A m. Q m P. piston or diaphragm

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1 Massachusetts Institute of echnology Department of Mechanical Engineering Modeling and Simulation of Dynamic Systems Assignment #3: GAS-CHARGED ACCUMULAOR he figure below (after Pourmovahed & Otis, 1984) is a schematic diagram of the hydraulic and mechanical system. motor Q s Q m P A m m F Q a P, v P oil gas piston or diaphragm y accumulator Our first goal is to model the thermofluid subsystem assuming the flow rate Qa as an input. he following bond graph is the simplest representation of the accumulator subsystem. P R w Q a (t): S C : f dv/dt ds/dt ds w /dt Se w hydraulic domain thermal domain he two-port capacitor on the left represents the reversible energy storage and work-to-heat transduction in the ideal gas in the accumulator. he two-port resistor on the right represents irreversible heat conduction through the wall of the accumulator with the concomitant entropy production. he outside of the wall is assumed to remain at a constant temperature (i.e., ambient temperature). he direction of the power bonds has been chosen to follow the usual convention that power is positive into the storage element. However, the sign convention for the two-port resistor follows a more intuitive power in = power out convention but assumes for convenience that positive heat flow is from the wall to the gas. Nonlinear equations Causal assignment indicates that both ports of the capacitor may be given the preferred integral causal form. Note that the two-port resistor assumes its preferred causal form with temperatures as inputs. hus we may be confident that the model will properly reflect the second law. We must choose state variables. here are many possible choices, but they are not equally simple. Gas-charged accumulator solution page 1 Neville Hogan

2 Hydraulic side: We could choose either pressure, P, or volume, V, as state variable but volume seems to be the most sensible choice as the corresponding state equation is simply V = Qa(t) he negative sign is due to the fact that positive flow rate compresses the gas. hermal side: We might choose the energy variable, total entropy, S, but temperature,, is more common. We could also use total internal energy, U, as a state variable. o choose between these, consider the available information. wo-port capacitor constitutive equations are derived assuming nitrogen may be modeled as an ideal gas. he ideal gas equation is PV = m R where m is the mass of the gas and R is the gas constant for nitrogen (not the universal gas constant). he second constitutive equation is derived by assuming du = m cvd where cv is the specific heat at constant volume. Integrating: U - Uo = m cv( o) where subscript o denotes reference state. he two-port resistor equations assume Fourier s-law (i.e., linear) heat conduction. Q = h Aw(w ) where h is a heat transfer coefficient and Aw and w are the area and temperature of the wall. Consider using S and V as state variables. Ṡ = Q = h Aw(w ) hence Ṡ = h Aw ( w 1) However, I need = (S,V) and P = P(S,V) to form state equations. From class notes, = o V R Vo cv exp S So mcv P = Po V R Vo cv + 1 exp S So mcv hese are output equations. State equations are Gas-charged accumulator solution page 2 Neville Hogan

3 Ṡ = h Aw R w o V cv exp S o S Vo mcv 1 V = Qa hese are highly nonlinear, coupled differential equations. Note that the thermal time-constant is far from obvious. Consider using U and V as state variables. Pressure and temperature are determined as follows by rearranging the expression for internal energy. = U m cv + o P = m R V = m R V U m cv + o hese are output equations. State equations may be found from an energy balance equation (the first law): U = Q P V V = Qa U = h Aw U m cv + o w + m R V U m cv + o Qa hese equations are also nonlinear and coupled. However, the thermal time constant 1 τ = h Aw m cv is clearly identified as may be seen by re-arranging. U = h Aw m cv U + h Aw(w o) + m R V U m cv + o Qa Note that choosing absolute zero for the reference state would simplify this equation to U = h Aw m cv U + h Aw w + U V R cv Q a Finally consider using and V as state variables. (Note that this is loosely analogous to choosing displacement and velocity the Lagrangian choice as the state variables to describe a mechanism.) U = Q P V U = m c v Ṫ = h A w ( w ) + m R V Q a Gas-charged accumulator solution page 3 Neville Hogan

4 Ṫ = h A w m c v + h A w m c v w + R c v V Q a V = Qa hese state equations are still coupled and nonlinear, but look a little simpler. Note that the thermal time constant is again obvious but now the choice of reference temperature is less important. he output equation for pressure is P = m R V he equations using energy as a state variable are similar to those using temperature as a state variable. hat shouldn t be surprising given the simple relation between temperature and internal energy for and ideal gas. he point to remember is that the integral causal form does not constrain your choice of state variables. State variables should be chosen for convenience and to maximize insight. I will use temperature and volume as state variables. Linearized equations he second task is to linearize the model and compare to P&O s linearized model. Linearizing the state equations is always helpful. In this case it can provide insight into the physical system and the model and parameters used by P&O. he two-port capacitor constitutive equation is derived from the ideal gas equation PV = m R. o linearize, we need to write this in a causal form any of the four causal forms will serve. Use the causal form with temperature and volume as (integrated) inputs, i.e. P = m R V P dv/dt C Denote the operating point by subscript o and linearize. P P o m R ( o ) m R o 2 (V ) Write V = V, = o and P = P P o ds/dt m R + m R o 2 ( V) Remember that the positive work compresses the gas, hence V is the appropriate displacement variable on the hydraulic side. his indicates that the hydraulic side of the linearized model may be represented by a transformer between thermal and hydraulic domains connected to a fluid capacitor by a 1- junction (common flow). transformer modulus: m R Gas-charged accumulator solution page 4 Neville Hogan

