Contents ELEC46 Power ystem Analysis Lecture ELECTRC POWER CRCUT BAC CONCEPT AND ANALY. Circuit analysis. Phasors. Power in single phase circuits 4. Three phase () circuits 5. Power in circuits 6. ingle phase equivalent circuits 7. The per-unit system Reference: Textbook (Glover et al), Chapter. CRCUT ANALY KL: KCL: voltage drops = around any loop currents = leaving any node Branch equations: branch relationship between & current v(t) i(t) branch voltage drop e.g. v Ri di v L dt dv i C dt i gv hv ystematic methods of writing KL, KCL and branch equations: node analysis mesh or loop analysis Node analysis: tep : choose a reference node (n=) assign unknown voltage v n to other nodes n,, N tep : for each node except n= and those connected to voltage sources, write KCL in terms of v n s tep : voltage sources (if any) tep 4: express answer as N N matrix equation and (if linear) solve using Gaussian elimination. ELEC46: Electric Power Circuits Basic Concepts and Analysis p. / ELEC46: Electric Power Circuits Basic Concepts and Analysis p. /
. PHAOR Power systems mostly operate in the sinusoidal steady state (). All voltages and currents are sinusoids. Oscillate at same frequency, differing only in amplitude and phase angle. A sinusoid xt Xcost value X and phase angle when referenced to cos t has an effective (rms). A sinusoid xt can be represented as a phasor X (rotating point in a complex plane) x t Re X j X Xe X X X X cos jx sin X sin e.g. vt cost 45 m X X Re X cos volts j45 its phasor: 7.745 7.7e 5 j5 Arbitrary choice of reference phasor arbitrary choice of clock time t= Phasors look like vectors but are not vectors. Properties of phasors: f: x t X cos tx X Xx XR jx yt Ycos ty YYy YR jy Then: dxt j X X 9 o x dt xt yt X Y X Y j X Y axt ax ax x XY. X. Y XY R R Also: x y x y MPEDANCE Linear time-invariant circuit elements: R, L, C inusoidal steady state (transients dissipated) vt cos t it cos t Y = impedance Y= admittance f R jx (R=resistance, X=reactance) Then and f Y Y G jb (G=conductance, B=susceptance) Then Y and Note: R Y jx R jx R X R X ELEC46: Electric Power Circuits Basic Concepts and Analysis p. / ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 4/
Resistor: v i nductor: R v Ri R R thus =R i.e. note: current and voltage are in phase. di i v L j L v L dt L j L i.e. 9 o note: current lags voltage by 9 o. Capacitor: v i C dv i C j C dt j i.e. C C 9 o note: current leads voltage by 9 o. mpedances in series: add ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 5/ thus Admittances in parallel: add thus Y Y Y. POWER N NGLE PHAE CRCUT. AERAGE (or REAL) POWER P Power is rate of change of energy with respect to time. vt it load cos vt t it cos t instantaneous power delivered to the load is: pt cos cos vtit t = dc component oscillating component define average power (real power) delivered to the load: P cos = rate of useful work (unit = watt = W) P = x = x component of resolved along component of resolved along ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 6/ cos Average power is also called real power or active power. Average power is important because it provides basis for
consumer billing by utility company. However, additional parameters are required to fully characterize the capacity requirements and efficiency of the power systems. Power factor is defined as: pf cos Lagging pf: Hence real power: P. pf. REACTE POWER inductive load lags i(t) peak occurs after v(t) peak Leading pf: capacitive load leads v(t) peak occurs after i(t) peak instantaneous power delivered to the load is: p t cos cos cos cos cos sin sin t v t i t t t cos sin R t X t cos nstantaneous power absorbed by resistive component sin p t R p X t nstantaneous power absorbed by reactive component cos sin pt P t Q t where: P cos = real power (W) Q sin = reactive power (Ar) Q = x component of orthogonal to = x component of orthogonal to sin ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 7/ ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 8/
