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Process Gain
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Each of t,hese lags forms one link in the chain from average coolant temperature to the measured reactor temperature. But since t,he manipulated variable is the flow of water added to the coolant stream, a suitable equation converting water flow to coolant temperature must be included. Adding a stream FTr at temperature TV to the coolant recycle stream F - Frr- at temperature Tc2 produces a mixture F at t emperature T,I, returning to the reactor. The heat balance is FT,] = FrTw + ( F - Fw)Tcx Rearranging, Tcz - T,I = (Te2 - Tr) T But, related to the heat load,
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(3.19)
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Since the response of average coolant temperature T, is sought, substitution is made for Te2: Tc2 zz , + Tc2 ; cl = T, +
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Combining Eqs. (3.19) and (3.20),
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Solving for T,, Tc= Tw+$(&-$ Process gain is the derivative of temperature with respect to flow: dTc Q - = -~ C3Fw2 dFw
(3.22) (3.21)
The adjustment of coolant temperature by water flow is demonstrably nonlinear. An equal-percentage valve should be used to deliver the water, to partially correct this situation.
example 3.4
If a reactor contains 40,000 lb of material of specific heat of 0.8 Btu/ (lb)(V), evolving 20000 Btu/min at 200 F aith a wall temperature of 170 F,
7, = (40,000)(0.8)(200 - 170) = 48 min 20000
72 can be estimated from the weight of the reactor wall, 8,000 lb, of specific heat 0.15 and a temperature gradient of 10 F:
72 = (8,000)(0.15)(10) = o 6 min 20000
Jacket contents of 500 gal (4,160 lb) of water at an average temperature of 140 F exhibits a time constant of
(4,160)(1.0)(160 - 140) = 4,2 min 20000
A typical value for lag in a temperature well is i-4 = 0.5 min. Finally, circulation through the jacket at a rate of 250 gpm yields a dead time
500 . ~~ = 250 = 2 mm
It happens that a reactor of this description will oscillate at a period of about 35 min in a closed loop. Even if all the secondary elements consisted of pure dead time, they could not cause the period to exceed 29 min. Therefore, some secondary element remains hidden, and the only place it could hide is in the reaction mass. The assumption has been made, in calculating its time constant, that the reaction mass was perfectly mixed -that it was all at the same temperature. This, of course, is a false premise, because it is impossible to transport fluid, hence heat, from the wall of the vessel to the temperature bulb in zero time. Heat is transferred both by convection and by conduction-conduction would be the mechanism if the fluid were motionless. It has been pointed out that
Analysis of Some Common Loops
heat transfer by conduction is a distributed process, involving some effective dead time. So it does not seem unreasonable that a small percentage of the 4%min primary time constant is dead time due to imperfect mixing. An examination of the mechanism of mixing will be taken up under composition control. example 3.5 The dynamic gain of the process is principally that of the primary time constant:
The response of average coolant temperature to water flow is plottedin Fig. 3.7 for values of F = 250 gpm (2,080 lb/min) and Tw = 80 F at a constant load of 20000 Btu/min. Because of the change in slope with flow, an equal-percentage valve characteristic is recommended. From Fig. 3.7, the required flow of water to produce an average coolant temperature of 140 F is found to be 37 gpm. Gain of the process is
Q dTc dFw --c3pw2
20000 Btu/min = - [l.O Btu/(lb)( F)][37 gpm12[8.33 lb/gal] = -1 75 F gpm
Gain of an equal-percentage valve is simply four times the flow being delivered :
G Y = 4 ~ = 1.5 37 gprn
100%
gpm/%
If a transmitter span of 200 F is selected, GT = 100 0/,/200 F, or 0.5 yO,/ F. The gain product can then be found: G = (0.116)(1.5 gpm/~)(-l.75 F/gpm)(0.5%/ F) (The negative sign indicates the sense of control action.) damping, P must be 200G, therefore P = 200(0.152) = 30% = -0.152 For >i-amplitude
:: -250
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