Exercise1 1.10.015 Informatik 7 Rechnernetze und Kommunikationssysteme
Review of Phasors Goal of phasor analysis is to simplify the analysis of constant frequency ac systems v(t) = max cos(wt + q v ) i(t) = I max cos(wt + q I ) Root Mean Square (RMS) voltage of sinusoid 1 T T 0 v() t max dt
Phasor Representation jq Euler's Identity: e cosq jsinq Phasor notation is developed by rewriting using Euler's identity v( t) cos( wt q ) j( wtq ) v( t) Re e (Note: is the RMS voltage) 3
Phasor Representation, cont d The RMS, cosine-referenced voltage phasor is: jq e q jwt v( t) Re e e cosq j I I cosq j I I jq sinq sinq (Note: Some texts use boldface type for complex numbers, or bars on the top ) I 4
Advantages of Phasor Analysis Device Time Analysis Phasor Resistor v( t) Ri( t) RI di() t Inductor v( t) L jw LI dt 1 1 Capacitor i( t) dt v(0) I C jwc 0 Z = Impedance R = Resistance X = Reactance t R jx Z X Z = R X =arctan( ) R (Note: Z is a complex number but not a phasor) 5
Example1 A 50-Hz, single-phase source with volts is applied to a circuit element. A)Determine the instantaneous source voltage. Also determine the phasor and instantaneous currents entering the positive terminal if the circuit element is B) a 0-ohm resistor, C) a 10-mH inductor, D) a capacitor with 5 ohm reactance. 3030 6
Complex Power Power p( t) v( t) i( t) v(t) = cos( wtq ) max i(t) = I cos( wt q ) max cos cos 1 [cos( ) cos( )] p( t) 1 max Imax[cos( q qi) cos( wt q q )] I I 7
Complex Power, cont d Average Power 1 p( t) max Imax[cos( q qi ) cos( wt q qi )] P avg 1 T T 0 1 I p() t dt I max max cos( q q ) cos( q q ) I I Power Factor Angle = = q q I 8
Complex Power S I cos( q q ) jsin( q q ) I * P jq I I (Note: S is a complex number but not a phasor) P = Real Power (W, kw, MW) Q = Reactive Power (var, kvar, Mvar) S = Complex power (A, ka, MA) Power Factor (pf) = cos If current leads voltage then pf is leading If current lags voltage then pf is lagging 9
Complex Power, cont d Relationships between real, reactive and complex power P S cos Q S sin S 1 pf Example: A load draws 100 kw with a leading pf of 0.85. What are (power factor angle), Q and S? S Q 1 -cos 0.85 31.8 100kW 117.6 ka 0. 85 117.6sin( 31.8 ) 6.0 kar 10
Power Consumption in Devices Resistors only consume real power P Resistor Resistor Inductors only consume reactive power Q Inductor IInductor XL Capacitors only generate reactive power Q Q Capacitor Capacitor C Capacitor I R I X X Capacitor X C C 1 wc (Note-some define X C negative) 11
Example A certain single phase load draws 5 MW at 0.7 power factor lagging. Determine the reactive power required from a parallel capacitor to bring the power factor of the parallel combination up to 0.9. 1
Example 3: A 8 MW/4 Mvar load is supplied at 13.8 k through a feeder with an impedance of (1+ j). The load is compensated with a capacitor whose output, Qcap, can be varied in 0.5 Mvar steps between 0 and 10.0 Mvars. What value of Qcap minimizes the real power line losses? What value of Qcap minimizes the MA power into the feeder? 13
Balanced 3 -- Zero Neutral Current I I I I I n a b c n Z (1 0 1 1 * * * * an a bn b cn c 3 an a S I I I I Note: means voltage at point x with respect to point y. xy 14
Three Phase - Wye Connection There are two ways to connect 3 systems: Wye (Y), and Delta (). Wye Connection oltages an bn cn 15
Wye Connection Line oltages cn ab ca - bn an bn (α = 0 in this case) bc (1 1 10 ab an bn bc 3 30 3 90 Line to line voltages are also balanced. ca 3 150 16
