How to Work Out Whether a Receiver Is Sized for the Demand

Why this matters

"Is the tank big enough?" is answered in most shops with a rule of thumb per horsepower, and that rule cannot answer the question, because it does not know what the plant does. A receiver is big enough or too small only with respect to a named demand event and a named usable pressure band. Change either one and the same vessel flips from generous to useless. A shop that can walk in, name the event, and hand the owner a required volume with the arithmetic attached is doing something the equipment supplier did not do when the package was quoted.

Before you touch anything in the air room

A receiver is a pressure vessel holding stored energy that does not dissipate when the compressor stops, through a power failure, a weekend or a lockout. Every reading in this procedure is taken from an existing gauge, an existing test port, or a display. If a test port has to be added, close the upstream isolation valve, open the section vent, confirm the section gauge reads zero with the vent still open, and lock and tag that isolation under 29 CFR 1910.147, which covers stored mechanical and pressure energy, before a wrench touches the joint.

If you drain a low point to clear a port, stand to the side of the discharge path with it directed into a container or a floor drain, wear eye protection and hearing protection, and never look into the path or point it at a person: air driven into a break in the skin can enter the bloodstream. Never clear a blocked port or a workbench with a blow gun. 29 CFR 1910.242(b) permits compressed air for cleaning only when reduced to less than 30 psi and then only with effective chip guarding and personal protective equipment. Air rooms commonly reach the 85 dBA eight-hour time-weighted average action level, at which point hearing protection and the rest of 29 CFR 1910.95 apply.

Step 1: Name the event, not the average

Storage covers events. It does nothing for averages, and a system short of air on average stays short with any vessel on it. So the first field on your sheet is a single demand event: the largest step change in flow the system has to ride through without the header falling below what the critical tool needs.

Write down three things about it. Its flow in scfm, from the tool's own data or a measurement. Its duration in minutes. Its repeat interval, which you will need in step 5 and which people leave blank.

Skipping this step is what produces a rule-of-thumb answer. Without a named event there is no number to size against, so the tank gets sized against the compressor, which is the one quantity that does not appear in the sizing relation at all.

Step 2: Subtract the compressor, because it does not stop

The vessel does not supply the whole event. The compressor keeps running through it. What storage has to cover is the deficit: event flow minus compressor output.

This is the single most common error in the procedure and it runs in the expensive direction. Size against the event's full flow rather than the deficit and you overstate the requirement by the ratio of the two, which on a tool drawing well above the machine's output is a factor of several.

If the deficit is negative, the compressor covers the event on its own and storage is not the constraint; you are looking at a pressure drop problem, which is a different measurement.

Step 3: Find the usable band, which is narrower than the switch differential

The band is the pressure you are allowed to spend during the event.

• The top is the header pressure the system actually sits at when the event starts, which is normal running pressure, not the compressor's cut-out.
• The bottom is the critical tool's minimum inlet pressure plus the distribution loss between the header and that tool at the event's flow.

That second term is the one that gets dropped. A tool needing 80 psig at its inlet behind a run that loses 9 psi at event flow cannot be fed from a header below 89 psig, so 89 is the floor, not 80. Measure the loss at the event's flow rather than at idle: friction loss scales with roughly the square of flow, so an idle reading understates it badly. The scaling itself and what sets it belong to the pipe-sizing article.

Step 4: Compute the required volume

Free air stored in a vessel between two pressures is the vessel volume times the pressure difference, divided by 14.7. That 14.7 is local barometric pressure, not a constant: at about 5,000 feet substitute 12.2 and the same vessel holds roughly 20 percent more free air per psi. Turn that around and the required volume is:

V = deficit flow x duration x 14.7 / usable band

with V in cubic feet, flow in scfm, duration in minutes and the band in psi. Multiply cubic feet by 7.481 to get gallons. This form assumes the air in the vessel stays near ambient temperature during the draw, which is close enough for a demand event lasting minutes and not for a fast blowdown lasting seconds.

Count only the volume on the right side of the treatment. A tank bolted under the compressor upstream of the dryer is a wet receiver, and air drawn from it during an event is undried air arriving at the dryer as a surge. It is real volume and it does real work on cycling; it is not event storage for a plant that needs dry air.

Step 5: Check recovery and cycling separately, and let the larger answer govern

Two more constraints sit on the same vessel, and each can demand more volume than step 4 did.

Recovery. After the event, the deficit has to be put back before the event repeats. Recovery time is the stored air spent, divided by compressor output minus background demand. Compare that against the repeat interval from step 1. An event that repeats faster than the system recovers is not an event, it is a sustained demand, and it needs capacity rather than storage.

Cycling. Total system volume, wet and dry together, sets how often the compressor starts. Cycle rate is inversely proportional to total volume, and the worst case sits at a demand near half of compressor output rather than at full demand. That relation and the starts-per-hour limits it has to satisfy belong to the sequencing article, which develops cycle time as T0 divided by d times one minus d; run it as a separate check and take whichever of the two answers is larger.

