Consider how every component interacts to design better air circuits.
Edited by Kenneth J. Korane
Too often, pneumatic circuits are sized based on the valve’s flow coefficient, Cv, without considering other system components. In the past, focusing on the valve alone often sufficed because air was considered “free” and designers tended to oversize valves. Today’s emphasis on better energy management and performance requires a more-exacting method to properly size pneumatic motion systems.
Sustainability conscious users can reduce air and energy consumption by using smaller, fewer, and less-expensive components and still ensure smooth and efficient operation.
The Cv equation
provides a standard means to predict flow through valves and other components. But designers must consider every component that could cause flow restrictions, pressure drops, or time delays when engineering pneumatic circuits.For instance:
Tips for optimizing air circuits
• Flow characteristics through air lines depend on tubing materials and construction.
• Air lines add volume that must be charged and discharged every cycle.
• Fittings that are not straight increase pressure losses and delays.
• Flow controls restrict airflow even in their full-open position.
Additional factors to consider:
• A programmable-logic controller (PLC) with a “scan time” typically controls the valve. Scan time is the time the processor takes to evaluate and execute a set of instructions before repeating the entire process.
• Response time for a valve to switch from fully closed to fully open once it receives a signal from the PLC.
• Loads cannot move until sufficient pressure difference develops across the cylinder piston to overcome gravity and friction caused by seals and guide bearings.
To put these factors into practical terms, let’s look at an example. Consider a typical pneumatic motion system with an air cylinder pushing a 10-lb mass vertically 3 in. in 0.170 sec. Supply pressure is 79.7 psia (65 psig) and exhaust pressure is 14.7 psia.
The valve shifts from closed to open in 0.032 sec and PLC scan time is 0.010 sec. Flexible tubing 14 in. long with no angled fittings connects the valve and cylinder, and there are no flow controls or external friction.
Cylinder motion is “bang-bang.” That is, velocity increases linearly with constant acceleration and the external structure stops the load abruptly without cushions or shock absorbers. And effects of temperature changes and air leaks are negligible, as are differences between static and dynamic friction.
These circuit calculations are iterative so a software program such as Excel or Math- CAD is recommended. Here are the key considerations.
Physics: First, calculate the required cylinder force. The maximum time air flows to the cylinder is 0.170 sec, minus the PLC scan time and half the valve-solenoid shift time. That leaves 0.144 sec for the actual move plus the delay time for enough pressure to build before the load moves. Start with an estimated move time of about two-thirds of 0.144 sec, or 0.096 sec. Average velocity va = 3 in./0.096 sec = 31.25 ips.
Assuming constant acceleration, peak velocity vp = 2va = 62.5 ips. Acceleration = vp/0.096 sec = 651 ips2 or 1.685 g. Thus, the required lifting force is 1.685 × 10 lb (due to acceleration) + 10 lb (due to gravity) = 26.85 lbf. If there were other forces, such as guide friction, they would be added here.
Cylinder: Size the cylinder to provide an extra 33 to 100% of the calculated required force to overcome pressure losses from other components in the circuit. In this case, design for a cylinder force between 36 and 54 lb at 65 psig.
Because F = PA, bore should be between 0.84 and 1.03 in. with no rod. Start with an oversized 1.50-in.- diameter cylinder with a 0.625-in.-diameter rod and 1 /8 -in. NPT ports. From the manufacturer’s catalog, friction is 13 psi. This is the minimum pressure needed to overcome internal friction and move the piston with no load. And start with 0.25-in.-ID flexible tubing to supply air to the cylinder.
Sum all the pressure drops around the circuit, starting at the valve’s exhaust port (atmospheric pressure) and ending at its supply port (regulated supply pressure). These pressure drops should add up to the pressure difference between supply and atmospheric pressures. (This is similar to Kirchhoff ’s Voltage Law used to solve electrical circuits.)
