Trouble With Your Bubbles

By Scott Baril, Director of Technical Services, Brett Admixtures

The pour is 234 yards. Your truck reached the job 3 minutes before scheduled, the slump was right on. Everything is perfect! Then the testing technician walks up and says your air is 4.6%. Problem is, the spec says 5-7 and they want 6.000% on this job. Been there?

In our mind, we quickly go over the mix design. We know we put 5 ounces of air per yard. Quick math says if 5 ounces equals 4.2%, then 1/3 more admixture should be 4.2 X 1/3 = 1.4% more air. That's 4.2 +1.4= 5.6 % right? A call to the plant, wait 3 loads, test. and air comes in at 4.8%.

You need to call your admixture supplier, there's got to be something wrong with that air. Wasn't it delivered just yesterday?

There's your answer, right? You're at 6.6 ounces now and it's still too low at 4.8%. Maybe the next increase will make it "kick-in", don't want to get too high. You go to 8 ounces per yard now. Little beads of sweat are beginning to collect on your forehead; it's only 72 degrees out!

The truck finally arrives and you stand over the tester with your fingers crossed as load number 7 is being tested. FIVE POINT THREE. Five point three? Finally within spec. but still low.

A call to the plant and 3 loads from now you'll know what 10 ounces per yard will do. That will be load number 10.

You know those days. If only there was a better way. Well, there is. Remember that you're only adding some air to what's already there (Let's call it "Base Air") to reach a total. There is entrapped air that is in all concrete. A typical value would be 1.2%. This is largely influenced by the shape and gradation of the aggregate. Then there is entrained air present before you ever add air-entraining solution. This is largely influenced by the water reducers being utilized along with the brand and type of cement. This value can and does vary with each brand of cement. Sometimes it varies with each load of cement. A typical value for entrained air would be also 1.2%. This means that a load of non-air concrete would contain 2.4% when tested.

Let's do some math on the mix we wrote of above at 5 oz. of admixture:

Tested Air Percent = 4.2 Less the

"Base Air" Percent of 2.4 Amount of air entrained by the air-entraining solution = 1.8%

The mix had 5 ounces per yard, so 1.8% divided by 5 ounces equals .36% air for every ounce we added. We will use 0.36 % air per oz. for the following calculations:

When we went to 6.6 ounces, we got 6.6 times .36 = 2.4% entrained air.

"Base Air" = 2.4 %

So, Total Air = 4.8 %

When we went to 8 ounces: 8 times .36% equals 2.9%

"Base Air" % 2.4 %

So, Total Air = 5.3 %

Many times I'm called with situations such as these. If we understand the math behind the dosage, many problems could be avoided. This math works exactly the same if you already work in ounces per hundredweight of cementitious materials (which is the best way).

Start by checking your air on non-air loads to find your "Base Air". Practice the math. Once in a while, the math doesn't work, then you actually do have a problem.

Did you figure what the air would be at 10 ounces from above problem? It was 6.0 %.

Air entrainment needed above base air: 6.0% total, minus 2.4% base air, equals 3.6 % needed.

3.6% divided by .36% entrained by each ounce equals 10 ounces.

Note: The above was a simplified example how air entrainment can be optimized under a very specific set of conditions. The reader should note that different conditions, such as a change in temperature, water content, and/or aggregate gradation, in addition to cement type and water reducer usage, could also affect the performance of a given air entraining admixture.