Evaporator
In this example, we will estimate a model for a four-stage evaporator to reduce the water content of a product, for example milk. The 3 inputs are feed flow, vapor flow to the first evaporator stage and cooling water flow. The three outputs are the dry matter content, the flow and the temperature of the outcoming product.
The example comes from STADIUS’s Identification Database
Zhu Y., Van Overschee P., De Moor B., Ljung L., Comparison of three classes of identification methods. Proc. of SYSID '94,
using DelimitedFiles, Plots
using ControlSystemIdentification, ControlSystemsBase
url = "https://ftp.esat.kuleuven.be/pub/SISTA/data/process_industry/evaporator.dat.gz"
zipfilename = "/tmp/evaporator.dat.gz"
path = Base.download(url, zipfilename)
run(`gunzip -f $path`)
data = readdlm(path[1:end-3])
# Inputs:
# u1: feed flow to the first evaporator stage
# u2: vapor flow to the first evaporator stage
# u3: cooling water flow
# Outputs:
# y1: dry matter content
# y2: flow of the outcoming product
# y3: temperature of the outcoming product
u = data[:, 1:3]'
y = data[:, 4:6]'
d = iddata(y, u, 1)
InputOutput data of length 6305, 3 outputs, 3 inputs, Ts = 1
The input consists of two heating inputs and one cooling input, while there are 6 outputs from temperature sensors in a cross section of the furnace.
Before we estimate any model, we inspect the data
plot(d, layout=6)
We split the data in two, and use the first part for estimation and the second for validation. A model of order around 8 is reasonable (the paper uses 6-13). This system requires the option zeroD=false
to be able to capture a direct feedthrough, otherwise the fit will always be rather poor.
dtrain = d[1:3300] # first experiment ends after 3300 seconds
dval = d[3301:end]
model,_ = newpem(dtrain, 8, zeroD=false)
Iter Function value Gradient norm
0 7.896730e+02 2.039439e+02
* time: 5.1975250244140625e-5
50 7.576979e+02 4.949210e+02
* time: 4.373308897018433
100 7.371685e+02 7.425396e+01
* time: 8.040960788726807
150 7.314174e+02 9.097227e+01
* time: 11.673417806625366
200 7.298209e+02 2.361471e+01
* time: 15.287195920944214
250 7.295993e+02 2.369459e+01
* time: 18.848236799240112
300 7.295811e+02 1.137475e+01
* time: 22.462119817733765
350 7.295730e+02 1.617648e+01
* time: 26.06983995437622
400 7.295718e+02 5.663065e-01
* time: 29.753087997436523
predplot(model, dval, h=1, layout=d.ny)
predplot!(model, dval, h=5, ploty=false)
The figures above show the result of predicting steps into the future.
We can visualize the estimated model in the frequency domain as well.
w = exp10.(LinRange(-2, log10(pi/d.Ts), 200))
sigmaplot(model.sys, w, lab="PEM", plotphase=false)
Let’s compare prediction performance to the paper
ys = predict(model, dval, h=5)
ControlSystemIdentification.mse(dval.y-ys)
3×1 Matrix{Float64}:
0.05761868858427213
0.15635501742095104
0.01926955677230884
The authors got the following errors: [0.24, 0.39, 0.14]