The pressure drop over a distance, L, of single-phase flow in a pipe can be obtained by solving the mechanical energy balance equation, which in differential form is
If the fluid is incompressible (ρ = constant), and there is no shaft work device in the pipeline (a pump, compressor, turbine, etc.), this equation is readily integrated to yield
for fluid moving from position 1 to position 2. The three terms on the right-hand side are the potential energy, kinetic energy, and frictional contributions to the overall pressure drop, or
7.2.3.1. ΔpPE, the Pressure Drop Due to Potential Energy Change
ΔpPE accounts for the pressure change due to the weight of the column of fluid (the hydrostatic head); it will be zero for flow in a horizontal pipe. From Equation (7-15), the potential energy pressure drop is given by
In this equation, Δz is the difference in elevation between positions 1 and 2, with z increasing upward. θ is defined as the angle between horizontal and the direction of flow. Thus, θ is +90° for upward, vertical flow, 0° for horizontal flow, and –90° for downward flow in a vertical well (Figure 7-4). For flow in a straight pipe of length L with flow direction θ,
Figure 7-4. Flow geometry for pipe flow.
Example 7-2. Calculation of the Potential Energy Pressure Drop
Suppose that 1000 bbl/d of brine (γw = 1.05) is being injected through 2 7/8-in., 8.6-lbm/ft tubing in a well that is deviated 50° from vertical. Calculate the pressure drop over 1000 ft of tubing due to the potential energy change.
Solution
Combining Equations (7-17) and (7-18),
For downward flow in a well deviated 50° from vertical, the flow direction is –40° from horizontal, so θ is −40°. Converting to oilfield units, ρ = (1.05)(62.4) lbm/ft3 = 65.5 lbm/ft3 and ΔpPE = −292 psi from Equation (7-19).
Shortcut Solution
For fresh water with γw = 1(ρ = 62.4 lbm/ft3), the potential energy pressure drop per foot of vertical distance is
For a fluid of any other specific gravity,
where γw is the specific gravity. Thus,
For the example given, γw = 1.05 and Δz = L sin θ, so ΔpPE = 0.433 γwL sin θ = −292 psi.
7.2.3.2. ΔpKE, the Pressure Drop Due to Kinetic Energy Change
ΔpKE is the pressure drop resulting from a change in the velocity of the fluid between positions 1 and 2.
It will be zero for an incompressible fluid unless the cross-sectional area of the pipe is different at the two positions of interest. From Equation (7-15),
or
If the fluid is incompressible, the volumetric flow rate is constant. The velocity then varies only with the cross-sectional area of the pipe. Thus,
and since A = πD2/4, then
Combining Equations (7-24) and (7-26), the kinetic energy pressure drop due to a pipe diameter change for an incompressible fluid is
Example 7-3. Calculation of the Kinetic Energy Pressure Drop
Suppose that 2000 bbl/d of oil with a density of 58 lbm/ft3 is flowing through a horizontal pipeline having a diameter reduction from 4 in. to 2 in., as illustrated in Figure 7-5. Calculate the kinetic energy pressure drop caused by the diameter change.
Figure 7-5. Flow with a reduction in pipe size (Example 7-3).
Solution
Equation (7-27) can be used if the fluid is incompressible. First, the volumetric flow rate must be converted to ft3/sec:
and from Equation (7-27),
For oilfield units of bbl/d for flow rate, lbm/ft3 for density, and in. for diameter, the constants in Equation (7-27) and unit conversions can be combined to yield
where q is in bbl/d, ρ in lbm/ft3, and D in in.
7.2.3.3. ΔpF, The Frictional Pressure Drop
The frictional pressure drop is obtained from the Fanning equation,
where ff is the Fanning friction factor. In laminar flow, the friction factor is a simple function of the Reynolds number,
whereas in turbulent flow, the friction factor may depend on both the Reynolds number and the relative pipe roughness, ε. The relative roughness is a measure of the size of surface features on the pipe wall protruding into the flow stream compared with the pipe diameter, or
where k is the length of the protrusions on the pipe wall. The relative roughness of some common types of pipes are given in Figure 7-6 (Moody, 1944). However, it should be remembered that the pipe
roughness may change with service, so that the relative roughness is essentially an empirical parameter that can be obtained through pressure-drop measurements.
Figure 7-6. Relative roughness of common piping material. (From Moody, 1944.)
The Fanning friction factor is most commonly obtained from the Moody friction factor chart (Figure 7-7; Moody, 1944). This chart was generated from the Colebrook-White equation,
Figure 7-7. Moody friction factor diagram. (From Moody, 1944.)
The Colebrook-White equation is implicit in ff , requiring an iterative procedure, such as the Newton-Raphson method, for solution. An explicit equation for the friction factor with similar accuracy to the Colebrook-White equation (Gregory and Fogarasi, 1985) is the Chen equation (Chen, 1979):
Example 7-4. Calculating the Frictional Pressure Drop
Calculate the frictional pressure drop for the 1000 bbl/d of brine injection described in Example 7-2.
The brine has a viscosity of 1.2 cp, and the pipe relative roughness is 0.001.
Solution
First, the Reynolds number must be calculated to determine if the flow is laminar or turbulent. Using Equation (7-7),
Note that the oilfield units used here are obviously not consistent; however, the constant 1.48 converts
the units to a consistent set. The Reynolds number is well above 2100, so the flow is turbulent. Either the Moody diagram (Figure 7-7) or the Chen equation [Equation (7-35)] can be used to determine the friction factor. Using the Chen equation,
and
Now using Equation (7-31), and noting that 2 7/8-in., 8.6-lbm/ft tubing has an I.D. of 2.259 in.,
then,
Notice that the frictional pressure drop is considerably less than the potential energy or hydrostatic pressure drop, which we calculated to be –292 psi in Example 7-2.