Consider every system component to ensure top performance with minimal losses.

#### What is in this article?:

- Right-sizing pneumatic motion systems
- Pneumatic circuit table
- Pressure calculations

I am disappointed every time I see the flow coefficient ## Pneumatic systemsThe provides a standardized means of gaging flow through a valve or other systems component. But because the valve is only one part of a complete system, every component that potentially causes a flow restriction or delay must be considered as part of the overall design. Additional factors to consider include: - A programmable-logic controller (PLC) with a "scan time" typically controls the valve. Scan time is the amount of time the processor takes to evaluate and execute a set of instructions before repeating the entire process.
- The response time for the valve's internal mechanism to switch from fully closed to fully open once it receives a signal from the PLC.
- Cylinder and valve ports can restrict airflow due to the fitting's internal orifices.
- Air lines between valve and cylinder have different flow characteristics depending on whether they are rigid pipe or flexible tubing.
- Air-line length adds volume to the system that must be charged and discharged every cycle.
- 90° fittings cause pressure drops and add delays.
- Flow controls can restrict airflow even in their full-open position.
- The load cannot move until sufficient pressure develops across the cylinder piston to overcome friction caused by seals and guide bearings.
## Practical considerationsTo put these notions into practical terms, let's look at the process involved in designing a typical system. Consider a pneumatic motion system consisting of: - A cylinder lifting a 10-lb mass vertically 3 in. in 0.170 sec or less.
- Nonlubricated supply pressure is 79.70 psia (65 psig) and exhaust is 14.70 psia.
- The PLC has a scan time of 0.010 sec.
- The valve solenoid shifts in 0.032 sec.
- Prior to the valve shifting, 79.70-psia pressure holds the load down with the bottom of the piston fully exhausted.
- A 14-in. length of flexible tubing with no right-angle fittings connects the valve and cylinder.
There is no external-guide friction and flow controls are not deemed necessary. Let's also make several assumptions: - The
*C*equation applies not only to valves but to all circuit components. The actual_{v}*C*constant may vary slightly but, on average, it is approximately the same._{v} - Exhaust flow is the same for all components in the exhaust-circuit branch and supply flow is the same for all components in the supply-circuit branch.
- Cylinder motion is "bang-bang" and can be represented by a constant acceleration where velocity increases linearly. The cylinder provides the acceleration force but external geometry stops the load abruptly without shock absorbers. This is used to calculate an average force the cylinder must generate. Actual motion forces and velocities can change significantly but the work done (effort X displacement) does not change much and does not particularly depend on the motion profile.
- The effects of temperature change are negligible.
- Air leakage is negligible.
## System sizingSolving this problem is iterative in nature so a spreadsheet format or MathCAD is recommended. Here are the key considerations.
Assuming a linear ramp (constant acceleration), peak velocity is twice this or 62.5 ips. Divide by time again to get acceleration, 651 ips
Because
An assumption is necessary at this point. Valve manufacturers recommend choosing between a 2 and 10-psi pressure drop across the valve exhaust port. Assume a 5-psi drop. Substituting C = 1.185. Choose a standard valve with _{v}C = 1.0 with 1/4-in. NPT ports. The significantly oversized cylinder makes this assumption possible. Because flow requirements are unchanged, calculate pressure drop based on the smaller valve. Rearranging the _{v}C equation,_{v}Valve exhaust-port pressure drop (Δ P) = 7.02 psi, well within the 2 to 10-psi requirement.
C = 1.411. Together with a downstream pressure of 21.72 psi (14.7 + 7.02) and 9.94 scfm flow, from the _{v}C equation exhaust-valve fitting pressure drop = 2.39 psi._{v}
