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The ability to calculate the volume of a space is critical to many applications in the field of well drilling, well abandonment / grouting, water pumping, and water treatment. Sometimes this information is needed to estimate the yield of the well, the well specific capacity, to properly purge the well, and to conduct chemical treatments to the well, such as: shock well disinfection.

First, a water well and a borehole are approximately the same general shape, i.e., cylinders. The volume of the cylinder is a function of the length of the cylinder and the area of the cylinder and the area of the cylinder is a function of the diameter of the cylinder and by default its radius.

What are we doing? We are calculating the volume of a wellbore or borehole and we not calculating that amount of water in the wellbore or borehole, we are calculating the total amount of volume that could be present.

What is the depth of the borehole or wellbore? 500 feet (H) tog (top of ground or grade)

Water is the diameter of the borehole or wellbore? 8 inches Diameter (D)

Volume (V) of a Cylinder is given by the following equation: V = πr^{2}H

π = pi is a the mathematical constant that is the ratio of the circumference of a circle to its diameter and it is equivalent to approximately 3.14159… (“Pi is an irrational number. An irrational number does not have a repeating pattern and does not end”).

H = 500 feet (given)

r = radius of the cylinder, which is equivalent to D/2, so if D equals 8 inches, r = 8/2 or 4 inches.

r ^{2} = r * r = 4 inches * 4 inches = 16 square inches (in^{2})

Therefore, all we need to do to calculate Volume is multiple π * 16 * 500 (RIGHT?) NO!

The units are not the same!

So we will convert inches into feet. The radius of the well is 4 inches. To convert to feet, we divide this number by 12, because there are 12 inches in 1 foot. Therefore, 4 inches equals 0.33 feet.

The Volume of the Borehole is

V = 3.14159 * 0.33 ft * 0.33 ft * 500 ft = 171.05 ft ^{3}(cubic feet)

1 cubic foot can contain or hold 7.480519 gallons of freshwater with a density of 1 g/cm^{3}

Therefore, the wellbore has a volume equivalent to:

Volume (gallons) = 171.05 * 7.480519 gal/ft = 1279.6 gallons

Short-Hand:

8 inch well that is 500 feet deep could contain or hold, if it did not leak, 1279.6 gallons of water or (1279.6/500) = 2.559 gallons per foot.

8 inch well that is 500 feet deep contains 171.05 ft ^{3} or (171.05/500) = 0.3412 cubic foot per foot of borehole.

Now, if after drilling the well before we return a few weeks later and found that the borehole had water at 50 feet from the top of the casing (toc) , how much water would we have to remove from the hole to purge 1 borehole volume?

The top of the well casing is 2 feet above current grade and since the well from the top of the original grade was 500 feet (tog), this means that water the actual depth below grade is only 8 feet. Therefore, there is 492 feet of water in the borehole.

492 feet of water * 2.559 gallons per foot = 1259.0 gallons (We need a pump)

For this project, we drilled a 15-inch borehole to a depth of 100 feet and then installed an 8 -inch diameter steel casing in the center of this borehole, the local agency is requiring the project to grout the annular space with neat cement grout. The annular space is the space between the 8-inch steel casing and the 15-inch borehole wall, how much grout might we use or need?

**Step 1:**

What is the volume (cubic feet) per foot for a 15-inch diameter borehole?

D = 15-inch, r = 7.5 inch or 0.625 feet

V = 3.14159 * 0.625 ft * 0.625 ft * 1 ft = 1.227 cubic per foot (15-inch diameter borehole)

Since we know the volume per unit foot for the 8-inch borehole is 0.3412 cubic foot, the difference and amount of volume that needs to be grouted between the borehole and the steel casing is (1.227 – 0.3412 cubic foot) multiplied by the depth of 100 feet or a total of 88.58 cubic feet of grout.

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