Modeling and Analysis of Dynamic Systems

Modeling and Analysis of Dynamic Systems Dr. Guillaume Ducard Fall 2017 Institute for Dynamic Systems and Control ETH Zurich, Switzerland G. Ducard c 1 / 34

Outline 1 Lecture 7: Recall on Thermodynamics Principles Thermodynamics Principles 1 & 2 Entropy Variation 2 3 G. Ducard c 2 / 34

Outline Lecture 7: Recall on Thermodynamics Principles Thermodynamics Principles 1 & 2 Entropy Variation 1 Lecture 7: Recall on Thermodynamics Principles Thermodynamics Principles 1 & 2 Entropy Variation 2 3 G. Ducard c 3 / 34

Thermodynamics Principles 1 & 2 Entropy Variation First principle of thermodynamics is about the conservation of energy. All system may be described by its total energy. Its energy variation is given by: For closed systems: The variation of the total energy of a closed system is given by du = δw +δq du: internal energy variation δw: mechanical energy: work of pressure forces at the system surfaces + work of gravity forces in the volume of the system (usually negligible). δq: thermal energy exchanged with the surrounding For a closed system (ex: cylinder): the work of the pressure forces is given by: δw = P ext dv G. Ducard c 4 / 34

Thermodynamics Principles 1 & 2 Entropy Variation First principle of thermodynamics In many industrial applications, a continuous flow of fluids goes through the (thermodynamic) system. 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: Enthalpy defined as H = U +PV, where the term PV takes into account the work (energy) of fluid transport. The total energy(e tot ) variation of an open system is : de tot = dh +dk = δτ +δq where δτ is the useful work (production of thrust, or torque, etc.) G. Ducard c 5 / 34

Second Principle Thermodynamics Principles 1 & 2 Entropy Variation Second principle: intuition The second principle of thermodynamics is about how (sense, direction) the energy exchanges (or transformation) take place during a process, the associated variable is called Entropy. Example of use: to predict the direction of heat flow (in between 2 medium at different temperatures), particles flow (in between systems with different pressures). G. Ducard c 6 / 34

Second Principle of Thermodynamics Thermodynamics Principles 1 & 2 Entropy Variation Second principle: statement All system is characterized by a state function S called Entropy. If in an isolated system, irreversible transformations (processes) take place, the Entropy of the system increases and is maximal at equilibrium. G. Ducard c 8 / 34

Outline Lecture 7: Recall on Thermodynamics Principles Thermodynamics Principles 1 & 2 Entropy Variation 1 Lecture 7: Recall on Thermodynamics Principles Thermodynamics Principles 1 & 2 Entropy Variation 2 3 G. Ducard c 9 / 34

Thermodynamics Principles 1 & 2 Entropy Variation Entropy variation w.r.t. Internal energy variation at constant volume ( ) S = 1 U T Entropy variation w.r.t. volume variation at constant Internal energy ( ) S = P V T Entropy variation w.r.t. variation in U and V V U ds = ( ) S U V du +( ) S V U dv = ( ( 1 T) du + P ) T dv dt = mc v T +mrdv V (ideal gas) dt = mc v T mrdp P (ideal gas) G. Ducard c 10 / 34

Thermodynamics Principles 1 & 2 Entropy Variation A process is said to be isentropic when ds = 0. It is the case for an adiabatic (no heat transfer) and reversible process. Isentropic expansion relations for a perfect gas 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 Isentropic processes will be considered later in the lecture, when modeling valves and gas turbines. G. Ducard c 11 / 34

Outline Lecture 7: Recall on Thermodynamics Principles 1 Lecture 7: Recall on Thermodynamics Principles Thermodynamics Principles 1 & 2 Entropy Variation 2 3 G. Ducard c 12 / 34

Valves Lecture 7: Recall on Thermodynamics Principles Figure: Example of a gate valve (top), butterfly valve (bottom) (http://www.ctgclean.com) G. Ducard c 13 / 34

