Since this site is about water and wastewater let’s start our discussion about math with some basic topics that are fundamental to the field of wastewater treatment. Our first topic will be a discussion of the density of fluids and the use of specific gravity. There are several constants associated with wastewater that you need to know, constants you will use over and over again.
The weight of water =
The specific gravity (S.G.) of water = 1.0
A cubic foot of water =
So, if water weighs 8.34 lbs/gal and there are 7.48 gallons of water in a cubic foot, a cubic foot of water weighs 62.4 lbs as shown in Equation 1.
Equation 1: Density of Water
The metric system is all around us and we need to have at least a basic understanding so that conversions can be made as needed. (You can also download a free unit converter to help you with this.) Here, then, are a few other constants that you may need:
The “lightness” or “heaviness” of an object is the way we typically describe what scientists refer to as the “density” of an object. More technically, the density of a substance is its “mass per unit volume.” A unit of mass, for example, is a pound, and a unit of volume is, for example, a cubic foot or a gallon. In water and wastewater treatment the density is most commonly measured as lbs/gal. The formula for calculating density is shown in Equation 2.
Equation 2: Density Defined
D = density
M = mass in lbs
V = volume as cubic feet or gallons
Figure 1: One Cubic Foot of Water
Figure 2: One Gallon of Water
The density of several other substances is shown in Table 1.
Table 1: Density Comparison
Specific gravity is the ratio of the density of a substance (other than water, such as a polymer, ACH, or sodium hydroxide) to that of a “standard.” The standard for solids and liquids is water with a density of 8.34 lbs/gal. Specific gravity is calculated as shown in Equation 3.
Equation 3: Specific Gravity Defined
We received a delivery of ACH and the “net weight” of product in the truck was shown to be 33,380 lbs. In order to convert this to gallons we need to know the density of ACH. Given that ACH has a specific gravity of 1.34 we calculate the density of ACH as shown in Equation 4.
Equation 4: ACH Density
We can now convert the pounds of ACH in the truck to gallons as shown in Equation 5.
Equation 5: ACH Gallons
Area and Circumference of a Circle
In water and wastewater treatment, many unit processes are circular (Figure 3). You will need to calculate the surface area (Equation 6) in order to determine the rise rate in circular water treatment clarifiers or the equivalent surface overflow rate (SOR) and the solids loading rate (SLR) in circular wastewater clarifiers. To compute the weir oveflow rate (WOR) you will need to know the circumference (Equation 7) of a circular clarifier
Calculation of the area and circumference of a circle requires the use of the mathematical constant π (pi) which is the ratio of a circle’s circumference to its diameter. The value of pi to 15 places is 3.141592653589790 as calculated in Excel using the function =PI(). For our calculations using a value of 3.14 for pi will be accurate enough.
Figure 3: Radius (r) and Diameter (d) of a Circle
Equation 6: Area of a Circle
Equation 7: Circumference of a Circle
The volume of a cylinder can be expressed in cubic feet (ft3) or gallons (see Figure 4). In equations 8 and 9, below, the volume of a cylinder will be calculated in units of cubic feet. Since each cubic foot of water contains 7.48 gallons, to convert cubic feet of volume to gallons of volume, multiply by 7.48 gal/ft3.
Figure 4: Dimensions of a Cylinder
Equation 8: Volume of a Cylinder (Diameter)
Equation 9: Volume of a Cylinder (Radius)
Let’s take a closer look at Equation 9 to see the various ways it can be used. When we first look at Equation 9 we see that it has three “unknown” variables: V, R2, and H. In order to solve this equation for volume we have two find the value of the two unknown variables, R2 and H. In any equation with three unknowns, finding the value of any two variables will allow you to solve for the third. So, if we know V and H, we can find (or solve for) R. And if we know V and R, we can find H. Here’s an example.
We have a chemical storage tank that holds 4,000 gallons. As we look at the tank we find that we have no other information about it in terms of its dimensions. And the tank is too tall to gain access to the top so that we might measure its diameter. But fortunately we can get a tape measure to the top of the tank to determine its overall height. The tank’s height, H, is found to be 130 inches or 10.83 feet tall. This was a rough field measurement so we are likely a little off in our estimation of the tank’s height. It is much more likely that the tank was made to a standard dimension of 132 inches or 11 feet.
We now have two of the three variables we need to solve the volume equation. But remember, volume is first determined in units of ft3. So we need to convert our volume in gallons to cubic feet as shown in Equation 10.
Equation 10: Convert Gallons to Cubic Feet
Now we need to algebraically “rearrange” our volume equation and then solve for the radius, R, as shown in Equation 11.
Equation 11: Algebraic Rearrangement
Again, we know from experience that a chemical storage tank is more likely to have a radius of 4-ft, or a diameter of 8-ft, than a radius of 3.9-ft or a diameter of 7.8-ft. So, just as we made an adjustment in our estimated height, going from 10.83-ft to 11-ft, we will now adjust our tank’s diameter from 7.8-ft to 8-ft. Therefore, the final dimensions of our chemical storage tank are 8-ft in diameter by 11-ft tall. We now need to recalculate our tank volume to see how it compares to the original assumption of 4,000 gallons, as shown in Equation 12.
Equation 12: Recalculation of Tank Volume
Our revised estimate of the tank’s volume as 4,136 gallons compares well with the 4,000 gallon volume listed on the side of the tank. We are confident at this point that the chemical storage tank is truly 8-ft in diameter by 11-ft tall and that its volume is greater than 4,000 gallons. But the 4,000 gallon volume stated on the tank is the tank’s “effective” volume, or, based on the location of the suction out of the tank, the useable volume available before we would lose suction to a pump. So we are coming to our final set of calculations which is to determine how many gallons the tank holds per foot and per inch of height as shown in Equation 13.
Equation 13: Gallons Per Foot and Per Inch