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Pressure drop across flow system: 100 psi pipe length: 200 fi, Inside pipe diameter: 1.0 in. C, = 4.18, mo = 0.03 x 4h Ihear psi 0 100 0 10 94.6 0.25 20 78.5 0.54 30 51.4 1.00
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* For the equal percentage valve, nzo = 0.03 when x = 0.
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FIGURE 20-6 Comparison of effective valve characteristics of a linear valve and an equal percentage valve from Example 20.3. I ideal linear characteristic, II linear valve, III equal percentage
valve.
Equal percentage valve. Calculation of the effective characteristic will now be made for an equal percentage valve having the same C, of 4.18 as calculated for the linear valve in the first part of this example. For mg = 0.03, the value of /3 is calculated to be /3 = ln(l/mo) = ln(U0.03) = 3.51 For a flow rate of 20 gpm, m = q/q- = 2Of37 = 0.54
Solving Eq. (20.9) for x and inserting the values for p, mg, -and m give n = (1/3.51)1n(0.54/0.03) = 0.82 For other values of flow, corresponding values of n are calculated and the results are shown in Table 20.2 and Fig. 20.6. BENEFIT OF AN EQUAL PERCENTAGE VALVE. It is often stated in the control
literature that the benefit derived from an equal percentage valve arises from its inherent nonlinear characteristic that compensates for the line loss to give an effective valve characteristic that is nearly linear. A study of Fig. 20.6 shows that in this example an equal percentage valve overcompensates for line loss and produces an effective characteristic that is not linear, but is bowed in the opposite direction to that of the effective characteristic of the linear valve. In summary,
neither valve in this example produces an effective characteristic that is linear.
One can show that as the line loss increases, the linear valve will depart more from
the ideal linear relation and the equal percentage valve will move more closely
toward the linear relation. In practice, a valve designated as linear will not give a linear characterisitic exactly as defined in this chapter. To achieve a truly linear characteristic would require very careful design and precision machining of the valve plug and seat. The same comment can be made for an equal percentage valve, as defined by Eq. (20.10). In order to know the effective characteristic of a valve, one must test it experimentally.
VALVE POSITIONER
The friction in the packing and guiding surfaces of a control valve causes a control valve to exhibit hysteresis as shown in Fig. 20.7, in which stem position is
CONTROL VALVES
pv, signal
to valve
FIGURE
Control valve hysteresis.
20-7
plotted against valve-top pressure. When the pressure increases, the stem position increases along the lower curve. When the pressure decreases, the stem position decreases along the upper curve. At the moment the air pressure signal reverses, the stem position stays in the last position until the dead band H is exceeded, after which the pressure begins to decrease or increase along the paths shown by the arrows. If the valve is subjected to a slow periodic variation in pressure, a typical path taken by the stem position is shown by the closed curve ABCDA in Fig. 20.7. The hysteresis described in the previous paragraph should be distinguished from the dynamic lag of a valve discussed in Chap. 10. The dynamic lag discussed in Chap. 10 is caused by the volume of space above the valve diaphragm, the resistance to flow of air to the valve top, and the inertia of the valve stem and plug; such a lag is expressed by a first-order or second-order transfer function. On the other hand, hysteresis, which is caused by the friction between the stem and the packing, is a nonlinear phenomenon and cannot be expressed by a transfer function. A valve can exhibit both dynamic lag and hysteresis. The presence of hysteresis in the valve can cause the controlled signal to exhibit an oscillation or ripple called a limit cycle. Since this limit cycle is usually considered objectionable and contributes to wear of the valve, a method is needed to eliminate it. Since the limit cycle is a nonlinear phenomenon related to the hysteresis, controller tuning is not a solution to the problem. To reduce the deleterious effect of hysteresis and to also speed up the response of the valve, one can attach to the control valve a positioner which acts as a high-gain proportional controller that receives a set-point signal from the primary controller and a measurement from the valve stem position. In this sense, the addition of a valve positioner introduces a form of cascade control, which was discussed in a previous chapter. A sketch of a control valve with a positioner attached is shown in Fig. 20.8. The positioner, bolted to the valve actuator, has an arm that is clamped to the valve stem to detect the stem position. Notice that the valve positioner shown in Fig. 20.8, has the usual connections for a controller: a set point that calls for a desired stem position in the form of a signal from the primary controller pc, a measurement in the form of stem position x, and a pneumatic output in the form of a pressure to the valve top pv. The mechanical details of an actual valve positioner involve a pneumatic mechanism functioning as a high-gain proportional controller. The gain is built
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