Valves Lecture 7: Recall on Thermodynamics Principles From a modeling point of view: this system can be viewed as the interface between 2 reservoirs (with associated level variables): fluid upstream (upstream pressure p in ) fluid downstream (downsteam pressure p out ) This is due to the pressure difference that the fluid flows through the orifice (valve). Two modeling approaches considered: 1 for incompressible fluids 2 for compressible fluids G. Ducard c 14 / 34

Valve and Incompressible Fluids Assumptions Flow frictions modeled through a correcting factor: determined experimentally c d : discharge coefficient Fluid assumed to be incompressible (constant density). This is the case for liquids and fluids at low Mach numbers. Mach number M M = u c [ ] where u is the local flow velocity [m/s], c is speed of sound in the medium [m/s]. Primary use: to determine whether a fluid flow can be treated as incompressible. If M < 0.2(upto0.3) and the flow is quasi-steady and isothermal, incompressible flow can be assumed. G. Ducard c 15 / 34

Valve and Incompressible Fluids Modeling objective Express the fluid mass flow m through the valve, with the following terms: c d : discharge coefficient (takes into account flow restrictions, flow friction and other losses); A: open area of the valve; ρ: density of the fluid (constant: incompressibility assumption); p in : pressure upstream of the valve; p out : pressure downstream of the valve. G. Ducard c 16 / 34

Valve and Incompressible Fluids If the fluid is incompressible, then the valve can be modeled using Bernoulli s law (conservation of energy), leading to m (t) = c d A 2ρ p in (t) p out (t) G. Ducard c 17 / 34

Valve and Compressible Fluids If the fluid is compressible, the valve can be modeled using the concept of Isenthalpic throttle. Why do we call it Isenthalpic throttle? Isenthalpic process: A fluid circulates in a tube with 1 no moving wall (no work from pressure forces), 2 no heat exchange dh = 0, no enthalpy variation in the system. G. Ducard c 18 / 34

Valve and Compressible Fluids The behavior of the flow may be separated into 2 distinct parts : 1) Valve upstream and 2) valve downstream. p in (t) p out (t) m in (t),ϑ in (t),p in (t) m out (t),ϑ out (t),p out (t) ϑ in (t) ϑ out (t) G. Ducard c 19 / 34

p in (t) p out (t) m in (t),ϑ in (t),p in (t) m out (t),ϑ out (t),p out (t) ϑ in (t) ϑ out (t) Valve upstream: up to the narrowest point Fluid accelerating, laminar flow, Pressure decrease, temperature decrease, potential energy stored in the flow (level variable = pressure) is converted isentropically (without losses) into kinetic energy. G. Ducard c 20 / 34

p in (t) p out (t) m in (t),ϑ in (t),p in (t) m out (t),ϑ out (t),p out (t) ϑ in (t) ϑ out (t) Valve downstream: from narrowest point onward Fluid decelerating, Turbulent flow Kinetic energy is dissipated in thermal energy (no pressure recuperation takes place). G. Ducard c 21 / 34

p in (t) p out (t) m in (t),ϑ in (t),p in (t) m out (t),ϑ out (t),p out (t) ϑ in (t) ϑ out (t) Consequences of assumptions made: 1 The pressure in the narrowest part of the valve is almost equal to downstream pressure. 2 The temperature of the flow before and after the valve is approximately the same (almost isothermal process). G. Ducard c 22 / 34

If we deal with gases, in conditions of perfect gases: p m in (t) (t) = c d A(t) Rϑin (t) Ψ(p in(t),p out (t)) where Ψ(.) is Ψ(p in (t),p out (t)) = and where ( κ 2 κ+1 ( pout p in ) 1/κ p cr = ) κ+1 κ 1 2κ κ 1 [ ] κ 2 κ 1 pin κ+1 [ ] )κ 1 1 ( pout κ p in for p out < p cr for p out p cr G. Ducard c 23 / 34

