardia has an inactivation requirement of 3 Log. If there was no treatment process you would be looking at the last column under pH of 7.5. However, referring back to Table 7.1 you can identify the Log credit and resulting disinfection inactivation requirements. You should come up with a required inactivation from disinfection of 1 Log. Using the first row for disinfectant concentration and the second column from the 7.5 pH portion of the table, you should come up with a required CT of 42 mg/L · min.

    Calculating Actual CT

    There can be multiple locations where a disinfectant is added to the water during the treatment process. Sometimes the water is pre-chlorinated in the raw water pipeline leaving a storage reservoir prior to entering the treatment facility. Sometimes the water is disinfected before the coagulation flocculation process and many times the water is disinfected after filtration prior to delivery to customers. Every time chlorine is added to the water supply it counts towards the inactivation of pathogens. Each step of the way CT will need to be calculated. The example information below will help illustrate this concept.

    Free chlorine is added at a concentration of 0.2 mg/L in a 12” diameter 5,000-foot long pipeline leaving a storage reservoir prior to entering the treatment plant. The flow through the pipeline is 2 MGD.

    Using this information, the time 0.2 mg/L of free chlorine is in contact with the water can be determined using the Detention Time formula.

    • Dt = Volume/Flow
    • Volume = 0.785 x D2 x L
    • Volume = 0.785 x 1’ x 1’ x 5,000’ x 7.48 gal/ft3 = 29,359 gallons
    • Flow = 2 MGD = 1,389 gpm
    • Dt = 29,359 gal/1,389 gpm = 21 minutes

    Multiply the detention time by the concentration and you get CT.

    • 0.2 mg/L x 21 minutes = 4.2 mg/L · min

    So, the actual CT through the pipeline is 4.2 mg/L · min.

    Tracer studies (T10) have determined that a free chlorine concentration of 1.2 mg/L through the treatment plant is 20 minutes.

    With this information, you can calculate the CT through the plant.

    \[1.2\, mg/L \times 20\, minutes = 24 \,mg/L \cdot min\]

    Since both sections are disinfected with the same chemical, the two CT values can be added together.

    \[4.2\, mg/L \cdot min + 24\, mg/L \cdot min = 28.2\, mg/L \cdot min\]

    The following table is used to organize the data.

    Location and Type of Disinfection

    Actual CT

    Required CT

    CT Ratio

    Pipeline + Plant

    (free chlorine)

    28.2 mg/L · min

    42 mg/L · min

    0.67

    Since the ratio of actual to required CT is less than 1.0, then CT is not met. If a treatment plant does not meet CT it can either increase the detention time through the pipeline or plant or it can increase the dosage.

    In a situation where two different disinfection chemicals are used, the required CT values would be different and you would not add the different disinfecting locations together. The next example illustrates this scenario.

    Example

    A conventional water treatment plant receives water with a 0.4 mg/L free chlorine residual from 9,000 feet of 3-foot diameter pipe at a constant flow rate of 10 MGD. The water has a pH of 7.5 and a temperature of 10°C. Tracer studies have shown a contact time (T10) for the treatment plant to be 30 minutes. The plant maintains a chloraminated residual of 1.2 mg/L. Does the plant meet CT compliance for Giardia?

    The first step should be setting up the table and identifying the CT Tables to use to find the required CT values. This particular problem uses CT Tables C-3 and C-10 (your instructor should hand these out in class). Remember to subtract out the 2.5 Log credit for conventional treatment.

    Location and Type of Disinfection

    Actual CT

    Required CT

    CT Ratio

    Pipeline (free chlorine)

    21 mg/L · min

    Plant (chloramines)

    310 mg/L · min

    Now, the actual CT needs to be calculated.

    Volume of the 9,000 feet of 3-foot diameter pipe.

    • Volume = 0.785 x 3ft x 3ft x 9,000ft x 7.48 gal/ft3 = 475,616 gallons
    • Flow rate = 10 MGD = 6,944 gpm
    • Dt = 475,616 gal/6,944 gpm = 68.5 minutes
    • 0.4 mg/L x 68.5 minutes = 27.4 mg/L · min (this is the CT through the pipeline)

    CT for the plant is:

    • 1.2 mg/L x 30 minutes = 36 mg/L · min (this is the CT through the plant)

    Now finish populating the table:

    Location and Type of Disinfection

    Actual CT

    Required CT

    CT Ratio

    Pipeline (free chlorine)

    27.4 mg/L · min

    21 mg/L · min

    1.3

    Plant (chloramines)

    36 mg/L · min

    310 mg/L · min

    0.12

    The sum of the CT ratios equals 1.42 mg/L · min. Therefore, CT is met. You may have noticed that CT was achieved through the pipeline only and the chloramination through the plant is not needed. This is true. So, when solving one of these problems, once you meet the ratio of 1.0 or greater, CT is met and you can stop solving the problem.

    Exercises


    7.1: How Much For How Long? is shared under a CC BY license and was authored, remixed, and/or curated by Mike Alvord.

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