Power Transfer Across a Power System Component as a Single Impedance

Transcription

1 1 Power Transfer Across a Power System Component as a Single Impedance Bus 1 S 1 =P 1 +jq 1 S =P +jq Z=jX + I + Bus V V V j 1 V1 e δ j V e δ 1 = V =

2 Power Transfer Across a Single Impedance Power system components, such as a high voltage (HV) lines and transformers, can often be represented by a single impedance or reactance. High Voltage (HV) lines and transformer impedances are predominantly reactive (Z = R+jX, where X dominates) (Loads are, however, usually resistive!) Typically, the reactance of HV lines, X 0.4 ohm/km/phase The resistance of HV lines, is typically an order of magnitude less

3 Power System Components as a Single Impedance 3 Consider a single impedance, Z connecting buses, both with a specified voltage. This impedance can represent a power line or a transformer. We define power flow, S 1 and S on each side. Evaluate S 1! I = V V 1 Z * * * V1 V 1 = = 1 = 1 * S P jq V I V Z Bus 1 S V 1 1 V S 1 =P 1 +jq I j 1 V1 e δ = Z=R+jX 1 1 V V V e R jx + - Bus V j V e δ 1 = V = The phase difference is: S =P +jq δ = δ1 δ j δ1 δ

4 4 Power transfer across a single reactance Bus 1 S 1 =P 1 +jq 1 Z=jX S =P +jq Bus + I + V 1 V - - V = 1 V = 1 1 j V e δ j V e δ

5 5 Power Transfer Across an Impedance S 1 = 1 1 V V V e R jx j δ1 δ δ = δ δ = + S R jx [ cosδ sinδ ] V V V + j R + X RV1 R V1 V cosδ + X V1 V 1 = P Q XV1 X V1 V cosδ + RV1 V 1 = R R + + X X sinδ sinδ

6 Power Transfer Across 6 an Impedance () Power systems are reactive: R << X R 0 P = P = 1 V V 1 X sin δ This is a very important, basic formula for real power flow between neighboring buses in an electrical power system. Similarly: Q 1 = 1 1 V V V X cosδ

7 Real Power Flow 7 Maximum real power flow occurs at phase angle difference of 90 between buses In normal operation phase angles in power systems are small (< ) Then: sinδ δ With voltage almost constant (within +/- 5% limits) and X constant, real power flow will depend only on phase angle (P ~ δ ) P increases with the square of the voltage!

8 8 Maximum Real Power Flow From: P(real power) V1 V P( ) sin δ δ = X For small angles: sinδ δ P(δ) P max δ(phase angle) The maximum power transfer corresponds to a phase angle difference of 90 between buses

9 Reactive Power Flow Power Through Engineering - Egill Benedikt Hreinsson 9 a Reactance Assumption: R = 0 Bus 1 S 1 =P 1 +jq 1 Z=jX S =P +jq Bus + I + P = P 1 V V V j 1 V1 e δ j V e δ 1 = V = Q 1 = 1 1 V V V X cosδ Q = V V cosδ V 1 X

10 Phasor Diagram for Power Power Transfer Engineering - Egill Benedikt Hreinsson Across 10 a Reactance Don t confuse δ with φ! Bus 1 S 1 =P 1 +jq 1 Z=jX + I S =P +jq + Bus δ V V 1 jxi P V 1 V V j 1 V1 e δ - - j V e δ 1 = V = φ Q V1 = V + jxi I Black lines: Phasor diagram Blue lines: Graphical representation of P and Q

11 Graphical Representation of Real Power Transfer in a Phasor Diagram Assume that V is constant fixed...but δ and V 1 is variable 11 V 1 δ V jxi φ I The length of this line is proportional to V1 sinδ P( δ ) = V1 V sin δ X and hence the real power transfer, P

12 Graphical Representation of Reactive Power Transfer in a Phasor Diagram Again V is constant fixed...but δ and V 1 is variable 1 V 1 δ V jxi I φ The length of this line is proportional to Q = V V cosδ V 1 X V cosδ V 1 and hence the reactive power transfer

13 Real Power Depends Heavily on Phase Angle 13 Assume that V is constant! Now increase the phase angle δ! δ f I V V 1 Δδ jxi ΔP, or change in real power, depends heavily on Δδ, or change in phase angle, but not so much on voltage, V

