barcode reading in asp.net wT~ = wsTis + w,(Ti - Ti,) + Ti,(W and in Software

Printer Code-128 in Software wT~ = wsTis + w,(Ti - Ti,) + Ti,(W and

wT~ = wsTis + w,(Ti - Ti,) + Ti,(W and
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(21.4) (21.5)
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wTo = wsTos + w,(To - To,) + To,(w - ws)
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Notice that for these cases the nonlinear terms are wT; and wT,. The first partial derivatives, evaluated at the operating point, are
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*The reader may refer to I. S. Sokolnikoff and R. M. Redheffer (1966) for further discussion of this expansion.
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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 = m-C% 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)
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where Kr = K2
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UA i- w,C = UA UA + w,C
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 first-order with a time constant rW. The steady-state 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 222-202 . 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.
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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
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