Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

3 Pages

lecture5

Course: EEL 4213, Fall 2011
School: FSU
Rating:

Word Count: 1304

Document Preview

Power The Flow Solution Most common and important tool in power system analysis also known as the Load Flow solution used for planning and controlling a system assumptions: balanced condition and single phase analysis Problem: determine the voltage magnitude and phase angle at each bus determine the active and reactive power flow in each line each bus has four state variables: voltage magnitude...

Register Now

Unformatted Document Excerpt

Coursehero >> Florida >> FSU >> EEL 4213

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
Power The Flow Solution Most common and important tool in power system analysis also known as the Load Flow solution used for planning and controlling a system assumptions: balanced condition and single phase analysis Problem: determine the voltage magnitude and phase angle at each bus determine the active and reactive power flow in each line each bus has four state variables: voltage magnitude voltage phase angle real power injection reactive power injection Power Systems I The Power Flow Solution Each bus has two of the four state variables defined or given Types of buses: Slack bus (swing bus) Regulated bus (generator bus, P-V bus) voltage magnitude and angle are specified, reference bus solution: active and reactive power injections models generation-station buses real power and voltage magnitude are specified solution: reactive power injection and voltage angle Load bus (P-Q bus) models load-center buses active and reactive powers are specified (negative values for loads) solution: voltage magnitude and angle Power Systems I Newton-Raphson PF Solution Quadratic convergence More efficient for large networks mathematically superior to Guass-Seidel method number of iterations required for solution is independent of system size The Newton-Raphson equations are cast in natural power system form solving for voltage magnitude and angle, given real and reactive power injections Power Systems I Newton-Raphson Method A method of successive approximation using Taylors expansion Consider the function: f(x) = c, where x is unknown Let x[0] be an initial estimate, then x[0] is a small deviation from the correct solution ( ) f x[ 0 ] + x[ 0] = c Expand the left-hand side into a Taylors series about x[0] yeilds () f x[0] Power Systems I 2 df [ 0 ] 1 d f [ 0] 2 + x + 2 2 x +L = c dx dx ( ) Newton-Raphson Method Assuming the error, x[0], is small, the higher-order terms are neglected, resulting in () fx [0] where df [ 0 ] df [ 0] [0] + x c c x dx dx () c[ 0 ] = c f x[ 0 ] rearranging the equations c [ 0 ] x[ 0] = df dx x[1] = x[ 0 ] + x[ 0] Power Systems I Example Find the root of the equation: f(x) = x3 - 6x2 + 9x - 4 = 0 Power Systems I Newton-Raphson Method 50 40 30 20 10 3 2 f(x) = x -6x +9x-4 0 -10 0 Power Systems I 1 2 3 x 4 5 6 Power Flow Equations KCL for current injection n n I i = Yij V j = Yij V j ij + j j =1 j =1 Real and reactive power injection Pi j Qi = Vi* I i Substituting for Ii yields: n Pi j Qi = (Vi ) Yij V j ij + j j =1 Power Systems I Power Flow Equations Divide into real and reactive parts n Pi = Vi V j Yij cos( ij i + j ) j =1 n Qi = Vi V j Yij sin ( ij i + j ) j =1 Power Systems I Newton-Raphson Formation Cast power equations into iterative form n ( Pi[ k ] = Vi[ k ] V j[ k ] Yij cos ij i[ k ] + [j k ] ) j =1 n ( Qi[ k ] = Vi[ k ] V j[ k ] Yij sin ij i[ k ] + [j k ] ) j =1 Matrix function formation of the system of equations sch Pinj c = sch Qinj Power Systems I x [k ] [ k ] = [ k ] V () fx [k ] ( ) ( ) Pinj x[ k ] = Qinj x[ k ] Newton-Raphson Formation General formation of the equation to find a solution c = f ( xsolution ) The iterative equation x x[ 0 ] = initial estimate of xsolution [ k +1] () () c f x[ k ] = x[ k ] + df x[ k ] dx The Jacobian - the first derivative of a set of functions () df x[ k ] a matrix of all combinatorial pairs dx Power Systems I The Jacobian Matrix P P df (x ) = Q dx Q P 1 P 1 1 M M P Pn 1 n1 = Q11 Q1 1 M M Qnm Qn m 1 Power Systems I P V V Q V L P 1 n1 P 1 V1 L O M M O L Pn1 n1 Pn1 n1 Q1 V1 Q1 V1 L M M Qnm n1 Qn m V1 O L L O L L P 1 Vnm 1 M M Pn1 n 1 Vnm Q1 V1 Vnm M M Qn m Vn m Vnm Jacobian Terms Real power w.r.t. the voltage angle Pi = Vi V j Yij sin ( ij i + j ) i j i Pi = Vi V j Yij sin( ij i + j ) i j j Real power w.r.t. the voltage magnitude Pi = 2 Vi Yii cos ii + V j Yij cos( ij i + j ) Vi j i Pi = Vi Yij cos( ij i + j ) i j V j Power Systems I Jacobian Terms Reactive power w.r.t. the voltage angle Qi = Vi V j Yij cos( ij i + j ) i j i Qi = Vi V j Yij cos( ij i + j ) i j j Reactive power w.r.t. the voltage magnitude Qi = 2 Vi Yii sin ii + V j Yij sin( ij i + j ) Vi j i Qi = Vi Yij sin( ij i + j ) i j V j Power Systems I Iteration process Power mismatch or power residuals difference in schedule to calculated power Pi [k ] = Pi sch Pi [k ] Qi[ k ] = Qisch Qi[ k ] New estimates for the voltages [ k +1] i = [k ] i + [k ] i Vi[ k +1] = Vi[ k ] + Vi[ k ] Power Systems I Bus Type and the Jacobian Formation Slack Bus / Swing Bus one generator bus must be selected and defined as the voltage and angular reference Generator Bus The voltage and angle are known for this bus The angle is arbitrarily selected as zero degrees bus is not included in the Jacobian matrix formation have known terminal voltage and real (actual) power injection the bus voltage angle and reactive power injection are computed bus is included in the real power parts of the Jacobian matrix Load Bus have known real and reactive power injections bus is fully included in the Jacobian matrix Power Systems I Newton-Raphson Steps 1. Set flat start For load buses, set voltages equal to the slack bus or 1.00 For generator buses, set the angles equal the slack bus or 0 2. Calculate power mismatch For load buses, calculate P and Q injections using the known and estimated system voltages For generator buses, calculate P injections Obtain the power mismatches, P and Q 3. Form the Jacobian matrix Use the various equations for the partial derivatives w.r.t. the voltage angles and magnitudes Power Systems I Newton-Raphson Steps 4. Find the matrix solution (choose a or b) a. inverse the Jacobian matrix and multiply by the mismatch power b. perform gaussian elimination on the Jacobian matrix with the b vector equal to the mismatch power compute and V 5. Find new estimates for the voltage magnitude and angle 6. Repeat the process until the mismatch (residuals) are less than the specified accuracy Pi[ k ] Qi[ k ] Power Systems I Line Flows and Losses After solving for bus voltages and angles, power flows and losses on the network branches are calculated Transmission lines and transformers are network branches The direction of positive current flow are defined as follows for a branch element (demonstrated on a medium length line) Power flow is defined for each end of the branch Example: the power leaving bus i and flowing to bus j Bus i Iij Vi IL Ii0 yi0 Power Systems I Bus j yij Vj Iji Ij0 yj0 Line Flows and Losses current and power flows: i j ji I ij = I L + I i 0 = yij (Vi V j ) + yi 0 Vi I ji = I L + I j 0 = yij (V j Vi ) + y j 0 V j * * * Sij = Vi I ij = Vi 2 ( yij + yi 0 ) Vi yij V j* S ji = V j I * = V j2 ( yij + y j 0 ) V j yij Vi * ji * power loss: * S Loss ij = Sij + S ji Bus i Iij Vi IL Ii0 yi0 Power Systems I Bus j yij Vj Iji Ij0 yj0 Example Using N-R method, find the phasor voltages at buses 2 and 3 Find the slack bus real and reactive power Calculate line flows j0.02 and line losses 100 MVA base 1 j0.025 j0.04 2 3 138.6 MW 45.2 MVAR Power Systems I Slack Bus V1 = 1.050 256.6 MW 110.2 MVAR
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

