Lecture 2 Single Phase Flow Concepts
Wellbore Performance Wellbore performance analysis involves establishing a relationship between tubular size wellhead and bottom-hole pressure fluid properties fluid production rate
Single-Phase Liquid Flow Single-phase liquid flow exists in an oil well only when the wellhead pressure is above the bubble-point pressure of the oil, which is usually not a reality. This is just a start point. Multiphase flow usually dominates.
Conservation of Momentum The rate of momentum out, minus the rate of momentum in, plus the rate of momentum accumulation in a given pipe segment must equal the sum of all forces on the fluids.
Conservation of Momentum Conservation of mass
Flow to the Surface - Liquids Taken as a macroscopic balance the changes at two selected points can be examined
Conservation of Momentum
Flow to the Surface - Liquids Flow of slightly compressible liquids is described by the steady state mechanical energy balance Steady state means that no properties are changing with time.
Kinetic Energy Term (KE) (acc) The change in kinetic energy is the energy change brought about by a change in velocity between points 1 & 2. For turbulent flow this is expressed as Velocity is related to volumetric throughput or flux by
Potential Energy Term (PE) (el) The change in potential energy is the change brought about by elevation change between point 1 & 2. The flow may be inclined at an angle, or be horizontal.
Friction Frictional energy loss is a function of velocity, viscosity, density, pipe size, & condition. Correlations developed by Moody are widely used to obtain a Fanning friction factor based on Reynolds Number Fanning friction factor Moody friction factor
Moody Friction Factor
Moody Friction Factor
Pipe Roughness
Fanning Friction Factor Many correlations exist. A good one is presented by Chen Fanning friction factor = (1/4) Moody friction factor
Friction in Fittings Friction loss in fittings is negligible in flow though long lines. However it may be important in short distance applications. If so it is added to the friction term as
Friction in Fittings
Mechanical Energy Balance for Liquids Note the condition of no flow. Calc an example.
Example 1 Calculate the pressure change in a water injection well. The following data are known.
Example 2 Suppose that 1,000 bbl/day of 40 API, 1.2 cp oil is being produced through 2 7⁄8-in., 8.6 lbm/ft tubing in a well that is 15 degrees from vertical. If the tubing wall relative roughness is 0.001, calculate the pressure drop over 1,000 ft of tubing.
Liquid Pressure Traverse A pressure traverse is a plot of the flowing pressure versus position in the pipe. For the case of a slightly compressible liquid the plot is a straight line from P1 to P2 across L1 to L2. You can verify this result by breaking the flow length into increments and computing and plotting the end point of each increment.
Production Well Example
Single-Phase Gas Flow The first law of thermodynamics (conservation of energy) governs gas flow in tubing. The effect of kinetic energy change is negligible because the variation in tubing diameter is insignificant in most gas wells. With no shaft work device installed along the tubing string, the first law of thermodynamics yields the following mechanical balance equation:
Single-Phase Gas Flow Since dz is Sin(θ).dL The gas density from the gas law can be expressed as The velocity can be written in terms of volumetric flow rate at the standard condition as
Main Equation for Gas Flow in Wellbores
Solution This equation contains Three variables that are functions of position Z (compressibility factor) Temperature Pressure A simple approach is to use average values of temperature and Z over a segment of the pipe. There are two options for temperature in a segment between points 1 and 2
Method to Calculate P in a Segment T_average can be obtained from T1 and T2 Z_average is calculated from T_average and P1 In the segment, we can calculate P2 Z_average can be updated with new P in the segment which is (P1+P2)/2 We repeat the calculate until convergence. P1, T1 P2, T2
Solution For the special case of horizontal well
Oil Field Unit P1 P2
Example 3