# 6.1: It's a Lag in Time

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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 in order 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. 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, etc). 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. Figure $$\PageIndex{1}$$

The chart above 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 Detention Time formula is:

$\text{Detention Time (Dt)} = \dfrac{\text{Volume}}{\text{Flow}}$

• When solving this equation make sure the units are correct

Take a look at the examples below.

• $$\text{Dt}$$ = Volume/Flow
• gallons/(gallons/minute)
• cubic feet/(cubic feet/second)
• gallons/(million gallons/day)

In the first two examples above, 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:

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.

100,000 gallons/(1,000,000 gallons/1 day) = 0.1 day

Converting to hours from days is easy, simply multiply by 24 hours per day.

0.1 day/x = 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.

1,000,000 gallons/1 day x 1 day/1,440 minutes = 694.4 gpm

Now solve for the Detention Time.

100,000 gallons/(694.4 gallons/min) = 144 minutes x 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.

## Exercises

Solve the following problems. Be sure you provide the answer in the correct units.

This page titled 6.1: It's a Lag in Time is shared under a CC BY license and was authored, remixed, and/or curated by Mike Alvord (ZTC Textbooks) .