Modeling and Analysis of Dynamic Systems

Transcription

1 Modeling and Analysis of Dynamic Systems by Dr. Guillaume Ducard c Fall 2016 Institute for Dynamic Systems and Control ETH Zurich, Switzerland 1/59

2 Outline 1 2 Introduction Example: Continuously Stirred Tank Reactor 3 2/59

3 Outline 1 2 Introduction Example: Continuously Stirred Tank Reactor 3 3/59

4 yourdictionnary.com 4/59

5 Recommanded Videos on gas turbine: GE 101 gas turbine: bbc link: 5/59

6 Thermodynamic cycles (Brayton cycles). 1 2 isentropic compression 2 3 isobar fuel combustion 3 4 isentropic expansion 4 1 isobar heat rejection into atmosphere 6/59

7 u vng ϑ 3 p 3 p 4 ω t turbine (fluid- and thermodynamics) ϑ t = ϑ 4 mt T t Figure: Causality diagram of a gas turbine. Inputs: ϑ 3 : Temperature before the turbine [K] P 3 : Pressure before the turbine [Pa] P 4 : Pressure after the turbine [Pa] ω t : Turbine speed [rad.s 1 ] u vnt Variable nozzle geometry: control input 7/59

8 u vng ϑ 3 p 3 p 4 ω t turbine (fluid- and thermodynamics) ϑ t = ϑ 4 mt T t Figure: Causality diagram of a gas turbine. Outputs: ϑ t (t) = ϑ 4 (t): Temperature after the turbine, [K] m (t) : gas mass flow [kg.s 1 ] T t : shaft torque [Nm] 8/59

9 u vng ϑ 3 p 3 p 4 ω t turbine (fluid- and thermodynamics) ϑ t = ϑ 4 mt T t Figure: Causality diagram of a gas turbine. Objectives Find a relationship between the outputs: ϑ 4 (t), m (t), T t and the inputs. 9/59

10 µ t (kg/s) 3 2 ω t = constant throttle approximation (A vnt < 1) throttle approximation (A vnt = 1) Π t Figure: Gas turbine massflow behavior. The flow behavior of gas turbines may be assumed to be similar to that of an isenthalpic valve or maps also are used. All variables are referenced to a nominal case ϑ 3,ref,p 3,ref Normalized mass flow : µ t = m t ϑ3 /ϑ 3,ref /(p 3 /p 3,ref ) Pressure ratio: Π t = p3 p 4 ( Π t > 1 for gas flow through the turbine) Turbine inlet area: A vnt (variable nozzle throttle area) 10/59

11 c u c us = = rtωt c us 2c p ϑ 3 [1 Π (1 κ)/κ t ] c u : turbine blade-tip speed ratio c us : speed of the gas resulting from an isentropic expension η t 3 Π t = π 1 2 simplified turbine Π t = π 2 > π cu Figure: Gas turbine efficiencies. knowing the parameter c u and the turbine pressure ratio Π t, the above map gives us the turbine efficiency η t 11/59

12 Once the efficiency of the turbine η t is known: The torque T t delivered by the turbine to the mechanical shaft T t = η t m t c p ϑ [ 3 ω t 1 Π (1 κ)/κ t with c p : mass gas specific heat at constant pressure, and κ = c p /c v, of the gases flowing through the turbine. ] Turbine output temperature: ϑ 4 [ ] ϑ 4 = ϑ 3 1 η t (1 Π (1 κ)/κ t ) Remarks: 1 See the influence of the efficiency on the real/actual temperature of the exhaust gas, compared to the isentropic temperature. 2 Remark the definition of the turbine pressure ratio: Π t = P t,in P t,out 3 Equations derived during the lecture. 12/59

13 Outline 1 2 Introduction Example: Continuously Stirred Tank Reactor 3 13/59

14 Compressor models are obtained using a simplification : to consider the compressor elements as static (see causality diagram). p 2 ϑ 1 p 1 ω c compressor (fluid- and thermodynamics) ϑ c = ϑ 2 mc T c Figure: Causality diagram of a gas compressor. 14/59

