Upward Leakage through Micro-Annulus of Injection Well
Art Palmer
The following paragraphs show how to estimate the rate of upward flow from the bottom
of a cased injection well back to the surface along a micro-annulus (for example, where
the casing has separated from the surrounding cement).
Assume laminar flow, which is most likely. Turbulent flow may develop later as the
opening enlarges by dissolution, erosion, or other stress, but its flow rate depends on
too many specific conditions to consider here. At the onset of turbulence the flow
actually becomes less efficient, even though its rate continues to increase gradually.
Variables (see diagram on next page):
v = flow velocity (cm/sec)
Q = discharge (cm3/sec) – this is what we want to calculate.
w = narrow dimension of micro-annulus (cm).
c = length of gap (cm) around the curved edge of the casing (example: if half the casing
has separated, c = half the circumference of the circle).
A = cross-sectional area of flow (cm2) = cw.
γ = specific weight of fluid (dynes/cm3) = density (ρ, grams/cm3) × gravitational field
strength (g, = 980 cm/sec2 = dynes/gram). Density of fluid is close to 1.00 and
does not vary significantly with temperature (for example, ρ rounds off to 1.00
between 0°C and 50°C). If the fluid density is not close to that of pure water,
here is the opportunity to change it.
μ = viscosity of fluid (dyne-sec/cm2). This varies a lot with temperature and type of fluid.
For water it is 0.0131 at 10°C, 0.0100 at 20°C, 0.00801 at 30°C, 0.00656 at
40°C, and 0.00549 at 50°C.
P = fluid pressure (dynes/cm2). Consider atmospheric pressure to be zero, since it is
essentially uniform. Pressure is artificially increased at the well head to force fluid
downward. This increases the pressure at the bottom of the well and in the
surrounding rocks. To convert raw pressure into hydraulic head, divide by γ.
Dwell = total depth of well (cm). We’re trying to keep the units as simple as possible,
even though the numbers may be quite large.
Dwater = depth of water in well (cm).
Z = elevation (cm). Use the top of the well as the datum (Z = 0). The bottom of the well
has a negative elevation = –D.
h = hydraulic head (cm) = elevation (Z) + pressure head (P ∕ γ) + velocity head (v2 ∕ 2g).
Velocity head is negligible in this situation so can be ignored.
I = hydraulic gradient (difference in head / flow distance) = cm/cm, so no dimensions.
[See next page for diagram, and last page for a detailed example.]
Calculations
Discharge = Q = vA (cm3/sec).
For laminar flow through the annulus, v = (w2 γ I) ∕ (12μ) cm/sec, and A = cw.
Therefore, Q = (cw3 γ I) ∕ (12μ)] cm3/sec.
Consider the elevation at the top of the well to be zero. The atmospheric pressure is
considered zero too. Artificial pressure (P) is applied to the well bore (dynes/cm2). The
pressure head at the bottom of the well is a combination of P/γ cm from the artificial
pressure, plus the additional P/γ caused by the weight of the fluid column, which simply
equals the water depth, Dwater, which typically = Dwell. The elevation at the bottom of well
= –Dwell. Head (h) at the bottom of the well = artificial P∕γ + Dwater – Dwell (cm).
The top of the well is the outlet for leakage through annulus, at zero pressure and zero
elevation), so head at top of well = 0 cm. So the difference in head = (artificial P∕γ +
Dwater – Dwell) – 0.
In this problem, the flow distance is the simply the well depth = Dwell. The hydraulic
gradient (I = head difference / flow distance = (artificial P∕γ + Dwater – Dwell) ∕ Dwell.
Example
A pressure of 2500 psi (lb/in2) is applied to the top of an injection well. The well is 3000
feet deep and is filled with water to the same depth during injection. These are typical
field units and need to be converted to cgs units for convenience. The water
temperature is 20°C, so its viscosity = 0.010 dyne-sec/cm2.
Dwell = 3000 ft x 30.48 cm/ft = 91,440 cm. Dwater = same.
Artificial P = 2500 psi x 68,944 dynes/cm2 per psi = 1.72 x 108 dynes/cm2.
Artificial pressure head = P ∕ γ = 1.72 x 108 dynes/cm2 / 980 dynes/cm3 = 175,500 cm.
Natural pressure head, from weight of water at bottom of well = Dwater = 91,440 cm
Elevation of bottom of well relative to datum at surface = –Dwell = – 91,440 cm.
Head (h) at bottom of well = artificial P ∕ γ + Dwater – Dwell = 175,500 + 91,440 –
91,440 cm = 175,500 cm. Head at top of well (top of annulus) = 0 cm.
I = hydraulic gradient = head difference (bottom vs. top) divided by flow distance
(Dwell) = (175,500 cm – 0) / (91,440 cm) = 1.92 (cm/cm, i.e., no dimensions).
Suppose that the micro-annulus is measured to have w = 0.081 cm and c = 14 cm.
Calculation of leakage velocity through annulus:
v = (w2 γ I) ∕ (12μ) = (0.081)2 (980)(1.92) ∕ [(12) (0.010)] = 102.9 cm/sec.
Q = vA = vwc = (102.9) (0.081) (14) = 116.7 cm3/sec.
116.7 cm3/sec = 0.117 liters/sec x 15.85 gal/min per L/sec = 1.85 gallons per
minute.
Testing for laminar flow: This flow equation is valid only for laminar flow, where the
Reynolds Number (Re) = (fluid density x fluid velocity x fissure width) / (viscosity) >
1100 in narrow fissures. In the above example, Re = (1)(102.9)(0.081)/0.010 = 833.5,
which is in the laminar flow range. Beyond 1100 the laminar flow begins to break down
into turbulent flow, for which other more complex equations are needed. It’s clear that
even the tiny aperture described here can yield a seriously large amount of leakage, so
anything beyond laminar flow would pose a very serious problem.
While injection is taking place, the pressure will be highest at the well bottom and in its
distributaries and will disperse outward into the surrounding rocks. When injection
ceases, the pressure will decrease rapidly in the vicinity of well, but will continue to
increase slightly in surrounding areas for a while, then finally disperse and return more
or less to the original pressure distribution.
Final note: Measuring the overall size of the micro-annulus is actually impossible.
Even if it can be measured at the top of the well, the dimensions farther down are
unknown. Arbitrary numbers can be inserted into the computation to show the
effect of any given annulus dimensions. Alternatively, measured leakage rates
can be used to estimate the effective dimensions.