In this presentation is shown the linearization of functions with two of more variables with an example for a thermal process.

1. LINEARIZATION OF FUNCTIONS OF TWO OR
MORE VARIABLES & THERMAL PROCESS
EXAMPLE
1
CONTROL PROCESS
FIRST-ORDERDYNAMICSYSTEMS
"Good, better, best. Never let it rest. 'Til your good is better and your better is best." - St. Jerome

2. LINEARIZATION OF FUNCTIONS OF TWO OR MORE
VARIABLES & THERMAL PROCESS EXAMPLE
IRIS BUSTAMANTE PÁJARO*
ANGIE CASTILLO GUEVARA*
ALVARO JOSE GARCÍA PADILLA *
KARIANA ANDREA MORENO SADDER*
LUIS ALBERTO PATERNINA NUÑEZ*
CHEMICAL ENGINEERING PROGRAM
UNIVERSITY OF CARTAGENA
2
CONTROL PROCESS
FIRST-ORDERDYNAMICSYSTEMS

3. 3
MATHEMATICALTOOLSFORCONTROLSYSTEMS
LINEARIZATION
FUNCTIONS OF TWO OR MORE VARIABLES
Rate of a chemical reactionEquation of state Raoult’s law
𝑘 𝑐 𝐴
2
𝑡 𝑐 𝐵(𝑡)
𝑟 𝑐 𝐴 𝑡 , 𝑐 𝐵(𝑡) ) 𝑦 𝑇 𝑡 , 𝑝 𝑡 , 𝑥(𝑡)
𝑝0
𝑇 𝑡
𝑝 𝑡
𝑥(𝑡)
𝜌 𝑝 𝑡 , 𝑇(𝑡)
𝑀 𝑝(𝑡)
𝑅 𝑇(𝑡)
Smith & Corripio, 2005

4. 4
MATHEMATICALTOOLSFORCONTROLSYSTEMS
LINEARIZATION
Smith & Corripio, 2005
LINEARIZATION OF FUNCTIONS OF TWO OR MORE VARIABLES
Taylor series expansion
𝑓 𝑥1 𝑡 , 𝑥2 𝑡 , … ≈ 𝑓 𝑥1, 𝑥2, … +
𝜕𝑓
𝜕𝑥1
𝑥1 𝑡 − 𝑥1 +
𝜕𝑓
𝜕𝑥2
𝑥2 𝑡 − 𝑥2 + ⋯
𝜕𝑓
𝜕𝑥 𝑘
=
𝜕𝑓
𝜕𝑥 𝑘 𝑥1, 𝑥2,…
Where,
𝑥1, 𝑥2, … basic values of each variable.

5. 5
MATHEMATICALTOOLSFORCONTROLSYSTEMS
LINEARIZATION
Smith & Corripio, 2005
EXAMPLE 2-6.2
FUNCTION
𝑎 𝑤 𝑡 , ℎ(𝑡) = 𝑤 𝑡 ℎ(𝑡)
Area of a rectangle
𝑤 𝑡
ℎ 𝑡
𝑎 𝑤 𝑡 , ℎ(𝑡) ≈ 𝑎 𝑤, ℎ +
𝜕𝑎
𝜕𝑤
𝑤 𝑡 − 𝑤 +
𝜕𝑎
𝜕ℎ
ℎ 𝑡 − ℎ
How to linearize?
𝑎 𝑤 𝑡 , ℎ(𝑡) ≈ 𝑎 𝑤, ℎ + ℎ 𝑤 𝑡 − 𝑤 + 𝑤 ℎ 𝑡 − ℎ
𝑎 𝑤, ℎ
𝑤
ℎ
𝑤 ℎ 𝑡 − ℎ
ℎ𝑤𝑡−𝑤
Error
Small
error

6. 6
MATHEMATICALTOOLSFORCONTROLSYSTEMS
LINEARIZATION
Smith & Corripio, 2005
EXAMPLE 2-6.3
T
p
Density of an ideal gas as function of
pressure and temperature:
𝜌 𝑝 𝑡 , 𝑇(𝑡) =
𝑀 𝑝(𝑡)
𝑅 𝑇(𝑡)
Linear approximation?
Additional information
𝑀 = 20
𝑘𝑔
𝑘𝑚𝑜𝑙 𝑇 = 300 𝐾
𝑃 = 101.3 𝑘𝑃𝑎 𝑅 = 8.314
𝑘𝑃𝑎 ∙ 𝑚3
𝑘𝑚𝑜𝑙 ∙ 𝐾

