2.7. RED VIARIA:
2.7.1. CAMINOS RURALES
We have considered here a few cases of junction power directions, and we have also explicitly written down the junction equations in each case.
Figure 5.2a shows a junction confi guration whose fl ow and effort equations
FIGURE 5.1 Tetrahedron of state.
edt
fdt e
f p
q
dq= f dt dp= e
dt
are written as Equation 5.7. Figure 5.2b shows a junction confi guration whose fl ow and effort equations are written as Equation 5.8. Similarly, Figure 5.2c–f shows different junction confi gurations whose equations are written as Equations 5.9 through 5.12, respectively.
1 Junction
e3 e2
f3 f2
f1 e1
1 (a)
e3 e2
f3 f2
f1 e1
1 1 Junction (b)
1 Junction
e3 e2
f3 f2
f1 e1
1 (c)
0 Junction 1 e3 e2
f3 f2
f1 e1
0 (d)
FIGURE 5.2
Some of the different possible joint confi gurations.
f1= f2= f3
e1+ e2+ e3= 0 (5.7)
f1= f2= f3
e1+ e3= e2
(5.8)
f1= f2= f3
e1= e2+ e3
(5.9)
f1= f2+ f3
e1= e2= e3 (5.10)
f1+ f2= f3
e1= e2 = e3 (5.11)
FIGURE 5.2 (Continued)
0 Junction 1 e3 e2
f3 f2
f1 e1
0 (e)
e3 e2
f3 f2
f1 e1
0 0 Junction 1 (f )
f1+ f2+ f3 = 0 e1= e2 = e3
(5.12)
We will now discuss the process of deriving the governing equations from a fully causalled bond graph representation. For the fi rst example (Example 5.1), we have chosen the well-known RLC circuit from the elec-trical domain (which is equivalent to the spring–mass–damper system in the mechanical domain).
EXAMPLE 5.1: RLC CIRCUIT
Figure 5.3 shows an RLC circuit along with its bond graph representation. The effort and fl ow associated with each bond are marked up in the fi gure.
Answers to Q1 (written in terms of the constitutive equations).
Q1. What do all the elements give to the system?
Answers: The answers are written for each bond.
Bonds:
1.
e1= V(t)
(5.13)
Bond 1 gives back effort to the system.
2. e2= R ⋅ f2 (resistor) (5.14)
Bond 2 gives back effort to the system.
3.
f3= p3
L (inductor) (5.15)
Bond 3 gives back fl ow to the system.
4.
e4= 1
Cq4 (capacitor) (5.16)
Bond 4 gives back effort to the system.
Answers to Q2: In writing the answers to this question, we need to use the answers from Q1 and the joint equations for effort and fl ow.
Q2. What does the system give back to the storage elements?
Bond 4 (for storage element C1)
q4= f4= f1= f2= f3= p3
L
(5.17) FIGURE 5.3
An RLC circuit and its corresponding bond graph.
f4 V
+−
R
L
C
f3 f2
e4
e3 e2
f1 e1
C C1
1
1 Junction I
I1 R
R1
Se Se1
Flows are equal in 1 junction and f3 brings the fl ow information to the
Summing of efforts at the 1 junction.
Final form is (by rewriting Equations 5.17 and 5.18):
Note a few important things in the fi nal form of these equations. These are important issues to keep in mind when the reader wants to develop the governing equations:
There are two fi rst order equations, one each for the two storage
•
elements with integral causality.
The two equations are coupled, that is, they have to be solved
•
simultaneously. The two state variables that appear in the equa-tions are the p and q associated with the storage elements that is, p 3 and q 4 , (and their fi rst derivatives).
The fi nal forms of the equations are written in terms of the input
•
to the system, system parameters, and state variables only and no other terms.
Although intermediate variables, such as effort and fl ow of
indi-•
vidual bonds, are used in the intermediate steps, they vanish from the fi nal equations.
This is a second order system because there are two equations
•
state space equations.
