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Power system modelling

Power system modelling. Janusz Bialek. What are these lectures about?. Power flow equations – you can find them in any textbook (e.g. mine: But: What’s the physical meaning of the variables (e.g. angles)? What is real and reactive power?

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Power system modelling

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  1. Power system modelling Janusz Bialek

  2. What are these lectures about? • Power flow equations – you can find them in any textbook (e.g. mine: • But: • What’s the physical meaning of the variables (e.g. angles)? • What is real and reactive power? • How can you simplify the equations without departing from reality? • Similarly for the dynamic power system model • When can you use them? • When do you have to use higher-order models?

  3. Syllabus • Phasor analysis of AC circuits • Power flow equations: AC and DC models • Physical meaning of real and reactive power. • Power system dynamics • Dynamic generator models • Steady-state stability problem • Transient stability problem

  4. Power system modelling • A network is a a planar graph with nodes (buses, vertices) and branches (lines, edges) • GB high-voltage transmission network consists of 810 nodes and 1194 branches • European network consist of several thousands nodes • Power stations are modelled by differential equations used to analyse power system dynamics • First static AC circuit analysis, then dynamic analysis

  5. How to manipulate AC voltages and currents? Circuit analysis consists of applying Kirchhoff’s Current and Voltage Laws: you need to add sinusoidal voltages and currents. Adding sine functions is cumbersome Steinmetz: represent a sine function by a rotating vector (phasor) - simple harmonic movement. Its projection on a vertical axis gives a sine function. Voltages and currents in a circuit have the same frequency so their phasors rotate with the same speed and are therefore stationary wrt each other. Apply Kirchhoff’s Laws to the phasors rather than sine functions. Charles Steinmetz (1865 –1923)

  6. Rotating Vector & Sine Wave AC voltage or current can be represented by a phasor that rotates anti-clockwise with a given frequency, amplitude (equal to rms value) and a phaseshift. In power circuits all the voltage and currents analysed are 50 Hz so all you need to specify is the amplitude and the phase shift. How to manipulate them?

  7. Phasors in AC circuit analysis As the phasorsrotate synchronously at the same speed (50 Hz), they are stationary with respect to each other Hence adding ACvoltages (or currents) can be represented by adding vectors (phasors) The angles between phasors = phase shifts between sine functions Phasor diagram is a snapshot of rotating phasors taken at any time instant That means that a reference is arbitrary and all the angles (i.e. phase shifts between sine functions) are relative. j V I Im V j I Re

  8. Phasors and complex numbers A complex number is a vector on the complex plane – so complexnumbers can be used to manipulate phasors Resistor: voltage and current are in phase V = RI Inductor: current lags the voltage by 900 V = j XL I ( XL is inductive reactance) Capacitor: current leads the voltage by 900 V = - jXC I ( XC is capacitive reactance) Im  Re

  9. Phase-shifting transformers (quad-boosters) • Phase voltages in 3 phase circuits are in quadrature with phase-to-phase voltages • Adding a voltage in quadrature changes the voltage angle and therefore power flows (power flow is proportional to sin(δ) - see next lecture) • Many PSTs installed in Europe to prevent loop flows around Germany (Poland, Czech Republic), Belgium, Switzerland etc. • Must be coordinated

  10. Loop flows in a meshed network • In meshed networks power does not flow directly from A to B but through all transmission lines • Only 38% of F-I trade flows directly • Loop flows via the neighbours from wind in Northern Germany • Overload the neighbours networks if dispatch is not coordinated (it isn’t) • Phase-shifting transformers (PST) – direct (partial) control of power flows

  11. Real and reactive power as complex numbers Power is a product of voltage and current. Real and reactive power as complex numbers: S = V I*= P + jQ (complex conjugate current as  is the angle between voltage and current: multiplying complex voltage by complex conjugate current means we are taking the difference between the angles) Real power P = V I cos (real component of S) Reactive power Q = V I sin (imaginary component of S)

  12. Real and reactive power • Real power P represents energy used to do work (dissipated) • What’s the physical meaning of reactive power?

