FIG 8.4. Three computing elements and a set station provide the steadgstate heat balance. in Visual Studio .NET

Generator QR Code in Visual Studio .NET FIG 8.4. Three computing elements and a set station provide the steadgstate heat balance.

FIG 8.4. Three computing elements and a set station provide the steadgstate heat balance.
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FIG 8.5. If the steady-state calculation is correct, temperature will eventually return to the set point following a flow change.
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Some unknown factors do exist, however. No allowance was made for losses. If they are significant, and particularly if they change, an offset in exit temperature will result. Steam enthalpy could also vary, as well as the calibration of the steam flowmeter, should upstream pressure change. But for the most part, these fact,ors are readily accountable, whereas heat and mass transfer coefficients may not be. Response to a set-point change will be exponential, appearing as if the loop were open. Since moving the set point causes steam flow to move directly to the correct value, the response is exactly what was sought with complementary feedback (see Fig. 4.11).
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APPLYING DYNAMIC COMPENSATION
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The transient deviation of the controlled variable depicted in Fig. 8.5 was attributed to a dynamic imbalance in the process. This characteristic can be assimilated from a number of different aspects. If the load on the process is defined as the rate of heat transfer, then increasing load calls for a greater temperature gradient across the heat transfer surface. Since the purpose of the control system is to regulate liquid temperature, steam temperature must increase with load. But the steam in the shell of the exchanger is saturated, so that temperature can be increased only by increasing pressure, which is determined by the quantity of steam in the shell. Before the rate of heat transfer can increase, the shell must contain more steam than it did before. In short, to raise the rate of energy transfer, the energy level of the process must first be raised. If no attempt is made to add an extra amount of steam to overtly raise the energy level, it will be raised inherently by a temporary reduction in energy withdrawal. This is why exit temperature falls on a load increase. Conversely, on a load decrease, the energy level of the process must be reduced by a temporary reduction in steam flow beyond what is required for the steady-state balance. Otherwise energy will be released as a transient increase in liquid temperature.
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The dynamic response can also be envisioned simply on the basis of the velocity difference between the two inputs of the process, although this is less representative of what actually takes place. The load change appears to arrive at the exit-temperature bulb ahead of a simultaneous steam-flow change. To correct this situation, steam flow must be made to lead liquid flow. The technique of correcting this transient imbalance is called dynamic compensation.
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Determining the Needs of the Process
Capacity and dead time can exist on both the manipulated and the load inputs to the process. There may also be some dynamic elements common to both, such as the lags in the exit-temperature bulb for the heat exchanger. The relative locations of these elements appear as shown in Fig. 8.6. A feedback controller must contend with g, X g,, which are in series in its closed loop. But the feedforward controller need only be concerned with the ratio g,Jg,, in order to make the corrective action arrive at the divider at the same time as the load. Recall the appearance of this ratio in both Eqs. (8.2) and (8.4). In some difficult processes, the manipulated variable enters at the same location as the load, e.g., in a dilution process where all streams enter at the top of a vessel. In this case, even though g, may be quite complex, g, and g, couId be nonexistent, making dynamic compensation unnecessary. Perhaps the easiest way to appreciate the need for dynamic compensation is to consider a process in which g, and g, are dead time alone. Let 7q and 7% represent their respective values. The response of the controlled variable as a function of time is
c(t) = K,
m(t - Tm)
a@ - 7,)
The division makes the process fundamentally nonlinear, which complicates dynamic analysis. To allow inspection of the transient response of the process, analysis must be made on an incremental basis, by differentiating both sides of the equation: dc (t) = K, dm (t - TV) _ m dq (t - T,) Q q2 1 (8.5)
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