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barcode reading in asp.net wT~ = wsTis + w,(Ti  Ti,) + Ti,(W and in Software
wT~ = wsTis + w,(Ti  Ti,) + Ti,(W and Code 128C Reader In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code128 Generator In None Using Barcode generation for Software Control to generate, create Code128 image in Software applications. (21.4) (21.5) Scan Code 128 Code Set C In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. Painting ANSI/AIM Code 128 In C#.NET Using Barcode drawer for VS .NET Control to generate, create Code 128A image in .NET framework applications. wTo = wsTos + w,(To  To,) + To,(w  ws) Printing Code 128 Code Set B In .NET Using Barcode creation for ASP.NET Control to generate, create Code 128C image in ASP.NET applications. Code 128C Printer In VS .NET Using Barcode printer for .NET framework Control to generate, create Code 128C image in VS .NET applications. Notice that for these cases the nonlinear terms are wT; and wT,. The first partial derivatives, evaluated at the operating point, are ANSI/AIM Code 128 Printer In Visual Basic .NET Using Barcode maker for .NET framework Control to generate, create Code 128B image in .NET applications. Bar Code Generator In None Using Barcode creator for Software Control to generate, create bar code image in Software applications. and so on.
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Painting Code 3/9 In None Using Barcode maker for Microsoft Word Control to generate, create ANSI/AIM Code 39 image in Office Word applications. Generate Data Matrix In None Using Barcode creation for Font Control to generate, create ECC200 image in Font applications. Introducing Eq. (21.4) and (21 S) into (21.1) gives the following linearized equation: [(Ti,  To,)(w  w,) + w,(Ti  T,)]C + UA(T,  To) = rnC% At steady state, dT,ldt = 0, and Eq. (21.1) can be written w,C(Ti,  To,) + UA(T,,  To,) = 0 Subtracting Eq. (21.7) from (21.6) and introducing the deviation variables T/ T; T; w and rearranging give the result C[(Ti$  T,,)W f w,(T/  Td)] + lJA(T:  TJ = mC% Taking the transform of Eq. (21.8) and solving for T;(s) give (21.8) = = = = Ti  Ti, To  To, T,,  Tys w  w , (21.7) (21.6) Print GTIN  12 In None Using Barcode generation for Online Control to generate, create UPCA image in Online applications. Generate Bar Code In Java Using Barcode creation for BIRT Control to generate, create barcode image in BIRT reports applications. where Kr = K2
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K = Wo,  Ti,) 3 UA + w,C mC rw = UA + w,C From Eq. (21.9), we see that the response of Td to T;, Ti, or W is firstorder with a time constant rW. The steadystate gains (KS ) in Eq. (21.9) are all positive. The following energy balance can be written for the steam side of the kettle: wvHv  w,H, = UA(T,  To) + vd:;Uv) + rnlCI% (21.10) Notice that we have made use of assumption 4 in writing the last term of Eq. (21. lo), which implies that the metal in the outer jacket wall is always at the steam temperature. A mass balance on the steam side of the kettle yields WV  WC =Vdp dt
(21.11) THEORETICAL
ANALYSIS
COMPLEX
PROCESSES
Combining Eqs. (21.10) and (21.11) to eliminate wc gives
W (H,  H,) = (U,  H,)Vfg + tTqCl% + UA(T,  To) dUv + VP,dt (21.12) The variables pv, Uy , H,, and H, are functions of the steam and condensate temperatures and can be approximated by expansion in Taylor series and linearization as follows: Pv = Pv, + 4Tv  TV,) uv = U , + W  TV,) Hv = Hv, + y(Tv  TV,) Hc = Hc, + o(Tc  Tc,) (21.13) where o = $$ IS Y
dHv Y =dTv dK a=dT, Is Is
The parameters (Y, 4, y, and g in these relationships can be obtained from the steam tables once the operating point is selected.* Introducing the relationships of Eq. (21.13) into Eq. (21.12) and assuming the condensate temperature T, to be the same as the steam temperature TV give the following result: *For example, if the operating point is at 212 F and the deviation in steam temperature is 10 F, we obtain the following estimate of y from the steam tables: TVs H,, At Ty H, At T, H, = 212 F = 1150.4 Btu/lb
= 222O F, = 1154.1 = 202' F, = 1146.6 1154.1  1146.6 = o 375 Y z 222202 . and H, = 1150.4 + 0.375(T,  212) In a similar manner, the properties of saturated steam can be used to evaluate LY, 4, and u. PROCESS
APPLICATIONS
b%,  H,, + (Y  u)(Tv  Tv,)Iw, .= i (U,,  H,,) L
Some of the terms in Eq. (21.14) can be neglected. The term (Y  M v  Tv,) , can be dropped because it is. negligible compared with (H,,  H,,). For example, for steam at atmospheric pressure, a change of lOoF gives a value of (y  a)(T,, T,,) of about 7 Btu/lb while (H,,  H,,) is 970 BtuIlb. Similarly, the term (24  (T)(T,  T,,) can be neglected. For example, this term is about 4 But/lb for a change in steam temperature of lOoF fbr steam at about 1 atm pressure; the term WV,  H,,) is 897 Btu/lb under these conditions. Also, the term c#~p,/ct is about 15 BtuAb and can be neglected. Discarding these terms, writing the remaining terms in deviation variables, and transforming yield (21.15) where Ti = T,  Tys