  13. What is reactive power? • The tandem analogy • The horse-boat analogy • The inclined-plane analogy • The trampoline analogy • The drying line analogy • The foam on the beer analogy • The snake analogy, etc. • Here: physical explanations, not silly analogies

  14. Inductive reactive power • Motor is an electromagnetic device and needs coils to produce magnetic field • Because current is AC (alternating), energy to supply the magnetic field (Em = L I2 /2 ) oscillates between the source and the coil. • When current is increasing, the energy flows from the source to the coil. When current is decreasing the coil returns energy to the source iL + + Power supplying magnetic energy to the coil: p(t) = iL(t) v(t) and the current lags the voltage by 900 - -

  15. The frequency of energy exchanges is 100 Hz as magnetisation is the same whether the current is negative or positive (Em = Li2/2) • That oscillating energy and power is termed reactive (imaginary, idle) as it cannot do any work (the average is zero). • However it is necessary for setting up of the magnetic fieldand therefore for the motor (inductor) to operate. • There is no reactive power in DC circuits as the current is constant. Once magnetic field is set up in a coil, it does not change

  16. Capacitive reactive power • Capacitors need energy to supply the electric field (Ee = C V2 /2) • When voltage is increasing, energy flows from the source to the capacitor. • When voltage is decreasing the capacitor returns energy to the source • Energy oscillations in a capacitor are in anti-phase with those for inductors • Inductive current is 900 behind voltage while capacitive current is 900 ahead of voltage • Capacitance is a “source” of reactive power, inductance is a “sink” • Capacitors and inductors compensate each other: when one needs energy, the other discards it. In other words one always supplies the other and no external source is required

  17. Real power is the average power unit: Watts [W] where φ is the phase shift between the voltage and current Reactive power: unit: VAr Apparent power: unit: VA Power factor Power factor (pf) is simply a measure of how reactive a circuit is. When pf = 0, the circuit is purely reactive and carries no real power. When pf=1, the circuit is purely resistive and carries no reactive power.

  18. Nasty effects of reactive power • Causes real power losses (because of oscillating power transfers) • Takes up capacity of wires: less is available for real power transmission • Causes voltage drops on transmission lines (see later): • You cannot transfer reactive power over long distances • Main “sources” of reactive power: synchronous generators, additional compensation by capacitors (voltage support) • Deficit of reactive power may lead to a voltage collapse

  19.  -equivalent model of a transmission line receiving end Sending end R series resistance causing heating I2R losses series reactance (inductive) due to magnetic field: “consumes” Q shunt susceptance (capacitive) due to electric field: “generates” Q (NB: reactance XC = 1/C = 1/BC) For high-voltage transmission lines X>>R and BC is small (large distance between conductors) For distribution lines R may be comparable to X and Bc may be neglected

  20. Valid for both transmission and distribution lines Valid for transmission lines (approximate, used for so-called DC model – see later) For underground cables, C is large (i.e. capacitive reactance XCis small) due to small distance between the phases of a cable and proximity of ground. This creates a high current IC flowing through BC (carrying reactive power) which loads up the cable reducing the cable’s capacity for real power transmission. Max length of ac cable is a few tens km. Longer cables must be DC

  21. Simplified power flow models • Power engineers tend to think in terms of power rather than current: AC current has no direction while real power is directional • Full AC power flow equations are complicated – how to use simplified models useful for cause-effect analysis (and mathematical models) • Generally: • Real power flow P is driven by the power angles δ difference • Reactive power flow is driven by the voltage magnitude V difference

  22. Simplified equivalent circuit of HV transmission line (R <<X): Vs I φ IX δ φ Vr IX is perpendicular and leading wrtI I Assume lagging power factor The angle δ, between V and Ef , is called the power (or load) angle and drives P Do not confuse with the power factor angle φ (between V and I).

