EEN320 - Power Systems I ( ) Part 6: Fundamentals of power - PowerPoint PPT Presentation

EEN320 - Power Systems I ( Συστήματα Ισχύος Ι ) Part 6: Fundamentals of power system operation Dr Petros Aristidou Department of Electrical Engineering, Computer Engineering & Informatics Last updated: April 10, 2020

Today’s learning objectives After this part of the lecture and additional reading, you should be able to . . . . . . describe and analyse the behaviour of a transmission line under 1 different operating conditions; . . . explain the Ferranti effect. 2 , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 2/ 50

Fundamentals of power system operation - Overview In this part of the lecture, we investigate the stationary current and voltage relations as well as the resulting active and reactive power flows on an AC power line For this purpose, we use the wave equation discussed in the previous part of the lecture (Part 5) Thereby, we focus on a series of practically relevant scenarios The analysis is performed under two assumptions: 1) The operating conditions are balanced → analysis is performed via single-phase equivalent circuits 2) The network is in steady-state (for assessment of dynamic phenomena other models are required) Furthermore, we consider all powers per phase . The corresponding three-phase power can be calculated using the conventions introduced in Part 2. , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 3/ 50

1 Decoupled quantities - Power flow on a power line P 1 + jQ 1 P 2 + jQ 2 − → − → V 1 ϕ 1 V 2 ϕ 2 Power line Several ways to mathematically describe power flow over a power line Usually, we use complex voltage together with active and reactive powers at each end of line This yields 8 real quantities V 1 , ϕ 1 , P 1 , Q 1 , V 2 , ϕ 2 , P 2 , Q 2 Which of the above quantities are decoupled (i.e. independent) of each other and which are not? , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 6/ 50

1 Decoupled quantities - Examples P 1 + jQ 1 P 2 + jQ 2 − → − → V 1 ϕ 1 Power line V 2 ϕ 2 Not all quantities in above graphic are independent of each other Examples: V 1 and V 2 are coupled via line characteristics (see previous lectures) → Therefore it is customary to take one angle, e.g. ϕ 2 , as reference; hence, one ”loses” one quantity in the formulas Power flows are also coupled; if P 1 and Q 1 are fixed, then P 2 and Q 2 can be computed if V 1 or V 2 is fixed, too If V 1 and V 2 are fixed, P 1 , P 2 , Q 1 and Q 2 are also fixed and can not be adjusted independently , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 7/ 50

1 Decoupled quantities - Common triples P 1 + jQ 1 P 2 + jQ 2 − → − → V 1 ϕ 1 Power line V 2 ϕ 2 V 1 , ϕ 1 , V 2 : powers result from line characteristics and given quantities; practical example: power line connects two bulk ”stiff” power networks V 1 , P 2 , Q 2 (or P 1 , Q 1 , V 2 ): By fixing voltage on one end of line and power on other end, remaining quantities follow; practical example: consumer with fixed power demand connected via power line to network V 1 , P 1 , Q 1 : By fixing quantities at sending end of line, voltage and powers at receiving end follow; practical example: power plant that feeds network over power line , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 8/ 50

2 Surge impedance loading - Meaning I 1 I 2 V 1 V 2 Z 2 x = 0 x = ℓ Surge impedance loading (SIL) = power delivered when line is loaded with its surge impedance, i.e. � R ′ + j ω L ′ Z 2 = Z w = G ′ + j ω C ′ SIL also called natural loading In the following, we consider two cases Lossless line ( R ′ = G ′ = 0) Lossy line ( R ′ � = 0 , G ′ � = 0) , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 10/ 50

2.1 SIL of lossless power line - Receiving end I 1 I 2 V 1 V 2 Z 2 x = 0 x = ℓ � Lossless power line: R ′ = G ′ = 0 → surge impedance Z w = L ′ C ′ Active power delivered at end of line P 2 = | V 2 | 2 = | V 2 | 2 Z 2 Z w Reactive power delivered at end of line ( Z 2 = Z w is real in lossless case) Q 2 = 0 Current at end of line I 2 = V 2 = V 2 Z 2 Z w , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 11/ 50

2.1 SIL of lossless power line - Sending end (1) I 1 I 2 V 1 V 2 Z 2 x = 0 x = ℓ √ L ′ C ′ = j β (see From solution of wave equation with x = 0 and γ = j ω Part 5, Sect. 4.2) V 1 = cosh( j βℓ ) V 2 + Z W sinh( j βℓ ) I 2 I 1 = V 2 sinh( j βℓ ) + cosh( j βℓ ) I 2 Z W With cosh( j β ) = cos( β ) and sinh( j β ) = j sin( β ) we obtain V 1 = cos( βℓ ) V 2 + jZ W sin( βℓ ) I 2 I 1 = j V 2 sin( βℓ ) + cos( βℓ ) I 2 Z W , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 12/ 50