Worked example: a blast cabinet against an 18 scfm machine

A body shop reports that its media blast cabinet loses cutting power after about a minute and the technician has to stop and wait.

The event. The cabinet draws 25 scfm at its nozzle and gets used in bursts of about 3 minutes, roughly every 5 minutes through the afternoon. Compressor output is 18 scfm.

The deficit. 25 minus 18 is 7 scfm. Over 3 minutes that is 21 standard cubic feet the storage has to supply.

The band. Header sits at 110 psig running. The cabinet's nozzle needs 80 psig at the cabinet inlet. Measured with one gauge moved between two ports at the cabinet's own flow, the loss from header to cabinet inlet is 9 psi. Floor is 80 plus 9, so 89 psig. Usable band is 110 minus 89, which is 21 psi.

Required volume. 7 x 3 x 14.7 / 21 = 308.7 / 21 = 14.70 cubic feet, which is 110 gallons.

What is installed. An 80-gallon tank under the compressor, which is 10.69 cubic feet, upstream of a refrigerated dryer. Downstream of the dryer there is no vessel at all. So against a 14.70 cubic foot requirement the plant has zero qualifying storage, and every cubic foot the cabinet pulls is dragged through a dryer while the header falls. The technician's minute is the header walking from 110 down through 89 with nothing holding it up.

The fix and what it moves. A 120-gallon vertical receiver downstream of the dryer is 16.04 cubic feet, which covers the 14.70 requirement with about 9 percent margin. Check recovery before accepting it: 21 standard cubic feet to replace, background demand between bursts measured at 6 scfm, so net refill is 18 minus 6 = 12 scfm, and recovery is 21 / 12 = 1.75 minutes against a 5 minute repeat interval. It recovers with time to spare. Had the cabinet been used every 90 seconds, no receiver in the catalogue would have fixed it and the answer would have been a larger machine.

Cycling improves as a side effect rather than as the goal: total volume goes from 10.69 to 26.73 cubic feet, a factor of 2.50, so the start rate falls to 40 percent of what it was.

The two ways this goes wrong. Size against the cabinet's 25 scfm instead of the 7 scfm deficit and you get 25 x 3 x 14.7 / 21 = 52.5 cubic feet, 393 gallons, which is 3.57 times too large and reads as authoritative because the arithmetic closes. Size against the pressure switch's 25 psi differential instead of the 21 psi usable band and you get 12.35 cubic feet, which is 16 percent light and produces a receiver that almost works, which is worse than one that obviously does not.

The lever nobody should reach for first. Widening the band by raising cut-out is arithmetically identical to adding volume and looks free. It is not. Every psi of extra discharge is compression work paid on every cubic foot the plant makes, on the order of four tenths to half of one percent of input power per psi in the neighbourhood of a 100 psig discharge from an atmospheric intake, a coefficient that does not travel far from that discharge. Volume is the lever without that penalty.

Fitting the vessel you specified

A compressed air receiver is a pressure vessel. 29 CFR 1910.169(a)(1) requires construction, installation and inspection in accordance with the ASME Boiler and Pressure Vessel Code, Section VIII, in the edition that rule incorporates, and your state's boiler and pressure vessel law names the edition and the inspection regime that actually bind the installation. 1910.169(b) requires a drain at the lowest point, an indicating pressure gauge, and one or more spring-loaded safety valves sized so pressure cannot exceed the maximum allowable working pressure by more than a small margin, and it forbids any valve between the receiver and its safety valve.

Tying it in opens the piping: stop the compressor, isolate and lock its energy sources under 29 CFR 1910.147, open the drain to vent the section to atmosphere, and confirm zero on the receiver's own gauge with the drain still open before a single fitting is broken.

How to verify you got this right

Re-read your own sheet and confirm four things. That the flow you divided by was a deficit and not a demand. That the band's floor came from a measured loss at the event's flow rather than from the tool's rated inlet pressure alone. That the volume you counted sits downstream of the treatment the critical tool requires. And that recovery time is shorter than the repeat interval you wrote in step 1.

Then prove it in the plant rather than on the sheet. Put a recording gauge on the header, run the event at the same time of day, and confirm the header's low point lands above the floor you computed with margin left. If it lands lower, the deficit was understated, which usually means the tool draws more than its published figure at the pressure it is actually running, and the fastest way to settle it is to measure the event rather than to add another vessel and hope.

References

• 29 CFR 1910.169, air receivers, including construction to the ASME Boiler and Pressure Vessel Code Section VIII at (a)(1) and the drain, gauge and safety valve requirements at (b)
• 29 CFR 1910.147, control of hazardous energy, for isolating and verifying depressurisation before opening a pressurised path
• 29 CFR 1910.242(b), compressed air used for cleaning, and 29 CFR 1910.95, occupational noise exposure, including the 85 dBA eight-hour time-weighted average action level
• Compressor, dryer and tool manufacturer data for output, rated inlet pressure and flow at pressure
• See related: What a Receiver Is For and What Happens Without One; Why Pipe Size Decides More Than Compressor Size; The Compressor Types and What Each One Is Suited To