Valve exhaust port: To size the valve, first calculate flow through the valve exhaust port. Calculate the volume being evacuated, divide by time, and multiply by the pressure ratio to convert to standard conditions. Air-cylinder exhaust volume = (3)(π /4)(1.52 – 0.6252) = 4.38 in.3 Airline exhaust volume = (14)(π /4)(0.252)(79.7-14.7)/79.7 = 0.56 in.3 (Note the air line does not empty completely — one atmosphere pressure remains.) Total exhaust volume = 4.94 in.3 Divide this by 0.096 sec to get 51.5 in.3/sec, or 1.787 cfm at full pressure. Multiply by 79.7/14.7 (pressure ratio) to convert to standard conditions, and get 9.69 scfm exhaust. In the same manner, the supply flow (without the rod) through the valve supply port is 11.50 scfm.
An assumption is necessary at this point. Valve manufacturers recommend designing for a 2 to 10-psi pressure drop across the valve’s exhaust port. We’ll assume a 6-psi drop. Substituting Q = 9.69, ΔP = 6, and P = 14.7 into the Cv equation, the recommended valve Cv = 1.055. Choose a standard valve with Cv = 1.0 and ¼-in. NPT ports. Because flow requirements are unchanged, calculate pressure drop based on the smaller valve. Rearranging the Cv equation, valve exhaust-port ΔP = 6.67 psi, well within the 2 to 10-psi requirement.
Each circuit component also has a critical pressure ratio, P2/P1, that must be ≥0.528. If P2/P1 = 0.528, airflow is sonic or “choked” and is at the maximum level possible. If P2/P1 < 0.528, flow is impossibly high and must be reduced. For this component P2/P1 = 14.7/21.37 = 0.688, so it’s subsonic.
Exhaust-valve fitting: A ¼-in. NPT straight fitting has a 0.28-in. ID. The flow coefficient equation for fittings is approximately Cv =18df2. This yields a fitting Cv = 1.411. With a downstream pressure of 21.37 psi (14.7 + 6.67) and 9.69-scfm flow, using the Cv equation gives an exhaust-valve-fitting pressure drop = 2.30 psi. P2/P1 = 21.37/(21.37+2.30) = 0.903, so it’s acceptable.
Exhaust-air line: Determine air line Cv using
The line-friction coefficient fl for flexible tubing is 0.02 (use 0.03 for rigid steel pipe). This yields an exhaust-air line Cv = 1.962. Note that if the line has 90° fittings, add 48ndf to the length of the line before calculating airline Cv. Line flow is based on the cylinder volume plus half the air-line volume or 9.14 scfm. Using 1.962 Cv, a downstream pressure of 23.68 psi (14.7 + 6.67 + 2.30) and 9.14 scfm from the Cv equation, exhaust air-line pressure drop = 0.96 psi. P2/P1 = 23.68/(23.68+0.96) = 0.961.
Exhaust and supply flow controls: Use the manufacturer’s Cv data, if available. Otherwise, calculate Cv for flow controls with the fitting equation but use the air-line ID. If a flow control were used here with a 0.25-in.-diameter line, Cv would be 1.125 (full open). Because flow controls typically mount in the cylinder port, calculate ΔP using flow based only on the cylinder volume or 8.59 scfm. A flow control is not in the circuit so we’ll move on to the next component.
Exhaust-cylinder fitting: A 1/8 -in. NPT straight fitting has a 0.19-in. ID. Using the fitting Cv equation, exhaust cylinder fitting Cv = 0.650. This gives a downstream pressure of 24.64 psi (23.68 + 0.96) and 8.59 scfm. From the Cv equation, exhaust-cylinder-fitting pressure drop = 7.41 psi and P2/P1 = 24.64/32.05 = 0.769.
Cylinder: At this point, piston exhaust-side pressure = 32.05 psia (24.64 + 7.41). Add the cylinder’s 13-psi internal friction to this to get 45.05 psia. Multiply this by the area to get a resistive force of 65.79 lb. Add the required cylinder output force of 26.85 lb to get a supply-side piston force requirement of 92.64 lb. Divide this by the area of a 1.50-in.-diameter piston with no rod to get supply side pressure = 52.42 psia.
Cylinder-supply fitting: Using Cv = 0.650 for 1 /8 -in. NPT and a downstream pressure of 52.42 psia and 10.40 scfm, cylinder supply fitting pressure drop is 5.10 psi. P2/P1 = 52.42/57.52 = 0.911. Supply air line: Cv = 1.962, downstream pressure = 57.52 psia, Q = 10.95 scfm, and supply air-line pressure drop = 0.57 psi. P2/P1 = 57.52/58.09 = 0.990.
