Casing Cement Slurry Volume & Weight Calculation

Adding Cementing additives is one of the variables influencing the base fluid density. This density ought to be gathered for additive concentration when deciding hydrostatic pressure. This part will provide you with many terms to assist you with density calculations. After completing the Cement Slurry Volume Calculation Article, you shall understand the following:

• The relationships among density, weight, and volume
• API gravity
• specific gravity
• absolute density vs. bulk density
• absolute volume
• cement slurry weight and volume calculations

Density

Density can be characterized as the weight of a substance per unit volume. In the English system, the unit of volume is one cubic foot, and the unit of weight is one pound. In the metric system, cubic centimeter (cc) is considered the unit of volume, and we use grams as the unit of weight. If you measured the weight of a particular volume of iron, lead, and copper, you would discover they have widely various weights. Accordingly, a term is expected to refer to the weight of a unit volume of any material — such term shall be called density. For instance, a cubic foot of gold weighs 1206.2 lb, so the density of gold is 1206.2 lb/ft^3.

If you have the density and the volume of an object, you can calculate its weight:

Weight = Density × Volume

Below, you will find some densities for your reference.

For practical purposes, the densities of gases are contrasted with air at atmospheric pressure rather than with water. Using air as a similar reference, the weights of the gases mentioned above are:

Specific Gravity

Specific gravity (abbreviated SG) can be defined as the weight of a volume of substance divided by the weight of the same volume of substance taken as a standard. For example, the standard for solids and liquids is water; the standard for gases is air. Another meaning of the term specific gravity is the proportion of the density of a material to the density of water or air. The weight of the water is 8.33 lb/gal. It is a straightforward issue to convert specific gravity to weight (or vice-versa).

Problem

Calculate the specific gravity of a 12 lb/gal brine.

Solution

SG = density of substance/density of standard

So, SG = 12 Ib/gal / 8.33 lb/gal

SG = 1.44

If it was given the specific gravity of the brine as 1.44, the density can be determined as follows:

1.44 × 8.33 lb/gal = 12 lb/gal

What Is The Definition Of API Gravity For Cement Slurry Volume Calculations?

Baume’s gravity can be defined as a scale that uses salt water instead of freshwater as a reference. It is mainly applicable in refineries to estimate the gravity of acids and alkalis only. API gravity is also utilized in most other materials in the oil field. Let’s say again the standard material is water. Water’s API gravity is considered to be 10 degrees.

The relation between specific gravity and API gravity is an Inverse relationship. As one increases, the other decreases. Crude oil with a 42° API gravity weight has an equivalent specific gravity of 0.82. American crude is commonly in between 0.768 and 0.966 SG, which is equivalent to 52.6° to 10.5° API.

API readings are typically taken at a temperature of 60°F. It is important to convert the readings to 60°F to be precise whenever we take at any other temperature. You can find tables for this conversion and conversion of API gravity to specific gravity in the API Standard 2500 bulletin and various drilling engineering data handbooks ( green book, …. ).

Conversion formulas are:

Degrees API Gravity = ((141.5 / SP GR)) – 131.5

Specific Gravity = 141.5 / ( API + 131.5)

What Is The Difference Between Absolute Density vs. Bulk Density?

We generally can define absolute density as the mass per one unit of volume. Absolute density uses only the actual volume occupied by a certain substance. In addition, we can define Bulk density as the mass per one unit of bulk volume – which incorporates the actual volume of the substance and the volume of trapped “air.”

Absolute Volume Calculations in Cement Slurry

Absolute volume can be defined as the volume per one unit mass. Here is a case of absolute volume. It is assumed that there is a container (Fig. 3) which have a volume of 1 cf and measures one cubic foot. This container will be filled with ping pong balls. There are empty spaces between such balls, filled with trapped air. Let’s calculate the volume in gallons, which is occupied by such ping pong balls only.

The volume occupied by the ping pong balls can best be estimated by first determining the volume occupied by the void spaces. To do this, we could add water to the box, and it will equal the volume of added water. After obtaining such volume of water required to fill the void spaces, we can subtract it from 7.4805 gal/ft^ 3 to get the volume occupied by the ping pong balls alone.

