Lesson - Detention Time

Recall how to calculate volume ....

To calculate the expected detention times in the flash-mix chamber, distribution channel, and flocculation basins, use the dimensions of the basin or pipe to calculate the volume in gallons and divide this volume in gallons by the flow. The calculation will be in the units of time that is used for the flow. These times are important when determining the optimum chemical dosage for the water that is being treated using jar tests. Also, these times are necessary for the desired chemical reactions to occur.

Formulas:

Use the following formula to calculate the volume of a circular tank or clarifier in cubic feet.

Cub Volume.JPG

Use the following formula to calculate the volume of a cylinder in cubic feet.

Cylinder Volume

Frequently, the volume is expressed in gallons rather than cubic feet. Remember that there are 7.48 gallons in each cubic foot. Therefore, the conversion factor to convert volume in cubic feet to volume in gallons is as follows.

$LaTeX: Volume\:\left(gal\right)=Volume\:\left(ft^3\right)\times\frac{7.48\:gal}{ft3}$

Use the following formula to calculate the detention time (Dt) of any chamber, tank, basin, or clarifier. Remember that each variable can be provided using a variety of units and all of the units must align in order to solve the equation.

$LaTeX: D_t=\frac{Volume}{Flow}$

Detention times are calculated by dividing the volume in gallons by the flow in gallons per day. This produces the detention time in days. Multiply by 24 hours per day to obtain the detention time in hours. To convert the detention time from hours to minutes, multiply by 60 minutes per hour. To convert the detention time from hours to minutes, multiple by 60 minutes per hour and minutes to seconds multiple by 60 seconds per minute. Many operators prepare curves of flow versus detention time for the basins in their plants. These curves allow for easy selection of stirring times when performing jar tests.

Application of Detention Time and Contact Time

Detention Time is an important process that allows large particles to “settle out” from the flow of water through gravity, prior to filtration. It is the time it takes a particle to travel from one end of a sedimentation basin to the other end. Conventional filtration plants require large areas of land to construct sedimentation basins and employ the detention time process. Not all treatment plants have the available land and may decide that direct filtration is suitable. Therefore, in direct filtration plants the sedimentation process is eliminated. However, in direct filtration plants, the filters have shorter run times and require more frequent backwashing cycles to clean the filters.

A term used that is interchangeable with detention time is contact time. Note that this should not be confused with CT, Concentration Time which will be discussed in another module. Contact times represent how long a chemical (typically chlorine) is in contact with the water supply prior to delivery to customers. For example, contact time can be measured from the time a well is chlorinated until it reaches the first customer within a community or, it could be how long the water mixes in a storage tank before it reaches a customer.

Calculating the Detention Time and Contact Time requires two elements, the volume of the structure holding the water (sedimentation basin, pipeline, and storage tank) and the flow rate of the water (gallons per minute, million gallons per day.) Since detention times and contact times are typically expressed in hours, it is important that the correct units are used. When solving Dt problems be sure to convert to the requested unit of time.

As with all water math-related problems, there are other parameters that can be calculated within the problem. For example, if the detention time and volume are known, then the flow rate can be calculated. Or, if the flow rate and detention time are known, the volume can be calculated. Sometimes the flow rate and the desired detention time is known and the size of the vessel holding the water needs to be designed. In this example, the area or dimensions of the structure can be calculated.

The pie wheel (below) shows a simple way of calculating the variables. If the variables are next to each other (Dt and Flow Rate) then multiply. If they are over each other (Volume and Dt or Volume and Flow Rate) then divide.

The following formula is used for calculating detention times.

$LaTeX: D_{t} = \frac{Volume}{Flow}$

You may also use the Pie Wheel to solve for Detention Time as shown below.

detention time circle.JPG

Units are extremely important when using this formula. There are three variables in this formula: Detention Time, Flow, and Volume. Each variable can be provided using a variety of units. You cannot calculate the solution unless all of the units align. If the units are similar (matching), then dividing volume by flow will yield a time (Detention Time). However, simply dividing a volume by a flow will not result in a time. For example, if you divide gallons by cubic feet per second there is no resulting answer. This is because “gallons” and “cubic feet” will not cancel each other.

Making sure the units are correct is important before solving this equation. Take a look at the examples below, starting with American Water Called video called "Problem Solved: Detention Time - Water Treatment Math" Please take notes as you watch.

Example:

Calculate the detention time.

$LaTeX: D_{t} = \frac{Volume}{Flow} = \frac{gallons}{gallons/minute}=minute$

$LaTeX: D_{t} = \frac{Volume}{Flow} = \frac{cubic feet}{second}=second$

$LaTeX: D_{t} = \frac{Volume}{Flow} = \frac{gallons}{million gallons/day}$

In the first two examples, the terms can be divided. However, in the third example they cannot. Dt should be expressed as a unit of time (i.e., sec, min, hours). If you divide the first two examples (gal/gpm and cf/cfs), you will end up with minutes and seconds respectively. However, in the third example, gallons and million gallons cannot cancel each other out. Therefore, if you had 100,000 gallons as the volume and 1 MGD as the flow rate:

$LaTeX: D_{t} = \frac{Volume}{Flow} = \frac{100,000 gallons}{1 MGD}$

Then you would need to convert 1 MGD to 1,000,000 gallons per day in order to cancel the unit gallons. The gallons then cancel leaving “day” as the remaining unit.

