THERMAL EXAMPLE: DAMPING
work in air sealed outlet
A BICYCLE PUMP WITH THE OUTLET SEALED. When the piston is depressed, a fixed mass of air is compressed. --mechanical work is done. The mechanical work done on the air is converted to heat. --the air temperature rises A temperature difference between the air and its surroundings induces heat flow. --entropy is produced The original work done is not recovered when the piston is withdrawn to the original piston. --available energy is lost
Mod. Sim. Dyn. Syst.
Thermal damping example
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MODEL THIS SYSTEM GOAL: the simplest model that can describe thermal damping (the loss of available energy) ELEMENTS: TWO KEY PHENOMENA work-to-heat transduction a two port capacitor represents thermo-mechanical transduction entropy production a two port resistor represents heat transfer and entropy production BOUNDARY CONDITIONS: For simplicity assume a flow source on the (fluid-)mechanical side a constant temperature heat sink on the thermal side
Mod. Sim. Dyn. Syst.
Thermal damping example
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A BOND GRAPH IS AS SHOWN.
P Tgas dSgas/dt thermal domain To dSo/dt
Q(t): Sf
dV/dt (fluid) mechanical domain
C
0
R
Se :To
CAUSAL ANALYSIS: The integral causal form for the two-port capacitor (pressure and temperature outputs) is consistent with the boundary conditions and with the preferred causal form for the resistor
Mod. Sim. Dyn. Syst.
Thermal damping example
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CONSTITUTIVE EQUATIONS: Assume air is an ideal gas and use the constitutive equations derived above.
T T o P P o
=
V c V v o R V c V v o
R
S S o exp mc v
=
+
1
S S o exp mc v
Assume Fourier's law describes the heat transfer process. kA = Q l (T1 - T2)
Mod. Sim. Dyn. Syst.
Thermal damping example
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ANALYSIS: For simplicity, linearize the capacitor equations about a nominal operating point defined by So and Vo
T T = o S o mcv P P = o S o mc v
T TR = o (- V ) o Vo c v
P P R = o + 1 (- V ) o Vo c v
Po To mc mc v v -1 Inverse capacitance: C = Po Po R + 1 mc v Vo c v equality of the off-diagonal terms (the crossed partial derivatives) is established using Po Vo = mRTo
Linearized constitutive equations
To T = mc v Po P mc v
where
S Po R + 1 (- V ) Vo c v Po mcv
S = S - So, V = V - Vo, T = T - To(So,Vo), P = P - Po(So,Vo)
Mod. Sim. Dyn. Syst. Thermal damping example page 5
NETWORK REPRESENTATION The linearized model may be represented using the following bond graph
P . -V T
This representation shows that in the isothermal case ( T = 0 ) the fluid capacitance is Cfluid = Vo Po
& in the constant-volume case ( V = 0 ) the thermal capacitance is C thermal = mc v To
the strength of thermo-fluid coupling is To Po This uses the convention that the transformer coefficient is for the flow equation with output flow on the output power bond
T & T & V = o S and hence T = o P Po Po
though causal considerations may require the inverse equations.
Mod. Sim. Dyn. Syst. Thermal damping example page 6
:
:
1 C : Vo/Po
TF
To/Po
0
. S
:
C: mcv/To
ALTERNATIVELY: It may be useful to express the parameters in term of easily-measured reference variables To and Vo as follows
P . -V T
:
This representation shows that the strength of the coupling is Vo mR proportional to the (nominal) gas volume inversely proportional to the mass of gas
V & V & V = o S and T = o P mR mR
:
1
TF
Vo/mR
:
0
. S
C : Vo2/mRTo
C: mcv/To
Mod. Sim. Dyn. Syst.
