Toolbox: Electrical Systems Dynamics

Toolbox: Electrical Systems Dynamics Dr. John C. Wright MIT - PSFC 05 OCT 2010

Introduction Outline Outline AC and DC power transmission Basic electric circuits Electricity and the grid Image removed due to copyright restrictions. Please see "How Electricity Gets from the Power Station to your Home." PowerWise Teacher's Center, 2007. 2 SE T-6 Electrical Systems

The war of the currents AC Pros and Cons One kills elephants One has simpler infrastructure Why do we have AC and not DC? Look at a simple transmission circuit to decide. Use Voltage=120 VDC and Power=1.2 GW Tesla Edison 3 SE T-6 Electrical Systems

Efficient transmission requires AC AC Goals of the analysis Find the generator voltage Find the power delivered by the generator Find the power dissipated by the transmission line Find the ratio P Trans /P Load 4 SE T-6 Electrical Systems

Simple circuit models A simple electric circuit R T + V T + I + V G V L R L The current I = V L R L The power to the load L P L = I 2 R L = V R L Equating currents (from Kirchhoff s Laws), the transmission line power L P T = I 2 R T = R T V 2 R L The power ratio is then the ratio of resistances: P T = R T P L R L Generator power P G = P L + P T = Generator voltage V G = P I G =1+ 2 V 1+ R 2 R T L R L L R R T L V L 5 SE T-6 Electrical Systems

Simple circuit models Efficiency requires most power is dissipated in the load Example for Al and household V If R T /R L 1, then P T P L Most of the voltage appears across the load. So, we have very small transmission losses. For P L =1.2 GW and V L = 120V Then R L = P L /V 2 =1.2 10 5 Ω L For transmission assume L = 50 km (a short distance) An Aluminum cable with A = 5 cm 2 (to minimize sag, Cu not used.) Resistivity of Al, η = 2.8 10 8 Ω-m R T = ηl/a = 2.8Ω R L Conclusion: not so good! 6 SE T-6 Electrical Systems

Simple circuit models AC can be used to increase the voltage With AC we can use transformers Step up the voltage at the generator Transmit power at high voltage, low current Step down the voltage at the load Transmitting at low current should reduce transmission losses 7 SE T-6 Electrical Systems

Simple circuit models An Ideal Transformer I 1 I 2 + V 1 + V 2 n 1 n 2 N =n 2 /n 1 = turns ratio V 2 =NV 1 I 2 =I 1 /N Physical process is conservation of magnetic flux/energy 8 SE T-6 Electrical Systems

Simple circuit models An Ideal Transformer Common examples of transformers: I 1 I 2 + V 1 + V 2 n 1 n 2 N =n 2 /n 1 = turns ratio V 2 =NV 1 I 2 =I 1 /N Physical process is conservation of magnetic flux/energy Bottom right: Photo by mdverde on Flickr. Bottom left: Image by Tau Zero on Flickr. Top right: Photo by brewbooks on Flickr. 8 SE T-6 Electrical Systems

Analysis AC transmission reduces losses + R T I 1 I0 + + V T I L V G V 1 V 2 R L N 1 N 2 As before, P L =1.2 GW, V L = 120 V, R L =1.2 10 5 Ω What are transformer and transmission requirements, Such that P T P L? 9 SE T-6 Electrical Systems

Analysis From the circuit: 2 P L =V L I L = R L I L 2 P T =V T I 1 = R T I 1 From the transformer relation, I L = N 2 I 1, it follows P T P L 1 R T = N2 2 R L 10 SE T-6 Electrical Systems

Analysis From the circuit: 2 P L =V L I L = R L I L 2 P T =V T I 1 = R T I 1 From the transformer relation, I L = N 2 I 1, it follows P T P L = 1 R T N 2 2 R L For N 2 1 a huge reduction in transmission losses Practical numbers: V 0 = 12kV, V 1 = 240kV (rms) This implies that N 1 = V 1 /V 0 = 20 Assume small voltage drop across the transmission line. Then V 2 V 1 Second turn ratio becomes N 2 = V 2 /V L = 2000 Our transmission loss formula gives P T /P L 6% 10 SE T-6 Electrical Systems

Reactive power The downside to AC: Reactive power A down side to AC: Reactive power Why? Load is not pure resistive Load usually has an inductive component Resistance absorbs power Inductor circulates power back and forth This oscillating power is the reactive power 11 SE T-6 Electrical Systems

Reactive power Resistors, inductors and capacitors, oh my! There are three basic circuit elements having different Ohm s laws. Element Resistor Inductor Capacitor Symbol R L C Ohm s Law V = RI V = L di dt dv dt = I /C I = sin ωt V = R sin ωt V = L cos ωt V = 1 C cos ωt Phase shift 0 π/2 π/2 Impedance Z [Ω] = V /I R jωl j ωc 12 SE T-6 Electrical Systems

Reactive power Phase lags increase reactive power Current, I = I m sin( Voltage, V = V m sin( 2πt T period ) 2πt T period ) 1 0.5 I /I m, V /Vm 0 0.5 1 0 1 2 3 4 t/t period 13 SE T-6 Electrical Systems