We can call the calculated volume of ping pong balls in this way, the absolute volume. While this example is overstated, it does assist in explaining the idea. In cementing jobs, we work with sand, cement, etc., instead of ping pong balls, but we can see the balls as an amplification of sand or cement particles. Void or empty spaces exist in sand; the volume added to fracturing fluids is simply the absolute volume of the sand.

Calculating absolute volume for cement can be streamlined by utilizing the tables for “Physical Properties of Cementing Materials and Admixtures” introduced in the Technical Data section of the Halliburton Cementing Tables (Figure. 4).

You can see four columns; the left-hand (first) column records the substance, and the second one lists the bulk weight in lb per ft^3. The third one provides the specific gravity, and the last gives absolute volume in gal/lb.

Note

When using fluids’ absolute volume calculations, you won’t see a gal/lb factor in the Red Book. Meanwhile, you can have this factor if you have the fluid density in lb/gal. In other words, just divide one by the fluid density. For instance, water is 8.33 lb/gal, so gal/lb = 1 ÷ 8.33 lb/gal = 0.12 gal/lb

Problem

Calculate the absolute volume (gallons) and weight (lb per gallon) of this slurry.

• Base Fluid is water at 8.33 lb/gal
• adding 4 lb of sand.

Solution

We will use the chart below to help calculate absolute volume. First, record the materials and their densities in the first two columns. After that, open the Red Book table and record the absolute volume factors (gal/lb) in the third column.

Multiply the substances (lb) by the factor (gal/lb) to get the absolute volume and then add such values to the above table. For the totals, add the substances (lb) together and then add together the absolute volumes:

Find the density of the mixed slurry using the following formula:

Total lb ÷Total abs gal = lb/gal

12.33 lb ÷1.1824 gal = 10.427943 lb/gal

Problem

Calculate the absolute volume (gallons) and weight (lb per gallon) of the following cement slurry?

• Base Fluid is water at 8.33 lb/gal
• The additive is 3% KC1
• 4 lb of sand is added

Solution

Total lb ÷ Total abs gal = lb/gal

12.5799 lb ÷1.1934705 gal = 10.5 lb/gal

Note: Generally, when working with sand and water (or base fluid) slurries, the calculations are based on 1 gallon of base fluid and the weight of that 1 gallon of fluid.

Sample Problem

Calculate this cement slurry volume’s absolute volume (gallons) and weight (lb per gallon). Class H Cement Water at 8.33 lb/gal

Solution

For the above cementing problem, we will add another column to the chart for mixing water requirements. Proceed as in the two previous example problems until you have to fill in the mixing water requirements for the Class H cement. You can find this quantity in the Red Book Technical Data section in the table for “Water Requirements.” For Class H cement, the requirements are 4.3 gal/sk. Enter this under the appropriate column:

Enter 4.3 gal for the absolute volume of water. Divide the water factor by the absolute volume to determine the materials (lb) for water. After that, calculate the totals:

Total lb ÷Total abs gal = lb/gal

129.82 lb ÷7.9 gal = 16.4 lb/gal

Using this formula, you can find the yield of cement (ft 3 per sack)

Total abs gal ÷7.4805 gal/ft 3 = ft /sk

7.9 gal/sk* ÷7.4805 gal/ft 3 3 = 1.06 ft /sk

NOTE: When dealing with cement slurries, the three calculations are usually based on one sack of cement and the weight of that sack.

Problem

Calculate the absolute volume (gallons) and weight (lb per gallon) of this cement slurry.

• Base Fluid is water at 8.33 lb/gal
• 4 lb of sand is added.

Solution

Use the chart below to assist in the calculation of absolute volume. At the beginning, record the materials and their densities in the first two columns. After that, using the Red Book table, record the absolute volume factors (gal/lb) in the third column.

Total mixing water shall be entered under absolute gallons before totaling.

Calculate the weight of the mixed cement by using this formula:

Total Pounds ÷ Total Absolute gallons = lb/gal

Calculate the cement yield in cubic feet per sack by using this formula:

Total Absolute gal ÷ 7.4805 gal/ft3 (constant) = ft3/sk

The mixing water per sack is the sum of the gallons in the far right column