  23. Real power P • Voltages in transmission network have to be kept ±10% of nominal values so we can assume that they are roughly constant • Hence the power flow P is roughly proportional to the sine of the angle difference δ between the voltages at the sending end and the receiving end • The maximum value of the angle is 900 – steady-state stability limit • It is a reasonably realistic model in the whole operating range – about 10% accuracy – recommended • It cannot be used in distribution networks as we neglect resistance

  24. DC linear network model • Power angles have to kept small (usually below 450 ) for stability reasons • For small angles the model becomes linear • In per-units, the voltages are close to 1: • Hence the (confusing) name DC model: • In actual DC circuit: I = V/R • In linear DC model of AC circuit: P = δ/X • Remember about simplifying assumptions (resistance and capacitance neglected) – only transmission network! • Only valid for small values of the angles – not valid for heavily loaded lines

  25. Approximate reactive power model (both X and R modelled) – valid also for distribution networks Vs Calculate voltage drop (Vs – Vr): difference in magnitude (the length of phasors) rather than the complex phasor difference IX Vr δ φ IR I

  26. (Vs – Vr) has two components: ΔVp and ΔVq φ e but hence Usually  is small (see earlier) so the vertical component ΔVq can be neglected when calculating the arithmetic difference between the voltages (difference between the length of the phasors):

  27. Due to flow of P Due to flow of Q The flow of both P and Q causes voltage drops but the sensitivity due to Q is larger because X >> R for a typical high-voltage transmission line Compensating for Q eliminates voltage drops and therefore improves voltage profile in the system (makes it flatter). Voltages have to be kept within limits ( 10% in UK for ). Long lines mean high reactance X , and that means high voltage drops due to Q transfers. Reactive power cannot be transferred over long distances as that would cause unacceptable voltage drops Distributed generation may reverse the power flow and cause voltage rise

  28. Ferranti effect If the power factor of the load is leading (due to e.g. overcompensation) then reactive power could flow in the opposite direction (negative Q): Q P Then│ΔV│<0 and therefore Vs < Vr pf lead Recall that underground cables have high capacitance and therefore produce Q hence increasing the receiving end voltage – Ferranti effect. Why 11 kV is a standard distribution voltage?

  29. How to calculate power flows in the whole network (rather than a single line)? • Assume we know real and reactive power injections, i.e. real and reactive power generation or demand in each node • Generation: the output of each generator is determined by the market , wind power is determined by a forecast • Demand: demand can be very accurately predicted based on history • Calculate voltage (magnitude and angle) at every node • Knowing nodal voltages, we can calculate branch currents from the Ohm’s Law: I = V/Z • Knowing voltages and currents we can calculate real and reactive power flows: S = P + jQ = V I* • Solution: nodal voltage analysis

  30. Nodal equations for a system with N nodes current injections or admittance matrix nodal voltages voltage at node i, mutual admittance between nodes i and j, equal to the negative of the branch series admittance ), self-admittanceof node iequal to the sum of all the admittances terminating on node i, including the shunt admittance Yi0 Pi-equivalent model of a transmission line

  31. Admittance matrix actual admittance elements of admittance matrix If shunt admittances are neglected, the diagonal elements become the sum of admittances (weights) and the off-diagonal elements are the negative admittances (weights). In graph theory terms, the admittance matrix is the Laplacian matrix of the weighted graph describing the network. It is symmetric, positive semidefinite and singular A is the (mxn) bus incidence matrix showing the incidence of lines to the buses in the system, m is the number of branches and y is the vector of m line admittances.

  32. Singularity of the admittance matrix The admittance matrix is singular. To deal with it, assume that the voltage at the reference node is known (usually nominal, i.e. 1 in per-units) and its angle is zero. Then remove the corresponding rows and columns from Y.

  33. Nodal equations in terms of real and reactive power As the inputs to the power flow problem, power injections (generations and demands) rather than current injections are used. For any node i, the current injection at the node can be extracted from I = Y V as Hence Separating the real and the imaginary part and changing admittances to polar coordinates gives :

  34. Linearising: Jacobian matrices Solution: Newton-Raphson method

  35. To obtain simpler and more symmetrical Jacobian submatrices, M and K are multiplied by voltage magnitudes

  36. DC power flow • Full power flow problem considers both real and reactive power and aims to calculate voltage magnitude and angle at every node • Simplifying assumptions • all the voltages are nominal (1 in per-units) • line resistance is much lower than reactance, (valid only in HV transmission networks but not in distribution networks) • The problem reduces by half as we are solving only real power-angle equations • Linear DC power flow assuming angle differences are small

  37. Voltage/reactive power flow In meshed transmission networks solving for voltages requires solving full AC equations. Distribution networks are radial (trees) so you can calculate voltages recursively starting from the ends (leafs) and going up the tree DistFlow – recursive calculations of both real and reactive flows taking advantage of the tree-structure of distribution network (Baran, Wu, 1989) Power demands Power flow from k to (k+1) k – consecutive node number

