As shown in this typical pneumatic circuit, designers must account for pressure drop across the valve, but also for other system components such as fittings and tubing. |

**Curtis Harvey**

Senior Design Engineer

Automation Div.**Remmele Engineering**

St. Paul, Minn.

I am disappointed every time I see the flow coefficient *C _{v}* used to size a valve without consideration given to the other components in a pneumatic circuit. In the old days, a focus on the valve alone often worked because designers tended to significantly oversize the valve. This approach is no longer acceptable. Given today's demands for optimized performance, lower costs, and minimal air consumption, a more-exacting method is needed to properly size pneumatic motion systems.

### Pneumatic systems

The*C*equationprovides 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.

_{v}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 considerations

To 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 sizing

Solving this problem is iterative in nature so a spreadsheet format or MathCAD is recommended. Here are the key considerations.

**Physics:** First, calculate the required cylinder force. The maximum time that the cylinder "sees" air is 0.170 sec, less the PLC scan time, and less half the valve-solenoid shift time (full flow for one-half the time or one-half the flow for full time). That leaves 0.144 sec for the actual move time plus the delay time for sufficient pressure to build up before the load moves. Start with a "guess" move time of about two-thirds of 0.144 sec, or in this case about 0.096 sec. Average velocity is 3 in./0.096 sec or 31.25 ips.

Assuming a linear ramp (constant acceleration), peak velocity is twice this or 62.5 ips. Divide by time again to get acceleration, 651 ips^{2} or 1.685 *g* (651 ips^{2}/386.4 ips^{2} = 1.685 *g*). Thus, the required force is 1.685 X 10 lb (due to acceleration) + 10 lb (due to gravity) = 26.85-lb force. If there are other forces, such as guide friction, they would be added here.

**Cylinder:** A good rule of thumb when first sizing a cylinder for speed is to make it large enough to provide approximately twice the calculated required force. In this case, design for a minimum cylinder force of 27 X 2 or 54 lb at 65 psig.

Because *F* = *PA*, the minimum bore requirement is 1.03 in. with no rod. Select a 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. Based on experience, make an initial guess that 0.25-in.-ID flexible tubing adequately supplies air to the cylinder.

**Valve exhaust port:** To size the valve, first calculate average flow during motion for the valve exhaust port. Calculate the volume being evacuated, divide by time, and then multiply by the pressure ratio to convert to standard conditions. Air-cylinder exhaust volume = (3π/4)(1.5^{2} - 0.625^{2}) = 4.38 in.^{3} Air-line exhaust volume = (14π/4)(0.25^{2}) = 0.69 in.^{3} Total exhaust volume = 5.07 in.^{3} Divide this by 0.096 sec to get 52.8 in.^{3}/sec, or 1.833 cfm. This is at full pressure. Multiply by 79.7/14.7 to convert to standard conditions, 9.94-scfm exhaust. In the same manner, calculate the supply flow during motion (without the rod) and get 11.74-scfm supply. These exhaust and supply flows will be used to estimate all other component pressure drops in the circuit.

*Q*= 9.94, Δ

*P*= 5, and

*P*= 14.7 into the

*C*equation, recommended valve

_{v}*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,Valve exhaust-port pressure drop (Δ

_{v}*P*) = 7.02 psi, well within the 2 to 10-psi requirement.

**Exhaust valve fitting:** A 1/4 -in. NPT straight fitting has a 0.280-in. ID. The flow coefficient for fittings can be approximated by *C _{v}* =18df2. This yields a valve fitting

*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}**Exhaust air line:**Determine air line

*C*usingThe line-friction coefficient

_{v}*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}**Exhaust and supply flow controls:** Calculate *C _{v}* for flow controls with the fitting equation, but use the air-line diameter. If a flow control were used here with a 0.25-in.-diameter line,

*C*would be 1.125. But because it is not in the circuit,

_{v}*C*is infinite and pressure drop would equal zero.

_{v}**Cylinder exhaust fitting:** A 1/8-in. NPT straight fitting has a 0.190-in. ID. Using the fitting *C _{v}* equation, exhaust cylinder fitting

*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}**Cylinder:** At this point, piston low pressure = 34.91 psia (25.22 + 9.69). Add to this the 13-psi friction to get 47.91 psia. Multiply this by the area to get a resistive force of 69.97 lb. Add the required cylinder output force of 26.85 lb for a high-side piston force of 96.82 lb. With a 1.50-in. diameter and no rod, supply side pressure = 54.79 psia.

**Cylinder-supply fitting:** Using *C _{v}* = 0.650 for 1/8-in. NPT, a downstream pressure of 54.79 psia, and 11.75 scfm, cylinder supply fitting pressure drop is 6.23 psi.

**Supply air line:***C _{v}* = 1.962, downstream pressure = 61.02 psia,

*Q*= 11.75 scfm, and supply air-line pressure drop = 0.61 psi.

**Supply-valve fitting:***C _{v}* = 1.411, downstream pressure = 61.63 psia,

*Q*= 11.75 scfm, and supply valve fitting pressure drop = 1.17 psi.

**Valve-supply port:***C _{v}* = 1.0, downstream pressure = 62.81 psia,

*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.

**Delay time:** When the valve shifts from closed to open, the pressure previously built up in the cylinder (rod side) begins to bleed out the high-volume exhaust side while, simultaneously, pressure builds on the low-volume supply side. Because cylinder volume on the supply side is at or near zero, supply pressure builds quickly. The time delay before motion begins is mostly due to the time it takes for exhaust pressure to fall below the supply pressure and develop the necessary pressure difference across the piston to overcome friction and move the load.

