CV ( t = 2 ) = K I 0 Edt + ∫ CV ( t = 1 )
CV ( t = 2 ) = K I 0 Edt + ∫ CV ( t = 1 )
t = = 2 K Et
I t = 1 + CV ( t = 1 )
05 . sec ) ( 10 %)( 21 −+ ) 75 %
Therefore, the new control variable output from t = 1 to t = 2 will be 80%. If the controller error drops to 5% for the next two seconds (from t = 2 to t = 4), the controller output will increase steadily from 80% (at t = 2) to 85% (at t = 4). After t = 4, the error is 0%, therefore, the controller output stays at 85%. Note that the family of curves shown in Figure 15-41b is the product of
K I E for a particular value of error. If the error stays constant for t seconds, then the change in the value of CV over that time period will follow the K I E
curve for that error value.
The inverse of the gain term K I is referred to as the integral time (T I ), or reset time, in seconds. The integral time is the time it takes for the control variable (CV) to change 1% for a 1% change in error. It is expressed as:
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The T I variable is used by some manufacturers to allow the user to indirectly enter the integral gain into the controller. If the integral time must be specified in minutes, as is required by some manufacturers, a simple conversion can
change T I from seconds to minutes:
(in seconds) or
K I () K I () min
60 sec ( in minutes )
So, for the previous example, the reset time is equal to:
= 1 − 05 1 . sec
= 2 seconds
The integral controller mode is also referred to as reset action, because it automatically resets the error to zero over time.
E X AM PLE 1 5 -6
Illustrate the transfer function of an integral controller with a gain of K I = 0.2 sec –1 over a process variable range of 100 °
F to 200 ° F. Plot the response of the controller’s output for an error due to a permanent load disturbance of +10% above the set point of 150 °
F over the full range. Two seconds after the controller increases its output, the error will drop by 5%. After 3 more seconds, the error will become zero. Find
the value of CV after 5 seconds.
S OLU T I ON
Figure 15-43 shows the integral controller’s transfer function. When the error is +10% above the set point, the process variable will be at 160 °
F, which will cause the controller’s output to increase at a rate of 2% per second:
dCV = KE I dt
= ( . )( ) 0 2 10 = 2 % per second
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S ECTION PLC Process Process Controllers C HAPTER 4 Applications
and Loop Tuning 15
10% Error PV = 160 ° F
PV 100 ° F 150 ° F 200 ° F
E =0
E (% Range) = 200 °
F – 100 ° F = 100%
F – 100 ° F
K = 10%/sec – (–10%/sec) I 100%
20%/sec
%/sec = 0.2 –1 = 0.2 sec
Figure 15-43. Integral controller’s transfer function.
As Figure 15-44 illustrates, after 2 seconds of integral action, the controller output will be 54%:
CV ( t = 2 ) = K I Edt
t = 0 + ∫ CV ( t = 0 )
After the 2 seconds have elapsed, the error will drop to 5% and the controller will integrate at a rate of:
dCV = KE I
dt = ( . )( %) 025
= 1 % per second
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S ECTION PLC Process Process Controllers C HAPTER 4 Applications
and Loop Tuning 15
E =10% PV = 160 ° F
PV = SP =150 ° F
Figure 15-44. Controller output.
Therefore, at the end of the next 3 seconds, the controller output will
At this point, the error will drop to zero, so the controller will stop changing the CV , maintaining its output at a new zero error value
of 57%.
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