Consider every system component to ensure top performance with minimal losses.
What is in this article?:
 Rightsizing pneumatic motion systems
 Pneumatic circuit table
 Pressure calculations
Curtis Harvey 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 moreexacting method is needed to properly size pneumatic motion systems. Pneumatic systemsThe C_{v} equation Additional factors to consider include:
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:
There is no externalguide friction and flow controls are not deemed necessary. Let's also make several assumptions:
System sizingSolving 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 valvesolenoid shift time (full flow for onehalf the time or onehalf 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 twothirds 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) = 26.85lb 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 1.50in.diameter cylinder with a 0.625in.diameter rod and 1/8in. 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 0.25in.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. Aircylinder exhaust volume = (3π/4)(1.5^{2}  0.625^{2}) = 4.38 in.^{3} Airline 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, 9.94scfm exhaust. In the same manner, calculate the supply flow during motion (without the rod) and get 11.74scfm supply. These exhaust and supply flows will be used to estimate all other component pressure drops in the circuit. An assumption is necessary at this point. Valve manufacturers recommend choosing between a 2 and 10psi pressure drop across the valve exhaust port. Assume a 5psi drop. Substituting Q = 9.94, ΔP = 5, and P = 14.7 into the C_{v} equation, recommended valve C_{v} = 1.185. Choose a standard valve with C_{v} = 1.0 with 1/4in. 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 C_{v} equation, Exhaust valve fitting: A 1/4 in. NPT straight fitting has a 0.280in. ID. The flow coefficient for fittings can be approximated by C_{v} =18df2. This yields a valve fitting C_{v} = 1.411. Together with a downstream pressure of 21.72 psi (14.7 + 7.02) and 9.94 scfm flow, from the C_{v} equation exhaustvalve fitting pressure drop = 2.39 psi. Exhaust air line: Determine air line C_{v} using Exhaust and supply flow controls: Calculate C_{v} for flow controls with the fitting equation, but use the airline diameter. If a flow control were used here with a 0.25in.diameter line, C_{v} would be 1.125. But because it is not in the circuit, C_{v} is infinite and pressure drop would equal zero. Cylinder exhaust fitting: A 1/8in. NPT straight fitting has a 0.190in. ID. Using the fitting C_{v} equation, exhaust cylinder fitting C_{v} = 0.650. This gives a downstream pressure of 25.22 psi (24.11 + 1.11) and 9.94 scfm. From the C_{v} equation, exhaust cylinder fitting pressure drop = 9.69 psi. Cylinder: At this point, piston low pressure = 34.91 psia (25.22 + 9.69). Add to this the 13psi 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 highside piston force of 96.82 lb. With a 1.50in. diameter and no rod, supply side pressure = 54.79 psia. Cylindersupply fitting: Using C_{v} = 0.650 for 1/8in. 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 airline pressure drop = 0.61 psi. Supplyvalve fitting:C_{v} = 1.411, downstream pressure = 61.63 psia, Q = 11.75 scfm, and supply valve fitting pressure drop = 1.17 psi. Valvesupply 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 highvolume exhaust side while, simultaneously, pressure builds on the lowvolume 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: 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 0.875P_{e} and 0.15P_{s}. 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 shifttime 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. 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 pneumaticmotion circuit in our lab using all the variables identified for the 1.50in.diameter cylinder, except without a PLC. Without the PLC's 0.01sec 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 1.0625in.diameter bore and 0.50in. rod diameter, would also work. It will lead to a smaller valve C_{v}. See the Pneumatic Circuit Table for tabulated results. Optimizing designsNote the 11.96psi pressure drop at the cylinder exhaust fitting and the 9.97psi drop at the valve exhaust. Recall the 11.96psi maximum allowable pressure drop for all components and the 10psi 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 airline 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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