We are more familiar with the second order ordinary differential equa-tion representing a second order system. It can be shown that the above two equations may be combined to give a second order ODE for the system.
Converting it into second order function by taking derivatives of Equation 5.19 and combining them, we get:
Equation 5.20 is the standard form that many are familiar with for an RLC circuit. A similar equation written for a mechanical system will be the well-known spring–mass–damper equation for a vibrating mass:
F(t) = mx+ Bx + kx (5.21) where x is the displacement, F is the external force applied, B is the damping coeffi cient, and m is the mass.
It is thus shown that a set of two state–space equations is equivalent to a single second order differential equation, and the two can be used to describe the same second order system. To reiterate, the system has two energy storage elements. Hence, it is a second order system.
EXAMPLE 5.2
Figure 5.4 shows the system and the corresponding bond graph. This is a mechanical system containing the basic components that make up typical mechanical systems. Based on what we learned in the previous example, the equations will be for the state variables associated with the energy storage ele-ments, that is, q 4 , p 1 , q 8 , and p 10.
This example has a transformer element. Before we start deriving the equa-tions, we need to obtain the transformer parameter.
Figure 5.5 shows how the transformer element could move. This means that f 6 and f 7 are proportional to each other in the following way:
f11
1 Junction 1 Junction 2
FIGURE 5.4
A generic mechanical system and its corresponding bond graph.
l2 l1
f7
f6
FIGURE 5.5
A schematic of the lever as a transformer.
The negative sign in the above relationship means that the two fl ow directions are opposite to one another. Since the transformer conserves power (i.e., e 7 f 7 = e 6 f 6 ) a similar relationship will exist between e 6 and e 7 .
e6l1= −e7l2⇒ e6 = −l2 l1e7
(5.23)
So the transformer factor is
TF= −l2
l1 .
Now we will attempt to derive the governing equations.
Answers to Q1 (written in terms of the constitutive equations).
Q1. What do all the elements give to the system?
Answers:
Bonds
1. f1= p1
m1
(5.24)
2. e2= F(t) (5.25) 3.
e3 = −m1g
(5.26)
4. e4 = k1q4 (5.27)
5. e5 = bf5 (5.28)
8. e8 = k2q8 (5.29)
10. f10 = p10
m2
(5.30)
11. e11= −m2g (5.31)
Answers to Q2: Use the answers from Q1 and the joint equations.
Q2. What does the system give back to the storage elements?
Bond 1
All four equations may be represented together as:
Four fi rst order equations are describing the behavior of this system. Each equation represents one of the energy storage devices. The state–space vari-ables representing the energy storage devices are displacement or momen-tum—the energy variables. This set of coupled ordinary differential equations
(5.36) can be solved simultaneously to obtain the system behavior. In order to solve these equations, four initial conditions will also be required.
EXAMPLE 5.3: AN ELECTROMECHANICAL SYSTEM, A PERMANENT MAGNET DC MOTOR
The bond graph shown in Figure 5.6 is that of a permanent magnet DC motor.
We will discuss the detailed derivation of this bond graph from the actual phys-ical system in a later chapter. For now we will try to use the method that we just learned to develop the governing equations for this model.
Before we start deriving the equations, let’s make sure the gyrator factor is determined. Let us assume the gyrator factor be m. Therefore, from the bond graph representation of the system, it is clear that:
e5= mf4
e4 = mf5
(5.37)
Now we will attempt to derive the governing equations.
Answers to Q1 (written in terms of the constitutive equations).
Q1. What do all the elements give to the system?
1 Junction 1 Junction 2
FIGURE 5.6
Bond graph representation of a permanent magnet DC motor.
2. f2 = p2
The two equations may be represented together as:
In Examples 5.1–5.3, the energy storage devices ended up with integral causality. We have, in a previous chapter, talked about both integral and dif-ferential causality. We also said that although integral causality is desired, it is not always possible to have integral causality, and, instead, a differen-tial causality may result. Example 5.4 in Section 5.3 discusses the governing equation derivation process in the presence of a differential causality.