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CS 367: Model-Based Reasoning
Lecture 16 (03/14/2002)
Gautam Biswas
5/23/2017
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Today’s Lecture
Last Lectures:
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Bond Graphs and Causality
State Space Equations from Bond Graphs
Today’s Lecture
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More Complex Examples
20-SIM
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Electrical Circuits: Example 2
Try this one:
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Mechanical Model: Example 2
Try this one:
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State Equations
Linear
x  A. x  B.u
x1  a11. x1  ....  a1n x n  b11u1  ....  b1m um
x 2  a 21. x1  ....  a 2 n x n  b21u1  ....  b2 m um
..
.
x n  a n1. x1  ....  a nn x n  bn1u1  ....  bnm um
Nonlinear
x   ( x , u)
x1  1 ( x1 ,...., x n , u1 ,...., um )
x 2   2 ( x1 ,...., xn , u1 ,...., um )
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..
.
x n   n ( x1 ,...., x n , u1 ,...., um )
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Causality for basic multiports
Note that a lot of the causal considerations are based on
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Causality Assignment: Example 3
Try this one:
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Equation Generation: Example 2
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Extending Modeling to other domains
Fluid Systems
e(t) – Pressure, P(t)
 f(t) – Volume flow rate, Q(t)
 Momentum, p = e.dt = Pp, integral of pressure
 Displacement, q = Q.dt = V, volume of flow
 Power, P(t).Q(t)
 Energy (kinetic): Q(t).dPp
 Energy (potential): P(t).dV
Fluid Port: a place where we can define an average pressure, P and a volume
flow rate, Q
Examples of ports: (i) end of a pipe or tube
(ii) threaded hole in a hydraulic pump
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Fluid Ports
Flow through ports transfers energy
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P – force/unit area
Q – volume flow rate
P.Q = power = force . displacement / time
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Moving fluid also has kinetic energy
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But it can be ignored if
1 Q 2
P   ( )
2 A
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Fluid Domain: Resistor
P1
Q
Q1, P1
Q2, P2
1
P
1 Q2
2
P3 Q3
l
R
Pipe, Porous plug or a
Constriction
P 3  R.Q 3
Q1  Q 2  Q 3   flow
P 3  P1  P 2 : pressure drop
How do we compute R?
For thin, long tubes with incompressible and laminar flow
128. .l
;  :viscosity, l : length, d : diameter
 .d 4
For turbulent flow :
R
3
4
P3  a t .Q3 Q3 ; a t :experiment al constant
Re 
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Resistance relationship changes
for orifices, etc. for turbulent
flow
4. .Q
; Re  2000  laminar
 .d.
above 4000  definitely turbulent
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Fluid Capacitor
C
P1
Q1
Q3
Q1
Q2
P1, P2, P3
For rigid pipe
P3 Q3
P
0 Q2
2
P1  P2  P3
  density
Q3  Q1  Q2
h  height
P3   .g.h 
Note : P3 
 . g.P3
A
g  gravitational const .
V3
A
; C 
V3  volume
C
 .g
A  area
V  .r .l
C 3  0  0 ; r0  radius
B
B
l  length; B  bulk modulus (gives us pressure change when
liquid is compressed)
2
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Fluid Inertia
P1
Q1
area, A
I
P2
Q2
P1
Q1
P3 Q3
P3  P1  P2
P
1 Q2
2
Q1  Q2  Q3
Linear System
I 3 .Q3  PP3  momentum
net force on slug of liquid
 P1. A  P2 . A ; fluid velocity  QA3
Mass of slug   . A.l

 F  P1. A  P2 . A  m.a   . A.l . QA3
As pressure difference
 .l 
 .l
 P3  P1  P2 
.Q3 ; I 3 
 fluid inertia increases, flow accelerates
A
A
quicker for larger pipes
Interesting point : A   I 3 
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Fluid effort sources
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Fluid flow sources
at rate Q(t)
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Modeling a fluid line
Simplest case: just a lumped resistance
Distributed Model (break pipe into shorter segments)
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Schematic of Secondary Sodium Cooling Loop
sodium flow stopping valve
(normally opened)
R2
Vin
main
motor
secondary
sodium
pump
I
IH X
intermediate
heat
exchanger
f11
e33
R1
e22
evaporator
sodium
overflow
tank
GY
C
E V
COFC
R4
R5
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e19
feed
water
loop
sodium flow
stopping valve
(normally opened)
e14
super heater
CSH
feed
water
loop
R3
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Nuclear reactor
Heat Transport from reactor core to turbine – two loops
primary and secondary loops.
We model the secondary loop – use liquid sodium to
transport heat from primary loop to evaporator and
super heater where steam is produced to drive the
turbine
Primary components: intermediate heat exchanger,
pump and motor system, super heater, evaporator,
pipes, valves, and overflow tank.
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Building Bond Graph Model
Step 1: Focus on energy and mass balance in fluid domain +
mechanical characteristics of main motor & pump
Assume motor is an AC synchronous motor; don’t model electric
field effects – assume it is present as soon as motor is turned on.
(dynamic electrical effects of motor not modeled)
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