The line-friction coefficient f for flexible tubing is 0.02. (For rigid steel pipe use 0.03.) This yields an exhaust air line _{l}C = 1.962. Note that if the line has 90° fittings, add 48_{v}nd to the length of the line before calculating air-line _{f}C. Using 1.962 _{v}C, a downstream pressure of 24.11 psi (14.7 + 7.02 + 2.39), and 9.94 scfm, from the _{v}C equation exhaust air-line pressure drop = 1.11 psi._{v}
C would be 1.125. But because it is not in the circuit, _{v}C is infinite and pressure drop would equal zero._{v}
C = 0.650. This gives a downstream pressure of 25.22 psi (24.11 + 1.11) and 9.94 scfm. From the _{v}C equation, exhaust cylinder fitting pressure drop = 9.69 psi._{v}
Q = 11.75 scfm, and supply air-line pressure drop = 0.61 psi.
Q = 11.75 scfm, and supply valve fitting pressure drop = 1.17 psi.
Q = 11.75 scfm, and valve supply port pressure drop = 2.29 psi. This adds up to a final supply pressure = 65.10 psia, with 14.60 psia to spare.
Calculate delay time, t = Δ_{d}P. Here, Δ_{e}gV_{e}k_{s}/q_{m}v_{s}^{2}Pe = 79.70 - 34.91 psia (the piston low side pressure). Density of air is approximately 0.075 lb/ft^{3} at 528°R. Velocity of sound v is approximately 1,127 fps at 528°R. Ratio of specific heats _{s}k for air = 1.4._{s}Therefore, This method has limitations. There is a maximum pressure drop that no component in this circuit can exceed. To determine that pressure, calculate the minimum of 0.875 P. These are 12.86 and 11.96 psi, respectively, so the maximum pressure drop for this example is 11.96 psi. There was extra pressure to spare when the last pressure (79.70 - 65.10 psi) was calculated, and no pressure drop exceeded 11.96 psi. This indicates that the initial shift-time guess can be reduced and all the above calculations repeated. This can be done 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._{s}Optimizing the shift time yields 0.146 sec (0.081 motion time + 0.039 delay time + 0.016 + 0.010) with a limiting maximum pressure drop of 11.96 psi at the cylinder exhaust port fitting. This means increasing the supply pressure will have little or no effect on speed. Now it is time for a reality check. We set up this actual pneumatic-motion circuit in our lab using all the variables identified for the 1.50-in.-diameter cylinder, except without a PLC. Without the PLC's 0.01-sec delay, cylinder actuation time was 0.134 sec. That is approximately 0.002 sec faster (0.136 sec without a PLC) than the initial estimate. The time for the actual cylinder to lower was 0.144 sec with no PLC, versus a calculated time of 0.158 sec. Thus, there is good agreement between theory and actual conditions. Selecting a smaller cylinder, say with a 1.0625-in.-diameter bore and 0.50-in. rod diameter, would also work. It will lead to a smaller valve Pneumatic Circuit Table for tabulated results.## Optimizing designsNote the 11.96-psi pressure drop at the cylinder exhaust fitting and the 9.97-psi drop at the valve exhaust. Recall the 11.96-psi maximum allowable pressure drop for all components and the 10-psi maximum pressure drop for the valve. These pressure drops prevent this cylinder from moving faster. Increasing supply pressure will have no effect on cylinder speed and may even increase delay time and slow it down. However, these limitations disappear with a larger fitting ID and a valve with a higher Also notice the 79.70 psia "Check Sum" value in the smaller cylinder example. This prevents the small cylinder from moving faster. All pressure drops are well within their allowable limits, so increasing supply pressure will increase the small cylinder's speed. Notice the small air-line pressure drop for the small cylinder. This indicates that the air line is possibly oversized. Using a smaller line diameter could speed up the cylinder by reducing the volume of air in the circuit. It also reduces air consumption. Compare flow values and move times for the two cylinders. The smaller cylinder and valve require less space, save money, and consume less air. If these effects were multiplied throughout a machine, space and cost savings could be significant. This method makes observations like these possible. Cylinder times are more predictable. Cylinders and valves can be sized to minimize air consumption. And because time is lost when a circuit does not operate as designed, savings can be realized in machine debug time and assembly rework. |

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