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Inertial Lag of a Flowing Liquid
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In the steady state, the velocity of flow in a pipe varies with pressure drop :
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AP IL2 = c2 2g P
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Analysis of Some Common Loops
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where u = velocity, ft/sec C = flow coefficient g = gravity, ft/sec2 Ap = pressure drop, lh/ftz p = density, lb/ft But velocity is proportional to flow: F ZL = A where F = flow, ft3/sec A = inside area, ftz Therefore the pressure drop due to flow in the steady stat e is AP=~= 2gA2C2 If the applied force A Ap exweds resistance to flow, acceleration takes place. An equation can he written for the unsteady state: net force equals mass times accelcrat,ion. AAp AF2p
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2gA2C2
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du M dF - a = x z
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where M = mass, slugs t = time, set The mass of fluid in the pipe is
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where L = length in feet. Rearranging,
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To find the time constant, the differential equation must be reduced to its standard form: 2gC2A2 Ap = pFThe time constant is then the coefficient of dF/dt: r2LAC2 F (3.1)
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1 Understanding
Feedback
Control
Flow coefficient C2 can be replaced by its steady-state equivalent:
= 2gA2 Ap
leaving (3.2)
example
To test the significance of the last expression, a numerical example is Consider a 200-ft length of l-in. Schedule 40 pipe, containing presented. mater flowing at 10 gpm with a 20-psi drop. L = 200 ft
F = 10 gpm = 0.0223 ft3/sec
g = 32.2 ft/seG il = 0.006 ft2 Ap = 20 lb/in2 = 2,880 lb/ftz = (32.2)(0.006)(2,880)
p = 62.4 Ib/fV
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Notice that the time constant varies with both flow and pressure drop, because of the square relation between the two. Nevertheless, the derivation permits evaluation of the dynamic response at a nominal flow and at least a qualitative indication of t,he response elsewhere. As may have been anticipated, the time con&ant, is small, but not, zero, except, at, zero
flOW.
Dynamic Elements Elsewhere in the Loop
This time constant is fundamentally the only dynamic element in the process. But its response is of the same order of magnitude as the instruments in the control loop, and therefore t,hc entire loop must) be analyzed. Figure 5.1 describes a pneumatic flow-control loop consistjing of t ransmit,ter (2), controller (4), valve (G), and two transmission lines (3,s). The flow knnsmittcr contains an amplifier wit,h certain dynamic prop-
FIG 3.1. At least six elements contribute to the dynamic response of the flow-control loop.
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FIG 3.2. Because of its velocity limit, a control valve appears to folloz small signals faster than large signals.
Small-signal response
Time
erties. Because of the amplifier, the lag of the transmission line is isolated from that of the flowing fluid. The transmission line is terminated by a controller, isolating it from the second trnnsn~ission line. The figure shows no isolating amplifier between the output line and the valve, however, allowing interaction there. Transmission lines can be conveniently represented by dead time plus a first-order lag. The vnluc of each is naturally a function of length and dianieter.1 The control valve cannot be so easily represented, however. If a valve motor were a constant-volume device, it would behave like a first-order lag. But every change in pressure is accompanied by a change in volunle of the motor. This property causes the motor to operate at a limited velocity, based on the n~asinnm~ rate of flow of air that can be delivered into the expanding volunw. In effect, a valve seems to exhibit a snlnller time constant for snlall changes than for large changes, because the velocity of stroke is not :I function of the nxlgnitude of the change. Consequcntly, :L valve cannot be adequately represented by a single time
collstnllt .
example
.\ clowtl about 6.5 nrinrd by l)linse and
lool) of this d(~scril)tion \Yill bc found to owillntc at n. period of sw. untlcr ~~~o~~o~tiol~nl-~~lns-~esrt control. I lir Iwriod is detcrthe I)h:tsc contribution of all the clcllwnts. Table 3.1 lists the g&iii cwitributioii of qwh at the 0.5~wc Iwiod. Thr loons ~)hnsr uxs found I)\- first srlrcting 6.5 see 3s the nntwal prriod; rcsct action nlust then contril)utc 2s t o bring the total t o 1SO . (.\ diffcrcnt value of rcsct tilllc I\-otlld challge 7,.) The gaili contribution of rcsetj at that l)hasr niik~lc is 1.11 . Sotiw that all rlrmcnts cwntributc s0111e l)llaSe lag, but, only thaw nhk~ phase lag approac~hcs -Go sffrcat the 1001) w i n notiwnbly.