How the power grid operation is modeled?𝒀𝒀•Step 4: Power flow calculationsto determine •how much power will be transmitted by each transmission line?•what the transmission line losses are?•Step 5: Calculate the losseson the transmission line and the power flowing through each line to verify the transmission lines are not overloaded.North Carolina State University Dr. Ning LuG2 (MW)100150…110G3 (MW)150200…140Total (MW)300420…310ECE 451 Lu Lec. 11
Power Flow Inputs and Outputs•Characteristics of power flow calculation•Model snap-shots of the steady state operation•Per unit value based•Admittance matrix based•Nodal based variables: basic variables 𝑃𝑃, 𝑄𝑄, 𝑉𝑉, 𝛿𝛿•Inputs •One swing/slack bus: 𝑉𝑉, 𝛿𝛿are known. 𝑃𝑃, 𝑄𝑄unknown•A few PV buses: 𝑃𝑃, 𝑉𝑉are known•Many PQ buses: 𝑃𝑃, 𝑄𝑄are known•Outputs:•Voltage magnitudeand angleat each node (bus)•Realand reactivepower on each line•Losseson each transmission line•Check violation of operational limits (line limits, bus voltage limits)North Carolina State University Dr. Ning Lu𝑦𝑦13𝑦𝑦23𝑉𝑉1= |𝑉𝑉1|∠δ1𝑉𝑉2= |𝑉𝑉2|∠δ2𝑉𝑉3= |𝑉𝑉3|∠δ3ECE 451 Lu Lec. 11
Step 1: Form the Admittance Matrix North Carolina State University Dr. Ning Lu𝐼𝐼1=𝑉𝑉1+𝑉𝑉1−𝑉𝑉2+ (𝑉𝑉1−𝑉𝑉3)𝑦𝑦1302+𝑦𝑦1202𝑦𝑦12𝑦𝑦13𝐼𝐼2=𝑉𝑉2+𝑉𝑉2−𝑉𝑉1+ (𝑉𝑉2−𝑉𝑉3)𝑦𝑦1202+𝑦𝑦2302𝑦𝑦12𝑦𝑦23𝐼𝐼3=𝑉𝑉3+𝑉𝑉3−𝑉𝑉1+ (𝑉𝑉3−𝑉𝑉2)𝑦𝑦1302+𝑦𝑦2302𝑦𝑦13𝑦𝑦23𝐼𝐼1=𝑉𝑉1𝑦𝑦1302+𝑦𝑦1202+𝑦𝑦12+𝑦𝑦13𝐼𝐼2=𝑉𝑉1+𝑉𝑉2+𝑉𝑉2+𝑉𝑉3+𝑉𝑉3−𝑦𝑦13−𝑦𝑦12𝑦𝑦1202+𝑦𝑦2302+𝑦𝑦12+𝑦𝑦23−𝑦𝑦12−𝑦𝑦23𝐼𝐼3=𝑉𝑉1−𝑦𝑦12+𝑉𝑉2+𝑉𝑉3−𝑦𝑦23𝑦𝑦1302+𝑦𝑦2302+𝑦𝑦13+𝑦𝑦23𝐼𝐼1𝐼𝐼2𝐼𝐼3=𝑉𝑉1𝑉𝑉2𝑉𝑉3𝑦𝑦1302+𝑦𝑦1202+𝑦𝑦12+𝑦𝑦13−𝑦𝑦12−𝑦𝑦13−𝑦𝑦12𝑦𝑦1202+𝑦𝑦2302+𝑦𝑦12+𝑦𝑦23−𝑦𝑦23−𝑦𝑦13−𝑦𝑦23𝑦𝑦1302+𝑦𝑦2302+𝑦𝑦13+𝑦𝑦23𝐼𝐼=𝑌𝑌𝑏𝑏𝑏𝑏𝑏𝑏𝑉𝑉⇒𝐼𝐼𝑖𝑖=�𝑗𝑗=1𝑛𝑛𝒀𝒀𝒊𝒊𝒊𝒊𝑉𝑉𝑗𝑗𝒀𝒀𝟏𝟏𝟏𝟏𝒀𝒀𝟏𝟏𝟏𝟏𝒀𝒀𝟏𝟏𝟏𝟏𝐼𝐼1=𝑉𝑉1𝒀𝒀𝟏𝟏𝟏𝟏+𝑉𝑉2𝒀𝒀𝟏𝟏𝟏𝟏+𝑉𝑉3𝒀𝒀𝟏𝟏𝟏𝟏𝐼𝐼2=𝑉𝑉1𝒀𝒀𝟏𝟏𝟏𝟏+𝑉𝑉2𝒀𝒀𝟏𝟏𝟏𝟏+𝑉𝑉3𝒀𝒀𝟏𝟏𝟏𝟏𝐼𝐼3=𝑉𝑉1𝒀𝒀𝟏𝟏𝟏𝟏+𝑉𝑉2𝒀𝒀𝟏𝟏𝟏𝟏+𝑉𝑉3𝒀𝒀𝟏𝟏𝟏𝟏ECE 451 Lu Lec. 11
