In this screencast we will take information about a process and put it together in order
to determine the ultimate gain and ultimate period of a process. For this example we'll
be looking at a temperature control loop which will be controlling a temperature via the use
of cooling water. We're given some information about the transfer function, and a time constant
for the process transfer function. Additionally we're given some information about the sensor/transmitter,
and the control valve. To find the ultimate gain and period requires the use of the direct
substitution method. Recall that the direct substitution method is a method to indicate
the stability of a process, and the ultimate gain and ultimate period represent the edges
of that stability between a stable process and an unstable process. So in order to use
the direct substitution method we need to know the characteristic equation. The characteristic
equation is one plus all the parts of the open loop gain, which will include the process,
the valve, the controller, and the transmitter. So what we'll now do is kind of go through each
of these to evaluate them. So our process is given that we have a transmitter gain of
negative five in degrees Celsius per liter per second, and a time constant of ten. We'll
get to the valve in just a second. The controller we use is recall that when we analyze using
direct substitution for the ultimate gain we always assume a p-only controller, so therefore,
here this will be kc and we're looking for a gain in percent controller output per percent
transmitter output. The transmitter has a time constant of three seconds, so therefore
this will be three s plus one in the denominator, and we know the fact that the range goes from
twenty degrees to a hundred degrees Celsius. So if we think about this for a second we
know that the transmitter takes in the value of whatever you're trying to measure and spits
out a signal. So therefore, if I want to find the gain here what's coming out is the signal
in percent transmitter output, and what's coming in is the temperature, in this case in degrees
Celsius. So we know the range of the temperature is a hundred to twenty, but we also know that
from transmitter output the lowest temperature will always represent zero percent transmitter
output, and the highest value represent a hundred percent transmitter output. So this will represent
a hundred over eighty, or one point two five over three s plus one. Now we'll analyze the
valve. So the valve is a little trickier because we have a couple pieces of information
here. We have that its linear, and the fact that the max flow is two liters per second,
and we also note the time constant for the valve is five seconds. So again the denominator
is quite easy, it's again another first-order transfer function, five s plus one, but now
the question is what is this gain? So the gain is dependent on a few things. One, is
if it's fail-open or fail-closed. Two, if it's linear equal percentage. Three, if the pressure
drop is constant or not. We know that the valve is linear. We are going to assume in
this analysis that delta p is a constant, if it's not we would use a slightly different
equation. The final part we have to figure out here is if it's fail-open or fail-closed.
So in the case of safety issue we would definitely want cold water flowing through the process,
particularly in the heat exchanger, where it would come in contact with a hot process
fluid. So therefore, we're going to want this valve to fail-open. The importance of the
fact that we're saying that it fails-open is that it dictates by convention that the
sign is negative. So for we have a linear valve with a constant pressure drop what that
tells us is that for the magnitude of the gain is simply the max flow rate over a hundred.
It's a hundred because again controller output here ranges from zero percent to a hundred
percent, if we think about a valve its input will be controller output, and its output would
be the flow rate. So here we'll have two over a hundred, or 0.02. So with this information
we can fill out a characteristic equation. We now find our least common denominator which
will be the three terms in the denominator of the second term, ten s plus one, five s
plus one, times three s plus one. We multiply negative five by negative 0.02 by one
point two five and kc to get point one two five kc. We can eliminate the denominator
now, and what we now have to do is factor out the first term. If we multiply the s out
what we now do is let s equal i omega to evaluate this using direct substitution. We
can take advantage of the fact that i squared equals negative one. We can now separate this
into the real and the imaginary parts. Since the whole term is zero that means that the real
part and the imaginary part must be zero. So if we look at our two equations, the real
equation has two variables, omega and kc, imaginary has just one, omega, so we'll start
there. The i's cancel, one of the omegas cancel, that would be an omega of zero which is immaterial
in this analysis. So we're left with here is that omega squared equals eighteen over
a hundred and fifty, or that omega is the square root of eighteen over a hundred and
fifty, which is approximately point three four six. Going back to the real equation
we can substitute in that value of omega squared. Kc equals eighty three point two, and this
is the assessment of our ultimate gain. In order to find the ultimate period we take
our ultimate frequency and use that, which we just found as point three four six, and
that goes in the denominator as the ultimate period is two pi over omega. And when we do
this we will have the fact that the ultimate period is eighteen point one five seconds.
As a note the units of the ultimate gain will be percent controller output per percent transmitter
output. So in this screencast we have shown how we can take the information from a process
and determine stability information from that.