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December 20, 2002 Start Presentation
Final Examination - Solution
• Electrical Circuit
• Mechanical System
• Chemical Reaction
• Multi-valued Function
• Tunnel diode
• System Dynamics
December 20, 2002 Start Presentation
Electrical Circuit I
- Given the following electrical circuit:
± U
I
December 20, 2002 Start Presentation
Electrical Circuit II
- Draw a bond graph for this circuit.
- Minimize the number of bonds needed (Hint: Apply the diamond rule).
- Add causality strokes.
- Decide whether the circuit contains any algebraic loops
and/or structural singularities.
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Electrical Circuit III
December 20, 2002 Start Presentation
Mechanical System II
- Draw a bond graph describing this system.
- Write a differential equation describing the motion of this
system in terms of the angular position θ.
- For F=0, determine the value of θ, for which the system is
in equilibrium. Call this value θ ss.
- Perform a variable transformation:
- and rewrite your differential equation in terms of ϕ.
- Create a state-space model for this system.
ϕ = θ - θss
December 20, 2002 Start Presentation
Mechanical System III
- Draw a bond graph describing this system.
F
Fm
F 1 τ 1 τ 2 Fk
τJ
τB m·g
v (^1)
v (^1)
v (^1)
v 1 ω v (^2)
ω
ω
ω
F = F(t) mg = m·g 0 = F + mg – Fm – F 1 0 = Fm – m · dv 1 /dt
0 = τ 1 – R · F 1
0 = v 1 – R · ω
0 = τ 1 – τJ – τB – τ 2
0 = τJ – J · d ω /dt
τB = B · ω
τ 2 = r · Fk
v 2 = r · ω
dFk /dt = k · v 2
December 20, 2002 Start Presentation
Mechanical System IV
- We have a structural singularity.
F = F(t) mg = m·g 0 = F + mg – Fm – F 1 0 = Fm – m · dv 1 /dt
0 = τ 1 – R · F 1
0 = v 1 – R · ω
0 = τ 1 – τJ – τB – τ 2
0 = τJ – J · d ω /dt
τB = B · ω
τ 2 = r · Fk
v 2 = r · ω
dFk /dt = k · v 2
F = F(t) mg = m·g 0 = F + mg – Fm – F 1 0 = Fm – m · dv 1
0 = τ 1 – R · F 1
0 = v 1 – R · ω
0 = τ 1 – τJ – τB – τ 2
0 = τJ – J · d ω /dt
τB = B · ω
τ 2 = r · Fk
v 2 = r · ω
dFk /dt = k · v 2
0 = dv 1 – R · d ω /dt
F = F(t) mg = m·g 0 = F + mg – Fm – F 1 0 = Fm – m · dv 1
0 = τ 1 – R · F 1
v 1 = R · ω
0 = τ 1 – τJ – τB – τ 2
0 = τJ – J · d ω /dt
τB = B · ω
τ 2 = r · Fk
v 2 = r · ω
dFk /dt = k · v 2
⇒ 0 = dv 1 – R · d^ ω^ /dt
December 20, 2002 Start Presentation
Mechanical System V
- We now have an algebraic loop.
F = F(t) mg = m·g 0 = F + mg – Fm – F 1 0 = Fm – m · dv 1
0 = τ 1 – R · F 1
v 1 = R · ω
0 = τ 1 – τJ – τB – τ 2
0 = τJ – J · d ω /dt
τB = B · ω
τ 2 = r · Fk
v 2 = r · ω
dFk /dt = k · v 2
0 = dv 1 – R · d ω /dt ⇒
F = F(t) mg = m·g 0 = F + mg – Fm – F 1 0 = Fm – m · dv 1
0 = τ 1 – R · F 1
v 1 = R · ω
0 = τ 1 – τJ – τB – τ 2
0 = τJ – J · d ω /dt
τB = B · ω
τ 2 = r · Fk
v 2 = r · ω
dFk /dt = k · v 2
0 = dv 1 – R · d ω /dt
F = F(t) mg = m·g F 1 = F + mg – Fm Fm = m · dv 1
τ 1 = R · F 1
v 1 = R · ω
τJ = τ 1 – τB – τ 2
d ω /dt = τJ /J
τB = B · ω
τ 2 = r · Fk
v 2 = r · ω
dFk /dt = k · v 2
dv 1 = R · d ω /dt
December 20, 2002 Start Presentation
Mechanical System VIII
ϕ = θ – θss ⇒ θ = ϕ + θss ; ϕ = θ ; ϕ = θ
⇒ (J + m·R^2 ) · ϕ·· = R · (F + m·g) – B · ϕ· – k·r · ( ϕ + m·g·R / k·r )
⇒ (J + m·R^2 ) · ϕ·· = R · F – B · ϕ· – k·r · ϕ
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Mechanical System IX
x 1 = ϕ ; x 2 = ϕ· ; u = F
(J + m·R^2 ) · ϕ·· = R · F – B · ϕ· – k·r · ϕ
J + m·R^2
k·r J + m·R^2
B
J + m·R^2
R
x· 1 = x (^2)
x· 2 = − · x 1 − · x 2 + · u
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Chemical Reaction I
- Dinitrogen-pentoxide (N 2 O 5 ) decays into nitrogen dioxide
(NO 2 ) and oxygen (O 2 ).
- Determine the stoichiometric coefficients necessary to
satisfy mass flow continuity.
2N 2 O 5 → 4NO 2 + O 2
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Chemical Reaction II
- The step reactions are:
- where N 2 O 5 *^ is an energized N 2 O 5 molecule.
N 2 O 5 + N 2 O 5 → N 2 O 5 *^ + N 2 O 5
k 1
N 2 O 5 *^ → NO 2 + NO 3
k 2
NO 2 + NO 3 → NO + NO 2 + O 2
k 3
NO + NO 3 → 2NO 2
k 4
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Multi-valued Function I
- Model the following multi-valued function in Modelica by
means of suitable if- and when-statements.
x
f(x)
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Once we operate inside the when-statement, we may assume
that x on the right hand side has assumed exactly one of the
three threshold values. Hence it is straightforward to test,
which of the events has occurred.
f = if x < xm /2 then fm else if x < x (^) p /2 then f (^) p else 0 ;
Multi-valued Function II
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We now can simulate.
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The value of the function f(x) can only change, when either x
becomes more negative than xm, or when x becomes more
positive than 0 , or finally, when x becomes smaller than x p.
In either of these three situations, an Iteration needs to take
place in order to determine the intersection accurately.
The event condition is formulated by means of a when-
statement.
when { x < xm , x > 0 , x < x (^) p } then ...
The iteration starts, when either one of these three conditions becomes true.
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We start by parameterizing the curve. A possible solution is
the following:
i
u u 1 u 2
i (^2) i (^1)
s = 0
s = 1 s = 2
s = 3 s = 4
s^ → ∞
- ∞ ← s
blocking conducting i2B
u1B i1B
u2B
December 20, 2002 Start Presentation
December 20, 2002 Start Presentation
System Dynamics I
- We wish to create a System Dynamics model that
describes the processes of osmosis and diffusion in sugar
cane:
Osmosis is the process of water intake by the plant.
Diffusion is the process of sucrose loss to the environment.
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System Dynamics II
- Both rates follow the same general rule:
- The gradient is the difference between the concentrations.
- The flow is always from the side of the higher
concentration to that of the lower concentration.
- The concentrations are computed as:
Rate = permeability · gradient
concentration =
amount volume
December 20, 2002 Start Presentation
December 20, 2002 Start Presentation
