How to Sequence Multiple Compressors Without Short Cycling

Why this matters

Sequencing gets treated as a controls exercise, and it is a storage exercise wearing a controls costume. You can arrange setpoints perfectly and still end up with a trim machine loading and unloading every ninety seconds, because the cycle rate is set by how much air the system can hold between two pressures, not by how cleverly the setpoints are stacked. Get the storage wrong and the only ways out are to widen the pressure band, which raises the top of the range and costs energy everywhere, or to accept the cycling and the wear that comes with it.

What you are producing at the end of this is one page: a setpoint schedule with eight fields, every one of them derived rather than chosen. Work them in order, because each field constrains the next.

Field 1: minimum acceptable pressure at the worst point of use

Not at the compressor, not at the header, at the inlet of the machine or tool that is least tolerant. Get it from that equipment's own documentation rather than from the plant's habit, because plant habit is usually a pressure somebody raised once and nobody lowered.

Skipping this field is what produces a schedule built around whatever pressure the plant currently runs, which imports every past overreaction into the new design.

Take 85 psig at the tool inlet for the worked schedule.

Field 2: measured distribution drop at peak flow

Measure it, at peak, between the compressor room and that worst point of use. The method belongs to the distribution pressure drop card in this library. Two things about it matter here: it must be measured at peak flow rather than at a convenient quiet moment, because drop through pipe in turbulent flow rises with roughly the square of flow, and it must be measured to the specific worst point rather than to a convenient tap.

Skipping this field means you will discover the drop later, as a pressure complaint, and the plant will solve it by raising the setpoint, which undoes the schedule.

Take 9 psi measured at peak.

Field 3: minimum header pressure

Field 1 plus field 2. Here, 85 plus 9 is 94 psig, so the bottom of the lowest machine's band goes at 95 psig with a psi of margin.

This is the number the whole schedule is anchored to. Everything above it is stacked on top, and every psi you add here is paid for continuously.

Field 4: the trim machine's band

The trim machine is the one that cycles: in a cascade the lead machine has the higher setpoints and ends up running loaded continuously once demand is high, while the trim machine loads and unloads to follow the swing above the lead machine's capacity.

Set the trim band's bottom at field 3. Its width is provisional at this stage because field 7 may force it wider, and that is the whole tension in the procedure. Start at 10 psi, so trim loads at 95 and unloads at 105.

Field 5: the lead machine's band, stacked above

The lead band sits directly above the trim band with no overlap, so the trim machine's unload point is the lead machine's load point. Lead loads at 105 and unloads at 115.

Skipping the no-overlap check gives you two machines responding to the same excursion together, which is a separate and expensive failure covered in this library under two compressors on one header. Note the cost this stacking carries: the total range is now 20 psi rather than 10, and the top of it is 115. A commonly published rule of thumb puts compressor power change at about 1 percent per 2 psi near 100 psig discharge on a lubricated rotary screw, so the top of the range is roughly 10 percent more power than the bottom, and it is paid whenever the lead machine is up there.

Field 6: storage, measured rather than calculated

You do not need a volume calculation. You need one timed test.

During a genuine no-production window, with the plant isolated at the main header valve and that valve tagged, and after confirming with the floor that nobody is on air-driven equipment that would stall mid-operation, load the trim machine alone and time the header rising through its band. Nobody opens a drain, a filter bowl or any joint during this test; if something has to be opened, the compressor's disconnect is locked open under 29 CFR 1910.147, the leg is isolated and bled, and the indicating gauge that 29 CFR 1910.169 requires the receiver to carry is confirmed at zero before a joint is broken.

Call the result T0. Take 24 seconds across the 10 psi band for the worked schedule.

T0 is the system's storage expressed in the only units the rest of this procedure needs: the time the storage buys at the trim machine's full output across that band.

Field 7: cycle rate, checked at two demands and against two different limits

With the trim machine's demand at a fraction d of its own capacity, the loaded time is T0 divided by (1 minus d) and the unloaded time is T0 divided by d, so the cycle time is T0 divided by d times (1 minus d).

That expression is worth reading rather than just using. Cycle time is longest at both ends of the range and shortest in the middle, so the fastest cycling happens when the trim machine's demand is near half its capacity, not near its limit. A plant that checks cycling at peak demand is checking the easy case.

Check A, at the worst-case demand for cycling, d equal to 0.5. Cycle time is 24 divided by 0.25, which is 96 seconds, so 37.5 cycles per hour. Loaded and unloaded are 48 seconds each. Against a published blowdown time of 40 seconds for this machine, the 48-second unloaded window clears it, so on load and no-load the schedule survives this check. If instead you intend to stop the trim motor rather than unload it, 37.5 starts per hour is far above the maximum starts per hour published for a motor of this size, and the schedule fails here.

