MIT22_081JF10_lec08a

MIT22_081JF10_lec08a - Toolbox: Electrical Systems Dynamics...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Toolbox: Electrical Systems Dynamics Dr. John C. Wright MIT - PSFC 05 OCT 2010 Introduction Outline Outline Image removed due to copyright restrictions. Please see "How Electricity Gets from the Power Station to your Home." PowerWise Teacher's Center, 2007. AC and DC power transmission Basic electric circuits Electricity and the grid 2 SE T-6 Electrical Systems AC "The war of the currents" 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 AC Efficient transmission requires AC Goals of the analysis Find Find Find Find the the the the generator voltage power delivered by the generator power dissipated by the transmission line ratio PTrans /PLoad 4 SE T-6 Electrical Systems AC Simple circuit models A simple electric circuit The current I = VL RL The power to the load 2 2 R = VL PL = I L RL RT I + VG - + VT - + VL - RL Equating currents (from Kirchhoff's Laws), the transmission line power 2 2 R = RT VL PT = I T RL The power ratio is then the ratio of resistances: PT = RT PL RL Generator power PG = PL + PT = 1 + 2 RT VL RL RL 5 Generator voltage VG = PIG = 1 + RT VL RL SE T-6 Electrical Systems AC Simple circuit models Efficiency requires most power is dissipated in the load Example for Al and household V For PL = 1.2 GW and VL = 120V For transmission assume L = 50 km (a short distance) An Aluminum cable with A = 5 cm2 (to minimize sag, Cu not used.) Resistivity of Al, = 2.8 10-8 -m Conclusion: not so good! 6 If RT /RL 1, then PT PL 2 Then RL = PL /VL = 1.2 10-5 Most of the voltage appears across the load. So, we have very small transmission losses. RT = L/A = 2.8 RL SE T-6 Electrical Systems AC 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 AC Simple circuit models An Ideal Transformer I1 + V1 - I2 + V2 - n1 V2 =NV1 I2 =I1 /N n2 N =n2 /n1 = turns ratio Physical process is conservation of magnetic flux/energy 8 SE T-6 Electrical Systems AC Simple circuit models An Ideal Transformer Common examples of transformers: I1 + V1 - I2 + V2 - n1 V2 =NV1 I2 =I1 /N n2 N =n2 /n1 = turns ratio Physical process is conservation of Bottom right: Photo by mdverde on Flickr. Bottom left: Image by Tau Zero on Flickr. magnetic flux/energy Top right: Photo by brewbooks on Flickr. 8 SE T-6 Electrical Systems AC Analysis AC transmission reduces losses RT I0 VG N1 + V1 - + V T I1 - IL RL N2 + V2 - As before, PL = 1.2 GW, VL = 120 V , RL = 1.2 10-5 What are transformer and transmission requirements, Such that PT PL ? 9 SE T-6 Electrical Systems AC Analysis From the circuit: 2 PL =VL IL = RL IL 2 PT =VT I1 = RT I1 From the transformer relation, IL = N2 I1 , it follows PT 1 RT = 2 PL N2 RL 10 SE T-6 Electrical Systems AC Analysis From the circuit: 2 PL =VL IL = RL IL 2 PT =VT I1 = RT I1 For N2 1 a huge reduction in transmission losses Practical numbers: V0 = 12kV , V1 = 240kV (rms) This implies that N1 = V1 /V0 = 20 Assume small voltage drop across the transmission line. Then V2 V1 Second turn ratio becomes N2 = V2 /VL = 2000 From the transformer relation, IL = N2 I1 , it follows PT 1 RT = 2 PL N2 RL Our transmission loss formula gives PT /PL 6% 10 SE T-6 Electrical Systems AC 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 AC Reactive power Resistors, inductors and capacitors, oh my! There are three basic circuit elements having different Ohm's laws. Element Symbol Ohm's Law I = sin t Phase shift Impedance Z = V /I Resistor R V = RI V = R sin t 0 R Inductor L dI V = L dt Capacitor C dV dt = I /C V = L cos t /2 jL 1 V = - C cos t -/2 -j C 12 SE T-6 Electrical Systems AC Reactive power Phase lags increase reactive power 2t Current, I = Im sin( Tperiod ) 2t Voltage, V = Vm sin( Tperiod ) 1 0.5 I /Im , V /Vm 0 -0.5 -1 0 1 2 t/Tperiod 3 4 13 SE T-6 Electrical Systems AC Reactive power Phase lags increase reactive power 2t Current, I = Im sin( Tperiod ) 2t Voltage, V = Vm sin( Tperiod ) Power, 1 0.5 I /Im , V /Vm 0 -0.5 -1 