15 p 2 ϑ 1 p 1 ω c compressor (fluid- and thermodynamics) ϑ c = ϑ 2 mc T c Figure: Causality diagram of a gas compressor. Inputs : Pressure at compressor output : P 2 [Pa] Pressure at compressor input : P 1 [Pa] Temperature at compressor input: ϑ 1 [K] Compressor turn speed: ω c [rad/s] 15/59

16 p 2 ϑ 1 p 1 ω c compressor (fluid- and thermodynamics) ϑ c = ϑ 2 mc T c Figure: Causality diagram of a gas compressor. Outputs : Temperature at compressor output ϑ 2 [K] Gas mass flow m c [kg/s] Torque absorbed by the compressor T c [N.m] 16/59

17 p 2 ϑ 1 p 1 ω c compressor (fluid- and thermodynamics) ϑ c = ϑ 2 mc T c Figure: Causality diagram of a gas compressor. Objectives Find an analytical expression of the Outputs ϑ 2, m c, T c as a function of the inputs. 17/59

18 In order to compute: 1 the gas temperature at the compressor outlet ϑ 2, the compressor efficiency η c is needed. 2 the compressor torque T c : compressor mass flow m c is needed. These 2 parameters are found using experimental maps, as follows: 18/59

19 Π c surge η c = constant ω c = constant µc (kg/s) 1 from the known compressor ratio Π c, 2 and the compressor turn rate ω c, the norm. mass flow µ c and thus m c and the efficiency η c of the compressor can be found. Remarks: Compressor pressure ratio: Π c = P 2 P 1 Normalized compressor mass flow: µ c= m ϑ1 P 1,ref c ϑ 1,ref P 1 19/59

20 Once the efficiency is known, the torque absorbed by the compressor can be found by the isentropic efficiency equation T c = m c c p ϑ [ ] 1 Π c (κ 1)/κ 1 η c ω c where c p is the specific heat at constant pressure and κ the ratio κ = c p /c v of the gases flowing through the compressor. Similarly, the temperature ϑ 2 = ϑ c is found to be [ ϑ 2 = ϑ ( ) ] Π c (κ 1)/κ 1 η c Note the definition of the compressor pressure ratio: Π c = P c,out P c,in 20/59

21 Remarks on power vs. efficiency Maximum power is not at maximum efficiency, usually industrial manufacturer prefer to optimize for maximal power and compromise on efficiency. 21/59

22 Outline 1 2 Introduction Example: Continuously Stirred Tank Reactor 3 22/59

23 Automotive Industry In most of cars nowdays there are turbine/compressor systems known as turbocompressor or turbocharger or simply turbo. 23/59

24 Standard turbocharger Figure: source: Honeywell 24/59

25 Standard turbocharger Figure: source: Honeywell 25/59

26 Standard turbocharger Figure: source: Honeywell 26/59

27 Turbocharger with Variable Geometry Turbine (VGT) with VGT, even at low engine speed (thus low gas exhaust flow) we can make the turbine spins fast,by accelerating the gas flow (reducing the in-between vane space), orientating the flow towards the turbo blades in order to increase its rotational speed. 27/59

28 Principle of VGT turbo: VGT explained: Audi VGT Turbo: Mechanical disassembly: 28/59

29 Outline Introduction Example: Continuously Stirred Tank Reactor 1 2 Introduction Example: Continuously Stirred Tank Reactor 3 29/59

30 Chemical Systems Introduction Example: Continuously Stirred Tank Reactor Definition A chemical reaction is a transformation during which reactants are transformed into products. Remark: during a chemical reaction, molecules may appear/disappear to be transformed in other molecules, but in all cases the atoms are conserved, and thus the total mass is preserved. Chemical reactions typically involve two reactants αa+βb γc +δd the integer coefficients {α, β, γ, δ} describe the stoichiometry of the reaction and the double arrow indicates that the reaction can evolve in both directions. 30/59

31 Chemical Systems: examples Introduction Example: Continuously Stirred Tank Reactor Combination sodium + chlorine sodium chloride 2 Na(s) + Cl 2 (g) 2 NaCl(s) Combustion burning of propane C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) burning of coal (carbon) gives carbon dioxide C(s)+O 2 (g) CO 2 (g) Other types of reactions include: Reduction Oxidation Precipitation reactions etc. 31/59

