were quick-opening, equal-percentage, or anywhere in between. All the positioner does is modify the valve stem position as per the desired characteristic function instead of proportionally follow the signal as it normally would.

This approach has the distinct advantage of convenience (especially if the valve is already equipped with a positioner) over changing the actual valve trim. However, if valve stem friction ever becomes a problem, its e ects will be disproportionate along the valve travel range, as the positioner must position the valve more precisely in some areas of travel than others when pressed into service as a characterizer.

It should be noted that not all process control loops benefit from control valves with equalpercentage inherent characteristics. There are some process applications, for example, where the pressure drop across a control valve holds relatively constant over a wide range of valve flow rates47, in which case an inherently linear control valve will indeed yield a linear installed characteristic. In other words, if the pressure drop across the control valve never changes much, what we have is an “ideal” scenario (i.e. no “droop” or “distortion”) that doesn’t need special valve characterization to behave linearly.

Some other process applications actually work quite well48 with a quick-opening installed valve characteristic, in which case an inherently linear valve with varying pressure drop at di erent flow rates su ces. Ultimately, what matters is the valve stem position’s e ect on the process variable and whether that relationship is consistent (linear) over a wide range of operating conditions.

27.14Control valve problems

Control valves are subject to a number of common problems. This section is dedicated to an exploration of the more common control valve problems, and potential remedies.

47Such applications are typically found when the purpose of the control valve is to regulate process fluid pressure. Consider, for example, a control valve regulating upstream gas pressure in a vessel by venting gas from that vessel to atmosphere. In such an application, the valve’s upstream pressure (P1) will be nearly constant due to the control loop’s action, and the valve’s downstream pressure (P2) will be constant due to it being atmospheric pressure.

48An example of such a process is temperature control through a heat exchanger where the controlled fluid flow regime happens to transition from laminar to turbulent as the control valve opens further: at low stem positions (nearly shut) where the flow is laminar and heat transfer is impeded, large changes in flow rate may be necessary to e ect modest changes in temperature; at high stem positions (nearly open) where the flow is turbulent and heat transfer is e cient, only small changes in flow rate are necessary to create modest changes in temperature. In such an application a quick-opening installed characteristic may actually yield more consistent behavior than a linear installed characteristic.

If we graph the mechanical response of a pneumatic actuator with substantial stem friction, we see something like this:

Stem motion

Increasing actuator pressure

What should be a straight, smooth line is reduced to a series of “stair-steps” as the combined e ect of static and dynamic friction, plus the dynamic e ects of a pneumatic actuator, conspire to make precise stem positioning nearly impossible. This e ect is commonly referred to as stiction.

Even worse is the e ect friction has on valve position when we reverse the direction of pressure change. Suppose that after we have reached some new valve position in the opening direction, we begin to ramp the pneumatic pressure downward. Due to static friction (again), the valve will not immediately respond by moving in the closed direction. Instead, it will hold still until enough pressure has been released to diminish actuator force to the point where there is enough unbalanced spring force to overcome static friction in the downward direction. Once this static friction is overcome, the stem will begin to accelerate downward because (lesser) dynamic friction will have replaced (greater) static friction. As the stem moves, however, air volume inside the actuating diaphragm or piston chamber will decrease, causing the contained air pressure to rise. Once this pressure rises enough that the stem stops moving downward, static friction will again “grab” the stem and hold it still until enough of a pressure change is applied to the actuator to overcome static friction.

What may not be immediately apparent in this second “thought experiment” is the amount of pressure change required to cause a reversal in stem motion compared to the amount of pressure change required to provoke continued stem motion in the same direction. In order to reverse the direction of stem motion, not only does the static friction have to be “relaxed” from the last movement, but additional static friction must be overcome in the opposite direction before the stem is able to move that way. We may clarify this concept by applying the problem-solving strategy of adding numerical quantities to the thought experiment. Suppose we have a control valve with a static packing friction of 50 pounds in either direction, and a diaphragm with a 12-inch diameter (113.1 square inches of area). According to the force/pressure/area formula (F = P A), an applied

pressure of 0.442 PSI will be required to overcome this static friction in either direction. That is to say, when moving in the upward direction, we must apply 0.442 PSI more pressure than is ideally required to achieve any given valve position. If after achieving this stem position we desire to move the valve downward to some new stem position, we must not only decrease the pressure by 0.442 PSI to relax the tension on the packing, but we must also decrease it another 0.442 PSI to overcome packing friction in the downward direction before the valve moves at all from its last position. Thus, a pressure reversal of 0.884 PSI (i.e. twice the equivalent value of packing friction) is required to make the valve reverse its direction of motion. This constitutes “deadband” in the control valve’s action, which degrades control behavior.

Thus, the e ects of friction on a pneumatic control valve actuator may be quantified by subjecting the valve to small reversals in applied actuator pressure and measuring the resulting stem position. The largest increment of actuator pressure reversal resulting in zero stem motion represents the total amount of friction within the valve mechanism.

The following graph shows such a test, plotting actuator pressure over time as well as valve stem position over time. As the actuator pressure is stepped up and down in successively smaller intervals, the control valve’s stem position is seen to respond with less and less motion until it fails to respond at all:

Short of rebuilding a “sticky” control valve to replace damaged or worn components, there is not much that may be done to improve valve stiction other than regular lubrication of the packing (if appropriate). Lubrication is applied to the packing by means of a special lubricator device threaded into the bonnet of the valve:

As one of the common sources of excessive packing friction is over-tightening of the packing nuts by maintenance personnel eager to prevent process fluid leaks, a great deal of trouble may be avoided simply by educating the maintenance sta on the “care and feeding” of control valve packing for long service life.