14 KiB
| title | chunk | source | category | tags | date_saved | instance |
|---|---|---|---|---|---|---|
| Bond graph | 8/11 | https://en.wikipedia.org/wiki/Bond_graph | reference | science, encyclopedia | 2026-05-05T14:13:40.582713+00:00 | kb-cron |
Place an 0-junction at each node Insert Sources, R, I, C, TR, and GY bonds with 1 junctions Ground (both sides if a transformer or gyrator is present) Assign power flow direction Simplify These steps are shown more clearly in the examples below.
=== Linear mechanical === The steps for solving a Linear Mechanical problem as a bond graph are as follows:
Place 1-junctions for each distinct velocity (usually at a mass) Insert R and C bonds at their own 0-junctions between the 1 junctions where they act Insert Sources and I bonds on the 1 junctions where they act Assign power flow direction Simplify These steps are shown more clearly in the examples below.
=== Simplifying === The simplifying step is the same regardless if the system was electromagnetic or linear mechanical. The steps are:
Remove Bond of zero power (due to ground or zero velocity) Remove 0 and 1 junctions with less than three bonds Simplify parallel power Combine 0 junctions in series Combine 1 junctions in series These steps are shown more clearly in the examples below.
=== Parallel power === Parallel power is when power runs in parallel in a bond graph. An example of parallel power is shown below.
Parallel power can be simplified, by recalling the relationship between effort and flow for 0 and 1-junctions. To solve parallel power, one will first want to write down all of the equations for the junctions. For the example provided, the equations can be seen below. (Please make note of the number bond the effort/flow variable represents).
f
1
=
f
2
=
f
3
e
2
=
e
4
=
e
7
e
1
=
e
2
+
e
3
f
2
=
f
4
+
f
7
e
3
=
e
5
=
e
6
f
7
=
f
6
=
f
8
f
3
=
f
5
+
f
6
e
7
+
e
6
=
e
8
{\displaystyle {\begin{matrix}f_{1}=f_{2}=f_{3}&&e_{2}=e_{4}=e_{7}\\e_{1}=e_{2}+e_{3}&&f_{2}=f_{4}+f_{7}\\&&\\e_{3}=e_{5}=e_{6}&&f_{7}=f_{6}=f_{8}\\f_{3}=f_{5}+f_{6}&&e_{7}+e_{6}=e_{8}\end{matrix}}}
By manipulating these equations one can arrange them such that one can find an equivalent set of 0- and 1-junctions to describe the parallel power. For example, because
e
3
=
e
6
{\textstyle e_{3}=e_{6}}
and
e
2
=
e
7
{\textstyle e_{2}=e_{7}}
one can replace the variables in the equation
e
1
=
e
2
+
e
3
{\textstyle e_{1}=e_{2}+e_{3}}
resulting in
e
1
=
e
6
+
e
7
{\textstyle e_{1}=e_{6}+e_{7}}
and since
e
6
+
e
7
=
e
8
{\textstyle e_{6}+e_{7}=e_{8}}
, we now know that
e
1
=
e
8
{\displaystyle e_{1}=e_{8}}
. This relationship of two effort variables equaling can be explained by an 0-junction. Manipulating other equations one can find that
f
4
=
f
5
{\displaystyle f_{4}=f_{5}}
which describes the relationship of a 1-junction. Once the relationships have been determineds, one can redraw the parallel power section with the new junctions. The result for the example show is seen below.
=== Examples ===
==== Simple electrical system ==== A simple electrical circuit consisting of a voltage source, resistor, and capacitor in series.
The first step is to draw 0-junctions at all of the nodes:
0
0
0
0
{\displaystyle {\begin{matrix}&0&&0&\\&&&&\\&&&&\\&0&&0&\end{matrix}}}
The next step is to add all of the elements acting at their own 1-junction:
R
|
0
−
1
−
0
|
|
S
e
−
1
1
−
C
|
|
0
_
−
−
−
0
{\displaystyle {\begin{matrix}&&&&R&&&&\\&&&&|&&&&\\&&0&-&1&-&0&&\\&&|&&&&|&&\\S_{e}&-&1&&&&1&-&C\\&&|&&&&|&&\\&&{\underline {0}}&-&-&-&0&&\end{matrix}}}
The next step is to pick a ground. The ground is simply an 0-junction that is going to be assumed to have no voltage. For this case, the ground will be chosen to be the lower left 0-junction, that is underlined above. The next step is to draw all of the arrows for the bond graph. The arrows on junctions should point towards ground (following a similar path to current). For resistance, inertance, and compliance elements, the arrows always point towards the elements. The result of drawing the arrows can be seen below, with the 0-junction marked with a star as the ground.
Now that we have the Bond graph, we can start the process of simplifying it. The first step is to remove all the ground nodes. Both of the bottom 0-junctions can be removed, because they are both grounded. The result is shown below.
Next, the junctions with less than three bonds can be removed. This is because flow and effort pass through these junctions without being modified, so they can be removed to allow us to draw less. The result can be seen below.
The final step is to apply causality to the bond graph. Applying causality was explained above. The final bond graph is shown below.
==== Advanced electrical system ==== A more advanced electrical system with a current source, resistors, capacitors, and a transformer
Following the steps with this circuit will result in the bond graph below, before it is simplified. The nodes marked with the star denote the ground.
Simplifying the bond graph will result in the image below.
Lastly, applying causality will result in the bond graph below. The bond with star denotes a causal conflict.
==== Simple linear mechanical ==== A simple linear mechanical system, consisting of a mass on a spring that is attached to a wall. The mass has some force being applied to it. An image of the system is shown below.
For a mechanical system, the first step is to place a 1-junction at each distinct velocity, in this case there are two distinct velocities, the mass and the wall. It is usually helpful to label the 1-junctions for reference. The result is below.
1
mass
1
wall
{\displaystyle {\begin{matrix}&&\\&&\\1_{\text{mass}}&&\\&&\\&&\\&&\\1_{\text{wall}}&&\end{matrix}}}
The next step is to draw the R and C bonds at their own 0-junctions between the 1-junctions where they act. For this example there is only one of these bonds, the C bond for the spring. It acts between the 1-junction representing the mass and the 1-junction representing the wall. The result is below.