32 Chemical Systems Introduction Example: Continuously Stirred Tank Reactor Chemical reactions typically involve two reactants αa+βb γc +δd Chemical reactions are best described on a molecular basis n A, n B, n C, n D with numbers of mol (quantity of element) n A = 1 mol is molecules of that species (Avogadro constant), Molar mass M A is the mass of 1 mol of A. m A = n A M A. Concentration [A] is the number of molecules n A in a given volume V: [A] = n A V 32/59

33 Introduction Example: Continuously Stirred Tank Reactor Chemical reaction kinetics We define the reaction advancement ξ, as the variation of the reactants quantity: dn A α = dn B β = dn C γ = dn D δ = dξ The reaction rate or speed of the reaction, or formation speed is thus v f = dξ dt v f = dξ dt = 1 dn A α dt = 1 β dn B dt = 1 γ dn C dt = 1 δ dn D dt If you divide by the volume V of the mixed reacting elements, the volumic speed of the reaction is v = 1 V dξ dt = 1 d[a] = 1 d[b] = 1 d[c] = 1 d[d] α dt β dt γ dt γ dt where [x] denotes the concentration of element x in [mol/m 3 ] 33/59

34 Introduction Example: Continuously Stirred Tank Reactor Chemical reaction kinetics v = 1 V dξ dt = 1 d[a] = 1 d[b] = 1 d[c] = 1 d[d] α dt β dt γ dt δ dt The reaction rate is also defined as v = r[a] p [B] q where p and q are called partial reaction orders. Remark: the coefficients p and q are often not equal to the stoichiometric coefficients, must be determined experimentally. 34/59

35 Case of First-order Reaction Introduction Example: Continuously Stirred Tank Reactor Consider the reaction A+B C Assume the reaction is first order in both reactants, then the reaction rate is: v = r [A][B] since the reaction rate is also it yields v = 1 d[a] α dt d[a] dt = α r [A][B] 35/59

36 Case of equilibrium or opposed reactions Introduction Example: Continuously Stirred Tank Reactor The forward reaction rate αa+βb γc +δd (causing element A to disappear ) is also defined as: v = r [A] α [B] β. and thus the concentration change rate of reactant A is obtained as : 1 d [A] = r [A] α [B] β α dt d [A] = α r [A] α [B] β dt Remark: in this formulation, the probability that the reaction takes place is proportional to the probability that the necessary number of molecules A and B are in contact (concentration). 36/59

37 Case of equilibrium or opposed reactions Introduction Example: Continuously Stirred Tank Reactor The backward reaction rate (causing element A to appear ) is also defined as: v + = r + [C] γ [D] δ and thus the concentration change rate of reactant A is obtained as : 1 d + [A] = r + [C] γ [D] δ α dt d + [A] = α r + [C] γ [D] δ dt Total rate of formation of the species A (in mol/(s m 3 )) d [A] = α ( r + [C] γ [D] δ r [A] α [B] β) dt 37/59

38 Introduction Example: Continuously Stirred Tank Reactor The rate of the reaction constants r +, and r, depend on: the pressure, and most importantly on the temperature. An Arrhenius model is used r + = k + (ϑ,p,...) e E+ /(Rϑ) R = J/mol K: universal gas constant. The constant k + is referred to as the the pre-exponential factor, E + is the activation energy. The Boltzmann term: exp{ E + /(Rϑ)} indicates the fraction of all collisions that have sufficient energy to start the reaction. 38/59

39 Introduction Example: Continuously Stirred Tank Reactor r + = k + (ϑ,p,...) e E+ /(Rϑ) r/k Rϑ/E Figure: Arrhenius function. Remark: reminds of probability function; see influence of temperature 39/59

40 Introduction Example: Continuously Stirred Tank Reactor Similarly, the reaction kinetic for the backward reaction r = k (ϑ,p,...)e E /(Rϑ) The four parameters: {k +,k,e +,E } must be determined experimentally. 40/59

41 Outline Introduction Example: Continuously Stirred Tank Reactor 1 2 Introduction Example: Continuously Stirred Tank Reactor 3 41/59

