An empty can, sized by a square root
knockout_sizing.py from the Souders-Brown correlation and the GPSA K table,
not quoted from a handbook page: the video and this page read the same module.
If you have walked past an empty tank in front of a compressor and wondered what it is for, this is it. It is the most boring vessel in the plant. It has no internals worth the name, no moving parts, and on a flowsheet it is one circle. It is there because a compressor is built for gas, and gas always arrives carrying liquid.
Gas squeezes. Liquid does not. A reciprocating compressor that takes a slug of liquid into the head end can bend a rod, blow a head or wreck a cylinder in a single revolution, because the piston arrives expecting something compressible and finds something that is not. A centrifugal machine trips on vibration and can shed blades and seals. The fix is not a clever machine. It is a big empty drum that slows the gas down until the droplets fall faster than it rises.
The whole design is one line, and this page works it through: the equation, the same equation for three real gases, what the mesh pad actually buys you, the residence-time table for all four vessels in the family, and the one number that changes when the pressure goes up.
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Why a slug is a mechanical event, not a process upset
A reciprocating compressor cylinder has a clearance volume at the end of the stroke, typically somewhere around a tenth of the swept volume. On a normal stroke the piston arrives, the gas in front of it compresses into that clearance, pressure rises, the discharge valve lifts and the gas leaves. The piston never touches anything solid.
Fill that clearance with liquid and the arithmetic stops working. Liquid is effectively incompressible at these pressures, so the piston has nowhere to put it. The load path that was carrying a gas pressure is now carrying a hydraulic one, and it is carried by the piston rod, the crosshead, the head bolts and the head itself. Something gives. Practitioners report bent rods, blown head gaskets and cracked cylinders, and they report them happening inside one revolution, which at 300 rpm is 200 milliseconds.
Centrifugal machines fail differently and more slowly. Liquid entering the eye erodes the blades, unbalances the rotor and drives vibration up until the machine trips. The damage is to blades, labyrinths and seals rather than to a rod. Either way the vessel in front of the machine is cheaper than the machine.
The whole design, in one line
A droplet in a rising gas stream is in a fight between drag pulling it up and weight pulling it down. Set those equal and you get the droplet's terminal velocity. Now turn it around: if the gas rises slower than the droplet falls, the droplet leaves at the bottom. So the design question is a velocity, and the answer is the Souders-Brown equation.
vmax = K √( (ρL − ρV) / ρV )
K is an empirical settling coefficient with units of velocity, and it carries everything the derivation threw away: droplet size distribution, drag coefficient, turbulence, how well the inlet is baffled. GPSA gives 0.107 m/s (0.35 ft/s) for a vertical drum with a mesh pad between 0 and 7 barg. Take the pad out and it halves to 0.0535 m/s.
Worked for air and water at 20°C, which is the case in the video:
- (998 − 1.2) / 1.2 = 830.7
- √830.7 = 28.82
- 0.107 × 28.82 = 3.08 m/s
The diameter falls straight out of it. On 2.4 actual m³/s of gas, the free area you need is 2.4 / 3.08 = 0.778 m², so D = √(4A / π) = 0.995 m, which is a 1.00 m drum. At that size the gas actually runs at 3.06 m/s, or 99% of the ceiling. Height is set by H/D of 2 to 3, so 2.5 × 1.00 = 2.50 m.
The same equation, three real gases
The equation does not change with pressure. The one number that does is ρV, the gas density, and it moves hard: from 1.2 kg/m³ at atmospheric to 35.6 kg/m³ at 50 barg. It sits under the square root, so the allowed velocity falls roughly as the root of the pressure ratio. K falls too, because GPSA fitted it against pressure on real separators.
| Service | P | ρ gas | K | √ratio | v max | D |
|---|---|---|---|---|---|---|
| air and water ambient separator |
atmospheric | 1.2 | 0.107 | 28.82 | 3.08 m/s | 1.00 m |
| steam and condensate saturated steam |
10 barg | 5.51 | 0.104 | 12.63 | 1.32 m/s | 0.71 m |
| natural gas and condensate high-pressure separator |
50 barg | 35.6 | 0.078 | 4.32 | 0.34 m/s | 0.55 m |
That is the counter-intuitive part worth taking away. High pressure does not mean a bigger knock-out drum. It means a much slower one, and a much smaller volumetric flow, and the second effect wins. The trap is the reverse: taking the atmospheric K of 0.107 m/s into a 50 barg separator undersizes it by about 37%.
