Modeling and Analysis of Dynamic Systems

Transcription

1 Modeling and Analysis of Dynamic Systems Dr. Guillaume Ducard Fall 2017 Institute for Dynamic Systems and Control ETH Zurich, Switzerland G. Ducard c 1 / 46

2 Outline 1 Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker 2 3 G. Ducard c 2 / 46

3 Outline Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker 1 Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker 2 3 G. Ducard c 3 / 46

4 Recalls Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker Lorentz force law F = I F = q ( l B ) ( v B ) ( is the vector product in R 3, l and B are vectors in R 3 ). Faraday induction law U = v ( l B ) Remark: right-hand rule G. Ducard c 4 / 46

5 Outline Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker 1 Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker 2 3 G. Ducard c 5 / 46

6 Loudspeaker Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker Figure: Loudspeaker mechanism: G. Ducard c 6 / 46

7 Loudspeaker: principle Recalls Case study: Loudspeaker G. Ducard c 7 / 46

8 Recalls Case study: Loudspeaker G. Ducard c 8 / 46

9 Recalls Case study: Loudspeaker a) b) S N S u(t) I(t) R L U ind (t) F(t) m d k x(t) Figure: Simplified loudspeaker: a) mechanical structure, b) electromechanical network. G. Ducard c 9 / 46

10 Recalls Case study: Loudspeaker Mechanical part m ẍ(t) = F(t) k x(t) d ẋ(t) Electric part L d dt I(t) = R I(t) U ind(t)+u(t) Motor law F(t) = B n d π I(t) = κ I(t) Generator law U ind (t) = B n d π v(t) = κ v(t) = κ ẋ(t) electro-mechanical coupling G. Ducard c 10 / 46

11 Loudspeaker: diagram Recalls Case study: Loudspeaker G. Ducard c 11 / 46

12 Outline Lecture 6: Electromechanical Systems 1 Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker 2 3 G. Ducard c 12 / 46

13 In this section, only the most important thermodynamics concepts are recalled, as a tool box for system modeling. Three examples will be considered: 1 Chemical reactor 2 Heat exchanger 3 Gas receiver and highlight differences between compressible and incompressible fluids. G. Ducard c 13 / 46

14 Lecture 6: Electromechanical Systems Definition To all systems, we associate a state function U, called internal energy. For a closed system (no matter transfer): during an arbitrary process, the variation of U is the sum of the work of external forces + thermal energy (heat) transferred by/to the system. du = δw +δq Types of thermodynamic process adiabatic process: no heat transfer, δq = 0, du = δw isochoric process: no volume change, δw = 0, du = δq isolated system: du = 0 G. Ducard c 14 / 46

15 U Lecture 6: Electromechanical Systems Work of external forces δw = P ext dv Heat, vibration energy of the molecules Assumptions: body with constant mass m, the heat transfer is 1 δq = m C v dt, if process with no change in Volume 2 δq = m C p dt, if process with no change in Pressure with C v : specific heat at constant volume [J (K kg) 1 ]. with C p : specific heat at constant pressure [J (K kg) 1 ]. G. Ducard c 15 / 46

16 Incompressible systems (solids and fluids) The internal energy U is approximated by : Remarks: U(T 1 ) = m C T 1, C = C v = C p independent of the temperature Assumptions also valid for compressible gases if temperature variations are not too large. Although a reference value for the internal energy can be chosen arbitrarily, we can consider that U = 0 at zero Kelvin. Example: C p for water is 4.18 J. g 1 K 1. Water can transport a lot of thermal energy used in cooling systems (nuclear power plant, cars, etc.) C p (T) is a function of temperature (usually increases with T). G. Ducard c 16 / 46

17 Outline Lecture 6: Electromechanical Systems 1 Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker 2 3 G. Ducard c 17 / 46