$LaTeX: D_{t} = \frac{Volume}{Flow} = \frac{\frac{100,000 gallons}{1,000,000 gallons} }{1 day}=0.1 day$

$LaTeX: D_{t} = \frac{Volume}{Flow} = \frac{0.1 MG}{1 MGD}=0.1 day = 2.4 hours$

Converting the days to hours is easy since there are 24 hours in one day.

$LaTeX: 0.1 day= \frac{24 hours}{1 day} = 2.4 hours$

Sometimes this can be the simplest way to solve detention time problems. However, people can be confused when they get an answer such as 0.1 days. There are other ways to solve these problems. One way is to convert MGD to gpm. Using the above example, convert 1 MGD to gpm.

$LaTeX: \frac{100,000 gallons}{1 day} \times \frac{1 day}{1,440 minutes} =694.4 gpm$

Now solve for the Detention Time

$LaTeX: \frac{\frac{100,000 gallons}{694.4 gallons} }{min}=144 minutes \times \frac{1 hour}{60 minutes} = 2.4 hours$

If the question is asking for hours there still needs to be a conversion. However, 144 minutes is more understandable than 0.1 days

Example:

What is the detention time in a circular clarifier with a depth of 50 ft and a 30 ft diameter if the daily flow is 2.2 MG. (Express your answer in hours:minutes.)

First you need to calculate the volume of the clarifier.

Clarifier Volume = 0.785 x D² x H = 0.785 x (30 ft)² x 50 ft = 35,325 ft³

In order to calculate the detention time, the units for Volume and the units for Flow must align. Since the flow rate is provided in MG, convert the cubic foot volume to MG.

$LaTeX: 35,325 ft^{3} \times \frac{7.48 gal}{1 cf}\times \frac{1 MG}{1,000,000 gal}=0.264231 MG$

Now you can calculate the detention time by substituting the volume calculated and the daily flow provided into the detention time formula.

$LaTeX: D_{t}=\frac{Volume}{Flow} =\frac{0.264231 MG}{2.2 MGD} = 0.120105 days\times \frac{24 hour}{1 day} = 2.88252 hours$

The problem asks for the detention time to be expressed in hours and minutes. Based on the calculation above, you know that you have 2 full hours and a portion of a third hour. To calculate the exact number of minutes, take the decimal amount and multiply by 60 min per hour.

$LaTeX: 0.88252 hours \times \frac{60 min}{1 hour}=52.95 min = 53 min$

D_t = 2 hours 53 minutes = 2:53

Example:

A water utility is designing a transmission pipeline collection system in order to achieve a chlorine contact time of 2 hours 10 mins once a 1,125 gpm well is chlorinated. How many feet of 30” diameter pipe are needed?

The problem statement provides both the detention time and the flow rate. However, in order to use the detention time provided, you need to convert it to minutes.

$LaTeX: \left(2\:hours\:\times\frac{60\:\min}{1\:hour}\right)+10\:\min\:=\:130\:\min$

Rearranging the terms in the detention time equation to solve for volume results in volume equals detention time times flow. Substitute the minutes calculated and the flow rate into this equation in order to solve for the total volume in gallons.

$LaTeX: Volume\:=\:D_{t} \times Flow = 130 min \times \frac{1,125 gal}{min}=146,250 gal$

Since the problem statement is asking for how many feet of pipe, convert the gallons to cubic feet.

$LaTeX: Volume = 146,250 gal \times \frac{1 cf}{7.48 gal}= 19,552.1390374 cf = 19,552.1 cf$

Now that you know the total volume of the pipe in cubic feet, you can use the formula for volume of a pipe to determine the pipe length. Substitute the pipe diameter and the pipe volume into the equation and rearrange the terms to solve for length.

Pipe Volume = 0.785 x D² x H =

$LaTeX: 0.785 x (30 in x \frac{1 ft}{12 in})^{2}\times Length = 19,552.1 ft^{3}$

$LaTeX: Length = \frac{19,552.1 ft^{3} }{0.785 \times (2.5 ft)^{2}} = \frac{19,552.1 ft^{3} }{4.90625 ft^{2} } = 3,985.1414 ft = 3,985 ft$

The information for this page is from the book titled "Advanced Waterworks Mathematics" version 1.6 dated 2021.Written by Michael Alvord, Regina Blasberg, Stephanie Anagnoson, and Ernesto.An Open Educational Resources Publication by College of the Canyons. 
The content in this book is licensed under a Creative Commons Attribution 4.0 International License.