reactive power sign convention: Lagging power factor: Q sin inductive loads absorb reactive power Leading power factor: Q sin capacitive loads generate reactive power Physical significance: The time average of the instantaneous power is called the real power P. t is the time-average rate of energy conversion from electrical to non-electrical form, and this is the power that costs you money on your electricity bill. Reactive power Q refers to maximum value of instantaneous power absorbed by the reactive component of the load. The instantaneous reactive power expresses the reversible flow of energy to and from the reactive component. t is alternately ve and ve and averages zero. For example, a bar heater is a rely resistive load that converts current directly into heat. The current is called the actual (real) current because it contributes directly to the production of actual power (heat). On the other hand, a motor represents a partially inductive load. The motor current consists of an actual current and a magnetizing current. The actual current produces real power which comprises: work performed by motor (lifting, moving), heat generated in ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 9/ motor winding resistance, heat generated in motor iron through eddy current and hysteresis losses, frictional loss in motor bearings, air friction loss in turning motor. The magnetizing (reactive) current is used to generate the magnetic field required for operation of motor. This current corresponds to an exchange of energy between power source and motor, but it is not converted into actual power.. COMPLEX POWER Complex power is defined as:. *.. j sin cos P jq P Q apparent power (unit = A) ang Q tan power angle P P cos power factor (lagging or leading) f load is impedance R jx ELEC46: Electric Power Circuits Basic Concepts and Analysis p. / *
power angle = ang Power flow direction and sign convention: * P R R jx Q X f load is admittance * P jq > (absorbed) inductive loads < (generated) capacitive loads Y G jb * * Y Y Y P G G jb Q B P jq jx R G > (absorbed) inductive loads < (generated) capacitive loads jb load source Load: current enters ve terminal. f P is ve then ve real power is absorbed. f Q is ve then ve reactive power is absorbed. f P (Q) is ve then ve real (reactive) power is delivered. ource: current leaves ve terminal. f P is ve then ve real power is delivered. f Q is ve then ve reactive power is delivered. f P (Q) is ve then ve real (reactive) power is absorbed. Relationship between P, Q and can be visualized in terms of a right triangle called power triangle: A P cos W AC power conservation principle: Q sin Ar Given an ac source driving a network of interconnecting impedances, the complex, real and reactive power of the source equals, respectively, the sums of the complex, real and reactive powers of the individual impedances. But be aware that the source apparent power is generally different from the sum of individual apparent powers. ELEC46: Electric Power Circuits Basic Concepts and Analysis p. / ELEC46: Electric Power Circuits Basic Concepts and Analysis p. /
4. Three phase () circuits Advantages: efficient use of copper: generator, tx & dx conductors power flow is constant (if balanced operation) motors exhibit efficient starting and constant torque. 4. CURRENT AND OLTAGE RELATONHP 4.. Y-connected loads and sources ELEC46: Electric Power Circuits Basic Concepts and Analysis p. / ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 4/
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5. POWER N THREE PHAE () CRCUT balanced source a b c i a i b i c v ab v bc balanced load Choose an as reference and let phase angle of single phase of load (e.g. angle of impedance of branch of Y-equivalent circuit) cos i t cost van t p t v t cos t o bn cn a p ib t Lcost cos o p ict Lcost v t t Total instantaneous power absorbed into load instantaneous power in each phase p t v t i t v t i t v t i t an a bn b cn c cos cos cos cos p L t t cos cost cos p L because: L o o t t t cos cos cos.77 (-.54j.6) (-.54-j.6) = thus (using p ) L L L p t cos P = total real power no oscillating component in total instantaneous power in balanced systems. however, even in balanced systems, oscillating components of instantaneous power exist in each phase. define complex power as sum of complex powers in each phase: for a balanced system thus: * * * an a bn b cn c o o * * * bn b an a an a o o * * * cn c an a an a j * an a p p cos p p sin cos j sin L L L L ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 9/ ELEC46: Electric Power Circuits Basic Concepts and Analysis p. /
Q LLsin = total reactive power tandard convention used when specifying the ratings for power system equipment: value of apparent power L L line (-to-line) value of voltage L power factor cos 6. NGLE PHAE EQUALENT CRCUT Define the a operator to be the complex number: Define: Thus: a e j.5 j.866 a 4.5 j.866 a 6 a 6.5 j.866 a* a a.5 j.866 a 6 a a.5 j.866 a a aa j aa For a balanced system with phase sequence a-b-c: bn cn ab a an bc a a a a a an imilar expressions for phase and line currents. Conclusion: Balanced system can solve equivalent -phase problem ca an an an a -a -a a a -a Also: a 6 a ELEC46: Electric Power Circuits Basic Concepts and Analysis p. / ELEC46: Electric Power Circuits Basic Concepts and Analysis p. /