FSU - EEL - 4213
ExampleUsing the Newton-Raphson PF,find the power flow solution10.01 + j0.03y12 = 10 j 20 puy13 = 10 j 30 puy23 = 16 j 32 puPower Systems I0.0125 + j0.0250.02 + j0.0423|V3| = 1.04200 MW400 + j 250S == 4.0 j 2.5 pu100200P3sch == 2.0 pu
FSU - EEL - 4213
there are many solution combinations for scheduling generationin practice, power plants are not located at the same distancefrom the load centerspower plants use different types of fuel, which vary in cost fromtime to timein the power flow analysis,
FSU - EEL - 4213
i = 1, 2, ku j ( x1 , x2 , xn ) 0j = 1,2, , mand the inequality constraintsgi ( x1 , x2 , xn ) = 0subject to the equality constraints)The Lagrange multiplier is extended to include theinequality constraints by introducing the m-dimensionalvector
FSU - EEL - 4213
Chapter 8: Transient Analysis ofSynchronous MachinesFAMU-FSU College of EngineeringSynchronous MachinesSteady state modelingrotor mmf and stator mmf are stationary with respect to each otherflux linkage with the rotor are invariant with timeno volt
FSU - EEL - 4213
Balanced 3-Phase Short CircuitConsider a synchronous generator operating at 60 Hzwith constant excitationExamine the impact on the stator currents when a threephase short circuit is applied to the generator terminalsThe initial currentsParkia (0+ )
FSU - EEL - 4213
power flow - evaluate normal operating conditionsfault analysis - evaluate abnormal operating conditionsthree-phasesingle-line to ground and double-line to groundline-to-line faultsunbalanced faultsbalanced faultsspecifying ratings for circuit brea
FSU - EEL - 4213
inversion of the bus admittance matrix is a n3 effortfor small and medium size networks, direct building of the matrixis less effortfor large size networks, sparse matrix programming withgaussian elimination technique is preferredDirect formation of
FSU - EEL - 4213
three-phasesingle-line to grounddouble-line to groundline-to-line faultsunbalanced faultsbalanced faults60-75%15-25%5-15%&lt;5%Percentage of total faultssymmetrical componentsaugmented component modelsUnbalance fault analysis requires new tools
FSU - EEL - 4213
TransformersEquivalent Series Impedance:Transformer bank of three single-phase transformersZ 0 = Z1 = Z 2 = Z Three-phase transformer with a three-leg coreZ1 = Z 2 = Z Z0 &gt; ZWye-Delta Wound TransformersWiring connection will always cause a phase s
FSU - EEL - 4213
Common Unbalanced Network FaultsSingle-line-to-ground faultsDouble-line-to-ground faultsLine-to-line faults March 2004Power Systems I1Single Line to Ground FaultVa = 0IaVaI f = IaIbVbIb = Ic = 0IcVcI a0 1 1 1 I a I = 1 1 a a 2 0 a1 3
FSU - EEL - 4213
StabilitylThe ability of the power system to remain in synchronismand maintain the state of equilibrium following adisturbing forceuSteady-state stability: analysis of small and slow disturbancesnugradual power changesTransient stability: analys
FSU - EEL - 4213
Steady State StabilityllllThe ability of the power system to remain in synchronismwhen subject to small disturbancesStability is assured if the system returns to its originaloperating state (voltage magnitude and angle profile)The behavior can be
FSU - EEL - 4213
Transient StabilityThe ability of the power system to remain in synchronismwhen subject to large disturbancesLyapunov energy functionsLarge power and voltage angle oscillations do not permitlinearization of the generator swing equationssimplified en
FSU - EEL - 4213
Solving Non-linear ODEObjectiveTime domain solution of a system of differential equations Given a function or a system of functions: f(x) or F(x) Seek a time domain solution x(t) or x(t) which satisfy f(x) or F(x)Integration of the differential equat
FSU - EEL - 4213