Reactive power Phase lags increase reactive power I /I m, V /Vm 1 0.5 0 0.5 Current, I = I m sin( Voltage, V = V m sin( Power, 2πt T period ) 2πt T period ) P =I V = I m V m sin 2 ( = 1 2 I mv m 1 cos(2 2πt T period ) 2πt T period ) «1 0 1 2 3 4 t/t period 13 SE T-6 Electrical Systems

Reactive power Phase lags increase reactive power I /I m, V /Vm 1 φ 0.5 0 0.5 Current, I = I m sin( Voltage, V = V m sin( Power, P = I V = I m V m sin( = 1 2 I mv m 2πt T period + φ) 2πt T period ) 2πt T period ) sin( cos(φ) cos(2 2πt + φ) T period «2πt T period + φ) cos(φ) is known as the power factor 1 0 1 2 3 4 t/t period 13 SE T-6 Electrical Systems

Reactive power Reactive power must be supplied. For parts of the AC cycle the instantaneous power is greater than the average power Generator must be able to deliver this higher power even though it is returned later Bottom line: generator must have a higher volt-amp rating than average power delivered: VARs and Watts. Higher rating bigger size higher cost 14 SE T-6 Electrical Systems

Reactive power Phase shifts are introduced by inductance A motor will have an inductance, L. L V cos(ωt) I R 15 SE T-6 Electrical Systems

Reactive power Phase shifts are introduced by inductance A motor will have an inductance, L. It will introduce a phase shift given by tan φ = ω L R L V cos(ωt) I R 15 SE T-6 Electrical Systems

Reactive power Phase shifts are introduced by inductance A motor will have an inductance, L. It will introduce a phase shift given by tan φ = ω R Amplitude of current will also be reduced. V I = (ω 2 L 2 + R 2 ) 1/2 L L V cos(ωt) I R 15 SE T-6 Electrical Systems

Reactive power Phase shifts are introduced by inductance A motor will have an inductance, L. It will introduce a phase shift given by tan φ = ω R Amplitude of current will also be reduced. V I = (ω 2 L 2 + R 2 ) 1/2 This all follows from adding up the voltages for a simple circuit: di L dt + RI = V cos(ωt) L L V cos(ωt) I R 15 SE T-6 Electrical Systems

Reactive power How to minimize the VA requirement? To minimize the VA requirements on the generator we want φ Assume the average power absorbed by the load is P L Calculate the peak generator power [P G (t)] max Note: The peak is 2 (rms volt-amp rating) Less generator power is cheaper. 0 as a function of P L 16 SE T-6 Electrical Systems

Reactive power Peak power from inductance The power dissipated in the load: P L = RI 2 RV 2 = ω 2 L 2 + R 2 The peak power delivered by the generator V 2 R P peak = VI (1 + cos φ) = 1+ (ω 2 L 2 + R 2 ) 1/2 (ω 2 L 2 + R 2 ) 1/2 Using the expression for P L, we get: P peak = ω 2 L 2 + R 2 1/2 + R 2 P L 2R 1 17 SE T-6 Electrical Systems

Reactive power Capacitance can balance out reactive power Recall from our table that the phase lags are opposite for inductance and capacitance tan φ L = ω L 1 R, tan φ C = ω C Short answer: there is a capacitance that will keep the current and voltage in phase (but not eliminate the power factor) L C = R 2 + ω 2 L 2 Long answer follows. 18 SE T-6 Electrical Systems

I V relation for a capacitor Iˆ 1 ()= t C Analysis dvg dt I V relation for the load di V R ˆ ˆ 2 G = I 2 + L dt Conservation of current It ˆ()= Iˆ() t + I ˆ () t 1 2 31

Solution Assume VG = V co s( t) ( all voltages now rms) Current in the capacitor branch Iˆ 1 ()= t CV sin( t) Current in the load branch (from before) V Iˆ 2 ()= t 1/2 2 2 ( R + L 2 ) cos( t ) 32

The total current The total current flowing from the generator It ˆ()= Iˆ() t + I ˆ () t 1 2 cos( t ) =V C sin( t) 2 2 2 1/2 ( R L ) + cos( t)cos sin = V 1/2 + sin( t) 2 2 2 C 2 2 2 ( R + L ) ( R + L ) 1/ 2 33

The value of C Choose C for zero reactive power Set sin(t) coefficient to zero sin C = 1/2 R L 2 2 ( + 2 ) Simplify by eliminating the power factor C = R L 2 2 ( + L 2 ) 34

Calculate the peak power Calculate the peak power to learn what has happened to the VA rating [ P ] [ ] P = () = peak G 2 VI max t max = 2V 2 cos ( R 2 2 + L 2 ) 1/2 RV 2 = 2 ( R 2 2 + L 2 ) = 2 P L 35

The Result It worked!! The VA requirement has been reduced Ppeak VA = = 2 P L 36

Reactive power Discussion AC is good for transmission Have to manage reactive power Other aspects: HVDC transmission lines AC losses from corona discharge Voltage and frequency tolerances Stability of the grid to perturbations, eg a power plant going offline or a transmission line going down. 19 SE T-6 Electrical Systems

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