  38. Optimal Power Flow (OPF) • Minimise the cost of dispatch subject to nodal power balance equations (i.e. power flow problem) and network constraints

  39. Security-Constrained Optimal Power Flow (SCOPF) • The secure operation must be maintained not only for a given operating state, but also immediately following "credible" contingencies (tripping of lines, generators, transformers). Almost universally adopted reliability criterion is (N-1), i.e. a system must withstand on its own a loss of any single element. That means that • there must be enough generations reserve that supplies can be quickly resumed if any of the generators trips • all the line flows must be below their limits not only for a given "normal" operating state, but also when any of the lines is disconnected and the flows reconfigure themselves.

  40. Example (2 systems connected by two parallel lines) Assume both lines are loaded up to maximum thermal capacity of 4 GW. What would happen if one of the lines tripped due to e.g. a short-circuit? What is the maximum secure loading of the transmission lines? SCOPF problem is similar to OPF but more complicated: not only system must be secure now but also following a set of given "credible" contingencies; usually (N-1) criterion SOs determine the actual dispatch always by solving (N-1) SCOPF, rather than just OPF. (N-1) is a static and conservative criterion. How to sail closer to the wind?

  41. Dynamic Model of the Power System Everything you need to know! You can find equations in this and other (e.g. Kundur) books – here I will concentrate on physical meaning and understanding

  42. Dynamic Model of the Power System Concentrate on traditional synchronous generators, rather than wind-powered Double-Fed Induction Generators or PV panels In the steady-state, synchronous generators are kept in synchronism by electromagnetic forces Rotors rotate at the same constant speed, i.e. they are stationary with respect to each other Following a disturbance (short-circuit, line tripping etc) the balance of torques is disturbed and the speed of machines is not constant or uniform Rotor motion is described by the swing equation: strings are elastic so approximately we have a system of connected masses on springs

  43. Rotor angle  • Imagine a radial line drawn on a rotor. Rotor angle  is the angle with respect to a synchronously rotating reference axis and is constant if the rotor speed is constant • The rotor angles are not unique (as the reference is arbitrary) but the relative rotor angles (i.e. 1-2 ) are unique • Following a disturbance the generators accelerate and decelerate as the balance of torques in not constant Speed deviation

  44. Generator swing equation (2nd Newton Law) Balance of torques (power): if turbine’s driving mechanical power is not equal to the electrical power developed by the generator, the rotor will accelerate Rotor angle Electrical power developed by the generator Mass moment of inertia Turbine mechanical power = approx. constant Damping coefficient Turbine’s mechanical power is assumed constant within the typical oscillations (swing) timeframe: a few seconds Pemay change almost instantaneously (electromagnetic phenomena) - hence creating imbalance

  45. Synchronous generator models • The key to dynamic power system modelling is modelling of the synchronous generator • The steady-state model is different than the dynamic model (i.e. shortly after a disturbance) • Each model will result in a different equation Peand a different value of D – butthe swing equation stays the same

  46. Construction of synchronous generator • Three-phase stator winding carries load current: armature • Three-phase stator winding produces a rotating magnetic field • Rotor: DC field (excitation) winding supplied by an external DC source (electromagnet) • DC rotor current creates a magnetic fieldthat rotates with the rotor and induces electromotive force (voltage) on the stator windings • There are also damping winding on the rotor • The rotor can be round or with salient poles

  47. Synchronous generator in the steady-state • The induced electromotive force,Ef , is the driving voltage and it is proportional to the rotor field current If • The internal resistance may be neglected. The internal reactance is referred to as the synchronous reactanceXs – its value depends on the path of the magnetic field • The electrical angle (phase shift) δbetween V and Ef , is also the mechanical rotor angle between the rotor and external rotating magnetic field • Hence we can link electrical power system model with the mechanical swing equation Generator Load Ef = f(If)

  48. Dynamic generator models • Following a short-circuit, the generator cannot any longer be modelled as a constant reactance as the path of the magnetic field changes with time due to the screening effect of the field and damper windings. Subtransientstate (a few cycles) Screening effect of field and damper windings High path reluctance via air-gap = Low reactance = High current

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