7. 7
MATHEMATICALTOOLSFORCONTROLSYSTEMS
LINEARIZATION
Smith & Corripio, 2005
How to linearize?
𝜌 𝑝 𝑡 , 𝑇(𝑡) ≈ 𝜌 𝑝, 𝑇 +
𝜕𝜌
𝜕𝑝
𝑝 𝑡 − 𝑝 +
𝜕𝜌
𝜕𝑇
𝑇 𝑡 − 𝑇
𝜕𝜌
𝜕𝑝
=
𝜕𝜌
𝜕𝑝 𝑝, 𝑇
=
𝜕
𝜕𝑝
𝑀 𝑝(𝑡)
𝑅 𝑇(𝑡) 𝑝, 𝑇
=
𝑀
𝑅 𝑇(𝑡) 𝑝, 𝑇
=
𝑀
𝑅 𝑇
𝜕𝜌
𝜕𝑇
=
𝜕𝜌
𝜕𝑇 𝑝, 𝑇
=
𝜕
𝜕𝑇
𝑀 𝑝(𝑡)
𝑅 𝑇(𝑡) 𝑝, 𝑇
= −
𝑀𝑝 𝑡
𝑅 𝑇2 𝑡 𝑝, 𝑇
= −
𝑀 𝑝
𝑅 𝑇2
𝜌 𝑝, 𝑇 =
𝑀 𝑝
𝑅 𝑇

8. 8
MATHEMATICALTOOLSFORCONTROLSYSTEMS
LINEARIZATION
Smith & Corripio, 2005
Linearized function
𝜌 𝑝 𝑡 , 𝑇(𝑡) ≈
𝑀 𝑝
𝑅 𝑇
+
𝑀
𝑅 𝑇
𝑝 𝑡 − 𝑝 −
𝑀 𝑝
𝑅 𝑇2
𝑇 𝑡 − 𝑇
Numerically
𝜌 𝑝 𝑡 , 𝑇(𝑡) ≈ 1.178 + 0.01163 𝑝 𝑡 − 101.3 − 0.00393 𝑇 𝑡 − 300
𝜌 =
𝑘𝑔
𝑚3
; 𝑇 = 𝐾; 𝑝 = 𝑘𝑃𝑎
Where,

9. 9
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
𝐹𝑖, 𝑇𝑖
𝐹𝑜, 𝑇𝑜
Assumptions
Control volume
Liquid is well mixed
Tank is well insulated
Energy input by the stirrer is
negligible
Constant and equal inlet and outlet
volumetric flow, liquid densities and
heat capacity
Question
Mathematical model, 𝑇𝑜 response
to changes in 𝑇𝑖
Case 1: Adiabatic

10. 10
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
Energy balance:
𝐹𝑖 𝜌𝑖ℎ𝑖 𝑡 − 𝐹𝑜 𝜌 𝑜ℎ 𝑜 𝑡 =
𝑑 𝑉 𝜌 𝑢 𝑡
𝑑𝑡
Rate of energy into
control volume
Rate of energy out
of control volume
Rate of change of
energy accumulated in
control volume
𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 = 𝑉 𝜌𝑐 𝑣
𝑑 𝑇 𝑡
𝑑𝑡
Replacing internal energy 𝑢(𝑡) and enthalpy ℎ(𝑡)
𝑢 𝑡 = 𝑐 𝑣 𝑇 𝑡 − 𝑇𝑟𝑒𝑓 ℎ 𝑡 = 𝑐 𝑝 𝑇 𝑡 − 𝑇𝑟𝑒𝑓
Eq. 1

11. 11
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
𝐹𝜌𝑐 𝑝 𝑇𝑖,𝑠𝑠 − 𝐹𝜌𝑐 𝑝 𝑇𝑠𝑠 = 0
Deviation
Subtracting Eq. 1 & 2
Stable State (SS): Eq. 2
𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝑇𝑖,𝑠𝑠 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 − 𝑇𝑠𝑠 = 𝑉 𝜌𝑐 𝑣
𝑑 𝑇 𝑡
𝑑𝑡
𝑇𝑖(𝑡) = 𝑇𝑖 𝑡 − 𝑇𝑖,𝑠𝑠 𝑇(𝑡) = 𝑇 𝑡 − 𝑇𝑠𝑠
𝐹𝜌𝑐 𝑝 𝑇𝑖(𝑡) − 𝐹𝜌𝑐 𝑝 𝑇(𝑡) = 𝑉 𝜌𝑐 𝑣
𝑑 𝑇 𝑡
𝑑𝑡
𝑇𝑖(𝑡) =
𝑉 𝜌𝑐 𝑣
𝐹𝜌𝑐 𝑝
𝑑 𝑇 𝑡
𝑑𝑡
+ 𝑇(𝑡)
𝝉