Power Flow Problem Formulation and 3 Node TypesNorth Carolina State University Dr. Ning LuType 1: Swing Bus Type 2. PV busType 3. PQ Bus𝑃𝑃𝑖𝑖=�𝑗𝑗=1𝑛𝑛𝑌𝑌𝑖𝑖𝑗𝑗𝑉𝑉𝑖𝑖𝑉𝑉𝑗𝑗cos𝜃𝜃𝑖𝑖𝑗𝑗+𝛿𝛿𝑗𝑗− 𝛿𝛿𝑖𝑖knownknownknownknownknownknown𝐼𝐼=𝑌𝑌𝑏𝑏𝑏𝑏𝑏𝑏𝑉𝑉𝐼𝐼𝑖𝑖=�𝑗𝑗=1𝑛𝑛𝒀𝒀𝒊𝒊𝒊𝒊𝑉𝑉𝑗𝑗𝑆𝑆𝑖𝑖=𝑉𝑉𝑖𝑖𝐼𝐼𝑖𝑖∗𝑃𝑃𝑖𝑖− 𝑗𝑗𝑄𝑄𝑖𝑖=𝑉𝑉𝑖𝑖∗𝐼𝐼𝑖𝑖=𝑉𝑉𝑖𝑖∗�𝑗𝑗=1𝑛𝑛𝒀𝒀𝒊𝒊𝒊𝒊𝑉𝑉𝑗𝑗𝑄𝑄𝑖𝑖=−�𝑗𝑗=1𝑛𝑛𝑌𝑌𝑖𝑖𝑗𝑗𝑉𝑉𝑖𝑖𝑉𝑉𝑗𝑗sin𝜃𝜃𝑖𝑖𝑗𝑗+𝛿𝛿𝑗𝑗− 𝛿𝛿𝑖𝑖𝑆𝑆𝑖𝑖∗=𝑉𝑉𝑖𝑖∗𝐼𝐼𝑖𝑖ECE 451 Lu Lec. 11
Step 2: Determine Bus Types and Known Nodal ValuesNorth Carolina State University Dr. Ning LuHour 12Hour 12Hour 12Step 2: Nodal injections = Generation – LoadIn this system, one of the bus, the real and reactive power of it cannot be fixed and need to be able to be adjusted. This is because _____________________________________________?Load 1Load 2Load 3G1G2G3𝑃𝑃1𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙+𝑗𝑗𝑄𝑄1𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑃𝑃2𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙+𝑗𝑗𝑄𝑄2𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑃𝑃3𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙+𝑗𝑗𝑄𝑄3𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑃𝑃3𝑔𝑔𝑔𝑔𝑛𝑛+𝑗𝑗𝑄𝑄3𝑔𝑔𝑔𝑔𝑛𝑛𝑃𝑃2𝑔𝑔𝑔𝑔𝑛𝑛+𝑗𝑗𝑄𝑄2𝑔𝑔𝑔𝑔𝑛𝑛𝑃𝑃1𝑔𝑔𝑔𝑔𝑛𝑛+𝑗𝑗𝑄𝑄1𝑔𝑔𝑔𝑔𝑛𝑛400 + j 500.4+j0.05 pu150 + j600.15+j 0.06 p.u.1200 + j2101.2 + j 0.21 p.u.1050 + j2001.05 + j0.2 p.u.600 + j80500 + j1000.5+j 0.1 p.uBus 1 𝑉𝑉1∠𝛿𝛿1Bus 2 𝑉𝑉2∠𝛿𝛿2Bus 3 𝑉𝑉3∠𝛿𝛿3P2+jQ2 = 1.2-1.05 + j(0.21-0.2)=0.15-j0.01P3+jQ3 = 0.15-0.4 + j(0.06-0.05)=-0.25+j0.01Swing BusECE 451 Lu Lec. 11
North Carolina State University Dr. Ning Lu𝑉𝑉1𝑉𝑉2𝑉𝑉3𝑦𝑦1302+𝑦𝑦1202+𝑦𝑦12+𝑦𝑦13−𝑦𝑦12−𝑦𝑦13−𝑦𝑦12𝑦𝑦1202+𝑦𝑦2302+𝑦𝑦12+𝑦𝑦23−𝑦𝑦23−𝑦𝑦13−𝑦𝑦23𝑦𝑦1302+𝑦𝑦2302+𝑦𝑦13+𝑦𝑦23𝐼𝐼1𝐼𝐼3𝐼𝐼2Bus 1Bus 2Bus 3𝑦𝑦1302𝑦𝑦1202𝑦𝑦12𝑦𝑦13𝑦𝑦1202𝑦𝑦1302𝑦𝑦23𝑦𝑦2302𝑦𝑦2302Swing BusPQ BusPQ Bus1 swing bus2 PQ busP2+jQ2 = ?𝟏𝟏.𝟎𝟎𝟏𝟏∠0°P2+jQ2 = 0.15-j0.01𝑉𝑉2∠𝛿𝛿2P3+jQ3 = -0.25+j0.01𝑉𝑉3∠𝛿𝛿3ECE 451 Lu Lec. 11
North Carolina State University Dr. Ning Lu𝑉𝑉1
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- Fall '11
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- Impedance, Electric power transmission, Transmission line, North Carolina State University, Dr. Ning Lu