2.1 SIL of lossless power line - Sending end (2) I 1 I 2 V 1 V 2 Z 2 x = 0 x = ℓ Using I 2 = V 2 Z w yields V 1 = cos( βℓ ) V 2 + jZ W sin( βℓ ) V 2 Z w = V 2 (cos( βℓ ) + j sin( βℓ ) = V 2 e j βℓ I 1 = j V 2 sin( βℓ ) + cos( βℓ ) V 2 Z W Z w = I 2 (cos( βℓ ) + j sin( βℓ ) = I 2 e j βℓ → Voltage and current are shifted by angle βℓ at end of line Thereby, their amplitudes remain unchanged , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 13/ 50

2.1 SIL of lossless power line - Active power I 1 I 2 V 1 V 2 Z 2 x = 0 x = ℓ For active power at both end of lines, we have that (as line is lossless) 2 = P 2 = | V 1 | 2 P 1 = V 1 I ∗ 1 = V 2 I ∗ Z w This particular loading of line is called surge impedance loading (SIL) P SIL = | V | 2 Z w For this loading we achieve optimal transmission conditions (amplitudes of voltage and current remain constant along whole line) In practice, loading usually differs from SIL , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 14/ 50

2.1 SIL of lossless power line - Reactive power I 1 I 2 V 1 V 2 Z 2 x = 0 x = ℓ For SIL, reactive power flow on line is zero → At each point on line, reactive power ”absorption” of line inductance equals reactive power ”production” of line capacitance V 2 I 2 = L ′ V 2 ω C ′ = I 2 ω L ′ Q ′ C = Q ′ C ′ = Z 2 ⇒ ⇒ L w , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 15/ 50

2.1 SIL of lossless power line - Comments on reactive power Surge impedance of overhead lines (OHLs) between 200 − 400 Ω OHL inductance significantly larger than OHL capacitance → Reactive power ”absorbed” by OHL inductance exceeds reactive power ”produced” by OHL capacitance even for small currents → OHLs often operated above their SIL; then they ”absorb” reactive power Compared to OHLs, cables have very low surge impedance ( ≈ 30 − 50 Ω ) → SIL usually above thermal limit of cable → Cables usually ”produce” reactive power , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 16/ 50

2.2 SIL of lossy power line - Sending end (1) I 1 I 2 V 1 V 2 Z 2 x = 0 x = ℓ Lossy line → Z w is complex As before, we consider the case Z 2 = Z w Current at receiving end of line I 2 = V 2 = V 2 Z 2 Z w Apparent power at receiving end of line 2 = | V 2 | 2 S 2 = P 2 + jQ 2 = V 2 I ∗ Z ∗ w , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 17/ 50

2.2 SIL of lossy power line - Sending end (2) I 1 I 2 V 1 V 2 Z 2 x = 0 x = ℓ From solution of wave equation with x = 0 and γ = α + j β (see Part 5) V 1 = cosh( γℓ ) V 2 + Z W sinh( γℓ ) I 2 = cosh( γℓ ) V 2 + Z W sinh( γℓ ) V 2 Z w � � = V 2 e γℓ = V 2 cosh( γℓ ) + sinh( γℓ ) I 1 = V 2 sinh( γℓ ) + cosh( γℓ ) I 2 Z W � � = I 2 e γℓ = I 2 cosh( γℓ ) + sinh( γℓ ) Note: To obtain the last equality, we have used cosh( x ) + sinh( x ) = e x , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 18/ 50

2.2 SIL of lossy power line - Sending end (3) I 1 I 2 V 1 V 2 Z 2 x = 0 x = ℓ Apparent power at sending end V ∗ e 2 αℓ = S 2 e 2 αℓ S 1 = P 1 + jQ 1 = V 1 I ∗ 2 1 = V 2 Z ∗ w → As in lossless case, phase angle between voltage and current remains constant along line; phase shift is proportional to β x → But now, active and reactive power decrease with line length; same applies to voltage and current , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 19/ 50

2.2 Typical values for SI and SIL of lossy power line Typical values for OHLs Rated voltage in kV 132 275 400 Z w [ Ω ] 373 302 296 P SIL [MW] 47 250 540 Source: B. M. Weedy et al., ”Electric Power Systems”, John Wiley & Sons, 2012 Typical values for cables Rated voltage in kV 115 230 500 Z w [ Ω ] 36.2 37.1 50.4 P SIL [MW] 365 1426 4960 Source: P . Kundur, ”Power System Stability”, McGraw-Hill, 1994 , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 20/ 50

3 The two extrema - Overview Next, we analyse the behaviour of a power line in two special cases No load Short circuit To simplify our calculations, we restrict ourselves to the lossless case , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 22/ 50

3.1 No load conditions - Setup I 1 I 2 V 1 V 2 x = 0 x = ℓ No load condition can occur if Voltage is applied to unloaded line Load at end of line is disconnected Main characteristic: I 2 = 0 , ΕΕΝ 320 — Dr Petros Aristidou — Last updated: April 10, 2020 23/ 50