Wye Connection, cont d We call the voltage across each element of a wye connected device the phase voltage. We call the current through each element of a wye connected device the phase current. Call the voltage across lines the line-to-line or just the line voltage. Call the current through lines the line current. 3 130 3 e I Line Phase Phase Line S 3 I 3 I Phase Phase * Phase j 6 17
Delta Connection I ca I b I bc I c I ab I a For Delta connection, voltages across elements equals line voltages For currents I I I a ab ca 3 3 I ab I I I b bc ab I I I c ca bc S 3 I Phase * Phase 18
Delta-Wye Transformation To simplify analysis of balanced 3 systems: 1) Δ-connected loads can be replaced by 1 Y-connected loads with ZY Z 3 ) Δ-connected sources can be replaced by Y-connected sources with Line phase 330 19
Example 4 A three-phase line, which has an impedance of ( + j4) per phase, feeds a balanced Y-connected threephase load that has an impedance of -4j. The line is energized at the sending end from a 50-Hz, threephase, balanced voltage source of 30 3 (rms, lineto-line). Determine: The current, real power, and reactive power delivered by the sending-end source. The line-to-line voltage at the load. 0
Example5 A industrial company has an average power consumption of 500 kw with average power factor of 0.7 and 4000 working hours. Assume that the company must not pay for reactive power if they maintain the power factor above 0.9. Calculate Annual electricity consumption (active power) The electricity cost for active and reactive power if 1 kwh costs 9 cents and 1 karh costs 1.5 cents Would it be profitable to install a 300 kar capacitor bank that costs 8000 euros? 1
Ideal Power Market Ideal power market is analogous to a lake. Generators supply energy to lake and loads remove energy. Ideal power market has no transmission constraints Single marginal cost associated with enforcing constraint that supply = demand buy from the least cost unit that is not at a limit this price is the marginal cost. This solution is identical to the economic dispatch problem solution.
Two Bus Example Total Hourly Cost : 8459 $/hr Area Lambda : 13.0 Bus A Bus B 300.0 MW 199.6 MW 400.4 MW AGC ON AGC ON 300.0 MW 3
Mathematical Formulation of Costs (C) and Incremental Cost (IC) Generator cost curves are usually not smooth. However the curves can usually be adequately approximated using piece-wise smooth, functions. Two representations predominate quadratic or cubic functions piecewise linear functions We'll assume a quadratic presentation i Gi i Gi Gi C ( P ) P P $/hr (fuel-cost) dci ( PGi ) ICi ( PGi ) PGi $/MWh =λ dp Gi In order to minimize the total operating cost IC 1 (PG 1 )=IC (PG )= IC N (PG N ) 4
Mathematical Formulation of Costs For the two bus example C(PGA) C(PGB) 399.8 11.69PGA 616.9 11.83PGB 0.00334PGA 0.00149PGB Euro Euro IC(PGA) IC(PGB) 11.69 0.00668PGA Euro/MWh 11.83 0.0098PGB Euro/MWh 5
Mathematical Formulation of Costs For the two generator system C(PGA) 399.8 11.69PGA 0.00334PGA Euro C(PGB) 616.9 11.83PGB 0.00149PGB IC(PGA) 11.69 0.00668PGA Euro/MWh IC(PGB) 11.83 0.0098PGB Euro/MWh PGA+PGB=600 λ =IC(PGA)=13.0 λ = IC(PGA)=13.0 PGA=00 C(PGA)=870.6 PGB=400 C(PGB)=5587.3 Total cost 8457.9 Euro 6
Example 6 The fuel-cost curves for two generators are given as follows: C1(P1)= 600 + 15 P1 + 0.05 P1 C(P) =700 +0P + 0.04 P Assuming the system is lossless, calculate the optimal dispatch values of P1 and P for a total load of 1000 MW, the incremental operating cost, and the total operating cost. 7