Here is the K table the middle column is interpolated from:
| Pressure | K | K, US units |
|---|---|---|
| atmospheric | 0.107 m/s | 0.35 ft/s |
| 7 barg | 0.107 m/s | 0.35 ft/s |
| 14 barg | 0.101 m/s | 0.33 ft/s |
| 28 barg | 0.091 m/s | 0.3 ft/s |
| 41 barg | 0.082 m/s | 0.27 ft/s |
| 55 barg | 0.076 m/s | 0.25 ft/s |
| 69 barg | 0.070 m/s | 0.23 ft/s |
What the mesh pad is actually for
Run the terminal-velocity balance backwards and you can ask what size of droplet the empty drum is sized to stop. At 3.08 m/s, in the Newton regime with a drag coefficient around 0.44, that droplet is about 385 microns, at a Reynolds number of 78.7. Note that Stokes' law is not valid here and gives the wrong answer by a wide margin: Re is far above the Stokes limit of about one.
So an empty drum catches the big drops and nothing else. Everything finer rides out of the top with the gas, and on a wet suction line there is a lot of it. A mesh pad is a few inches of knitted wire that the fines cannot follow the streamlines around. They hit the wire, coalesce into drops big enough to fall, and drain back down. That is the whole mechanism, and it is worth exactly a factor of two on K, which is a factor of √2 on the diameter for the same duty.
It is also the part that fouls. A pad in a dirty service plugs, the pressure drop across it climbs and the velocity through the open area goes up, at which point it stops coalescing and starts re-entraining. A knock-out drum that used to work and now passes liquid is usually a pad problem, not a sizing problem.
Four vessels, one idea, four residence times
Once you see the shape of the argument, most of the vessels on a flowsheet turn out to be the same vessel. Make the section big, let the velocity drop, and give the mixture long enough for the denser phase to separate under gravity. All that changes is how long "long enough" is, and that is set by how far apart the two densities are and how big the dispersed particles are.
| Vessel | Separates | Residence time |
|---|---|---|
| flash drum | vapour / liquid | 30 to 60 s |
| knockout drum | droplets from gas | 5 to 30 s |
| settler | liquid / liquid | 10 to 60 min |
| clarifier | solids from liquid | 30 to 120 min |
A knock-out drum gets seconds because a water droplet is a thousand times denser than the gas around it. A clarifier gets hours because a floc is barely denser than the water it is in. Same physics, four orders of magnitude of time.
One number to keep straight: the 5 to 30 seconds is liquid holdup in the boot, not the gas's trip through the shell. Gas crosses a 1.96 m³ shell at 2.4 m³/s in 0.82 seconds. At 2% liquid carry-over by mass, the boot sees 0.059 litres a second, so 5 to 30 seconds of holdup is 0.3 to 1.8 litres of surge. That is the volume the level controller has to work in before the compressor sees anything.
What to check on a drum that is already there
- Is the K you are using the K for your pressure? This is the most common error in the calculation, and it is always in the unsafe direction.
- Is there a pad, and is the K assuming one? Halving K without halving the assumption is how a drum that calculates fine passes liquid continuously.
- Is the inlet baffled? A bare nozzle fires a jet across the vessel and the gas never sees the full section, so the effective velocity is much higher than Q/A says. The baffle is not decoration.
- Is the boot sized for the surge, not the average? Slugs are not steady, which is the entire reason the vessel exists.
- Has the duty changed since it was sized? A debottleneck that raises throughput raises Q linearly and the drum's ceiling does not move.
Next time you walk past an empty tank in front of a compressor, the question worth asking is not what it does. It is which of those five someone checked, and when.