How is the air consumption of a pneumatic cylinder worked out?
From the swept volume and the pressure. The piston area is A extend = pi x D x D / 4, the area on the rod side A retract = pi x (D x D - d x d) / 4. A double acting cylinder fills both sides each cycle, so the swept volume is (A extend + A retract) x L; at the working pressure it is brought to free air by (p + pa) / pa. For D 63 mm, a 20 mm rod and a 200 mm stroke: 3117.2 mm2 and 2803.1 mm2, 0.6234 l + 0.5606 l = 1.1841 l swept; at 6 bar(g) the ratio is (6 + 1.013) / 1.013 = 6.923, so 8.20 l of free air per cycle. At 10 cycles/min that is 82.0 l/min, 4.92 m3/h or 2.89 cfm.
What is free air, and why multiply by (p + pa) / pa?
Free air is the volume the air takes at the ambient pressure. The cylinder fills at the working pressure, so the swept volume holds more air than it would at the ambient pressure; the ratio of the absolute pressures, (p + pa) / pa, gives the free air. The calculator takes pa as 1.013 bar(a), the standard atmosphere rounded, and says so; a barometer reading can be entered instead (atm 0.99 bar).
What changes for a single acting cylinder?
Only one stroke takes air: the extend stroke, while a spring makes the return. The free air per cycle is A extend x L x (p + pa) / pa, and the rod does not change it. For a 50 mm piston and a 100 mm stroke at 6 bar(g): 1963.5 mm2 x 100 mm = 0.1963 l, times 6.923 is 1.36 l per cycle, or 27.2 l/min at 20 cycles/min. The tubing counts one line instead of two.
How is the force on the rod worked out?
As the pressure times the area: F = p x A, with p the gauge pressure. 6 bar(g) is 0.6 N/mm2, so a 63 mm piston gives 0.6 x 3117.2 = 1870 N on the extend stroke, and with a 20 mm rod 0.6 x 2803.1 = 1682 N on the retract stroke. This is the theoretical force. On a single acting cylinder the spring force is not taken off.
Why count the tubing between the valve and the cylinder?
Because the tube is filled to the working pressure and vented every cycle, and that air does no work. A double acting cylinder has a line to each port, a single acting one a single line. Two lines of 2 m with a 6 mm inside diameter hold 2 x pi x 6 x 6 / 4 x 2000 mm = 0.1131 l; times 6.923 that is 0.783 l of free air per cycle, 8.7 % of the 8.98 l the cylinder and tubing take together in the example above.
Which inputs, units and range does the calculator take?
The piston diameter D, the rod diameter d and the stroke L in mm or inches (cm and m are read too); the working pressure in bar, psi, MPa or kPa, gauge or absolute, read as gauge when it has no mark; the cycles per minute (or per hour) and the number of cylinders; the tube length per line and its inside diameter, if wanted; the ambient pressure, 1.013 bar(a) unless entered. It gives no recommended pressures and no sizing margins.