The production rate of a well can be described using scientific principles which include Darcy’s law and the conservation of mass 1Matthews, C. S., & Russell, D. G. (1967). Pressure buildup and flow tests in wells (Vol. 1, p. 27). New York: Henry L. Doherty Memorial Fund of AIME.. A simple analytical equation can be derived to describe flow into a vertical wellbore from a horizontally layered reservoir by making some mathematical and physics-based simplifications, including the following:
- the reservoir properties, such as permeability and porosity, are constant (don’t vary with time), homogeneous (the same everywhere within the reservoir) and isotropic (the same magnitude in all directions)
- the fluid is a single phase oil that is slightly compressible, described with
the following properties
- a Newtonian viscosity of \(\mu_o\) (subscript o stands for oil), which describes the flow resistance of the fluid (high viscosity is “thick like molasses” and low viscosity is “thin like water”)
- a formation volume factor of \(B_o\), which accounts for the volume change of the oil from downhole reservoir conditions (quantified in reservoir barrels, RB) to surface, stock tank conditions (quantified in stock tank barrels, STB)
- the wellbore is vertical, with a fixed radius, \(r_w\), and fully penetrates the entire thickness of the reservoir
- the reservoir geometry consists of a uniform thickness and either a circular or square drainage area of finite extent
- the wellbore is located in the center of the reservoir, so flow moves radially from the reservoir into the wellbore
- the flow is steady state, which means there are no changes with time (pressures and rates do not change)
Under the above idealized conditions, the radial inflow equation for an oil well using common oilfield units is given by:
where
- \(q_{oil}\) is the oil production rate in STB/day,
- \(k\) is the reservoir permeability in md (millidarcies),
- \(h\) is the reservoir thickness in ft,
- \(\Delta P = (P_e - P_w)\), the pressure difference in psi between the fluid pressure in the formation at the reservoir boundary, \(P_e\), and in the wellbore, \(P_w\) (\(\Delta P\) is often called drawdown pressure in a production well),
- \(\mu_o\) is the fluid viscosity in cp (centipoise),
- \(B_o\) is the formation volume factor in STB/RB,
- \(r_e\) is the radius to the reservoir boundary in ft,
- \(r_w\) is the radius of the wellbore in ft,
- and 141.2 is the constant to reconcile the equation to the chosen units.
This flow rate equation, although highly idealized, is well suited for quick ballpark calculations of well performance that can be utilized for demonstration purposes and sensitivity analyses, among other things. As we move through this lesson, we will introduce slightly more comprehensive versions that can address more complex downhole conditions.
Equation 2.1 shows that production rate is directly proportional to formation permeability, reservoir thickness, and the pressure drawdown of the well. Production rate is inversely proportional to fluid viscosity, formation volume factor and the ratio of the boundary radius to the well radius. The only properties in this equation that can be routinely controlled by the operator are \(\Delta P\) (through the wellbore pressure, \(P_w\)) and the ratio of \(r_e / r_w\) (through the wellbore radius, \(r_w\)). Lowering the wellbore pressure increases \(\Delta P\) which will increase flow rate. Increasing the radius of the wellbore will also increase the rate. The rest of the parameters in equation 2.1 are fixed natural properties of the subsurface, with the possible exception of the oil viscosity. Oil viscosity can be reduced by injecting gas or heat into the reservoir during enhanced oil recovery (EOR), which increases production rate.
Example Calculation:
In conventional reservoirs, a common drainage area size (also known as well spacing) is 40 acres, which corresponds to a circle with a radius of 745 ft or a square with sides of half-dimension 660 ft (we are assuming a circular drainage area for this example). If the reservoir fluid is normally pressured, it is defined to have a gradient of 0.44 psi/ft (overpressure and underpressure are also possible). Assuming a reservoir depth of 5,000 ft with a normal pressure gradient, the reservoir pressure is \(P_{rsv} = (5{,}000\ \text{ft}) \times (0.44\ \text{psi/ft}) = 2{,}200\ \text{psi}\). Other reasonable values for a conventional oil well are the following:
- reservoir thickness, \(h = 50\) ft
- permeability, \(k = 100\) md
- viscosity, \(\mu_o = 1\) cp
- wellbore pressure, \(P_w = 500\) psi
- wellbore radius, \(r_w = 0.25\) ft
- formation volume factor, \(B_o = 1.3\) RB/STB
The assumption of steady state flow (unchanging with time) is an assumption made to simplify the math for equation 2.1, but it is not an unreasonable assumption for oilfields undergoing waterflooding. During waterflooding, the amount of liquid produced (oil and water) out of production wells is typically balanced by the amount of water injected at injection wells, which maintains reservoir pressure at a constant value, fixing \(\Delta P\) as constant and thus sustaining a constant rate.
Given the above specified parameters, calculate the flow rate in STB/day for this conventional well.
Answer:
Example extension of this problem:
Calculate the rate given \(k = 0.1\) md, which is at the high end of the permeability range for unconventional reservoirs. Assume the same other data and conditions as the previous calculation.
Answer:
This unconventional permeability is only 1/1000th of the conventional example. Since flow rate is directly proportional to permeability, the unconventional flow rate is 1/1000th of the conventional case, or 5.8 STB/day. Such a well would be uneconomic without taking steps to enhance production through wellbore or reservoir modification. Such operations are termed stimulation, and the typical stimulation method chosen for such a well is hydraulic fracturing.