12. 12
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
Use of Laplace transform yields
ℒ 𝑇𝑖(𝑡) = 𝜏 ℒ
𝑑 𝑇 𝑡
𝑑𝑡
+ ℒ 𝑇(𝑡) ℒ
𝑑 𝑦 𝑡
𝑑𝑡
= 𝑠𝑦 𝑠 − 𝑦(0)
𝑇𝑖(𝑠) = 𝜏 𝑠 𝑇 𝑠 − 𝑇 0 + 𝑇(𝑠) 𝑇 0 = 𝑇 0 − 𝑇𝑠𝑠 = 0
𝑇 𝑠 =
1
(𝜏𝑠 + 1)
𝑇𝑖 (𝑠) Transfer function first-order processes
Assuming inlet
temperature increases of
M degrees in magnitude
𝑇𝑖 𝑡 = 𝑇𝑖,𝑠𝑠
𝑇𝑖 𝑡 = 𝑇𝑖,𝑠𝑠 + 𝑀
𝑡 < 0
𝑡 ≥ 0

13. 13
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
𝑇𝑖 𝑡 = 𝑀 𝑢(𝑡) 𝑇𝑖 𝑠 =
𝑀
𝑠
Unit step function
𝑇 𝑠 =
1
(𝜏𝑠 + 1)
𝑀
𝑠
Partial fractions method
𝑇 𝑠 =
𝑀
𝑠 𝜏𝑠 + 1
=
𝐴
𝑠
+
𝐵
𝜏𝑠 + 1
𝐴 𝜏𝑠 + 1 + 𝐵𝑠 = 𝑀
𝑇 𝑠 =
𝑀
𝑠
+
−𝑀𝜏
𝜏𝑠 + 1
ℒ−1
1
𝑠 + 𝑎
= 𝑒−𝑎𝑡
ℒ−1
1
𝑠
= 𝑒−0𝑡
= 1ℒ−1
𝑇 𝑠 = ℒ−1
𝑀
𝑠
+ ℒ−1
−𝑀𝜏
𝜏𝑠 + 1
Laplace transform inverse

14. 14
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
𝑇 𝑡 = 𝑀 (1 − 𝑒−𝑡/𝜏) 𝑇 𝑡 = 𝑇𝑠𝑠 + 𝑀 (1 − 𝑒−𝑡/𝜏)or
0
M
𝑇 𝑡 , °C
𝑇
𝑇 + 𝑀
𝜏 Time
0.632 𝑀
Figure. Response of a first-order process to a step change in input variable

15. 15
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
Case 2: Non adiabatic
𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 − 𝑞 𝑡 = 𝑉 𝜌𝑐 𝑣
𝑑 𝑇 𝑡
𝑑𝑡
𝑞 𝑡 = 𝑈𝐴 𝑇 𝑡 − 𝑇𝑠(𝑡)Assumption: U constant
𝐹𝜌𝑐 𝑝 𝑇𝑖,𝑠𝑠 − 𝐹𝜌𝑐 𝑝 𝑇𝑠𝑠 − 𝑈𝐴 𝑇𝑠𝑠 − 𝑇𝑠,𝑠𝑠 = 0
Stable State (SS):
Eq. 2
Eq. 1𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 − 𝑈𝐴 𝑇 𝑡 − 𝑇𝑠(𝑡) = 𝑉 𝜌𝑐 𝑣
𝑑 𝑇 𝑡
𝑑𝑡
Subtracting Eq. 1 & 2

16. 16
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝑇𝑖,𝑠𝑠 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 − 𝑇𝑠𝑠 − 𝑈𝐴 𝑇 𝑡 − 𝑇𝑠𝑠 − (𝑇𝑠 𝑡 − 𝑇𝑠,𝑠𝑠) = 𝑉 𝜌𝑐 𝑣
𝑑 𝑇 𝑡
𝑑𝑡
𝑇𝑖(𝑡) = 𝑇𝑖 𝑡 − 𝑇𝑖,𝑠𝑠
𝑇(𝑡) = 𝑇 𝑡 − 𝑇𝑠𝑠
𝑇𝑠(𝑡) = 𝑇𝑠 𝑡 − 𝑇𝑠,𝑠𝑠
𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 − 𝐹𝜌𝑐 𝑝 𝑇 𝑡 − 𝑈𝐴 𝑇(𝑡) − 𝑇𝑠(𝑡) = 𝑉 𝜌𝑐 𝑣
𝑑 𝑇 𝑡
𝑑𝑡
𝑉 𝜌𝑐 𝑣
𝑑 𝑇 𝑡
𝑑𝑡
+ 𝐹𝜌𝑐 𝑝 + 𝑈𝐴 𝑇 𝑡 = 𝐹𝜌𝑐 𝑝 𝑇𝑖 𝑡 + 𝑈𝐴 𝑇𝑠(𝑡)

