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Course: EEL 4213, Fall 2011
School: FSU
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nodal the voltages for a given load and generation schedule known real (P) and reactive (Q) power injections known real (P) power injection and the voltage magnitude (V) known voltage magnitude (V) and voltage angle () must have one generator as the slack bus takes up the power slack due to losses in the network slack bus (swing bus) generator bus load bus Types of network buses The...

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nodal the voltages for a given load and generation schedule known real (P) and reactive (Q) power injections known real (P) power injection and the voltage magnitude (V) known voltage magnitude (V) and voltage angle () must have one generator as the slack bus takes up the power slack due to losses in the network slack bus (swing bus) generator bus load bus Types of network buses The utility wants to know the voltage profile Power Systems I Power Flow Solution n j =1 * ii ji Pi jQi Ii = Vi* ji n n Pi jQi = Vi yij yijV j * Vi j =1 j =0 Pi + jQi = V I Power Law j =0 = Vi yij yijV j n = ( yi 0 + yi1 + yi 2 + + yin )Vi yi1V1 yi 2V2 yinVn I i = yi 0Vi + yi1 (Vi V1 ) + yi 2 (Vi V2 ) + + yin (Vi Vn ) KCL Power Systems I Power Flow Equations find an iterative improvement of x[k], that is: x[k+1] = g( x[k] ) a solution is reached when the difference between two iterations is less than a specified accuracy: x[k+1] - x[k] make an an initial estimate of the variable x: x[0] = initial value take a function and rearrange it into the form x = g(x) {there are several possible arrangements} x[k+1] = x[k] + ( g( x[k] ) - x[k] ) can improve the rate of convergence: > 1 modified step: the improvement is found as acceleration factors method of successive displacements iterative steps: A non-linear algebraic equation solver Power Systems I Gauss-Seidel Method 4 x = 1 x3 + 6 x2 + 9 = g ( x) 9 9 9 x = x3 + 6x2 + 4 Step 1. Cast the equation into the g(x) form. Find the root of the equation: f(x) = x3 - 6x2 + 9x - 4 = 0 Power Systems I Gauss-Seidel Example x[8] = 4.0000 x[ 7 ] = 3.9988 x[ 6 ] = 3.9568 x[5] = 3.7398 x[ 4 ] = 3.3376 6 4 x[3] = g ( x[ 2 ] = 2.5173) = 1 (2.5173)3 + 9 (2.5173) 2 + 9 = 2.8966 9 6 4 x[ 2 ] = g ( x[1] = 2.2222) = 1 (2.2222)3 + 9 (2.2222) 2 + 9 = 2.5173 9 6 4 x[1] = g ( x[ 0 ] = 2) = 1 (2)3 + 9 (2) 2 + 9 = 2.2222 9 Step 2. Starting with an initial guess of x[0] = 2, several iterations are performed. Power Systems I Gauss-Seidel Example 0 Power Systems I 0 0. 5 1 1. 5 2 2. 5 3 3. 5 4 4. 5 0. 5 1 Initial Value 1. 5 Iterations 3 2 x 2. 5 x 3 3. 5 x = g(x) Solution Points 2 g(x) =-1/9x +6/9x +4/9 Matlab Results Gauss-Seidel Example 4 4. 5 x[ 2 ] = 2.2778 + 1.25 [2.5902 2.2778] = 2.6683 6 4 g (2.2778) = 1 (2.2778) 3 + 9 (2.2778) 2 + 9 = 2.5902 9 x[1] = 2 + 1.25 [2.2222 2] = 2.2778 4 g (2) = 1 (2) 3 + 6 (2) 2 + 9 = 2.2222 9 9 x[ 0 ] = 2 Starting with an initial guess of x[0] = 2. Find the root of the equation: f(x) = x3 - 6x2 + 9x - 4 = 0 with an acceleration factor of 1.25 Power Systems I Gauss-Seidel Example x[8] = 4.0005 x[ 7 ] = 3.9978 x[ 6 ] = 4.0084 x[5] = 3.7238 x[ 4 ] = 3.1831 x[3] = 3.0801 Additional iterations Power Systems I Gauss-Seidel Example 0 Power Systems I 0 0. 5 1 1. 5 2 2. 5 3 3. 5 4 4. 5 0. 5 1 1. 5 2 x 2. 5 x 3 3. 5 x = g(x) Solution Points 2 with acceleration factor: 1.25 Initial Value Iterations 3 g(x) = -1/9x +6/9x +4/9 Matlab Results Gauss-Seidel Example 4 4. 