Approximation (air and many other gases OK): Ψ(p in (t),p out (t)) = 1 2 [ ] 2p out p in 1 pout p in for p out < 0.5p in for p out 0.5p in G. Ducard c 24 / 34

Approximation (unrealistically large threshold: π tr = 0.9) Ψ(Π) (-) Ψ exact (solid) and approximated (dashed)(κ = 1.4), laminar part (dash-dot) 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Π (-) G. Ducard c 25 / 34

A laminar flow condition can be assumed for very small pressure ratios Π tr := p out p in < 1 If larger pressure ratios occur, then use a smooth approximation with Ψ(Π) = a (Π 1) 3 +b (Π 1) a = Ψ tr (Π tr 1) Ψ tr 2 (Π tr 1) 3, b = Ψ tr 3 a (Π tr 1) 2 is used (Ψ tr is the value of Ψ and Ψ tr Ψ at the threshold Π tr ). the value of the gradient of G. Ducard c 26 / 34

Model of Gas Turbines yourdictionnary.com http://www.aptech.ro G. Ducard c 27 / 34

Thermodynamic cycles (Brayton cycles). 1 2 isentropic compression 2 3 isobar fuel combustion 3 4 isentropic expansion 4 1 isobar heat rejection into atmosphere G. Ducard c 29 / 34

Simplified model of a gas turbine ω 1 Θ ω 1 Θ 2 2 k ϕ T 1 T 2 d 1 d 2 rotor 2: the turbine stage driving torque T 2, Moment of inertia: Θ 2 rotor 1: the compressor stage breaking torque T 1, Moment of inertia: Θ 1 shaft elasticity constant: k friction losses: d 1 and d 2 [Nm.(rad/s) 1 ] G. Ducard c 30 / 34

Step 1: Inputs and Outputs Inputs: Torques T 1 and T 2 Outputs: Rotor speed at the compressor stage: ω 1 Step 2: Reservoirs and level variables reservoir 2: kinetic energy of the turbine E 2 (t) a. Level: ω 2 reservoir 1: kinetic energy of the compressor E 1 (t). Level: ω 1 reservoir 3: potential energy stored in the elasticity of the shaft U shaft (t). Level: ϕ What is the energy associated with each reservoir? E 2 (t) = E 1 (t) = U shaft (t) = a the energies are noted E1 and E 2 to avoid confusion with the torques T 1 and T 2. G. Ducard c 31 / 34

Simplified model of a gas turbine Step 3: Dynamics equation - Mechanical power balance de 2 (t) dt de 1 (t) dt du shaft (t) dt = = = G. Ducard c 32 / 34

Step 3: Dynamics equation - Mechanical power balance P mech,1 = compressor power = T 1 ω 1 P mech,2 = friction loss in bearing 1 = d 1 ω 1 ω 1 P mech,3 = power of the shaft elasticity at rotor 1 = kϕ ω 1 P mech,4 = power of the shaft elasticity at rotor 2 = kϕ ω 2 P mech,5 = friction loss in bearing 2 = d 2 ω 2 ω 2 P mech,6 = turbine power = T 2 ω 2 ( ) d 1 dt 2 Θ 1ω1(t) 2 ( ) d 1 dt d dt = P mech,1 (t) P mech,2 (t)+p mech,3 (t) 2 Θ 2ω2 2 (t) = P mech,4 (t) P mech,5 (t)+p mech,6 (t) ( ) 1 2 kϕ2 (t) = P mech,3 (t)+p mech,4 (t) G. Ducard c 33 / 34

Simplified model of a gas turbine Step 4: Dynamics equations of the level variables Θ 1 d dt ω 1(t) = T 1 (t) d 1 ω 1 (t)+k ϕ(t) Θ 2 d dt ω 2(t) = T 2 (t) d 2 ω 2 (t) k ϕ(t) d dt ϕ(t) = ω 2(t) ω 1 (t) G. Ducard c 34 / 34