# 5.1: Units


“Units” are the single most important component of many mathematical problems especially in the waterworks industry. If you are reporting a flow rate of 1,000, what does this mean? 1,000 gallons per minute? 1,000 cubic feet per second? Or, 1,000 acre feet per year? It is strongly recommended that you get in the habit of writing down the units for all the problems you are solving. Most of the problem solving in waterworks mathematics requires the conversion of units.

Cfs gpm MGD AFY

Therefore, Unit Dimensional Analysis (UDA) is the key to solving a majority of water related mathematical problems. UDA is the process of converting units of one type (e.g. gallons) to another similar type (e.g. cubic feet). Setting up your UDA problem is the most important step conversion problems. By “stacking” your units you create a visual representation of the problem where units can be canceled. This cancellation of units is very similar to cross canceling or reducing of fractions.

Example $$\PageIndex{1}$$

Convert seconds to days using UDA

$\dfrac{\text {sec} }{1} \times \dfrac{\text {min} }{\text {sec} } \times \dfrac{\text{hour}}{\min } \times \dfrac{\text { day }}{\text { hour }} \nonumber$

Solution

If we break the problem up into segments you can see how the canceling of units occurs.

$\dfrac{\not {\text {sec}}}{1} \times \frac{\min }{\not {\text {sec}}}=\min \nonumber$

In the above example, if we just do the first step you can see that canceling seconds will yield minutes as the result. By continuing the same process you can end up with the unit the question is asking.

$\dfrac{\not {\sec}}{1} \times \dfrac{\not {\min}}{\not {\sec}}=\dfrac {\not {\text {hour}}}{\not {\min }} \times \dfrac{\text {day}}{\not {\text {hour}}} = \text {day} \nonumber$

The reason seconds is set up over a “1” is because it is the units you are wanting to convert from and remember anything over 1 is that value.

Multiple units can be converted in a single problem too. However, convert one unit at a time. A common waterworks mathematics problem is the conversion from cfs to gpm. Cubic feet needs to be converted to gallons and seconds needs to be converted to minutes.

Example $$\PageIndex{2}$$

Convert cfs to gpm.

$\dfrac{\text{cf}}{\text{sec}} \rightarrow \dfrac{\text{gal}}{\text{min}} \nonumber$

Solution

An important thing to note is that you will be converting “like” units. This means that you will be converting cubic feet to gallons which are both a measurement of volume and converting seconds to minutes which are both a measurement of time.

There are two conversion factors that need to be used to perform this conversion. You need to ask yourself…How many seconds are in a minute? How many gallons are in a cubic foot?

There are 60 seconds in every minute.

There are 7.48 gallons in every cubic foot.

$\dfrac{\text{lcf}}{\sec } \times \dfrac{60 \text{sec}}{\min } \nonumber$

The expression 60 sec per minute is a conversion factor that is an equality. The conversion factor is deliberately written with the 60 sec on top of the equality because it needs to cancel the unit sec from the 1 cfs.

$\dfrac{1 \text{cf}}{\not {\sec}} \times \dfrac{60 \not{\sec} }{\min }=\dfrac{60 \text{cf}}{\min } \nonumber$

Since the sec cancel we end up with cubic feet per minute (cfm).

Now let’s look at converting cubic feet to gallons. Remember that there are 7.48 gallons in every cubic foot.

$\dfrac{1 \text{cf}}{\sec } \times \dfrac{7.48 \text{gal}}{\text{cf}} = \dfrac{1\not{\text{cf}}}{\sec} \times \dfrac{7.48 \text{gal}}{\not{\text{cf}}}=\dfrac{7.48 \text{gal}}{\sec} \nonumber$

If you combine both of the examples above you can see how cfs is converted to gpm.

$\dfrac{1\not{\text{cf}}}{\not{\sec}} \times \dfrac{60 \not {\sec}}{\min } \times \dfrac{7.48 \text{gal}}{\not{\text{cf}}}=\dfrac{448.8 \text{gal}}{\min } \nonumber$

## Exercise 5.1

Solve the following problems using UDA

1. Convert 5 cfs to gpm.
2. Convert 1,500 gpm to cfs.
3. Convert 4 MGD to gpm.
4. Convert 2.5 cfs to MGD.
5. Convert 7 fps to mph.
6. A hose is flowing at a rate of 1.5 cfs. How many gallons will flow through it in one minute?
7. A well pumps 2,000 gallons per minute. How many million gallons of water will it pump in one day?
8. A water utility sold 10 million gallons of water. They need to report this in one hundred cubic feet (CCF or HCF). How many HCF is this?
9. A water utility operator is filling up a storage tank at a rate of 1,000 gallons per minute. How long will it take until the tank has 1.5 MG of water?
10. Water is moving through a pipe at a velocity if 5.75 feet per second. Express the velocity in miles per hour.
11. Last year a water utility delivered 2.35 million gallons per day. How many acre feet did they deliver in one year?
12. A water tank was being drained at a rate of 750 gallons per minute. If the tank holds 5 MG of water, how long will it take for the tank until the tank is empty? (Express your answer as Days, Hours, Minutes – ex 2 days, 3 hours, and 10 minutes)

5.1: Units is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.