5 xn = cn + g n ( x1 , x2 , , xn ) x2 = c2 + g 2 ( x1 , x2 , , xn ) x1 = c1 + g1 ( x1 , x2 , , xn ) Rearrange each equation for one of the variables f n x1 ( , x2 , , xn ) = cn f 2 ( x1 , x2 , , xn ) = c2 f1 ( x1 , x2 , , xn ) = c1 Consider a system of n equations Power Systems I Gauss-Seidel for a System of Equations 0 2 0 n k +1] [ [ , x2k +1] , , xnk +1] ) in the Gauss-Seidel method, the updated values of the variables calculated in the preceding equations are used immediately in the solution of the subsequent equations 1 (x[ find the results in a new approximate solution 0 1 (x[ ], x[ ],, x[ ] ) assume an approximate solution for the independent variables s t e ps Power Systems I Gauss-Seidel for a System of Equations Power Systems I j =0 y n ij Pi jQi n + yijV j * Vi j =1 j i Vi[ k +1] = The Gauss-Siedel form ji Pi jQi Ii = Vi* n n Pi jQi = Vi yij yijV j * Vi j =1 j =0 Pi + jQi = V I * ii T h e e q u a t io n Vi = The Power Flow Equation j =0 y n ij Pi jQi n + yijV j[ k ] Vi*[ k ] j =1 ji ji ji for generation the powers are positive for loads the powers are negative the scheduled power is the sum of the generation and load powers the real and reactive powers are scheduled for the load buses that is, they remain fixed the currents and powers are expressed as going into the bus n *[k ] [k ] n [k +1] [k ] Qi = Vi Vi yij yij V j j =0 j =1 n *[k ] [k ] n [k +1] [k ] Pi = Vi Vi yij yij V j j =0 j =1 Rewriting the power equation to find P and Q Power Systems I Power Injections j =0 y n ij ji n *[k ] [k ] n [k +1] [k ] Qi = Vi Vi yij yij V j j =0 j =1 n *[k ] [k ] n [k +1] [k ] = Vi Vi yij yij V j Pi j =0 j =1 Vi [k +1] = Pi [sch ] jQi[sch ] n + yij V j[k ] Vi*[k ] j =1 ji ji The complete set of equations become: Power Systems I Solution by Gauss-Seidel n *[k ] [k ] [k +1] [k ] = Vi Vi Yii + Yij V j Pi j =1 j i n Qi[k +1] = Vi*[k ] Vi [k ]Yii + Yij V j[k ] j =1 j i n Pi [sch ] jQi[sch ] Yij V j[k ] Vi*[k ] j =1, j i Vi [k +1] = Yii Rewriting the equations in terms of the Y-Bus Power Systems I Solution by Gauss-Seidel the voltage magnitude and angle must be estimated in per unit, the nominal voltage magnitude is 1 pu the angles are generally close together, so an initial value of 0 degrees is appropriate Since both components (V & ) are specified for the slack bus, there are 2(n - 1) equations which must be solved iteratively For the load buses, the real and reactive powers are known: scheduled System characteristics Power Systems I Solution by Gauss-Seidel e [ k +1] i = Vi [ sch ] 2 ( fi ) [ k +1] 2 Vi = ei + j f i the real power is scheduled the reactive power is computed based on the estimated voltage values the voltage is computed by Gauss-Seidel, only the imaginary part is kept the complex voltage is found from the magnitude and the iterative imaginary part For the generator buses, the real power and voltage magnitude are known Power Systems I Solution by Gauss-Seidel V1 = 1.050 0.01 + j0.03 0.0125 + j0.025 0.02 + j0.04 0.452 pu 1.386 pu 1.102 pu 2.566 pu Using the Gauss-Seidel method, determine the phasor values of the voltage at the load buses 2 and 3, accurate to 2 decimal places Power Systems I Example
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FSU - EEL - 4213
The Power Flow SolutionMost common and important tool in power systemanalysisalso known as the Load Flow solutionused for planning and controlling a systemassumptions: balanced condition and single phase analysisProblem:determine the voltage magnit
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
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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