Multi-machine SystemsEach synchronous machine is represented by a constantvoltage source behind the direct axis transient reactanceThe input powers are assumed to remain constantUsing the pre-fault voltage, all loads are converted toequivalent admitt
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 2CHAPTER 1.8 and 21. A three phase wye-connected unbalanced load is supplied by a balanced three-phasedelta connected source. The three source voltages and two of the line currents are:Vab = 13.2 kV 90
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 3CHAPTER 3(Problems 3-1, 3-2, 3-3, 3-4, 3-6, 3-7, 3-8, pp 154-157 in Textbook)1. The secondary winding of a transformer has a terminal voltage ofvs(t) = 282.8 sin 377t V. The turns ratio of the transfo
FSU - EEL - 3216
Life Long LearningProfessional development: seminars, workshops, conferences,summer courses, PE licenseLiterature resourcesReference books, i.e. McGraw Hill Standard Handbook for ElectricalEngineersMagazines, i.e. IEEE Power &amp; Energy, CIGRE Electra
FSU - EEL - 3216
Transmission LinesOverhead ConductorOverhead Spacer CableUnderground CableThree-Conductor CableService CablesFundamentals of Power SystemsLecture 141Overhead ConductorsACSRAluminum Conductor withinner Steel ReinforcedstrandsACARAluminum Con
FSU - EEL - 3216
Induced Voltage in Stator Coil P-PolesFor a P-pole machine the flux per pole remains the same(integrating d =rlBmcos(P/2 )d over one pole pitch - /P to + /P )2rlBMThe electrical frequency becomesP2elmTherefore, the induced voltage per pole-pair (
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 1SOLUTIONS1. In the 60Hz circuit shown in Figure 1, VS = 240.0VR1 = 2.530 and I1 = 38.8A-36.2.L1 = 0.015 Ha) Determine the phasor currents, IS and I2, and the impedance Z2.b) Calculate the apparent
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 2SOLUTIONS1. A three phase wye-connected unbalanced load is supplied by a balanced three-phasedelta connected source. The source voltages and load impedances are:Vab = 13.2 kV 90Vbc = 13.2 kV 210Vca
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 3SolutionsCHAPTER 3(Problems 3-1, 3-2, 3-3, 3-4, 3-6, 3-7, 3-8, pp 154-157 in Textbook)1. The secondary winding of a transformer has a terminal voltage ofvs(t) = 282.8 sin 377t V. The turns ratio of t
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 4SOLUTIONS1. A Y-connected bank of three identical 100-kVA, 7967/480-V transformers is suppliedwith power directly from a large constant-voltage bus. In the short-circuit test, therecorded values on th
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 5, SOLUTIONS4-2. A three-phase four-pole winding is installed in 12 slots on a stator. There are 40turns of wire in each slot of the windings. All coils in each phase are connected inseries, and the thr
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 6SOLUTIONS4-6. The flux density distribution over the surface of a two-pole stator of radius r andlength l is given by)B BM cos( m tProve that the flux density under each pole face is2rlBMSolve fir
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 7 Solution1. A 3-phase, 4-pole synchronous machine rotates CCW with 1500 rpm and produces arated torque of 668.45 Nm at the shaft in CW direction. The total rated losses are5 kW and the rated stator (co
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 8A balanced 3-phase impedance type load is rated 3 MVA at 4.16 kV, with a laggingpower factor of 0.75. It is supplied by a generator via a 2 km long overhead transmissionline. The GMD of the conductors
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 8A balanced 3-phase impedance type load is rated 3 MVA at 4.16 kV, with a laggingpower factor of 0.75. It is supplied by a generator via a 2 km long overhead transmissionline. The GMD of the conductors
