Flow Through Channels & Fractures PDF

# Channels and Fractures in Parallel Only the matrix permeability has been discussed in the analysis to this point. In some sand and carbonate reservoirs the formation frequently contains solution channels and natural or artificial fractures. These channels and fractures do not change the permeabil...

# Channels and Fractures in Parallel Only the matrix permeability has been discussed in the analysis to this point. In some sand and carbonate reservoirs the formation frequently contains solution channels and natural or artificial fractures. These channels and fractures do not change the permeability of the matrix but do change the effective permeability of the flow network. In order to determine the contribution made by a fracture or channel to the total conductivity of the system, it is necessary to express their conductivity in terms of the darcy. ## Channels Recalling Poiseuille's equation for fluid conductivity of capillary tubes, $ Q = \frac{\pi r^4 \Delta P}{ 8 \mu L} \\ $ The total area available to flow is $ A = \pi r^2\\ $ So that the equation reduces to $ Q = \frac{A \Delta P}{ 8 Lμ} \\ $ From Darcy's law it is also known that $ \frac{\Delta P}{Lμ} = Ak \\ $ Equating Darcy's and Poiseuille's equations for fluid flow in a tube, $ k = \frac{r^2}{8} \\ $ where k and r are in consistent units. If r is in centimeters, then k in darcys is given by $ k = \frac{r^2}{8(9.869)(10^{-9})} = 12.50 \times 10^6 r^2 \\ $ where 9.869 × 10^-9 is a conversion factor Then if r is in inches, $ k = 12.50 \times 10^6(2.54)^2r^2 \\ $ $ = 80 \times 10^6 = 20 \times 10^6d^2\\ $ where d is the diameter of the opening in inches. Therefore, the permeability of a circular opening 0.005 in. in radius is 2,000 darcys or 2,000,000 millidarcys. ## Consider a cube of reservoir rock Consider a cube of reservoir rock 1 ft on the side and having a matrix permeability of 10 millidarcys. If a liquid of 1-centipoise viscosity flows linearly through the rock, under a pressure of gradient 1 psi per ft, the rate of flow will be 0.011271 bbl per day. If a circular opening 0.01 in. in diameter traverses the same rock, then the rate of flow can be considered to be the above value plus the rate of flow Q' through the circular opening. Then $ Q' = 1.1271\frac{kA \Delta P}{\mu L}\\ $ $ = 1.1271(2,000)\frac{\frac{\pi d^2}{4}(1)}{1(1)}\\ $ $ = 0.785(2254.2)\frac{ (0.0001)}{144}\\ $ $ = 0.785(0.22542)\frac{1}{144}\\ $ $ = 0.00122 \text{ bbl/day}\\ $ Therefore the combined rate is 0.012491, or an increase of about 11 percent. If the cube matrix has 1-millidarcy permeability, the increase would be 108 per cent. ## 1. Case of circular hole = 0.01" dia *Qt=0.00122+ 0.011271 bbl/day = 0.012491 bbl/day* *% increase =( 0.00127 / 0.0122) × 100 = 10.82% ~ 11% [For K=10 md] % increase = 108% if Kmatrix= 1md* ## Fractures For flow through slots of fine clearances and unit width Buckingham reports that $ \Delta P = \frac{12\muυL}{h^2}\\ $ where h is the thickness of the slot. By analogy to Darcy's law where $ \Delta P = \frac{\mu\υL}{k}\\ $ then $ k = \frac{h^2}{12}\\ $ where h is in centimeters and k in darcys. The permeability of the slot is given by $ k = \frac{h^2}{12(9.869)(10^{-9})} \\ $ When h is in inches and k is in darcys, $ k = 84.4 \times 10^6h^2 \\ $ $ k = 54.4 \times 10^6h^2 \\ $ The permeability of a fracture 0.01 in. in thickness would be 5,440 darcys or 5,440,000 millidarcys. Under the same flowing conditions used for the circular opening, a fracture 0.01 in. in thickness across the width of the block would contribute a flow rate Q", computed as $ Q" = 1.1271 \frac{kA \Delta P}{\mu L} \\ $ $ = 1.1271(5,440)\frac{ (0.01/12)(1)(1)}{1(1)} \\ $ $ = 1.1271(5,440)(0.00083)\\ $ $ = 5.1095 \text{ bbl/day} \\ $ The combined rate is 5.12077 bbl per day, or an increase of 45,433 percent. It is obvious that in situ fractures and solution cavities contribute substantially to the productivity of any reservoir. ## 2. Case of a Linear fracture *Qt = Q" + Q Qt= 5.1095+0.011271 bbl/d = 5.120771 bbl/d* *% increase = 5.1095 / 5.01127 × 100 = 45333% against circular hole case of 11%*