7. THE PER UNT () YTEM (Textbook.) Express quantities (voltage, current impedance, power, ) as a fraction of () reference values Advantages include: reduce incidence of in calculations automatic checking of answers: ball park feel simplified transformer calculations. 7. NGLE PHAE CRCUT Choose two (scalar, i.e. not phasor) quantities; usually: voltage ( or k) (apparent) power P Q (A or ka or MA) Then other s are: current impedance (A or ka) R X () admittance Y G B () quantities are: x x x e.g. if MA, P MW then P =. examples of quantities: Y Y Y P P thus: i.e. Q Q. P P cos cos i.e. 7. THREE PHAE CRCUT reference quantities: P cos. ELEC46: Electric Power Circuits Basic Concepts and Analysis p. / ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 4/
= apparent power (A) L, = line-to-line voltage line current is:, L L, ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 5/ other quantities are then: (note: phase quantity ; L line quantity) Y-connected -connected, L,,, L,, L, L,,, L,, L,,, L,, L, L,, L, L,, L,, L, L, L,, thus, for either or Y connection: L,, and L,, L, L, L, complex power: L ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 6/ * L L, L,. * L, L, * L, L, Note: single phase equivalent, does not need 7. TRANFORMER must use: one A throughout circuit one voltage between transformers consistent voltage s across transformers i.e. related by nominal turns ratio circuit in actual quantities: ideal N: circuit
N and N N and s:,, * * N and,, example: ideal N: impedance s: N,,, N, i.e. quantities: N.,, N N,,,,, i.e.,,. in physical quantities, referred to : in quantities: N N N,,,, N similarly:,, ;,, ;,,,, in calculations: can eliminate ideal transformers note: N = turns ratio, rated, rated,,,,, ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 7/ ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 8/
7.4 CHANGNG BAE old : ( old ) ; ( old ) new : ( new) ; ( new) change of for current quantities: ( new) ( old ) ( old ) ( new) ( old ) ( new) ( old ) ( new) ( old ) change of for impedance quantities: ( old ) ( old ) ( new) ( new) ( new) ( old ) ( old ) ( old ) ( new) 7.5 EXAMPLE Example : A Y-connected load consists of equal impedances of each. Load voltage is 4.4k line-to-line. mpedance in each of the lines connecting the load to the substation bus is.475. Find line-to-line voltage at substation bus by working in per unit on values of: L, 4.4k and L, 7A L, 44 For Y-connection:,, L, 7 A, 44/ Base impedance:,, 7 oltage across each load is 44/. Choose voltage across the load on a phase as reference, i.e. 44/ Hence: Load impedance: Load current: Line impedance: an 44/ an an,, 44/, an, an, an,.475,.7 75 Thus, line-to-neutral voltage at the substation is: ELEC46: Electric Power Circuits Basic Concepts and Analysis p. 9/ ELEC46: Electric Power Circuits Basic Concepts and Analysis p. /
...7 75.5.7 and line-to-line voltage at the substation is:.5.54.4k 4.6k Example : L, The schematic diagram of a radial transmission system is shown below. Calculate terminal voltage of the synchronous machine using a of MA for all circuits. k k 5MA X=% Line j Base impedance of the line: 74 6 k k 5MA X=% k 5MW.8pf lagging..4 ( new) ( old ) j j ( old ) 5 Load current (using formula P LLcos ): 6 5 A.8 Base current for the k line: 6 75 A Hence, per-unit load current is:.687 75 Per-unit voltage of load busbar:.9 The equivalent circuit is shown below: j. j.575 j.4 E R =.9.687.8pf lagging Load Per-unit reactance of the line: j j.575 74 Per-unit reactance of sending-end transformer: ( new) ( old ) j. j. ( old ) 5 Per-unit reactance of receiving-end transformer: Hence,.687.8 j.6 j. j.575 j.4.9 j..8 j.558.44 or.44k 5.84k ELEC46: Electric Power Circuits Basic Concepts and Analysis p. / ELEC46: Electric Power Circuits Basic Concepts and Analysis p. /
Example : A balanced Y-connected voltage source with Eag 77 volts is applied to a balanced-y load in parallel with a balanced- load, where Y j and 45 j5 ohms. The Y load is solidly grounded. Using per-phase values of, ka and, 77volts, calculate the source current a in per-unit and in amperes. 77 7.67 6.A 77 j Y,.9 j. 4. 8.4 7.67 Convert load into equivalent Y-load: 45 j5,.955 j.86.6 9.5 7.67. Y,, 4.8.4.69.5.9 j..955 j.86.669.74.669.74 6..99.74 ELEC46: Electric Power Circuits Basic Concepts and Analysis p. /