Solution of Homework 1: Interacting Tank-in-Series System

President University Erwin Sitompul SMI 2/1

Dr.-Ing. Erwin SitompulPresident University

Lecture 2System Modeling and Identification

http://zitompul.wordpress.com


Solution of Homework 1: Interacting Tank-in-Series System

v1

qi

h1

Chapter 2 Examples of Dynamic Mathematical Models

h2

v2

q1 a1 a2

A1 : Cross-sectional area of the first tank [m2]

A2 : Cross-sectional area of the second tank [m2]

h1 : Height of liquid in the first tank [m]

h2 : Height of liquid in the second tank [m]

qo

The process variable are now the heights of liquid in both tanks, h1 and h2.

The mass balance equation for this process yields:1 1

i 1( )d Ah q qdt

2 21 o

( )d A h q qdt


Solution of Homework 1: Interacting Tank-in-Series SystemChapter 2 Examples of Dynamic Mathematical Models

Assuming ρ, A1, and A2 to be constant, we obtain:

After substitution and rearrangement,

1 1 1q v a 1 2 12 ( )g h h a

o 2 2q v a 2 22gh a

1 i 11 2

1 1

2 ( )dh q a g h hdt A A

2 1 21 2 2

2 2

2 ( ) 2dh a ag h h ghdt A A


Homework 2: Interacting Tank-in-Series SystemChapter 2 Examples of Dynamic Mathematical Models

Build a Matlab-Simulink model for the interacting tank-in-series system and perform a simulation for 200 seconds. Submit the mdl-file and the screenshots of the Matlab-Simulink file and the scope of h1 and h2 as the homework result.Use the following values for the simulation.

a1 = 210–3 m2

a2 = 210–3 m2

A1 = 0.25 m2

A2 = 0.10 m2

g = 9.8 m/s2

qi = 510–3 m3/stsim = 200 s


Homework 2: Triangular-Prism-Shaped TankChapter 2 Examples of Dynamic Mathematical Models

Build a Matlab-Simulink model for the triangular-prism-shaped tank and perform a simulation for 200 seconds. Submit the mdl-file and the screenshots of the Matlab-Simulink file and the scope of h as the homework result.Use the following values for the simulation.

NEW

a = 210–3 m2

Amax= 0.5 m2

hmax = 0.7 mh0 = 0.05 m (!)g = 9.8 m/s2

qi1 = 510–3 m3/sqi2 = 110–3 m3/stsim = 200 s



Heat ExchangerConsider a heat exchanger for the heating of liquids as

shown below.Assumptions: • Heat capacity of the tank is small

compare to the heat capacity of the liquid.

• Spatially constant temperature inside the tank as it is ideally mixed.

• Constant incoming liquid flow, constant specific density, and constant specific heat capacity.



Heat ExchangerConsider a heat exchanger for the heating of liquids as

shown below.

Tj

qTl

qT

Vρ Tcp

Tl : Temperature of liquid at inlet [K]

Tj : Temperature of jacket [K]T : Temperature of liquid inside

and at outlet [K]

q : Liquid volume flow rate [m3/s]V : Volume of liquid inside the

tank [m3]ρ : Liquid specific density [kg/m3]cp : Liquid specific heat capacity

[J/(kgK)]



Heat ExchangerThe heat balance equation becomes:

l j

( )( )p

p p

d V c Tq c T q c T A T T

dt

A : Heat transfer area of the wall [m2] : Heat transfer coefficient [W/(m2K)]

Rearranging:

l j( )p p pdTV c q c T q c A T ATdt

l jp p

p p p

V c q cdT AT T Tq c A dt q c A q c A

The heat exchanger will be in steady-state if dT/dt = 0, so the steady-state temperature at outlet is:

steady-state l jp

p p

q c AT T Tq c A q c A


τ : Space variable [m]Ti : Liquid temperature in the

inner tube [K] Ti(τ,t)To : Liquid temperature in the

outer tube [K] To(t)q : Liquid volume flow rate

in the inner tube[m3/s]ρ : Liquid specific density in the

inner tube [kg/m3]cp : Liquid specific heat capacity

[J/(kgK)]A : Heat transfer area per

unit length [m]Ai : Cross-sectional area of

the inner tube [m2]


Double-Pipe Heat Exchanger

L

To,ss

τ dτ

A single-pass, double-pipe steam heat exchanger is shown below. The liquid in the inner tube is heated by condensing steam.

Ti,ss

τ q



To(t)

Ti(τ,t)

dτ

ii

TT d

The profile of temperature Ti of an element of heat exchanger with length dτ for time dt is given by:

i ii

T TT d dtt

(taken as approximation)

i ii i i o i( ) ( )p p p

dT TAd c q c T q c T d Ad T Tdt

The heat balance equation of the element can be derived as:





i i io i

p pAd c q cdT T T TAd dt A

The equation can be rearrange to give:

The boundary condition is Ti(0,t) and Ti(L,t).

