8/2/2019 Correction in Solution of Book Prob on Bond Graph

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Corrections in Example 3.11 and 3.12

Example 3.11

Step 3a:

Given in book corrected bond graph

1 3 5 1 3 5F 0 1 I F 0 1 I

2 4 2 4

C R C R

Step 3b:

Given in book corrected bond graph1 3 5 1 3 5

F 0 1 I F 0 1 I

2 4 2 4

C R C R

OR

In steps 3a and 3b

Bond graphs given instep 3a and 3b will be correct if I is replaced by Ci.e. correct bond graphs are

Step 3a:

1 3 5

F 0 1 C2 4

C R

Step 3b:1 3 5

F 0 1 C2 4

C R

8/2/2019 Correction in Solution of Book Prob on Bond Graph

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Example 3.12Corrected bond graph of figure 3.33(a) and (b) are

1 3 5 6 8

F 0 1 TF 0 I2 4 7

C R RFig. 3.33(a)

1 3 5 6 8

F 0 1 TF 0 I2 4 7

C R RFig 3.33 (b)

Corrected equations

Eq. 2a should be ( )534

4

1ee

Rf =

Equations 4a and 4b are correctUsing constraints equations, eqs 4a and 4b can be written as

( )

( )84778

82

4

4

1

fnfRee

neeR

f

==

=

Solving these two equations we get 84 and ef in terms of state variables 82 and fe

87

7

2

4

872

78

7

2

4

872

4fR

RnR

fnRenRe

RnR

fnRef

++

=++

=

Substituting 84 and ef in eqs 1a and 1b, we get resulting equations in state and sourcevariables

( )

( )01

011

8

0

87

7

2

4

872

78

2

0

87

7

2

4

872

72

4

2

fdtfRRnR

fnRenR

If

edtfRRnR

fnRenRne

RF

Ce

t

t

+

++

=

+

++

=

These are state equations in integral formWe can get differential equations by differentiating these equations

++

=

++

=

87

7

2

4

872

7

8

87

7

2

4

872

72

4

2

1

11

fRRnR

fnRenR

Idt

df

fRRnR

fnRenRne

RF

Cdt

de