SME
Process Dynamics & Control · Problem type
FOPDT model from a step test
Use this for “After a step, the output makes 28.3% of its change by 4 min and 63.2% by 10 min. Find τ and θ.”
Chart
y(t)
=
(1
−
exp(−
t − θ
τ
)),
τ
=
9
min
Dead time θ = 1 min: nothing moves until then.
τ = 1.5(t₆₃ − t₂₈) and θ = t₆₃ − τ.
After θ + τ the response has made 63.2% of its change; after 5τ, 99.3%.
This is an example. Type your own values to start a new problem.
τ (min)
=
1.5
×
(
t₆₃ (min)
−
t₂₈ (min)
)
θ (min)
=
t₆₃ (min)
−
τ (min)
Formulas
τ
=
1.5
×
(
t₆₃
−
t₂₈
)
9 = 1.5 × (10 − 4)
θ
=
t₆₃
−
τ
1 = 10 − 9
Assumptions
•
The process is first order plus dead time (FOPDT), and the test starts at steady state.
•
The curve 1 − e−(t − θ) ÷ τ passes 28.3% at θ + τ ÷ 3 and 63.2% at θ + τ.
•
Times are read from the moment the input steps; the gain K is the output change ÷ the input step.
Step-by-step
WE KNOW
t₂₈ = 4 min, t₆₃ = 10 min
FIND
time constant (τ), dead time (θ)
Step 1 · Find time constant (τ)
Use τ = 1.5 × (t₆₃ − t₂₈)
The
two
points
are
τ
−
τ
3
=
2
τ
3
apart,
so
τ
is
1.5
times
the
gap.
τ
=
1.5
×
(10
−
4)
τ
=
1.5
×
6
τ
=
9
min
Step 2 · Find dead time (θ)
Use θ = t₆₃ − τ
The
63.2%
point
comes
one
τ
after
the
dead
time
ends.
θ
=
10
−
9
θ
=
1
min
Check
Put the numbers back in. Both sides should match.
9 = 1.5 × (10 − 4) ✓
1 = 10 − 9 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.