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.