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barcode reading in asp.net THEORETICAL in Software
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Bar Code Generator In Java Using Barcode drawer for Android Control to generate, create bar code image in Android applications. Painting Code 128 Code Set A In Java Using Barcode printer for BIRT reports Control to generate, create Code128 image in BIRT applications. (21.55) An energy balance is next written for the metal in the wall, over the volume element of length Ax. This can be stated as follows: Expressing each term in this balance by symbols gives rDohoAX(Ty  Tw)  nDihiAx(Tw  T) = A,AxP,C,$ Simplifying this expression gives (21.56) 8TW  &TV  T,,,)  &(T,  T) dt  952 where 1
TDihi
&pwC, rD,ho
 = &p&v 722
We now have obtained the differential equations that describe the dynamics of the system. As in previous problems, the dependent variables will be tmnsformed to deviation variables. At steady state, the time derivatives in Eqs. (21.55) and (21.57) am zero, and it follows that PROCESS
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0 = vz + L(T,,  Ts) (21.58) 0 = $(T,,  Tw,)  $v,  Td
where the subscript s is used to denote the steadystate value. Note that to determine the steadystate values of the temperature requires the solution of two simultaneous equations, the first of which is an ordinary differential equation. Thus, the steadystate temperature T, is a function of x and may be obtained by solution of Eqs. (21.58) and (21.59) as Ts = TV, + K,  T&XP [ ~71 x/(l+:)l
where Tso is the normal entrance temperature. All equations for T ' to be derived below should be recognized as deviations from this expression. Subtracting Eq. (21.58) from (21.55) and Eq. (21.59)from (21.57) and introducing deviation variables give , dT' $ + +T; T') dt= 71
(21.60) dT,: $T;  T;) $(T;  T') 1 dt where T'= (21.61) TT, T; = T,  T,, T; = TV  TVs I S ; = $g + L(T:,  ; ) Equations (21.60) and (21.61) may be transformed with respect to t to yield (21.62) (21.63) where T = ; (x ,s) T; = ;;(x,s> ;I = T:(s) In Eqs. (21.62) and (21.63) it has been assumed that the exchanger is initially at steady state, so that T(x,O) = T,, T,(x,O) = T,,, and T,(O) = Tys. THEORETICAL ANALYSIS OF COMPLEX PROCESSES
Eliminating ;k from Eqs. (21.62) and (21.63) gives, after considerable simplification (21.64) where a(s) = s + ! 71 722 71(7127228 712 71(7127229 + 712 + 722) + 712 + 722) b(s) = Equation (21.64) is an ordinary firstorder differential equation with boundary condition ; (x,s) = ; (O,s) at x = 0. It can be readily solved to yield T (x, s) = ; (O, s) + [l  c(a )x] $(s)  ; (O, s) i where ; (O,s) is the transform of the fluid temperature at the entrance to the pipe and F:(s) is the transform of the steam temperature. From Eq. (21.65), the transfer functions can be obtained as follows: If the steam temperature does not vary, T:(s) = 0; the transfer function relating temperature at the end of the pipe (X = L) to temperature at the entrance is a+ $1 _ e (alv)L T (0, s) (21.66) Setting l/r1 to zero in the expression for a(s) [Eq. (21.64)] shows that a(s) = s and hence the response is simply that of a transport lag. This is in agreement with the physical situation where hi approaches zero [Eq. (21.55)], for which case the wall separating cold fluid from hot fluid acts as a perfect insulator. We saw in Chap. 8 that this situation is represented by a transport lag. If the temperature of the fluid entering the pipe does not vary, the transfer function relating the exit fluid temperature to the steam temperature is T (L, s) = $[I  ewv)q (21.67) m In principle, the response in the temperature of the fluid leaving the exchanger can be found for any forcing function, T(O,t) or T,(t), by introducing the corresponding transforms into Eq. (21.66) or (21.67). However, the resulting expression is very complex and cannot be easily inverted. For the case of sinusoidal inputs, the substitution rule discussed in Chap. 16 can be used to determine the AR and phase angle of the frequency response. Cohen and Johnson give a Bode diagram corresponding to Eq. (21.67) for a specific set of heatexchanger parameters. This diagram is shown in Fig. 21.10.