17. 17
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
𝑉 𝜌𝑐 𝑣
𝐹𝜌𝑐 𝑝 + 𝑈𝐴
𝑑 𝑇 𝑡
𝑑𝑡
+ 𝑇 𝑡 =
𝐹𝜌𝑐 𝑝
𝐹𝜌𝑐 𝑝 + 𝑈𝐴
𝑇𝑖 𝑡 +
𝑈𝐴
𝐹𝜌𝑐 𝑝 + 𝑈𝐴
𝑇𝑠 (𝑡)
𝝉 𝑲 𝟏 𝑲 𝟐
Use of Laplace transform yields
𝜏 𝑠 𝑇 𝑠 − 𝑇 0 + 𝑇 𝑠 = 𝐾1 𝑇𝑖 𝑠 + 𝐾2 𝑇𝑠 𝑠
𝑇 𝑠 =
𝐾1
𝜏𝑠 + 1
𝑇𝑖 𝑠 +
𝐾2
𝜏𝑠 + 1
𝑇𝑠 𝑠
𝐾 =
∆𝑂
∆𝐼
=
∆ output variable
∆ input variable

18. 18
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
𝑇 𝑠 =
𝐾2
𝜏𝑠 + 1
𝑇𝑠 𝑠
Constant: surrounding temperature
Constant: liquid temperature
𝑇 𝑠 =
𝐾1
𝜏𝑠 + 1
𝑇𝑖 𝑠 𝑇𝑠 𝑡 = 𝑇𝑠,𝑠𝑠; 𝑇𝑠 𝑠 = 0
𝑇𝑖 𝑡 = 𝑇𝑖,𝑠𝑠; 𝑇𝑖 𝑠 = 0
Assuming inlet
temperature increases of
M degrees in magnitude
𝑇𝑖 𝑡 = 𝑇𝑖,𝑠𝑠
𝑇𝑖 𝑡 = 𝑇𝑖,𝑠𝑠 + 𝑀
𝑡 < 0
𝑡 ≥ 0

19. 19
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
𝑇𝑖 𝑡 = 𝑀 𝑢(𝑡) 𝑇𝑖 𝑠 =
𝑀
𝑠
Unit step function
𝑇 𝑠 =
𝐾1
(𝜏𝑠 + 1)
𝑀
𝑠
Partial fractions method
𝑇 𝑠 =
𝐾1 𝑀
𝑠 𝜏𝑠 + 1
=
𝐴
𝑠
+
𝐵
𝜏𝑠 + 1
𝐴 𝜏𝑠 + 1 + 𝐵𝑠 = 𝐾1 𝑀
𝑇 𝑠 =
𝐾1 𝑀
𝑠
+
−𝐾1 𝑀𝜏
𝜏𝑠 + 1
ℒ−1
1
𝑠 + 𝑎
= 𝑒−𝑎𝑡
ℒ−1
1
𝑠
= 𝑒−0𝑡
= 1ℒ−1
𝑇 𝑠 = ℒ−1
𝐾1 𝑀
𝑠
+ ℒ−1
−𝐾1 𝑀𝜏
𝜏𝑠 + 1
Laplace transform inverse

20. 20
FIRST-ORDERDYNAMICSYSTEMS
THERMAL PROCESS EXAMPLE
Smith & Corripio, 2005
THERMAL PROCESS
𝑇 𝑡 = 𝐾1 𝑀 (1 − 𝑒−𝑡/𝜏) 𝑇 𝑡 = 𝑇𝑠𝑠 + 𝐾1 𝑀 (1 − 𝑒−𝑡/𝜏)or
0
M
𝑇 𝑡 , °C
𝑇
𝑇 + 𝐾1 𝑀
Time
𝐾1 𝑀
Figure. Response of a first-order process to a step change in input variable

21. THANKS !
21