When a sudden air demand occurs at a point distant from the compressor room, the pressure drop signal must travel back through the distribution piping to the compressor before the compressor can respond. During this transit time \(t_{transit} = d_{pipe} / v_{air}\) (in seconds), the receiver must supply all the required air. The formula converts the transit time to an equivalent tank volume using the air demand flow rate, pressure ratio, and the gallon-to-cubic-foot conversion.

\begin{equation}\label{eq:receiver-tank-bridging-size} V_{required} = \frac{d_{pipe}}{v_{air}} \cdot \frac{Q_{demand}}{60} \cdot \frac{P_{atm}}{\Delta P} \cdot k_{gal} \end{equation}

Symbols

\(V_{required}\)	Required receiver tank size \([\unit{ \gallon}]\)
\(d_{pipe}\)	Distance from demand event to compressor room \([\unit{ \foot}]\)
\(v_{air}\)	Speed of compressed air in distribution piping \([\unit{ \foot\per\second}]\)
\(Q_{demand}\)	Air demand at the event location \([\unit{ \cubicFoot\per\minute}]\)
\(60\)	Seconds per minute conversion \([\unit{ \second\per\minute}]\)
\(P_{atm}\)	Atmospheric pressure \([\unit{ \psi}]\)
\(\Delta P\)	Allowable pressure drop at the demand event \([\unit{ \psi}]\)
\(k_{gal}\)	Gallons per cubic foot (7.48) \([\unit{ \gallon\per\cubicFoot}]\)

Note

The factor \(d_{pipe} / v_{air}\) yields the signal transit time in seconds. Dividing \(Q_{demand}\) by 60 converts from cfm to ft³/s so the product of transit time × flow rate gives a volume in ft³, which is then scaled to gallons.