Check B, at the real high-demand condition. Take the plant at 180 units against a lead machine of 100 units, so the trim machine is carrying 80 of its own 100 units and d is 0.8. Loaded time is 24 divided by 0.2, which is 120 seconds; unloaded time is 24 divided by 0.8, which is 30 seconds; cycle time is 150 seconds, so 24 cycles per hour. The unloaded window is now 30 seconds against a 40-second blowdown, so the machine never reaches its low unloaded power. It is paying something close to loaded power while delivering nothing, and calling itself a load and no-load machine while doing it.

Two checks, two different gates, failing at two different demands. That is normal, and it is why both get run.

Field 8: rotation and failure fallback

Rotate lead and lag on a fixed interval so run hours stay comparable, and record the rotation in the schedule rather than leaving it to whoever is on shift.

Then set the local pressure switches so that if the sequencing controller fails or is bypassed, what the machines fall back to is a working cascade rather than two machines racing. The fallback settings are part of the schedule, not an afterthought, because a controller that fails at 2am with both machines set to the same band is exactly the fight this procedure exists to prevent.

The worked schedule, and the decision it forces

Field 7 failed, so something has to move. There are two ways to fix it and they are not equivalent.

Widen the band. Going from 10 psi to 16 psi scales T0 proportionally, from 24 seconds to 38.4 seconds. At d equal to 0.8 the unloaded window becomes 38.4 divided by 0.8, which is 48 seconds, clearing the 40-second blowdown. But the trim band is now 95 to 111, the lead band has to stack above it at 111 to 121, and the top of the plant's range has gone from 115 to 121. At the rule of thumb above that top-of-range excursion is worth roughly another 3 percent of compressor power whenever the lead machine is up there, and every leak and unregulated point of use flows more at 121 than at 115 because those restrictions pass choked flow and their mass flow tracks absolute upstream pressure.

Add storage. T0 scales with total pressurised volume, receiver plus header plus every branch, not with the receiver alone. Getting T0 from 24 to 38.4 seconds means raising total volume by a factor of 1.6, so the vessel you add is 0.6 times the whole current volume. Estimate the piping before you size it, or you will buy a receiver that moves T0 by half what you expected, and it costs nothing in setpoints. The band stays at 10 psi, the top of the range stays at 115, and the cycle problem is gone. At d equal to 0.8 the cycle becomes 240 seconds, 15 cycles per hour, with a 48-second unloaded window.

Storage wins whenever the top of the range is already where you want it, which is almost always, and that is the general result: given a fixed maximum allowable pressure, storage is the only free variable in a sequencing schedule.

And storage is what buys you the right to stop the lag machine. A trim machine that is merely unloaded still draws real power, and this library's card on two machines on one header shows that stopping it is where the saving actually lives. To stop the motor you have to clear the published starts-per-hour limit at the worst-case demand, which is d equal to 0.5. Cycle time there is four times T0, so bringing 23.4 starts per hour at T0 of 38.4 seconds down to the machine's published maximum, which on this motor is 10 per hour and which you take from its own data rather than from a habit, needs a cycle of 360 seconds, which needs T0 of 90 seconds, which is 3.75 times the storage the plant started with.

That is a real number to put in front of a customer, and it is a far better conversation than a third compressor. Any receiver added has to be a code vessel: ASME Boiler and Pressure Vessel Code Section VIII, in the edition your state's boiler and pressure vessel program has adopted, which binds the owner and reaches you through the jurisdiction's inspection, and it carries the drain, indicating gauge and spring-loaded safety valve that 29 CFR 1910.169 requires of every air receiver.

Verifying the schedule after it is set

Run the plant a full shift and log four things: the header pressure trace at the worst point of use, the trim machine's cycles per hour, its unloaded window at the busiest hour, and the lead machine's loaded fraction.

The header at the worst point of use must never touch field 1. The cycles per hour must clear the gate that applies to how the trim machine actually unloads, blowdown time if it idles and published starts per hour if it stops. The unloaded window is the one people forget to log and it is the one that failed in the example. And the lead machine's loaded fraction tells you whether the lead is genuinely acting as a base machine or whether the split is wrong and the roles should swap.

If the trim machine's cycle rate is highest at moderate demand and lowest at both extremes, the schedule is behaving as designed. If it cycles fastest at peak, your load split is not what you think it is, and the lead machine is probably not delivering what its nameplate says.

References

• Manufacturer documentation for each machine's rated delivery, published blowdown time, maximum motor starts per hour, and permitted stop and restart behaviour
• ASME Boiler and Pressure Vessel Code Section VIII, in the edition adopted by your state's boiler and pressure vessel program, which binds the vessel owner; 29 CFR 1910.169 for the receiver's required drain, indicating gauge and spring-loaded safety valve
• 29 CFR 1910.147 for lockout and stored air energy before opening any pressurised component during the storage test
• See related: Why Two Compressors Fight Each Other on the Same Header; What a Compressor Control Strategy Is Choosing Between; What Storage Buys You That Horsepower Cannot; What Pressure Drop Through Distribution Actually Costs