how to use barcode scanner in asp.net c# Inertial Lag of a Flowing Liquid in Visual Studio .NET

Paint Quick Response Code in Visual Studio .NET Inertial Lag of a Flowing Liquid

Inertial Lag of a Flowing Liquid
QR Code JIS X 0510 Reader In .NET Framework
Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in .NET framework applications.
QR Code JIS X 0510 Creator In .NET
Using Barcode generation for .NET Control to generate, create Denso QR Bar Code image in .NET framework applications.
In the steady state, the velocity of flow in a pipe varies with pressure drop :
QR Code ISO/IEC18004 Decoder In Visual Studio .NET
Using Barcode scanner for .NET framework Control to read, scan read, scan image in .NET framework applications.
Barcode Generator In Visual Studio .NET
Using Barcode maker for Visual Studio .NET Control to generate, create barcode image in Visual Studio .NET applications.
AP IL2 = c2 2g P
Recognizing Bar Code In VS .NET
Using Barcode recognizer for VS .NET Control to read, scan read, scan image in .NET applications.
QR Code JIS X 0510 Printer In C#.NET
Using Barcode creation for VS .NET Control to generate, create QR Code ISO/IEC18004 image in .NET applications.
Analysis of Some Common Loops
Generate QR Code In .NET
Using Barcode drawer for ASP.NET Control to generate, create QR Code image in ASP.NET applications.
QR Code ISO/IEC18004 Generation In VB.NET
Using Barcode encoder for Visual Studio .NET Control to generate, create QR Code JIS X 0510 image in VS .NET applications.
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
Matrix 2D Barcode Creator In .NET
Using Barcode generator for Visual Studio .NET Control to generate, create Matrix 2D Barcode image in VS .NET applications.
Create UPC Code In Visual Studio .NET
Using Barcode generation for .NET framework Control to generate, create Universal Product Code version A image in .NET applications.
2gA2C2
Barcode Creation In .NET Framework
Using Barcode printer for .NET Control to generate, create bar code image in VS .NET applications.
RM4SCC Creation In .NET Framework
Using Barcode creation for VS .NET Control to generate, create RoyalMail4SCC image in VS .NET applications.
du M dF - a = x z
EAN128 Generator In VB.NET
Using Barcode encoder for .NET framework Control to generate, create EAN128 image in .NET applications.
Code 3/9 Creation In Objective-C
Using Barcode drawer for iPhone Control to generate, create USS Code 39 image in iPhone applications.
where M = mass, slugs t = time, set The mass of fluid in the pipe is
Print UPC-A Supplement 5 In Objective-C
Using Barcode creation for iPad Control to generate, create UPC-A image in iPad applications.
Data Matrix 2d Barcode Decoder In Visual Basic .NET
Using Barcode scanner for VS .NET Control to read, scan read, scan image in .NET framework applications.
where L = length in feet. Rearranging,
Creating Code 128 Code Set B In Visual Basic .NET
Using Barcode maker for VS .NET Control to generate, create Code-128 image in .NET applications.
GS1 128 Printer In None
Using Barcode creation for Microsoft Word Control to generate, create GS1-128 image in Word applications.
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)
Reading Code 128B In C#
Using Barcode recognizer for .NET framework Control to read, scan read, scan image in .NET framework applications.
Print Bar Code In Java
Using Barcode generation for Eclipse BIRT Control to generate, create barcode image in BIRT reports applications.
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
@fWW2W(@.4)
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.
Analysis of Some Common Loops
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.
Copyright © OnBarcode.com . All rights reserved.