42 Introduction Example: Continuously Stirred Tank Reactor Example: continuously stirred tank reactor (CSTR) Assumptions A+B C The concentration [B] can be assumed to remain constant. The dissociation A+B C is negligible. The mass m and the density ρ of the fluid in the CSRT are constant. The CSTR is perfectly insulated, the only heat transfer occurs through the controllable heat exchanger. 42/59

43 Introduction Example: Continuously Stirred Tank Reactor Example: continuously stirred tank reactor (CSTR) [A i (t)],ϑ i (t), m i Q (t) m ρ c ϑ(t) [A(t)],[C(t)] LC [C(t)],ϑ(t), m o Figure: Continuous chemical reactor. 43/59

44 Modeling of the Chemical Reactor Introduction Example: Continuously Stirred Tank Reactor Step 1: Define Inputs and Outputs Control input = Q(t): rate of heat transferred by the heat exchanger Outputs = concentration of C and temperature ϑ(t) Disturbances = [A i (t)] and ϑ i (t) Notice that and m i = m o = m V i = V o = V= m /ρ 44/59

45 Modeling of the Chemical Reactor Introduction Example: Continuously Stirred Tank Reactor Step 2: Energy reservoirs Using the assumptions mentioned, three reservoirs must be modeled: n A : the amount of species A in the CSTR, level variable [A]; n C : the amount of species C in the CSTR, level variable [C]; U: the internal energy in the CSTR, level variable ϑ. 45/59

46 Modeling of the Chemical Reactor Introduction Example: Continuously Stirred Tank Reactor Step 3: Conservation laws For species A, the conservation laws yield d dt n A(t) = V [A i (t)] V [A(t)] V k [B] e E/(Rϑ(t)) [A(t)] for species C d dt n C(t) = V [C(t)]+V k [B] e E/(Rϑ(t)) [A(t)] and for the CSTR energy d dt U(ϑ(t),n A(t),n B (t),n C (t)]) = Hi(ϑ i (t)) H(ϑ(t))+ Q(t) Notice that the concentration [B] is (assumed to be) a constant. 46/59

47 Modeling of the Chemical Reactor Introduction Example: Continuously Stirred Tank Reactor Step 4: Express of internal energy as a function of temperature and composition du(ϑ,n A,n B,n C ) = U ϑ dϑ + U n A dn A + U n B dn B + U n C dn C = ρvc v dϑ +H A dn A +H B dn B +H C dn C where H A, H B, and H C are the enthalpies of formation for the corresponding species. Remark: Neither c v nor H A, H B, and H C are assumed to depend on the temperature ϑ. 47/59

48 Introduction Example: Continuously Stirred Tank Reactor Step 5: Inserting the last equation in the conservation laws yields: Consider: c p c v for liquids, τ d dt [A(t)] = [A i(t)] ( 1+τ k e E/(Rϑ(t))) [A(t)] τ d dt [C(t)] = [C(t)]+τ k e E/(Rϑ(t)) [A(t)] τ d dt ϑ(t) = ϑ i(t) ϑ(t)+ 1 Q(t) k ρc v +τ H 0 c vρ e E/(Rϑ(t)) [A(t)] V where the residence time τ, the overall reaction rate k and the reaction enthalpy H 0 are defined by τ := V/ V, k := k [B], H 0 = H A +H B H C Remark: H x > 0 if energy is needed to form species x. 48/59

49 Introduction Example: Continuously Stirred Tank Reactor Static behavior of the CSTR: i.e., Q = 0, rate of heat transferred by the heat exchanger, ϑ i = constant, and [A i ] = constant. Heat removed by the in- and out-flowing mass flow Hflow(ϑ)+ Q chem (ϑ) = 0 where Hflow(ϑ) = m c p (ϑ i ϑ) Vke E/(Rϑ) Q chem (ϑ) = H 0 1+τke E/(Rϑ)[A i] 49/59

50 Introduction Example: Continuously Stirred Tank Reactor P 3 Hflow Q chem P 2 P 1 Rϑ/E Figure: Steady-state points as a function of temperature. 50/59

51 Distributed Parameter Systems Assessments Many physical systems are best described by partial differential equations (PDE), representing for instance, mechanical continua, fluids, electromagnetic waves, etc. 51/59

