SME
Process Dynamics & Control
Process dynamics
Use this for “A process with gain 2 and time constant 5 min gets a step of 3 in its input. Find the output after 10 min.”
Chart
y(t) = 6(1 − exp(−t/τ)), τ = 5 min
After τ the response has made 63.2% of its change; after 5τ, 99.3%.
At t = 10 min: y = 5.188.
This is an example. Type your own values to start a new problem.
y∞
=
K
×
Δu
y
=
y∞
×
(1
−
e
−
min
÷
min
)
Formulas
y∞
=
K
×
Δu
6 = 2 × 3
y
=
y∞
×
(1
−
e
−
t
÷
τ
)
5.188
=
6
×
(1
−
e
−10 ÷ 5
)
Assumptions
•
First order, τ dy ÷ dt + y = Ku, with no dead time.
•
The process starts at steady state; y and u are deviations from it (deviation variables).
•
After one τ the output has made 63.2% of its change; after 5τ, 99.3%.
Step-by-step
WE KNOW
K = 2, Δu = 3, τ = 5 min, t = 10 min
FIND
final value (y∞), response (y)
Step 1 · Find final value (y∞)
Use y∞ = K × Δu
Once
dy
dt
is
0
the
equation
leaves
y
=
KΔu:
the
gain
times
the
step.
y∞ = 2 × 3
y∞ = 6
Step 2 · Find response (y)
Use y = y∞ × (1 − e−t ÷ τ)
Solving
τ
dy
dt
+
y
=
KΔu
from
y
=
0
gives
this
curve:
the
share
1
−
e
−
t
÷
τ
of
the
change.
y
=
6
×
(1
−
e
−10 ÷ 5
)
y
=
6
×
(1
−
0.1353)
y
=
6
×
0.8647
y
=
5.188
Check
Put the numbers back in. Both sides should match.
6 = 2 × 3 ✓
5.188
=
6
×
(1
−
e
−10 ÷ 5
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.