5 (inverse) fluid capacitance: m R o 2 he second constitutive equation is in the form S = S(,V) It may be derived from the internal energy equation and the first law. du = m c v d = ds P dv ds = m c v Ṡ = m c v o d + m R V dv Ṫ m R ( V) Again, remember that V is the appropriate flow variable of the hydraulic side given the power sign we have assumed. his indicates that the thermal side of the linearized model may be represented by a thermal capacitor connected to a transformer between thermal and hydraulic domains by a -junction (common effort). transformer modulus: m R thermal capacitance: m c v (as before) A bond graph of the linearized two-port capacitor is as follows. dv/dt 1 F C m R/ : 2 /m R o C : m c v / o Note that this linearized model indicates that the strength of the coupling between the thermal and mechanical domains is proportional to the mass of gas and inversely proportional to the volume it occupies a useful insight. he two-port resistor equations describe entropy flow as a function of temperature. For the gas Ṡ = Q = h Aw w 1 Linearizing Ṡ Ṡ o h A w w o 2 ( o ) Ṡ h A w w 1 h A w w o o 2 ( o ) Gas-charged accumulator solution page 5 Neville Hogan

6 Assume the operating point (and also the reference thermal state) is at equilibrium with the wall temperature. o = w Ṡ h A w ( o ) o Under the same operating conditions, the entropy flow rate from the wall is identical Ṡ w = Q w Ṡ w h A w ( o ) o A bond graph of this linearization of the two-port resistor is as follows. ds/dt 1 S e : w R : o /h A w Note that this linearized model does NO reflect the second law of thermodynamics linearization has eliminated entropy production due to heat transfer. his is due to the particular choice of operating point. By choosing o = w we make the entropy flow from the wall identical to the entropy flow to the gas. Ṡ net = Ṡ Ṡ w = However, entropy production is not an essentially nonlinear phenomenon it s the choice of operating point that matters. Linearization about equilibrium eliminates entropy production; a model linearized about a non-equilibrium operating point will describe entropy production. A bond graph of the complete linearized system is as follows. dv/dt Q a : S f 1 F m R/ V 2 o /m R o : C m c v / o : C ds/dt 1 S e R : o /h A w Using the causal assignment shown, linearized equations may be read from the graph. Ṫ = o m c Ṡ m R v V V o : w Gas-charged accumulator solution page 6 Neville Hogan

7 Ṡ = h A w o State equations V = Q a Ṫ = h A w m c v + R c v o Q a Output equation P = m R m R o 2 V o compare with P&O s linearized equations, take the Laplace transform. s + 1 τ = R c v o Q a V = Q a s P = m R o Substitute m R o 2 = P o R c + s + 1 v τ (s + 1 Q a τ s) 2 R c v + 1 = R + c v c v = γ P Q a = P o γ τ s + 1 s (τ s +1) his is the same as derived by P&O. he complete hydraulic system Adding the rest of the system is straightforward. he piston can be modeled as a transformer linking the fluid domain with the mechanical domain. he piston itself is simply a mass with a force acting on it (though with an unusual sign convention). A bond graph is shown below. Gas-charged accumulator solution page 7 Neville Hogan

8 Q s :Sf P F 1 R dv/dt C I : 1/M ds/dt A m v m Se : F Se : w he translational inertia is an ideal linear element. It makes little difference whether momentum or velocity is used as its state variable. For clarity I will use velocity with temperature and volume as the gas state variables. State and output equations for the complete system are as follows. V = A m v m Qs Ṫ = h A w m c v ( w ) + R c v V ( A m v m + Qs ) v m = 1 M m R A m V F P = m R V where A m is the piston area, v m is the velocity of the inertia and M its mass. Gas-charged accumulator solution page 8 Neville Hogan

9 Parameters for simulation For P&O fig. 1: Po = psia (operating pressure in the accumulator) Vo = 13 cu. in. (operating volume of the accumulator) m =.293 lbm (mass of gas) V/ = ±.25 For P&O fig. 3: Po = psia Vo = 122 cu. in. m =.15 lbm V/ = ±.5 For P&O fig. 4: Po = psia Vo = 122 cu. in. m =.149 lbm V/ = ±.1 In all cases: w = 54 R (room temperature) R = 66 in-lbf/lbm R (gas constant) c v = 19 in-lbf/lbm R τ = m c v h Aw = 15.3 seconds For the complete hydraulic system o estimate the effective mass I assumed isothermal conditions and used the linearized model. Isothermal conditions keeps things simple because it essentially eliminates the thermal domain from participating in the dynamics. he undamped natural frequency is given by ωn = Po Am 2 M Vo hence the estimated mass is M = P o Am 2 Vo ωn 2 Gas-charged accumulator solution page 9 Neville Hogan