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 9Problems from book 9-6, 9-11, and 9-15 as shown below.Problem 9-6 refers to a single-phase, 8 kV, 50 Hz, 50 km long transmission line consisting of twoaluminum conductors with a 3 cm diameter separated
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 9Problems from book 9-6, 9-11, and 9-15 as shown below.Problem 9-6 refers to a single-phase, 8 kV, 50 Hz, 50 km long transmission line consisting of twoaluminum conductors with a 3 cm diameter separated
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 10Problem from book 5-29 as shown below.
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 10SOLUTION
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 11Problem from book 7-5, 7-7, 7-11 as shown below.7-57-77-11
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 11 SOLUTIONSProblem from book 7-5, 7-7, 7-11 as shown below.7-5Attention: this solution, as provided by the author of the book, is inconsistent with the text of thebook. The core losses are supplied by
FSU - EEL - 3216
EEL 3216 Fundamentals of Power SystemsHomework # 12You are encouraged to solve the two problems below in a group effort. Therefore, amaximum of 5 students may submit one assignment together. If you do so, clearly writethe names of the 5 students who h
FSU - EEL - 3216
Electric MotorsThree basic types of motorsSynchronous, SMInduction (asynchronous), IMDC motorsIn the past: preferred for variable speed applicationsDecreasing in popularity; replaced by combination of IM and powerelectronic variable speed drive (VS
FSU - EEL - 3216
U.S. Electric Power Industry ExistingNet Summer Capacity by State, 2003Fundamentals of Power SystemsLecture 11U.S. Electric Power TransmissionSystem in 2000Fundamentals of Power SystemsLecture 1Page 12Power Systems ComponentsPrimary power sour
FSU - EEL - 3216
Electric Powersingle-phase (1~) systemsOne sourceTwo energy-carrying wiresPulsating instantaneous power (2f)three-phase (3~) systemsThree sources, 120 phase shiftedLecture 2Fundamentals of Power Systems1Single Phase Powerptv, i, pPVvtptDe
FSU - EEL - 3216
Notes on Phasor Notation andDiagramsMathematically correctComplex quantity(underlined)VmMagnitudeV Vm e jvttVm cos( t )Re(V) Vm cosvtorjVm sin( t )tIm(V) Vm sintPractical notation and usageComplex quantity, butI = 10 Aunderlining omitt
FSU - EEL - 3216
Ideal 1~ Transformer ModelFlux linkages (winding flux)Ideali1i2N1N1 ,1N22N2SourceV2av1e1 , v2e2aV1N1N2e1e2ZLN1N2Loadv1v2i2i1Dot conventionCurrent into doted end produces positive MMF (NI) orampere-turnsTherefore, orientat
FSU - EEL - 3216
FSU - EEL - 3216
Three-phase transformersCore arrangementsTransformer bank =&gt; three 1~ transformers (core typeor shell type)3~ transformer (3, 4, 5 legged core-type)Winding arrangementsY-Y-YYWith tertiaryFundamentals of Power SystemsLecture 61Core Type vs. She
FSU - EEL - 3216
Changing BasesThe impedances of generators, transformers,transmission lines, and loads are supplied by themanufacturers in per unit based on theequipments own rating.When placing the equipment into a system, theimpedance must be converted to the sy
FSU - EEL - 3216
AC Machine FundamentalsGeneratorsMotors3-phase1-phaseSynchronousInductionFundamentals of Power SystemsLecture 91AC Machines SystematicFundamentalsPrinciple of generating a rotating magnetic fieldMagnetomotive force and induced voltagePrincip
FSU - EEL - 3216
SM No-Load CharacteristicRotor flux magnitudeInduced stator voltage magnitudeEAkCharacteristic of theiron (stator and rotor)Rotor field currentRotor field currentEffect of saturation: for large field currents the no-loadarmature voltage magnitud
FSU - EEL - 3216
AmpacityLecture 15Fundamentals of Power Systems1Geometric Mean Radius (GMR)'L102r4lnDr02Inductance of conductorup to a distance of 1 unit(e.g. 1 foot)'L102lnDGMRr4L'GMR1Dlnr110lnGMR2lnInductance of conductor in Hpe
