Customer Questions: Calculating Secondary Containment Needs

Of course you want to protect the environment. And you would never want your employees exposed to leaks and spills of hazardous waste. That’s why you’ve very diligently investigated the purchase of a secondary containment system. But how much secondary containment do you actually need? Here are some easy (promise!) formulas for figuring that out. Federal containment regulations [40 CFR 264.175] require those storing hazardous waste to be able to “contain 10% of the volume of the containers, or the volume of the largest container, whichever is greater.” And what, exactly, does that mean? For example, you are:  

Secondary containment units range in sump capacity from 11 gallons and up. Custom secondary containment units are common if you need a higher sump capacity.

If you’re considering a custom containment system, the sump capacity of the system is a Very Important Number. (Alert: More math ahead.) First, you have to figure out the number of gallons you wish to contain, including the volume displaced by vehicles, containers or other items. Use the following formula to determine the amount of liquid your system will contain:

Length (L’) x Width (W’) x Height (H’) x 7.48 = Sump Capacity (Gallons)

So now you know how much sump capacity you need. Next, you need to figure out how big a system you need — that is, the dimensions. Of course we have an example for you! Let’s say you are purchasing a PIG Custom Collapse-A-Tainer Containment System and you need 100% containment for a 520-gallon tank that has dimensions of 5.5’L x 3.5’W. Using the formula above, you can easily figure out the dimensions your Collapse-A-Tainer has to be in order to have enough sump capacity. Two things to remember: 1) 2’ is the highest side for a Collapse-A-Tainer, and 2) you need to leave extra room around the container you are surrounding. Still with us? For our example, you’d have to solve for L and W:

L x W x 2’H x 7.48 = 520 gallons

L x W x 14.96 = 520 gallons

L x W = 520/14.96 = 34.76

So any combination of L x W >= 34.76, where L > 5.5 and W > 3.5 will work. Let’s use L=7’ and W=5’.

7’L x 5’W x 2’H x 7.48 = 523.6 gallons

Now that you know how much secondary containment you need for your primary containers, see the PIG Spill Containment Pallets and Decks and PIG SpillBlockers, Berms, Dikes and Barriers to protect your facility and the environment from releases.