18 Lecture 6: Electromechanical Systems In many industrial applications, a continuous flow of fluid goes through the (thermodynamic) system (transfer of matter). This is no longer a closed system, we talk about open systems (valve, turbine, compressor, etc.) For open systems: we rather use an other state function: defined as H = U +PV, where the term PV takes into account the work (energy) of fluid transport. G. Ducard c 18 / 46

19 Lecture 6: Electromechanical Systems In order to work out the internal energy of a body w.r.t. the variables: T, P, V, a new function is introduced as : H H = U +P V [J] Any variation in may be captured with: dh = du +P dv +V dp = δw +δq+p dv +V dp = P dv +mc p dt +kdp +P dv +V dp = mc p dt +(k +V) dp Remark: the term k dp is inserted to encompass non-isobaric processes as well. G. Ducard c 19 / 46

20 For a isobaric process we get: dh = mc p dt +V dp dh = mc p dt = du +P dv Heat energy flows are denoted as enthalpy flows H For a isochoric process we get: H (t) = m (t)c p (T(t))T(t) [W or J s 1 ] du = δq v = mc v dt G. Ducard c 20 / 46

21 Ideal gases (= at relatively high temperature and low pressures) An ideal gas behaves according to the 2 laws of Joule: 1 its internal energy U only depends on the temperature T, 2 its enthalpy H only depends on the temperature T. For an ideal gas, we have: du = dh = δq = δq = mc v dt mc p dt mc v dt +P dv mc p dt V dp G. Ducard c 21 / 46

22 Ideal Gases Lecture 6: Electromechanical Systems P V = nr i T Ideal gas constant: R i = 8.31 J K 1 mol 1, Quantity: n in [mol], Pressure: P in [Pa], Volume: V in [m 3 ], Temperature: T in K. P V = nr i T m = M R it = m R T mr i P = V M T = ρ RT with the gas-specific constant R = R i /M gas, gas molar mass M gas G. Ducard c 22 / 46

23 Gas constant: R = C p (T) C v (T) constant for all temperatures R = C p C v in [J (kg K) 1 ] Adiabatic process (no heat transfer) PV γ = cst γ = Cp,m C v,m = Cp C v, dimensionless γ = 5/3 for mono-atomic gas, γ = 7/5 for di-atomic gas (P,V): P V γ = cst (T,V): T V γ 1 = cst (P,T): P 1 γ T γ = cst Isothermal process P V = cst Remarks: the term γ is always > 1 G. Ducard c 23 / 46

24 Outline Lecture 6: Electromechanical Systems 1 Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker 2 3 G. Ducard c 24 / 46

25 Heat Transfer Lecture 6: Electromechanical Systems Figure: 3 ways for heat transfer (picture from processtechacademy.com) G. Ducard c 25 / 46

26 Heat Transfer by Conduction Heat conduction: molecules agitation, vibration energy of molecules Fourier s law: one-dimensional case: thin cylinder of cross section area A and length l: Q= κ A (T 1 T 2 ) [W or J s 1 ] l κ: thermal conductivity (depends on material) in [W K 1 m 1 ] l A A l G. Ducard c 26 / 46

27 Heat Transfer by Convection Heat convection: Transport of matter Newton s law: heat (thermal energy) transfer between a solid body with contact surface A and the surrounding fluid is : Q= k A (T 1 T 2 ) [W or J s 1 ] k: heat transfer coefficient (depends on the surface + fluid flow properties) in [W K 1 m 2 ] G. Ducard c 27 / 46

28 Heat Transfer by Radiation Heat radiation No transport of matter, can go through vacuum. Example: heat transfer from Sun to Earth Heat radiation: Stefan-Boltzmann s law Heat radiation from a body whose surface A at temperature T 1 and its surrounding at T 2. Q= ǫ σ A (T 4 1 T4 2 ) [W or J s 1 ] Emissivity ǫ (dimensionless, < 1, depend on the material). Example of pure water: ǫ = 0.96 Stefan-Boltzmann constant σ = Wm 2 K 4 G. Ducard c 28 / 46

