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In this screencast we will take a look at
how we can use process data in order to determine a transfer function. In this case our focus
will be on developing a first-order plus dead time transfer function. First-order plus dead
time transfer functions are quite important for a couple of reasons. One, in order to
analyze a process it is much easier due to its mathematical simplicity in comparison
to using a higher order transfer function, so FOPTDs are commonly used to approximate
higher order transfer functions, and a second reason is the fact that many controlled tuning
parameter methodologies are based on the knowledge of the FOPDT process transfer function. So
here we are looking at how the change in the cooling water flow rate impacts the temperature
of a process fluid. So therefore the transfer function we’re looking for here, our output
over our input, is equal to the controlled variable, which is the temperature of the process fluid,
and we’re interested in developing a transfer function that analyzes how it handles adjustments
in the cooling water flow rate. If we’re trying to model this as a FOPDT the transfer function
for a FOPDT is K e to the negative theta s divided by tau s plus one. So in order to
develop a FOPDT we need to be able to determine three variables: the steady state gain,K;
the dead time, theta; and the time constant, tau. Finding the steady state gain is probably
the easiest of the bunch, because what the steady state gain represents is the change
in the output at steady state, divided by the change in the input, so in this case our
step change. So that information we can easily get from the data. In order to find tau and
theta there are a number of different correlations that have been developed by different researches,
here we will use the correlations developed in the late seventies by Cinder Essen and
Christian Swanning, which stated the fact that the dead time can be found by one point
three multiplied by the time corresponding to thirty five point three percent of the
change, this will be explained shortly, minus point two nine multiplied by the time it takes
for eighty five point three percent of the change to occur, and tau can be defined as point
six seven multiplied by the difference between the eighty five point three percent time mark,
and the thirty five point three percent time mark. Here, the eighty five point three percent
and the thirty five point three percent, these times represent the time it takes for either
thirty five point three percent or eighty five point three percent of the change in
the output variable to occur, so for example, if a function went from ten to twenty we would
be interested in the time it took for thirty five point three percent of that change to
occur, so the difference between twenty and ten is ten, so we would be interested in the
change of three point five three of that, or thirteen point five three, and similarly
we would be interested in where eighty five point three percent of the change happened,
which is eighteen point five three. The first thing we would want to do here, although the
question asks us to model this as a FOPDT, we first would want to graph the data to see
whether or not a FOPDT fit would be a possibility, and as we look at the graph which is provided
here, we can see that a FOPDT relationship is reasonable, although it may look a little
awkward in comparison to usual because in this case the gain is negative, not positive,
and the reason why is we think about the process with our heat exchanger is that if you increase
the cooling water flow rate what your’re going to do is cause a decrease in the temperature
of the other fluid. So to devaluate that gain we just use the definition of gain, and if
we go back to our problem statement we’ll see the fact that our input change, in terms
of the change in our flow rate, was one point two gallons per minute, and that the change
in temperature went from fifty five to fifty three point two. So when we substitute those
all in we end up with negative one point five gallons per minute, so there we have found
our gain. Now we want to determine the time it takes to reach thirty five point three
percent, and eighty five point three percent of the change, but in order to do that we
need to find what those temperatures are. So the temperature that corresponds to thirty
five point three is going to be our starting value, fifty five, plus thirty five point
three percent, or point three five three, of the way from fifty three point two minus
fifty five. When this is done we end up with about fifty four point four degrees Fahrenheit,
so what we want to do here is determine the time. If we look at our fifty four point four
graph we find a time that corresponds to it we end up with that t thirty five point three
is approximately six point three minutes or so. Note for better accuracy we would have
more lines on the graph, but for this analysis this will work OK. Now that we have found
the time corresponding to thirty five point three we now need to find the time corresponding
to eighty five point three, and we do this using a very similar method. First, we find
a temperature that corresponds to that value, and when we solve for this we end up with
something thats about fifty three point five Fahrenheit. So now we can attempt to find
the t eighty five point three, and when we do that we end up with a t eighty five point
three of about nine and a half minutes. Now that we have the t thirty five point three
and eighty five point three we can now use the correlations to find our dead time and
our time constants to finish out our FOPDT analysis. So we have the relationship for
theta, and when we do that we end up with five point four minutes. Similarly we can
find tau and we end up with our time constant of two point one four minutes. We now have
determined all three of our parameters, so therefore we know the fact that our final
FOPDT transfer function which relates the temperature of the process fluid to the flow
rate of the cooling water equals negative one point five, our gain, e to our dead time,
negative five point four s, divided by two point one four s plus one. So an important
point to note here is the fact that these were all done approximately using the graph
that we have. So these evaluations at thirty point three and eighty five point three could
have been improved if the graph had a bit more graduated lines on it, but still the
process that we have done here is unchanged.

Reynold King

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