14 Reactive Power Depends Heavily on Voltage Magnitude 14 Assume that V is fixed. Now increase voltage magnitude, V 1! ΔV 1 δ φ V V 1 jxi I ΔQ, or change in reactive power, depends heavily on ΔV 1 ) or change in voltage magnitude (but not so much on change in phase angle, Δδ )

15 15 Strong and Weak Coupling P Strong coupling δ weak coupling Q Strong coupling V

16 Strong and Weak Coupling of Quantities in Power Systems 16 Phase angle and real power are connected by a strong coupling. Voltage magnitude and reactive power are connected by a strong coupling Phase angle and reactive power are connected by a weak coupling Voltage magnitude and real power are connected by a weak coupling

17 A Mechanical Analogy of Power Flow Across an Elastic Rotational Coupling 17 Elastic coupling To generator To load P R Rsinδ 1 Maximum power transfer at 90 angle

18 18 Multimachine stability The power system consists of a network of elastic couplings. We can imagine a set of weights hanging from a fixed ceiling and tied together by rubber bands. Originally all the weights are at rest until a disturbance occurs. We can cut any of the elastic couplings (to simulate a line outage) or poke the weights to simulate a drop in load/generation. The system will go into a damped oscillation or a cascaded breakdown. A damped oscillation will occur if all the ties hold without breaking, while a cascaded breakdown may occur if one link breaks

19 19 Summary of Power Transfer Electrical power system Formula Diagram AC system P = V X V 1 sinδ Bus 1 S 1 =P 1 +jq 1 + I Z=R+jX S =P +jq + Bus V 1 V - - V R V DC system P = 1 ( V ) V 1 = V = 1 1 j V e δ 1 P1 I Z=R(+jX) j V e δ V 1 V

20 0 Voltage Magnitude Power systems must supply electric power within a narrow voltage range, typically with 5% of a nominal value. For example, wall outlet should supply 30 volts, with an acceptable range from 18 to 4 volts. Voltage regulation is a vital part of system operations.

21 Voltage Regulation 1 A number of different types of devices participate in system voltage regulation Generators: reactive power output is changed to keep terminal voltage constant. Capacitors: switched either manually or automatically to keep voltage within a range. Load-tap-changing (LTC) transformers: vary tap ratio to keep voltage within a range. Static var compensators (SVCs): electronic devices instantaneously change reactive power output to keep voltage within range.

22 Voltage Control Voltage control is necessary to keep system voltages within an acceptable range. Because reactive power does not travel well, it would be difficult for it to be supplied by a third party.

23 Series capacitors with electronic control 3 Electronically controlled series capacitors can be used in a power system to increase stability against Voltage stability Uload/Ugen Voltage collapse Ugen Uload Stability limits P The computer and electronic control of power system is called FACTS (=Flexible AC transmission systems) Angular stability U/α1 U/α P α1 α

24 The Formula for Calculating Total 3 Phase Power in a Component 4 P = 3 I V cosφ 3 f f L V L is the voltage between phases (line to line voltage) I f is the current in each phase

25 5 Voltage stability We consider the flow at bus in our bus system We can eliminate δ by using Euler s equation and rearrange Q Bus 1 V V 1 S 1 =P 1 +jq j 1 V1 e δ 1= cos δ + sin I Z=jX S =P +jq + - Bus 1 = V = V1 V cosδ V = X X δ V j V e δ Q Q P = = V V cosδ V 1 V 1 V V V + = X X V 1 X cosδ X sinδ Q tanφ = P and we get... V V1 V P tanφ P + = X X

26 6 Voltage stability () We rearrange the previous eq. and get: ( φ ) 4 1 φ V + tan P X V V + (1+ tan ) P X = 0 This is a nd order equation with a solution: Introduce the following simplifying notation and we get: 1 4 V 1 V 1 = tanφ ± + 1 ( tanφ ) V P X P X P X V 4 4 E E V = QX ± P X XE Q 4 E= V 1 V = V Q = Q P = P