29 Outline Lecture 6: Electromechanical Systems 1 Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker 2 3 G. Ducard c 29 / 46

30 Example 1: Stirred Reactor System Q (t) m c k,a ϑ i (t) ϑ o ϑ i,ϑ o = temperatures inside and outside of the reactor K m = mass in the reactor kg c = specific heat of the reactor liquid J/(kg K) A = active heat exchange surface of the reactor m 2 k = heat transfer coefficient of the active surface W/(m 2 K) G. Ducard c 30 / 46

31 Example 1: Stirred Reactor System Assumptions 1 The reactor fluid has a uniform temperature ϑ i distribution, and the temperature of the environment ϑ o is constant. G. Ducard c 31 / 46

32 Example 1: Stirred Reactor System Assumptions 1 The reactor fluid has a uniform temperature ϑ i distribution, and the temperature of the environment ϑ o is constant. 2 The heat exchanger can impose an arbitrary heat flux (positive or negative) to the reactor s liquid. G. Ducard c 31 / 46

33 Example 1: Stirred Reactor System Assumptions 1 The reactor fluid has a uniform temperature ϑ i distribution, and the temperature of the environment ϑ o is constant. 2 The heat exchanger can impose an arbitrary heat flux (positive or negative) to the reactor s liquid. 3 Heat flows through the reactor s poorly insulated wall. G. Ducard c 31 / 46

34 Example 1: Stirred Reactor System Assumptions 1 The reactor fluid has a uniform temperature ϑ i distribution, and the temperature of the environment ϑ o is constant. 2 The heat exchanger can impose an arbitrary heat flux (positive or negative) to the reactor s liquid. 3 Heat flows through the reactor s poorly insulated wall. 4 The only relevant reservoir is the thermal (internal) heat stored in the reactor s liquid. G. Ducard c 31 / 46

35 Example 1: Stirred Reactor System Assumptions 1 The reactor fluid has a uniform temperature ϑ i distribution, and the temperature of the environment ϑ o is constant. 2 The heat exchanger can impose an arbitrary heat flux (positive or negative) to the reactor s liquid. 3 Heat flows through the reactor s poorly insulated wall. 4 The only relevant reservoir is the thermal (internal) heat stored in the reactor s liquid. 5 The reaction taking place inside the reactor is assumed to be neutral, i.e., no heat is generated or absorbed by the reactions inside the reactor. G. Ducard c 31 / 46

36 Example 1: Stirred Reactor System Q (t) m c k,a ϑ i (t) ϑ o G. Ducard c 32 / 46

37 Example 1: Stirred Reactor System Step 1: Input & Output Input: controlled input heat flow u(t) = Q in (t) [W] Output : internal reactor temperature (normalized to the external temperature), i.e., ϑ(t) = ϑ i (t) ϑ o Step 2: Relevant reservoir and associated level Internal energy stored is given by U(t) = m c ϑ(t) G. Ducard c 33 / 46

38 Example 1: Stirred Reactor System Step 3: The energy balance yields the (only) differential equation d dt U(t) = mc d dt ϑ(t) = Q in (t) Q out (t) Step 4: The heat flows are given by Q in (t) = u(t), controlled internal heat exchanger and according to Newton s law: Q out (t) = k A ϑ(t) poorly insulated reactor walls Step 5: Final formulation mc dϑ(t) dt = u(t) kaϑ(t) G. Ducard c 34 / 46

39 Exercise mc dϑ(t) dt = u(t) kaϑ(t) 1 Bring this system into the form of a transfer function 2 What is the characteristic time constant τ? (for a step input, after 3τ, the system reaches 95% of its final steady-state value). 3 What is the static gain of the system? 4 Check the consistency of the units for τ. G. Ducard c 35 / 46