Supply-valve fitting: Cv = 1.411, downstream pressure = 58.09 psia, Q = 11.50 scfm, and supply-valve-fitting pressure drop = 1.19 psi. P2/P1 = 58.09/59.28 = 0.980.
Valve-supply port: Cv = 1.0, downstream pressure = 59.28 psia, Q = 11.50 scfm, and valve-supply-port pressure drop = 2.33 psi. P2/P1 = 59.28/61.61 = 0.962. This adds up to a final supply pressure = 61.61 psia, with 18.09 psi to spare.
Delay time: When the valve opens, pressure in the cylinder (rod side) begins to bleed out the exhaust side while, simultaneously, pressure builds on the supply side. Because cylinder supply-side volume is small, supply pressure builds quickly. The delay before motion begins is mostly due to the time it takes exhaust pressure to fall below supply pressure and develop the necessary pressure difference across the piston to overcome gravity and friction, and initiate motion.
Calculate delay time td = ΔPegVeks/qmvs2. Here, ΔPe = 79.70 – 32.05 psia (the piston low-side pressure). Air density is approximately 0.075 lb/ft3 at 528°R. Therefore, td = 0.059 sec. So the initial time estimate = 0.096 shift time + 0.059 delay time + 0.032/2 valve solenoid time + 0.010 PLC scan time = 0.181 sec.
Optimizing the circuit
Besides the critical pressure ratio already discussed, experts recommend that no component in a circuit should exceed 15% of the absolute upstream pressure or 11.96 psi in this example. There was extra pressure to spare in the first iteration (79.70 – 61.61 psi) and no pressure drop exceeded 11.96 psi. This indicates we can improve the initial move-time estimate and repeat all the above calculations until the last pressure equals the supply, one component is at the maximum pressure drop, or the pressure drop across the valve exhaust port exceeds 10 psi.
Optimizing the shift time yields 0.146 sec (0.078 motion time + 0.042 delay time + 0.016 + 0.010) with a limiting pressure drop of 10 psi across the valve-exhaust port. Increasing supply pressure has little effect on speed and will increase delay time before motion begins.
As a test, we built this pneumatic-motion circuit using all the variables identified for the 1.50-in.-diameter cylinder, except without a PLC and its 0.01-sec delay. Cylinder actuation time was 0.134 sec, approximately 0.002 sec faster (0.136 calculated without a PLC) than the initial estimate. The time for the actual cylinder to lower was 0.144 sec versus a calculated time of 0.158 sec. Thus, there is good agreement between theory and actual conditions.
Assuming this cylinder operates 24/7, extending and retracting every 3 sec, operating costs will run about $182/yr based on air costing $0.50/1,000 scf. The energy to generate compressed air produces about 620 lb/yr of CO2 (based on 1.7 lb of CO2/1,000 scf of air.)
Reducing cylinder pressure 5.5 psi to 74.2 psia (59.5 psig) improves raise time from 0.146 to 0.138 sec and the lower time from 0.168 to 0.157 sec. It also saves about 7% in air. Reducing supply pressure further provides additional savings while staying within the 0.170-sec time frame. Further gains are possible by repeating this analysis with dual air pressures and quick exhaust valves.
Likewise, selecting smaller cylinders and reducing Cv in other components offers additional savings. For instance, adding a regulator to the 1.50-in.-bore cylinder circuit reduced costs 32%. Adding regulators with quick exhaust valves (Cv = 0.72) to the same circuit reduced costs 54%. And reducing cylinder size to 1.06-in. bore, with regulators and quick-exhaust valves, results in savings of 76%. Further reducing the cylinder bore to 0.87 in. doesn’t appear to have as many advantages, but there could be some space savings and component cost reductions. (See the “Optimized circuit” table for details.)
Benefits of an optimized circuit include lower air and electricity consumption with a corresponding drop in CO2 production. It also lets OEMs start with smaller, fewer, and less-expensive components that consume fewer resources and generate less CO2. If we all start to think a little greener today, we should all breath a little cleaner tomorrow.