Changes

:<math>\mbox{V}_{\rm h}  = [(8.0^{2})/1029.4] \times 10{,}000 = 621.7 \mbox{ bbl}</math>

:<math>\mbox{V}_{\rm h}  = [(8.0^{2})/1029.4] \times 10{,}000 = 621.7 \mbox{ bbl}</math>

:<math>&\\\mbox{V}_{\rm a} &= [(8.0^{2} - 5.0^{2})/1029.4] \times 10{,}000 = 378.9 \mbox{ bbl}\\&\\\mbox{V}_{\rm d} &= [(5.0^{2} - 4.0^{2})/1029.4] \times 10{,}000 = 87.4 \mbox{ bbl}\\&\\\mbox{V}_{\rm c} &= [(4.0^{2})/1029.4] \times 10{,}000 = 155.4 \mbox{ bbl}</math>

:<math>\mbox{V}_{\rm a} = [(8.0^{2} - 5.0^{2})/1029.4] \times 10{,}000 = 378.9 \mbox{ bbl}</math>

:<math>\mbox{V}_{\rm d} = [(5.0^{2} - 4.0^{2})/1029.4] \times 10{,}000 = 87.4 \mbox{ bbl}</math>

:<math>\mbox{V}_{\rm c} = [(4.0^{2})/1029.4] \times 10{,}000 = 155.4 \mbox{ bbl}</math>

Although this solution is correct mathematically (that is, the sum of annular volume, displacement, and capacity equals the hole volume), the basic formula results in an underestimate for displacement of drill pipe. Since drill pipe (not collars) has tool joints that are welded onto the pipe, calculation of drill pipe displacement involves an additional step to account for the additional volume displaced by the tool joints. For most calculations with commonly used sizes and weights of drill pipe (4.5 and 5 in.), the addition of the quantity 0.001 to the barrel per foot volume calculation for displacement accounts for the additional tool joint volume with sufficient accuracy. Using the data from the previous example, we obtain the following the displacement calculation:

Although this solution is correct mathematically (that is, the sum of annular volume, displacement, and capacity equals the hole volume), the basic formula results in an underestimate for displacement of drill pipe. Since drill pipe (not collars) has tool joints that are welded onto the pipe, calculation of drill pipe displacement involves an additional step to account for the additional volume displaced by the tool joints. For most calculations with commonly used sizes and weights of drill pipe (4.5 and 5 in.), the addition of the quantity 0.001 to the barrel per foot volume calculation for displacement accounts for the additional tool joint volume with sufficient accuracy. Using the data from the previous example, we obtain the following the displacement calculation:

:<math>\mbox{V}_{\rm d}  = ([5.0^{2} - 4.0^{2})/1029.4] + 0.001) \times 10{,}000</math>

:<math>\mbox{V}_{\rm d}  = ([5.0^{2} - 4.0^{2})/1029.4] + 0.001) \times 10{,}000</math>

:<math>&=97.4 \mbox{ bbl}</math>

:<math>=97.4 \mbox{ bbl}</math>

Notice that when drill pipe tool joints are considered, the displacement is 10.0 bbl greater. Again, this additional step is not necessary when calculating the displacement of collars since collars do not have welded-on tool joints.

Notice that when drill pipe tool joints are considered, the displacement is 10.0 bbl greater. Again, this additional step is not necessary when calculating the displacement of collars since collars do not have welded-on tool joints.