2t P =I V = Im Vm sin2 ( ) Tperiod ,, 1 2t = Im Vm 1 - cos(2 ) 2 Tperiod 0 1 2 t/Tperiod 3 4 13 SE T-6 Electrical Systems AC Reactive power Phase lags increase reactive power 2t Current, I = Im sin( Tperiod + ) 2t Voltage, V = Vm sin( Tperiod ) Power, 1 0.5 I /Im , V /Vm 0 -0.5 -1 2t 2t P = I V = Im Vm sin( ) sin( + ) Tperiod Tperiod ,, 1 2t = Im Vm cos() - cos(2 + ) 2 Tperiod cos() is known as the power factor 0 1 2 t/Tperiod 3 4 13 SE T-6 Electrical Systems AC 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 AC Reactive power Phase shifts are introduced by inductance A motor will have an inductance, L. L I V cos(t) R 15 SE T-6 Electrical Systems AC Reactive power Phase shifts are introduced by inductance A motor will have an inductance, L. It will introduce a phase shift given by tan = L R L I V cos(t) R 15 SE T-6 Electrical Systems AC Reactive power Phase shifts are introduced by inductance A motor will have an inductance, L. It will introduce a phase shift given by tan = Amplitude of current will also be reduced. V I = 2 2 ( L + R 2 )1/2 L R L I V cos(t) R 15 SE T-6 Electrical Systems AC Reactive power Phase shifts are introduced by inductance A motor will have an inductance, L. It will introduce a phase shift given by tan = L R Amplitude of current will also be reduced. V I = 2 2 ( L + R 2 )1/2 This all follows from adding up the voltages for a simple circuit: dI L + RI = V cos(t) dt L I V cos(t) R 15 SE T-6 Electrical Systems AC Reactive power How to minimize the VA requirement? To minimize the VA requirements on the generator we want 0 Assume the average power absorbed by the load is PL Note: The peak is 2 (rms volt-amp rating) Less generator power is cheaper. Calculate the peak generator power [PG (t)]max as a function of PL 16 SE T-6 Electrical Systems AC Reactive power Peak power from inductance The power dissipated in the load: PL = RI 2 RV 2 = 2 2 L + R2 The peak power delivered by the generator Ppeak = VI (1 + cos ) = V2 ( 2 L2 + R 2 )1/2 1+ R ( 2 L2 + R 2 )1/2 Using the expression for PL , we get: Ppeak = 2 PL 2 L2 + 2R 17 2 1/2 R +R 1 SE T-6 Electrical Systems AC Reactive power Capacitance can balance out reactive power Equivalent cir Recall from our table that the phase lags are opposite for inductance and capacitance tan L = L R , tan C = -1 C Short answer: there is a capacitance that will keep the current and voltage in phase (but not eliminate the power factor) L C= 2 R + 2 L2 Long answer follows. Goal: Find C so there is no re 18 SE T-6 Electrical Systems Analysis I V relation for a capacitor dV ^ I 1(t ) = C G dt I V relation for the load ^ dI 2 ^ VG = RI 2 + L dt Conservation of current ^ ^ ^ I (t ) = I 1(t ) + I 2 (t ) 31 Solution Assume VG = V cos( t) (all voltages now rms) Current in the capacitor branch ^ I 1(t ) = CV sin( t) Current in the load branch (from before) ^ I 2 (t ) = V (R 2 + 2 2 1/ 2 L ) cos( t ) 32 The total current The total current flowing from the generator ^ ^ ^ I (t ) = I 1(t ) + I 2(t ) =V cos( t ) 2 2 1/ 2 (R 2 + L ) C sin( t) sin =V cos( t)cos (R 2 + 2 2 1/ 2 L ) + (R 2 + 2 2 1/ 2 L ) C sin( t) 33 The value of C Choose C for zero reactive power Set sin(t) coefficient to zero C = sin (R 2 + 2 2 1/ 2 L ) Simplify by eliminating the power factor C = L (R 2 + 2L2 ) 34 Calculate the peak power Calculate the peak power to learn what has happened to the VA rating Ppeak = [PG (t )]max = 2 [VI ]max = 2V 2 cos (R 2 + 2L2 )1/ 2 RV 2 =2 2 (R + 2L2 ) = 2 PL 35 The Result It worked!! The VA requirement has been reduced Ppeak VA = = PL 2 36 AC 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 MIT OpenCourseWare http://ocw.mit.edu 22.081J / 2.650J / 10.291J / 1.818J / 2.65J / 10.391J / 11.371J / 22.811J / ESD.166J Introduction to Sustainable Energy Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
View Full Document

This note was uploaded on 02/24/2012 for the course MECHANICAL 2.650J taught by Professor Johnc.wright during the Fall '10 term at MIT.

Ask a homework question - tutors are online