27 7 Voltage stability and collapse (3) The figure shows dimensionless variables on the 3 axes These are p = PX/E and q = QX/E and finally v = V/E Normal operation is in the upper part of the surface Operation on the lower part is unacceptable The points at the equator correspond to maximum power (where the inner square root vanishes in the last equation) The previous equation is then: 1 1 v = q± p q 4 The projections of the equator curve on the horizontal surface corresponds to a parabola between P and Q as part of the inner square root of the last equation

28 Voltage Collapse and the Nose Curves (4) 8 With increased load we eventually get Voltage collapse The Nose Curves correspond to the solid lines on the surface of the previous slide Each curve corresponds to a given load factor, cosφ (or tanφ) A negative tanf corresponds to a capacitive circuit and we get increased voltage with increasing load, to start with

29 References 9 T. V. Cutsem: Voltage Instability: Phenomena, Countermeasures and Analysis Methods, Proceedings of the IEEE, Vol. 88, No., Feb. 000 B.M. Weedy, B.J. Cory: Electric Power Systems, 4th ed. John Wiley, 1998 J.J. Grainger, W.D. Stevenson: Power System Analysis, McGraw-Hill, 1994 J.D. Glover, M.S. Sarma: Power System Analysis and Design, 3rd ed., Brooks/Cole Thomson Learning, 00

30 30 Example (4a) - Power factor correction Example (4a) 3 fasa hreyfill notar 0 kva og er hann tengdur við 380 volta 3 fasa spennu. Aflstuðullinn cosφ 1 = 0.7 er spankenndur. Finnið stærð Y-tengds þéttis sem hliðtengja þarf við hreyfilinn til að aflstuðullinn aukist í cosφ = 0.9. Finnið einnig strauminn í hverjum fasa sem hreyfill+þéttir dregur fyrir og eftir tengingu þéttisins. [A 3-phase motor uses 0 kva and is connected to a 380 volt 3-phase supply. The lagging power factor is cosφ 1 = 0.7. Determine the size of a capacitor to be Y-connected in parallel with the motor to increase the power factor to a new value, cosφ = 0.9. Also determine the phase current drawn by the motor+capacitor before and after connecting the capacitor]

31 31 Example (4a) - solution

32 3 Example (4b) Combined load Example (4b) Álag A og álag B eru samhverf Y-tengd og samsíðatengd. Álag A tekur til sín 15 kw raunafl með spankenndum aflstuðli 0,6. Álag B dregur 8 kw raunafl með rýmdarkenndum aflstuðli 0,8. Samhverf 3-fasa spennulind er tengd við samsett álag A og B með spennu milli fasa 480 V A. Finnið heildar raunafl, launafl og sýndarafl sem samsetta álagið dregur B. Finnið aflstuðul samsetta álagsins og segið til hvort hann er rýmdarkenndur ( leading eða spankenndur ( lagging ) C. Finnið stærð straumsins í línunni frá spennulindinni. [Load A and load B are balanced Y-connected loads and are connected in parallel. Load A draws the active power 15 kw at 0.6 power factor lagging (inductive). Load B draws the active power 8 kw at 0.8 power factor leading (capacitive). The loads are supplied by a balanced, three phase source with line to line voltage 480 V. A. Find the total active, total reactive and total apparent power absorbed by the combined load B. Determine the power factor of the combined load and state whether inductive (lagging) or capacitive (leading) C. Determine the magnitude of the line current from the source]

33 33 Example (4b) - solution

34 34 Example (4c) power in a series reactance The voltage at the sending end of a 3 phase transmission line is kept constant at j0 V = V 0 = V e [kv]. The line has a pure reactive impedance of Z L = jx L [ohm/phase]. The voltage at the receiving end is V = V δ where it is connected to a load impedance Z D = Z D ϕ [ohm/phase]. a) Determine the real power in the receiving end of the line, P, as a function of V1, X, Z and ϕ b) Determine what conditions ZD has to meet so the maximum value for the real power output at the receiving end of the line P,max is reached, and determine if or how this condition depends on V, 1 X and/or ϕ. L c) When V = kv, cos ϕ =0.8 and P = 00 MW. Determine in this case the load impedance ZD = ZD ϕ and the current I in each phase. L D

35 Example (4c) solution, p1 35

36 Example (4c) solution, p 36

37 Example (4c) solution, p3 37 MVA MVar