40 Outline Lecture 6: Electromechanical Systems 1 Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker 2 3 G. Ducard c 36 / 46

41 m 2,o (t) ϑ 2o (t) m 1,i (t) ϑ 1i (t) m 1,o (t) ϑ 1o (t) Figure: Heat exchanger. m 2,i (t) ϑ 2i (t) G. Ducard c 37 / 46

42 ϑ 1o,j (t) m 1,c 1 Q j ϑ 1o,j (t) ϑ 1i,j (t) m 1 (t) m 2 (t) ϑ 2i,j (t) ϑ 2o,j (t) m 2,c 2 A k ϑ 2o,j (t) Figure: Heat exchanger element, massflows m 1, m 2, specific heat constants c 1, c 2, active heat exchange contact area A, heat transfer coefficient k. Assumptions: the temperature of the fluids leaving the heat-exchanger element j: ϑ 1o,j/2o,j is the same as inside the element. G. Ducard c 38 / 46

43 Main equations governing a cell s behavior and d dt U dϑ 1o,j (t) 1,j = m 1 c 1 = m 1 c 1 (ϑ 1i,j (t) ϑ 1o,j (t)) Q dt j (t) d dt U dϑ 2o,j (t) 2,j = m 2 c 2 = m 2 c 2 (ϑ 2i,j (t) ϑ 2o,j (t))+ Q dt j (t) and the heat transfer equation Q j (t) = ka(ϑ 1o,j (t) ϑ 2o,j (t)). Remark: Alternative formulation Q j (t) = ka( ϑ 1i,j +ϑ 1o,j 2 ϑ 2i,j +ϑ 2o,j ) 2 G. Ducard c 39 / 46

44 m 2 c p2 ϑ 2o,j (t) ϑ 1i,j (t) m 1 c p eq. (2.94) ϑ 1o,j (t) ka(ϑ 1o,j ϑ 2o,j ) + Q j + eq. (2.95) m 2 c p2 - + ϑ 2i,j (t) ϑ 1o,j (t) m 1 c p1 G. Ducard c 40 / 46

45 Outline Lecture 6: Electromechanical Systems 1 Lecture 6: Electromechanical Systems Recalls Case study: Loudspeaker 2 3 G. Ducard c 41 / 46

46 More complex situation when fluids are compressible V,R,c p,c v constant m in (t), Hin (t),ϑ in (t) m(t),p(t) U(t),ϑ(t) m out (t), Hout (t),ϑ(t) Step 1: Inputs/Outputs The inputs and outputs are the mass m in/out (t) and the enthalpy flows H in/out (t). The outflowing gas has the same ϑ(t) as the gas inside the receiver ( lumped parameter assumption). G. Ducard c 42 / 46

47 Step 2: Energy reservoir, level variables Two reservoirs are to be considered: 1 mass m(t) 2 internal energy U(t) Level variables: pressure p(t), temperature ϑ(t). Step 3: Energy balance and d dt U(t) = Hin(t) Hout(t) d dt m(t) = m in (t) m out (t) G. Ducard c 43 / 46

48 Step 4: Differential equation in the level variables d dt ϑ = ϑr { c p min ϑ in c p mout ϑ ( m in } m out )c v ϑ pvc v and d κr { min p(t) = (t)ϑ in (t) m out (t)ϑ(t)} dt V G. Ducard c 44 / 46

49 Remark: So far adiabatic (no heat-exchange) receiver was assumed. The other extreme situation is to assume isothermal conditions d dt ϑ(t) = 0 and using ideal gas law and isothermal assumption ( d dtϑ(t) = 0) d Rϑ { min p(t) = (t) m out (t)} dt V G. Ducard c 45 / 46

50 Next lecture + Upcoming Exercise Next lecture Fluid-dynamic Systems : Valves, Gas Turbines, Compressors Next exercises: Heating/Ventilation System + hot-air balloon G. Ducard c 46 / 46