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how to use barcode scanner in asp.net c# T+ UA + FpC VPC  H,Fx,,(dy/aT),dt = Tr in .NET framework
T+ UA + FpC VPC  H,Fx,,(dy/aT),dt = Tr QR Recognizer In .NET Using Barcode Control SDK for .NET framework Control to generate, create, read, scan barcode image in .NET framework applications. Printing Quick Response Code In Visual Studio .NET Using Barcode creator for .NET Control to generate, create QR image in Visual Studio .NET applications. The thermal time constant is rT = UA + FpC  H,Fx,,(ay/c T). QRCode Decoder In VS .NET Using Barcode recognizer for .NET framework Control to read, scan read, scan image in VS .NET applications. Bar Code Generation In Visual Studio .NET Using Barcode encoder for Visual Studio .NET Control to generate, create barcode image in Visual Studio .NET applications. (10.19) Scan Bar Code In Visual Studio .NET Using Barcode recognizer for .NET Control to read, scan read, scan image in VS .NET applications. Draw QR Code In Visual C# Using Barcode printer for VS .NET Control to generate, create QR Code ISO/IEC18004 image in .NET applications. If T, is the manipulated variable, t,he steadystate process gain turns out to be UA UA + FpC  H,Fxo(ay/aT), (10.20) Generate Quick Response Code In Visual Studio .NET Using Barcode encoder for ASP.NET Control to generate, create Denso QR Bar Code image in ASP.NET applications. Encoding QRCode In Visual Basic .NET Using Barcode drawer for .NET framework Control to generate, create QRCode image in VS .NET applications. P66 [ A p p l i c a t i o n s
Generating Data Matrix 2d Barcode In Visual Studio .NET Using Barcode creation for .NET Control to generate, create DataMatrix image in .NET applications. GS1 DataBar14 Encoder In .NET Using Barcode creator for .NET Control to generate, create GS1 DataBar Stacked image in .NET framework applications. If the reactor is unstable, bbth the gain and the time constant will be negative. The denominator in both expressions is the difference between the slopes of the heatremoval and heatevolution curves, as in Fig. 10.3. If both denominators are positive, the reactor behaves as a simple firstorder lag. If both are negative, positive feedback dominates; the dynamic gain is the same as a simple lag, but the phase angle goes from 90 at zero period to 180 at an infinite period: 4~ = 7r + tan 2n z 70 (10.21) 2D Barcode Generator In .NET Using Barcode printer for Visual Studio .NET Control to generate, create Matrix 2D Barcode image in VS .NET applications. Standard 2 Of 5 Printer In .NET Using Barcode drawer for Visual Studio .NET Control to generate, create Code 2 of 5 image in .NET framework applications. The  r indicates a negative steadystate gain, while the plus sign in front of the tan indicates a negative time constant. Both the time constant and the steadystate gain can also approach infinity, in which case the reactor acts as an integrator whose dynamic gain at period ~~ is To .+EE2nVpC/ CIA 277TT (10.22) Bar Code Decoder In C# Using Barcode recognizer for Visual Studio .NET Control to read, scan read, scan image in .NET applications. EAN13 Creator In Visual C#.NET Using Barcode encoder for .NET Control to generate, create GTIN  13 image in Visual Studio .NET applications. Equation (10.22) defines the dynamic asymptote of process gain for all conditions of stability, exhibiting an effective time constant of VpC/UA. A stable reactor can operate without temperature control; regulation of T, alone is ordinarily sufficient. But an unstable reactor will drift away from the control point in either direction at an everincreasing rate, unless feedback control is enforced. Unfortunately, it may not always be possible or economical to design for stability. Enough heat transfer area must be provided so that only about 50 to 60 F differential is required across it to remove the rated flow of heat. (This is an estimate of the T  T, ordinarily required to exceed dT/dy, such as that given in Fig. 10.2.3) Stability will be assured if heat is removed by boiling one or more of the ingredients in the reaction, since this makes the system almost isothermal. On the other hand, if heat is removed by a mechanism like evaporation of liquid into a dry gas stream, its flow may change very little with temperature. In this case the slope of the heatremoval curve would be slight, and the reactor could be expected to be unstable. All of the foregoing statements on stability were used to describe openloop situations. Some unstable reactors can be given steadystate stability by applying enough negative feedback from the control system to overcome the positive feedback of the reaction. To visualize how this is possible, consider the proportional control loop of Fig. 10.4 for steadystate conditions only. Figure 10.4 can be represented mat.hematically by Make Universal Product Code Version A In None Using Barcode encoder for Font Control to generate, create GTIN  12 image in Font applications. UPC  13 Decoder In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. T c = (, Linear Generator In Visual Basic .NET Using Barcode generation for .NET Control to generate, create 1D Barcode image in VS .NET applications. Generating EAN / UCC  13 In VB.NET Using Barcode maker for .NET Control to generate, create EAN 13 image in VS .NET applications. T)=x
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FIG 10.4. Negative feedback of the controller must overcome positive feedback in the reactor in order to attain steadystate stability. FIG 10.5. The dynamic gain varies almost linearly with the amount of dead time in the loop.
The closedloop steadystate gain is found by solving for T/T,: 1 1 ;, = 1 + P/100& Steadystate stability is identified by positive gain. to be positive, (10.23) 111 order for T/T, i&P If KT were posit ive, P could have any value, because the reactor would be stable with the loop open. But with KT negative, P cannot be greater than 100 KT: if KT = 2, P < 2007& This sets an upper limit on P. The dynamic properties of the rest of the loop set a lower limit on P. The nat ural period of the temperaturecontrol loop is found by equating the sum of the phase lags of all dynamic elements to 180 . Since the phase of a negative lag is between 90 and 180 , there is little room for other elements. This clearly rules out integral control. If all the other dynamic elements in the loop can be lumped toget her as dead time rd, t he period of oscillation can be found by equating the sum of the phase lags to 180 : r = 242  r + tan1 2=721 70 To Having found 70, determined : the dynamic gain of the unstable reactor can be A plot of dynamic gain vs. the ratio of Td/rT is given in Fig. 10.5. dynamic gain of a stable react,or is included for comparison.