Calculate delay time, *t _{d}*, as follows:

*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, *t _{d}* = 0.054 sec. So the initial time estimate = 0.096 shift time + 0.054 delay time + 0.032/2 valve solenoid time + 0.010 PLC scan time = 0.176 sec.

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 _{e}* and 0.15

*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 *C _{v}*. See the

*Pneumatic Circuit Table*for tabulated results.

### Optimizing designs

Note 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 *C _{v}*. These changes should increase speed with the capability for faster response with increasing supply pressure.

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.

## Pneumatic circuit table

**System parameters**

**First**

iteration

iteration

**Last**

iteration

iteration

**Smaller**

cylinder

and valveCylinder diameter, in.1.501.501.0625Rod diameter, in.0.6250.6250.500Exhaust pressure, psia14.7014.7014.70Supply pressure, psia79.7079.7079.70PLC scan time, sec0.0100.0100.010Valve

cylinder

and valve

*C*1.01.00.7Air-line length, in.141414Air-line ID, in.0.250.250.25Valve-solenoid shift time, sec0.0320.0320.013Target total move time, sec0.1700.1700.170Initial estimate: move + delay time, sec0.144----Move time, sec0.0960.0810.075Supply flow, scfm11.7413.998.40Exhaust flow, scfm9.9411.846.92Average velocity, ips31.2537.240.0Acceleration,

_{v}*g*1.6852.3922.761

## Pressure calculations

Valve exhaust ΔP, psi7.029.976.95Pressure 1, psia21.7224.6721.65Valve-fitting exhaust ΔP, psi2.392.981.16Pressure 2, psia24.1127.6522.82Air-line exhaust ΔP, psi1.111.380.57Pressure 3, psia25.2229.0323.39Flow-control exhaust ΔP, psiPressure 4, psia25.2229.0323.39Cylinder-fitting exhaust ΔP, psi9.6911.965.07Pressure 5, psia34.9140.9828.46Pressure 6, psia54.7963.8174.70Cylinder-fitting supply ΔP, psi6.237.592.34Pressure 7, psia61.0271.4077.04Flow-control supply ΔP, psi0.000.000.00Pressure 8, psia61.0271.4077.04Air-line supply ΔP, psi0.610.740.25Pressure 9, psia61.6372.1477.29Valve-fitting supply ΔP, psi1.171.420.48Pressure 10, psia62.8173.5777.76Valve supply ΔP, psi2.292.781.94### Results

Check Sum pressure, psia65.1076.3579.70Maximum allowable pressure drop, psi11.9611.9611.96Cylinder force output, lb26.8533.9237.61Low-pressure force + friction force, lb69.9778.8328.62High-pressure force, lb96.82112.7566.23Calculated delay time, sec0.0540.0390.048Estimated total time, sec0.1760.1460.140Tabulated results are for the example circuit lifting a 10-lb mass 3 in. in 0.170 sec or less. Cylinder ports are 1/8 NPT, valve ports are 1/4 NPT, and there are no flow controls in the system. Note that the 11.96-psi cylinder fitting exhaust pressure drop is the limiting factor with the larger cylinder, and the 79.70-psia Check Sum pressure limits the small-cylinder circuit.

### Optimizing circuit design

Here are a few suggestions for optimum circuit design:

- Make air-line lengths as short as possible.
- Make air lines between the valve and cylinder as straight as possible with minimal bends.
- Select cylinder bore sizes to handle the expected load plus a reasonable safety factor. Larger-than-necessary cylinders cost more money, waste energy, and add cycle time.
- Cylinder stroke should be no more than required. Longer-stroke cylinders cost more, waste energy, and add cycle time.
- A valve can be oversized without appreciably wasting energy. However, cycle time will increase if solenoid shift time increases.
- Overpressurizing a circuit beyond a certain point -- the maximum pressure drop -- does not increase cylinder speed but does waste air and can increase delay time and total cycle time.
- If the application calls for different loads or speeds for extend and retract motions, consider using different pressures or add flow controls.
- Consider using quick-exhaust dump valves. The dumped air bypasses the exhaust circuit, possibly reducing cycle time and increasing speed.
- Each application has an optimum air-line ID. Increasing the air-line diameter increases
*C*but also increases the volume that must be filled and evacuated each cycle._{v} - Components with the smallest
*C*s and largest pressure drops limit circuit performance. Increase these_{v}*C*s first to have the greatest impact on circuit performance._{v} - Components with the largest
*C*s and smallest pressure drops are possibly oversized. Decreasing these_{v}*C*s could improve circuit performance._{v}

### Nomenclature

A= Air-line cross-sectional area, in._{l}^{2}d= Cylinder diameter, in.d= fitting ID, in._{f}d= tubing ID, in._{t}f= line-friction coefficient_{l}l= line length, in.Q= Flow rate at 1 atmosphere, 68°F, and 36% relative humidity, in scfmC= Flow coefficient_{v}G= Specific gravity of air at one atmosphere, 68°F, and 36% relative humidity. Usually,G= 1.g= Acceleration of gravityk= Specific-heat ratio;_{s}k= 1.4 for air._{s}n= Number of fittingsP= Exhaust pressure, psi_{e}P= Supply pressure, psi_{s}P= Upstream pressure at T1, psia_{1}P= Downstream pressure, psia_{2}q= Mass flow rate_{m}T= Upstream temperature, °R. Usually,_{1}T_{1}= 528°R.V= Exhaust volume, in._{e}^{3}