Topic 1: Basics of Power Systems

1 Topic 1: Basics of Power Mohsenian Rad (U of T)1 Networking and Distributed SystemsECE 5332: Communications and Control for SmartSpring 2012 Power SystemsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid2 The Four Main Elements in Power Systems : Power Production / Generation Power Transmission Power Distribution Power Consumption / Load Of course, we also need monitoring and control SystemsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid3 Power Production: different Types: Traditional Renewable Capacity, Cost, Carbon Emission Step up TransformersPower SystemsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid4 Power Transmission: High Voltage (HV) Transmission Lines Several Hundred Miles Switching Stations Transformers Circuit BreakersPower SystemsDr.

2 Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid5 The Power Transmission Grid in the United SystemsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid6 Major Inter connections in the United SystemsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid7 Power Distribution: Medium Voltage (MV) Transmission Lines (< 50 kV) Power Deliver to Load Locations Interface with Consumers / Metering Distribution Sub stations Step Down Transformers Distribution TransformersPower SystemsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid8 Power Consumption: Industrial Commercial Residential Demand Response Controllable Load Non ControllablePower SystemsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid9 GenerationTransmissionDistributionLoadPo wer SystemsDr.

3 Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid10 Power system Control: Data Collection: Sensors, PMUs, etc. Decision Making: Controllers Actuators: Circuit Breakers, Grid Graph RepresentationDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid11 Nodes: BusesLinks: Transmission LinesGeneratorLoadPower Grid Graph RepresentationDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid12 Nodes: BusesLinks: Transmission LinesGeneratorLoadBuses(Voltage) Power Grid Graph RepresentationDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid13 Nodes: BusesLinks: Transmission LinesGeneratorLoadTransmission Lines ( Power Flow, Loss) Power Grid Graph RepresentationDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid14 Nodes: BusesLinks: Transmission LinesGeneratorLoadConsumersPower Grid Graph RepresentationDr.

4 Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid15 Nodes: BusesLinks: Transmission LinesGeneratorLoad10 MW3 MW7 MWTransmission Line AdmittanceDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid16 Admittance y is defined as the inverse of impedance z: z = r + j x (r: Resistance, x: Reactance) y = g + j b(g: Conductance, b: Susceptance) y = 1 / z Parameter g is usually positive Parameter b: Positive: Capacitor Negative: InductorTransmission Line AdmittanceDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid17 For the transmission line connecting bus ito bus k: Addmitance: yik Example:yik= 1 j 4 (per unit) Note that, yiiis denoted by yiand indicates: Susceptance for any shunt element (capacitor) to ground at bus i. Y-Bus MatrixDr.

5 Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid18 We define: Ybus= [ Yij] where Diagonal Elements: Off diagonal Elements: Note that Ybasmatrix depends on the Power grid topology and the admittance of all transmission lines. N is the number of busses in the grid. NikkikiiiyyY,1ijijyY Y-Bus MatrixDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid19 Example: For a grid with 4 buses, we have: After separating the real and imaginary parts: 4342414434241343432313323124232423212211 413121413121yyyyyyyyyyyyyyyyyyyyyyyyyyyy YbusBjGYbus Bus VoltageDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid20 Let Videnote the voltage at bus i: Note that, Viis a phasor, with magnitudeand angle. In most operating scenarios we have:iiiVV jijiVV Power Flow EquationsDr.

6 Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid21 Let Sidenote the Power injectionat bus i:Si= Pi+ j Qi Generation Bus: Pi> 0 Load Bus: Pi< 0(negative Power injection)Active Power Reactive PowerPower Flow EquationsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid22 Using Kirchhoff laws, AC Power Flow Equationsbecome: Do we know all notations here? If we know enough variables, we can obtain the rest of variables by solving a system of nonlinear equations. NjjkkjjkkjjkkNjjkkjjkkjjkkBGVVQBGVVP11)c os()sin()sin()cos( Power Flow EquationsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid23 The AC Power Flow Equations are complicated to solve. Next, we try to simplify the equations in three steps. Step 1: For most networks, G << B.

7 Thus, we set G = 0: NjjkkjjkkNjjkkjjkkBVVQBVVP11)cos()sin( Power Flow EquationsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid24 Step 2: For most neighboring buses: . As a result, we have: 15 to10 ji 1)()(jkjkjkCosSin NjkjjkkNjjkkjjkkBVVQBVVP11)( Power Flow EquationsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid25 Step 3: In per unit, |Vi| is very close to ( to ). As a result, we have: . Pkhas a linear model and Qkis almost jiVV kNkjjkjkkNjkjkNjjkkjkbBBBQBP ,111)( Power Flow EquationsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid26 Step 3: In per unit, |Vi| is very close to ( to ). As a result, we have: . Pkhas a linear model and Qkis almost jiVV kNkjjkjkkNjkjkNjjkkjkbBBBQBP ,111)( DC Power Flow EquationsPower Flow EquationsDr.

8 Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid27 Given the Power injection valuesat all buses, we can useto obtain the voltage anglesat all buses. Let Pijdenote the Power flowfrom bus ito bus j, we have: NjjkkjkBP1)( )(jiijijBP Power Flow EquationsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid28 Example: Obtain Power flow values in the following grid:puPg21 puPg22 puPg14 puPl43 puPl12 1014jy 1013jy 1034jy 1023jy 1012jy Power Flow EquationsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid29 First, we obtain the Y bus matrix: 4443424134333231242322211413121143424144 3424134343231332312423242321221141312141 3121 BBBBBBBBBBBBBBBBjBjjbbbbbbbbbbbbbbbbbbbb bbbbbbbbjYbusPower Flow EquationsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid30 Next, we write the (active) Power flow equations: This can be written as: 4434241343242141443433432312321313424323 2242321121241431321211413121 BBBBBBPBBBBBBPBBBBBBPBBBBBBP 4321434241434241343432313231242324232121 1413121413124321 BBBBBBBBBBBBBBBBBBBBBBBBPPPPP ower Flow EquationsDr.

9 Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid31 From the last two slides, we finally obtain: Therefore, the voltage angles are obtained as: 4321 14321 Power Flow EquationsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid32 However, the last matrix in the previous slide is singular! Therefore, we cannot take the inverse. The system of equations would have infinite solutions. The problem is that the four angles are not independent. What matters is the angular/phase difference. We choose one bus ( , bus 1) as reference bus: .01 Power Flow EquationsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid33 We should also remove the corresponding rows/columns: The angular differences (with respect to ): 43212010010103010100102010101010301412 4322010010301001020141 1 Power Flow EquationsDr.

10 Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid34 Finally, the Power flow values are calculated as:puPg21 puPg22 puPg14 puPl43 puPl12 ) (10)( ) (10)( ) (10)( ) (10)( ) (10)(433434322323411414311313211212 Flow EquationsDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid35 What if the generator connected to bus 1 is renewable? What if the capacityof transmission link (1,3) is 1 pu? What if we can apply demand responseto load bus 3? What if one of the transmission lines fails? Economic Dispatch ProblemDr. Hamed Mohsenian-RadTexas Tech UniversityCommunications and Control in Smart Grid36 In the example we discussed earlier, we had: In particular, we had: However, generation levels and assumed given. Q: What if the generators have different generation costs?