Separator Jetting – Movement of Solids in Separators (B-FSM-114)

How then do we calculate the amount and size of solids that settle in the production separator?

As a first-order basis I use a mechanistic approximation detailed as follows.

Calculate residence time of liquid (water) in vessel (weir height and inlet to weir distance)
Calculate fall (capture) velocity as vertical distance divided by residence time
Use Stokes Law to calculate size of particle matching capture velocity
Particles with diameter greater than capture size are collected within the separator – smaller particles pass through with produced water
Sum the total of all particles greater than cut off size gives total amount of solids settled in separator

An example is provided as detailed below.

First calculate the cut-off size of solids that is captured in the vessel:

Water volume in vessel = ½(π/4)d²L = 0.393(2.5)²(8)=19.63m³
Water flow rate = 0.0828 m³/s, ∴ RT=V/Q=19.63/0.0828=237 sec=3.95 min
Horizontal velocity = Q/A = 0.0828m³/2.45m² = 0.0336 m/s
What size particle falls in 1.25m/237s = 0.00527 m/s?
Stokes law: 𝒗_𝑽=(𝒈𝒅^𝟐 (𝝆_𝒔−𝝆_𝒍 ))/𝟏𝟖𝝁→𝐝=√(𝟏𝟖𝝁𝒗/𝒈(𝝆_𝒔−𝝆_𝒍 ) ) = 71 micron
Particles greater than ~70 microns captured in separator, with the rest passing into the produced water treatment system

The second step is to calculate the amount of captured solids in the vessel.

Assume all solids are in the water phase at 100 ppmv
Qw=45,000 BPD, ∴ Qs=4.5 BPD=0.715 m³/d=1896 kg/d
Calculate total weight of particles greater than capture size (use a spreadsheet)

References:

Rawlins, C.H., “Design of a Cyclonic Solids Jetting Device and Slurry Transport System for Production Systems”, paper 166118, presented at the SPE Annual Technical Conference and Exhibition, New Orleans, LA, 30 September – 2 October, 2013. https://doi.org/10.2118/166118-MS

Next week I will cover the behavior of settled solids.