## Liquid Desander – Example Calculations for Desander Sizing (B-FSM059)

In this article I’ll plan to merge the information from posts B-FSM-054 through -058 in order to provide detailed examples of how to use the information previously detailed. The example calculations below are taken from exercise #3 which is worked through in my Facilities Sand Management training course.

A desander with the following characteristics is operating on a well with the following characteristics:

- Nominal diameter: 10 inch
- Solids concentration: 120 ppmv
- Pressure drop: 18 psi
- Solid, oil, and water density: 2710 kg/m³, 910 kg/m³, and 1050 kg/m³, respectively
- Watercut: 78%
- Oil and water viscosity: 13.59 cP and 0.82 cP, respectively

- What is the separation size (D98) for the desander in this operation?

Determine Base D_{50}, corrections, then D_{98}:

- D
_{50(base)}= 24 microns (from chart in post B-FSM-056) - Inlet volume % solids ~0%, C1 = 1.0
- Pressure drop = 18 psi, C2 = 0.83
- G. solids corrected to water
- Liquid mixture density = 0.78*1050+0.22*910 = 1019 kg/m
^{3}(1.019 s.g.) - Effective solids density = 2710*1000/1019 = 2659), C3 ~ 1.0

- Liquid mixture density = 0.78*1050+0.22*910 = 1019 kg/m
- Liquid mixture viscosity (0.78*0.82+0.22*13.59 = 3.62 cP), C4 = 1.91
- D
_{50}= 24*1.0*0.83*1.0*1.91 = 38 microns - D
_{98}~ 2*D_{50}= 76 microns (assume a = 4 from post B-FSM-050)

- The flow rate drops such that the desander now operates at 4 psi pressure drop. What is resulting D98?

Perform same calculation with new pressure drop

- D
_{50(base)}, C1, C3, and C4 are the same - C
_{2}= 1.35 - D
_{50}= 62 microns, D_{98}= 124 microns

- A well is added to the flow that reduces the watercut to 18% and brings pressure drop back to 18 psi. What is the resulting D98?

Repeat calculation with revised watercut and original pressure drop

- D
_{50(base)}, C_{1}, C_{2}are the same - G. solids corrected to water
- Liquid density = 0.18*1050+0.82*910 = 935 kg/m
^{3} - Effective solids density = 2710*1000/935 = 2898), C
_{3}= 0.9

- Liquid density = 0.18*1050+0.82*910 = 935 kg/m
- Liquid viscosity (0.18*0.82+0.82*13.59 = 11.29 cP), C
_{4}= 3.36 - D
_{50}= 60 microns, D_{98}= 120 microns

- Another well is added in that changes the mixture to 25% watercut and 63% gas void fraction (gas actual density and viscosity is 38 kg/m³ and 0.011 cP, respectively). Assuming same pressure drop, what is the resulting D98?

(Note: we have not covered multiphase flow yet – that is the topic of module M5 to be covered in the future. However you can do these calculations with what you know already. Just realize that free gas greatly reduces the effective fluid density and viscosity).

Repeat calculation with multiphase flow (new viscosity and density)

- D
_{50(base)}, C_{1}, C_{2}are the same - Calculate multiphase mixture density and correction:
- Liquid density = 0.25*1050+0.75*910 = 945 kg/m
^{3} - Mixture density = 0.37*945+0.63*38 = 374 kg/m
^{3} - G. solids corrected to water = 2710*1000/374 = 7253, C
_{3}= 0.5

- Liquid density = 0.25*1050+0.75*910 = 945 kg/m
- Calculate multiphase mixture viscosity and correction:
- Liquid viscosity = 0.25*0.82+0.75*13.59 = 10.4 cP
- Mixture viscosity = 0.37*10.4+0.63*0.011 = 3.85 cP
- C
_{4}= 1.96

- D
_{50}= 24*1.0*0.83*0.5*1.96 = 19.5 microns, D_{98}= 39 microns

Free gas significantly improves performance of desander and this effect will be covered in detail in future posts.

The next article will start the discussion of hydraulic throughput model(s).

**References**

- Rawlins, C.H., and Wang, I. I. 2001. “Design and Installation of a Sand Separation and Handling System for a Gulf of Mexico Oil Production Facility,” SPE Production and Facilities, paper 72999, Vol. 16, No. 3, pp. 134-140.