The initial condition is Ti(τ,0).


Heat Conduction in a Solid Body

L

q(0) q(L)q(x) q(x+dx)

xdx

Consider a metal rod of length L with ideal insulation.Heat is brought in on the left side and withdrawn on the

right side.Changes of heat flows q(0) and q(L) influence the rod

temperature T(x,t).The heat conduction coefficient, density, and specific

heat capacity of the rod are assumed to be constant.




t : Time variable [s]x : Space variable [m]T : Rod temperature [K] T(x,t)ρ : Rod specific density [kg/m3]A : Cross-sectional area of the rod [m2]cp : Rod specific heat capacity [J/(kgK)]q(x) : Heat flow density at length x [W/m2]q(x+dx) : Heat flow density at length x+dx [W/m2]

L


x dx




The heat balance equation of at a distance x for a length dx and a time dt can be derived as:

L


x dx

in out

( )pd mc TQ Q

dt

( ) ( )pdTAdx c A q x q x dxdt

( )pdT dq xAdx c A dxdt dx

( )( ) ( ) dq xq x dx q x dxdx

x dxx

( )q x

( )q x dx ( )( ) dq xq x dxdx

0dxfor



Heat Conduction in a Solid BodyAccording to Fourier equation:

( ) dTq xdx

λ : Coefficient of thermal conductivity [W/(mK)]

Substituting the Fourier equation into the heat balance equation:

2

2pdT d TAdx c A dxdt dx

2

2pdT d Tcdt dx

2

2p

dT d Tdt c dx

: Heat conductifity factor [m2/s]pc



The boundary conditions should be given for points at the ends of the rod:


0(0, ) ( )T t T t( , ) ( )LT L t T t

The temperature profile of the rod in steady-state Ts(x) can be dervied when ∂T(x,t)/∂t = 0.


The initial conditions for any position of the rod is:0( ,0) ( )T x T x

2

2

( )0sd T x

dx ( )sT x ax b

(0)sT b 0,sT( )sT L aL b

0,saL T ,L sT , 0,L s sT T

aL


Thus, the steady-state temperature at a given position x along the rod is given by:

Chapter 2


, 0,0,( ) L s s

s s

T TT x x T

L

Examples of Dynamic Mathematical Models


A suitable model for a large class of continuous theoretical processes is a set of ordinary differential equations of the form:

11 1 1 1

( ) , ( ), , ( ), ( ), , ( ), ( ), , ( )n m sdx t f t x t x t u t u t r t r tdt

State EquationsChapter 2 General Process Models

22 1 1 1

( ) , ( ), , ( ), ( ), , ( ), ( ), , ( )n m sdx t f t x t x t u t u t r t r tdt

1 1 1( ) , ( ), , ( ), ( ), , ( ), ( ), , ( )n

n n m sdx t f t x t x t u t u t r t r tdt

t : Time variable x1,...,xn : State variablesu1,...,um : Manipulated variablesr1,...,rs : Disturbance, nonmanipulable variablesf1,...,fn : Functions


A model of process measurement can be written as a set of algebraic equations:

1 1 1 1 1( ) , ( ), , ( ), ( ), , ( ), ( ), , ( )n m m mty t g t x t x t u t u t r t r t

Output EquationsChapter 2 General Process Models

2 2 1 1 1( ) , ( ), , ( ), ( ), , ( ), ( ), , ( )n m m mty t g t x t x t u t u t r t r t

1 1 1( ) , ( ), , ( ), ( ), , ( ), ( ), , ( )r r n m m mty t g t x t x t u t u t r t r t

t : Time variable x1,...,xn : State variablesu1,...,um : Manipulated variablesrm1,...,rmt : Disturbance, nonmanipulable variables at outputy1,...,yr : Measurable output variablesg1,...,gr : Functions


State Equations in Vector FormChapter 2 General Process Models

If the vectors of state variables x, manipulated variables u, disturbance variables r, and the functionsf are defined as:

1 1 1 1

, , ,

n m s n

x u r f

x u r f

x u r f

Then the set of state equatios can be written compactly as:

( ) , ( ), ( ), ( )d t t t t tdt

x f x u r


Output Equations in Vector FormChapter 2 General Process Models

If the vectors of output variables y, disturbance variables rm, and vectors of functions g are defined as:

1 1 1

, , m

m

r mt r

y r g

y r g

y r g

Then the set of algebraic output equatios can be written compactly as:

( ) , ( ), ( ), ( )mt t t t ty g x u r

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Solution of Homework 1: Interacting Tank-in-Series System