52 Distributed Parameter Systems Assessments Many physical systems are best described by partial differential equations (PDE), representing for instance, mechanical continua, fluids, electromagnetic waves, etc. The underlying physical laws (Navier-Stokes, Maxwell, etc.) are relatively simple to formulate but extremely difficult to solve, especially for non-trivial boundary conditions. 51/59

53 Distributed Parameter Systems Assessments Many physical systems are best described by partial differential equations (PDE), representing for instance, mechanical continua, fluids, electromagnetic waves, etc. The underlying physical laws (Navier-Stokes, Maxwell, etc.) are relatively simple to formulate but extremely difficult to solve, especially for non-trivial boundary conditions. Typically, PDE are obtained by analyzing an infinitesimally small lumped parameter cell. This is shown in the following example of a heat exchanger. 51/59

54 Distributed Parameter Systems x x dx ϑ 1 (x,t) dm 1,c 1 d Q (x,t) ϑ 1 (x,t) da k F 1 ϑ 1 (x dx,t) ρ 1 w 1 ρ 2 w 2 ϑ 2 (x,t) F 2 dm 2,c 2 ϑ 2 (x dx,t) Figure: Infinitesimally small heat exchanger element. 52/59

55 The internal energy balance for the upper part yields t [dm 1c 1 ϑ 1 (x,t)] = m 1 c 1 [ϑ 1 (x dx,t) ϑ 1 (x,t)] kda[ϑ 1 (x,t) ϑ 2 (x,t)] (1) Introducing the fluid flow speeds w 1 and w 2, the fluid densities ρ 1 and ρ 2, the flow cross sections F 1 and F 2, and the total heat exchanger length L, equation (1) can be written as a series expansion F 1 dxρ 1 c 1 ϑ 1 (x,t) t ϑ 1 (x,t) F 1 w 1 ρ 1 c 1 dx x k A L dx[ϑ 1(x,t) ϑ 2 (x,t)] 53/59

56 Collecting terms and dividing by dx the final formulation is obtained ϑ 1 (x,t) t = w 1 ϑ 1 (x,t) x ka Lc 1 F 1 ρ 1 [ϑ 1 (x,t) ϑ 2 (x,t)] The lower part can be analyzed in the same way which yields ϑ 2 (x,t) t = w 2 ϑ 2 (x,t) x + ka Lc 2 F 2 ρ 2 [ϑ 1 (x,t) ϑ 2 (x,t)] 54/59

57 Special case k = 0 (no heat transfer between the two streams) dϑ o,j (t) mc p = m c p (ϑ i,j (t) ϑ o,j (t)) (2) dt behavior is modeled by stacking together n such first-order systems. m = ρf L n, m= ρfw with m representing the mass in one cell, F the constant area and L the total length of the heat-exchanger tube, and w the velocity of the flow, equation (2) is τ dϑ o,j(t) dt = ϑ i,j (t) ϑ o,j (t), τ = L nw, j = 1,...,n 55/59

58 The PDE (2) in this simplified case reads ϑ(x, t) t = w ϑ(x,t) x This hyperbolic PDE is known as the (simplified) advection equation. (3) A closed-form solution is ϑ(x,t) = f(t x w ). It is easy to see that the function f(.) satisfies the PDE (3) f(t x w ) t = f(t x w ) (t x (t x w ) w ) = g(t x t w ) and, f(t x w ) x = f(t x w ) (t x (t x w ) w ) = g(t x x w ) 1 w respectively. 56/59

59 Obviously, equation (3) is satisfied because w 1 w g(t x w ) = g(t x w ) Therefore, taking x at the input (x = 0) and at the output (x = L) extremes, the input/output dynamics u(t) = ϑ(t,0) y(t) = ϑ(t,l) of the simplified heat-exchanger system is described by y(t) = u(t L ) = u(t T) (4) w which is, of course, a simple delay element with delay T = L w. 57/59

60 1 0.8 input u(t) output ODE temperatures (K) output PDE time t (s) 58/59

61 Next lecture + Upcoming Exercise Next lecture System Parameter Identification Least squares Next exercise: Diesel Engine Injector 59/59