FSU - EEL - 3216
ExampleM1TST = 1.5 MVAX = 3%NH/NL = 5.4586BusV = 4.16 kVM2M150 Hz, 6 pole, Y-connectedPR = 75 kW, Xm = 7.2R1 = 0.082 , R2 = 0.07X1 = 0.19 , X2 = 0.18Friction and windage losses 1.3 kWCore losses 1.4 kWMiscellaneous losses 150 WM2Rated 10
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 1CHAPTER 1.8 and 21. In the 60Hz circuit shown in Figure 1, VS = 240.0VR1 = 2.530 and IS = 35 A15.L1 = 0.015 Ha) Determine the phasor currents, I1 and I2, and the impedance Z2.b) Calculate the appa
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 71. A 3-phase, 4-pole synchronous machine rotates CCW with 1500 rpm and produces arated torque of 668.45 Nm at the shaft in CW direction. The total rated losses are5 kW and the rated stator (copper) los
FSU - EEL - 3216
EEL 3216 Introduction to Power SystemsHomework # 3CHAPTER 31. (Problem 3-10 in Textbook)2. (Problem 3-22 in Textbook)3. In a certain single-phase system the per unit values of the impedance, current, voltage,and apparent power are:Z = 0.65 pu I = 1
FSU - EEL - 3216
Power Flow in a SM - GeneratorPout3V I A cos( )PoutReactive QoutpowerNote: is the phase anglebetween armature voltage V andarmature current IA. It is NOT theangle between VT and IL!Qout3 VT I L cos( )3V I A sin( )3 VT I L sin( )Lecture 13Fu
FSU - EEL - 3216
Magnetomotive Force and FluxDistributionThe magnetic fieldused to inducesinusoidal voltages(in the stator) has tobe distributedsinusoidal along thecircumference of thestator/rotor (air gap)Snapshot of oneinstantaneousvalue in timeThis spatial
FSU - EEL - 3216
T-Line Equivalent CircuitsThree general models for equivalent transmissionline circuitsChoice influenced by the line length, type (cable oroverhead line), and operating voltage levelChoice based on the analysis (e.g., short circuit orvoltage drop)M
Drexel - BLAW - 201
Milos RujevicEthics In BusinessMy approach to the topic ethics in business will start with the slogan on the wallof the main factory hall of the paper factory of Mr. Vapa in Belgrade in the years 1930-44.It read: Live, work and give the other to live
Drexel - BLAW - 201
Chapter 1: Terms of EndearmentDefining Information Technology12Chapter 1: Terms of EndearmentDefining Information TechnologyB. megabytesC. gigabytesD. terabytesChapter 1 Test BankTrue/False1.Most of the computers you would see in an airport term
Drexel - BLAW - 201
12Chapter 7 Test Bank True/False 1. Debugging means to solve IT problems. 2. The purpose of debugging is to improve systems. 3. Computers dont make mistakes. 4. 0 is not allowed in an email address. 5. 0 and 1 are allowed in URLs and email addresses. 6.
Drexel - BLAW - 201
CS 103, Fall 2008Midterm 2Prof. NakayamaFamily (or Last) Name_Given (or First) Name_Student ID_Instructions1. This exam has 7 pages in total, numbered 1 to 7. Make sure your copy has all the pages.2. Note the number written on the upper right -han
Drexel - BLAW - 201
Answer Key Testname: MT2-08F1) TRUE 2) FALSE 3) FALSE 4) TRUE 5) TRUE 6) C 7) A 8) 9) 10) FALSE 11) TRUE 12) FALSE 13) FALSE 14) TRUE 15) B 16) B 17) B 18) A 19) D 20) A 21) B 22) B 23) byte 24) nibble 25) 256 26) Hexadecimal 27) TRUE 28) FALSE 29) TRUE
Drexel - PSY - 101
Psychology 101 Dr. D. L. ChuteSample Quiz 11.The inability to recognize familiar faces as a result of brain damage is calledPROSOPAGNOSIA2. What are the primary functions of the frontal lobes?PLANING AND EXECUTIVE FUNCTIONS3.Damage to which lobe o
Drexel - PSY - 101
The principle reason, reduce anxietyThis potential long term effect of alcoholism is caused by a deficiency in thiamine and ischaracterized by encephalitis and psychosis: FETAL ALCOHOL SYNDROMEBlood Alcohol Content (BAC) depends on which of the followi