10 Assuming a piston diameter of.75 inches, Am is.44 inches squared. Using the pressure and volume of P&O fig.1 and an undamped natural frequency of.1 Hertz yields M = x.44 2 lbf in x (2 π x.1) 2 = sec 2 in 2 in 3 rad 2 Convert to more meaningful units: 1 lbf sec2 in = lbm M = 1.88 x 1 5 lbm. hus the mass required to yield an undamped natural frequency that low is unbelievably large 84 tons! However, the mass required to yield an undamped natural frequency of.1 Hz is a little more reasonable 1,88 pounds. o choose an initial velocity I assumed undamped oscillations and used vm = V Am V = V sin ωnt vm() = V ω n Am Simulations Nonlinear simulations of P&O figs. 1, 3 and 4 are as follows..5 P&O fig. 1, nonlinear.4.3 normalized pressure normalized volume Gas-charged accumulator solution page 1 Neville Hogan

11 47 P&O fig. 3, nonlinear pressure (psi) normalized volume 5 P&O fig. 4, nonlinear pressure (psi) normalized volume Note that in all cases these simulations are quite close to P&O s even though those authors used a much more elaborate model of the gas. For 1% volume changes, the ideal gas model is essentially identical to their data. For 5% volume changes the ideal gas model is essentially linear. Gas-charged accumulator solution page 11 Neville Hogan

12 ransient responses Simulations of the transient responses for the large mass assuming thermally damped, adiabatic and isothermal conditions are as follows. 4 ransient response of complete nonlinear system velocity (inch/sec) time (seconds).4 normalized pressure normalized volume he plot shows the velocity of the piston mass. Note that the oscillation is steadily losing amplitude as time increases. Also note that no mechanical losses such as friction have been included in our model. his loss in energy is due to the irreversible entropy production due to conduction in the gas-charged accumulator. Assuming only a reversible work storage in the gas accumulator would not predict this behavior. Just like a bicycle pump, compression of the gas generates heat that flows through the walls of the chamber. Not all of this heat energy is recovered when the gas expands, so energy is lost and we have thermal damping. he adiabatic case was modeled by adding an ideal flow source on the thermal side of the twoport capacitor. Setting this flow source to zero sets the entropy flow to zero, i.e. zero entropy flows which in this case means no heat flow. he result of this simulation shows oscillations as in the previous case. However, the amplitude of the oscillations does not decrease. We have essentially removed the heat lost through the wall leaving only the reversible work storage in compressing the gas. he system without thermal damping behaves like an undamped springmass system. Gas-charged accumulator solution page 12 Neville Hogan

13 4 Adiabatic transient response velocity (inch/sec) time (seconds).4 normalized pressure normalized volume he isothermal case was modeled by setting d/dt = and setting the initial gas temperature equal to w. his is equivalent to assuming no thermal resistance across the wall of the accumulator. Hence there is no entropy production and again, no thermal damping. 4 Isothermal transient response velocity (inch/sec) time (seconds).4 normalized pressure normalized volume he isothermal and adiabatic cases are undamped, as expected. Note that the isothermal frequency of oscillation is lower than the damped case which in turn is lower than the adiabatic case. his is also as expected. Gas-charged accumulator solution page 13 Neville Hogan

14 Finally, a simulation of the transient response for the smaller (more reasonable) mass is as follows. 4 ransient response for reduced piston mass velocity (inch/sec) time (seconds).4 normalized pressure normalized volume Note that even though the normalized pressure-volume plot shows very little evidence of energy dissipation (i.e., the area inside the loop is small) the effect on the transient response is substantial. I conclude that thermal damping is likely to be a real phenomenon in practical dynamic systems. Gas-charged accumulator solution page 14 Neville Hogan

15 Not used: U = mcvo V R cv exp S So Vo mcv 1 + Uo Choose an operating point at thermal equilibrium with the environment. Assume no forcing, i.e., Q a = o = w For P&O fig. 4: Po = bar = psia (operating pressure in the accumulator) Vo = 1687cc = 13 cu. in. (operating volume of the accumulator) Gas properties of 3 K (room temperature): cp = 141 Joules/kg K (specific heat at constant pressure) γ = (polytropic index) R = 297 Joules/kg K (gas constant) Remember: cv = c p γ c p = 256 in-lbf/lbm R γ = 256/19 = 1.35 = c p - R Gas-charged accumulator solution page 15 Neville Hogan