Thermal damping example
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RESISTOR EQUATIONS The two-port resistor constitutive equations are
& & 1 = Q = kA T1 - T2 S T1 l T1 & & 2 = Q = kA T1 - T2 S T T2 l 2
Linearize the resistor constitutive equations about a nominal operating point defined by T1o and T2o
T2o & S1 2 kA = T1o l 1 & S2 T 2o
1 T T1o 1 T1o - 2 T2 T2o -
This is in conductance form, f = Ge Note that this conductance matrix is singular:
G =-
T1oT2o 1 - =0 2 T12 T2o T1o T2o o
this is because both entropy flows are associated with the same heat flow
Mod. Sim. Dyn. Syst.
Thermal damping example
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LINEARIZE ABOUT ZERO HEAT FLOW If the two nominal operating temperatures are equal, T1o = T2o = To , the linearized constitutive equations are
1 & S1 = kA To l 1 & S2 T o
1 T To 1 1 - T2 To -
This simple form can be represented by an equally simple bond graph
T1 . S1
1
T2 . S2
R : Tol/kA
This follows the usual convention of writing the resistor parameter in resistance form
Mod. Sim. Dyn. Syst.
Thermal damping example
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ASSEMBLE THE PIECES... LINEARIZED BOND GRAPH
Pgas . -Vgas
:
Tgas
:
. So
:
Sf
1 C
TF
Po/To
:
0 C
1 R
Se: To
:
Vo/Po
mcv/To
:
Tol/kA
Note the sign change on the capacitor thermal port (to avoid a superfluous 0-junction) Causal assignment indicates a first-order system Time-constant is determined by thermal (conduction) resistance and thermal capacitance Gas pressure is determined by fluid capacitance and (reflected) thermal capacitance and resistance
Mod. Sim. Dyn. Syst.
Thermal damping example
:
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INCLUDE PISTON INERTIA BOND GRAPH
. Vgas Tgas
:
. -Sout_of_gas
: :
:
mpiston 1/Apiston
:
I
TF
:
1 C
TF
To/Po
:
0 C
1 R
Se
Tambient
:
Vo/Po
mcv/To
:
Tol/kA
Causal analysis indicates a third-order system capable of resonant oscillation In this model the only damping is in the thermal domain heat transfer, entropy flow
Mod. Sim. Dyn. Syst.
Thermal damping example
:
page 11
SUMMARIZING THE GAS STORES ENERGY. It also acts as a transducer because there are two ways to store or retrieve this energy --two interaction ports energy can be added or removed as work or heat. The "energy-storing transducer" behavior is modeled as a two-port capacitor. --just like the energy-storing transducers we examined earlier.
Mod. Sim. Dyn. Syst.
Thermal damping example
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IF POWER FLOWS VIA THE THERMAL PORT, AVAILABLE ENERGY
IS REDUCED
--the system also behaves as a dissipator. The dissipative behavior is due to heat transfer. Gas temperature change due to compression and expansion does not dissipate available energy. If the walls were perfectly insulated, no available energy would be lost, but then, no heat would flow either. Without perfect insulation temperature gradients induce heat flow Heat flow results in entropy generation. Entropy generation means a loss of available energy. THE SECOND LAW.
Mod. Sim. Dyn. Syst.
Thermal damping example
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DISCUSSION ALL MODELS ARE FALSE. It is essential to understand what errors our models make, and when the errors should not be ignored. It is commonly assumed that modeling errors become significant at higher frequencies. --not so! Compression and expansion of gases is common in mechanical systems. Hydraulic systems typically include accumulators (to prevent over-pressure during flow transients). The most common design uses a compressible gas. Compression and expansion of the gas can dissipate (available) energy.
Mod. Sim. Dyn. Syst.
Thermal damping example
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This dissipation requires heat flow, but heat flow takes time. For sufficiently rapid compression and expansion, little or no heat will flow, and little or no dissipation will occur. The simplest model of a gas-charged accumulator may justifiably ignore "thermal damping". That is an eminently reasonable modeling decision but that model will be in error at low frequencies not high frequencies.
THIS IS A GENERAL CHARACTERISTIC OF PHENOMENA DUE TO THERMODYNAMIC IRREVERSIBILITIES.
Mod. Sim. Dyn. Syst.
Thermal damping example
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