### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# ELECTROMECHL DYNMICS EEL 4220

FSU

GPA 3.89

### View Full Document

## 423

## 0

## Popular in Course

## Popular in Electrical Engineering

This 195 page Class Notes was uploaded by Tracy Okuneva Sr. on Thursday September 17, 2015. The Class Notes belongs to EEL 4220 at Florida State University taught by Staff in Fall. Since its upload, it has received 423 views. For similar materials see /class/205574/eel-4220-florida-state-university in Electrical Engineering at Florida State University.

## Similar to EEL 4220 at FSU

## Reviews for ELECTROMECHL DYNMICS

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/17/15

Chapter 4 DC Generators 982003 Electromechanical Dynamics Armature Reaction rotation A Current owing in the armature coils torque due tor creates a owerful ma netomotive force p g in F 0 that d1storts and weakens the ux com1ng o F p 9 from the poles F x p armature flux neutral zone 39owquot 5 51 or Q N load Considering the armature along the armature current produces a magnetic eld that acts at a right angle to the eld produced by the poles neutral zone The ux intensity depends on the current 982003 Electromechanical Dynamics 2 Armature Reaction rotation Contrary to the eld ux the armature ux Ir neutral is not constant but varies With the load zone r The ux in the neutral zone is no longer zero and a ux is induced in the coils resultant shorted by the brushes 9 3933 mm The armature mmf distorts the ux produced by the poles the neutral zones have shifted in the direction of rotation I I Y Flux is concentrated at the far end of new the poles positions 2 amp 3 in g zone increase in ux causes saturation to set in the far ends the total ux produced by the poles is less than When the generator runs at noload 982003 Electromechanical Dynamics 3 Improving Commutation The shift in the neutral zone causes an increase in arcing we can move the brushes in the direction of rotation to reduce the arcing For time varying loads the uctuating current raises and lowers the armature magneticmotive force and the neutral zone shifts back and forth it is not practical to continuously move the brushes to minimize the arcing for small machines the brushes are set in an intermediate position to ensure reasonably good commutation at all loads 982003 Electromechanical Dynamics Commutating Poles In larger machines a set of commutating poles are placed to counter the effect of armature reaction narrow poles carry windings that are designed to develop a MMF equal and opposite to the N MMF of the armature as load current varies the two MMF s rise and fall together the vertical component of the eld is nulli ed and the neutral 20116 is I GStOI Gd pole quot O mmt c 982003 Electromechanical Dynamics 5 Separately Excited Generators 0 Instead of using permanent magnets to create the magnetic eld pairs of electromagnets called eld poles are employed 39 Separately excited eld poles are supplied by an independent current source q batteries or another generator the current source is referred to as the exciter 982003 Electromechanical Dynamics Machine Saturation Curve rated flux 5 In a separately exclted noload 5 125 generator a change in excitation b current causes a corresponding gt Flux 100 z 075 2 change in the induced voltage 050 the saturation curve relates the 025 ux produced to the current 0 0 1 2 for small currents the ux is 1quot linearly proportionate 10 tad XO39tage at higher currents the ux E 120 output decreases due to iron 90 saturation i so the segment from a to b is the 30 saturation knee 0 4E o 1 2 the 1nduced voltage curve 1s Ix identical to the ux curve No39load saturatm curve 982003 Electromechanical Dynamics Equivalent Circuit Model Circuit model development armature circuit windings containing a set of identical coils and possessing a certain resistance Which can be modeled as a series resistance Wrt the terminals total armature resistance RO is measured between the terminals When the machine is at rest resistance is in series With the induced voltage Which is represented by a voltage source EO eld winding circuit Ra J 1 F W1nd1ng conta1n1ng a set of identical coils in series F total eld resistance Rf 2 02 982003 Electromechanical Dynamics Shunt Generators A shuntexcited generator is a machine With the eld Winding in parallel With the armature terminals this eliminates the need for an external source of excitation the generator becomes selfexciting this results in a small mmf acting in Starting the S 1f XCita On the same d1rectlon as the remanent ux and causmg the ux per pole to remanent ux 1n the pole 1nduce a 1ncrease small armature voltage When there the increased ux raises E Which 1s rotatlon 0 feeds back to increase I the voltage produces a small X exciting current 1X EO 1ncreases untll Rf and the saturation limits the feedback 982003 Electromechanical Dynamics lO Voltage Control 0 The induced voltage of the shunt generator is easily controlled by varying the excitation current by means of a rheostat connected in series With the shunt eld coil n J field rheostat R 982003 bx 4 V Electromechanical The noload value of E0 is determined from the saturation curve and Rf it is the intersection of the Rf line and the voltage curve I V I 1 5 0 1quot f 1 410 W Er i 39139 1 20 Q I I Q I 100 ED 52 T 30 f If 139quot 50 f 1 x 40 20 D 11 Shunt Generator Under Load v The terminal voltage of a self 18 excited shunt generator falls off 90 1 tr more sharply With increasing E12 I load than that of a separately 50 I I excited generator I i the eld current in a separately I excited generator remains 0 1 l constant under any load 0 I 5 10 A 982003 the eld current in a shunt generator is a function of the terminal voltage For a selfexcited shunt generator the voltage drop from noload to fullload is about 15 of the fullload voltage for separately excited generators it is less than 10 Electromechanical Dynamics increased loading causes a drop in terminal voltage and consequently a drop in excitation current 12 Compound Generator The compound generator prevents the terminal voltage of 11quot a shunt generator from A A decreasing with increasing load i i i a compound generator is similar to a shunt generator except that it has additional 1 eld coils connected in series with the armature circuit these series eld coils are series field shunt field AAA H composed of a few turns of heavy gage wire for carrying the armature load current the total resistance of the series coils is very small 982003 Electromechanical Dynamics l3 Equivalent Circuit At noload the current in the series coils is zero the shunt coils carry the excitation current X to produce the eld ux As load increases the terminal voltage tends to drop a but the load current IC now ows through the series eld coils the mmf developed by the series eld coils acts in the same direction as the mmf of the shunt eld coils AA lll VUV b shunt the ux increases under rising load 982003 Electromechanical Dynamics series field shunt field AAA H l4 Differential Compound Generator In a differential compound generator the mmf of the series eld acts opposite to the shunt eld under load the terminal voltage falls drastically with increasing load the series field circuit is reversed in polarity to make a compound generator into a differential compound generator useful in welding applications limits shortcircuit currents 982003 Electromechanical Dynamics 1 5 Loading Characteristics loading characteristics of several generator types Wovercompound 100 39 compound 8 separate excitation 3 80 a I shunt 2 E r 39 73 60 differential compound C lt quot I g I I 20 I a I 0 O 50 100 gt Load current 982003 Electromechanical Dynamics 16 Chapter 2 Fundamentals of Magnetism 8282003 Electromechanical Dynamics Magnetic Field Intensity Whenever a magnetic ux Q exist in a conductor or component it is due to the presence of a magnetic eld intensity H given by U H Where H magnetic eld intensity Am U magnetomotive force A or ampere turns l length of the magnetic circuit m 50 A example nd the magnetic eld intensity at the circle U 50A 159Am p 5 cm 2 27r005m 8282003 Electromechanical Dynamics Magnetic Flux Density For a magnetic ux Q there exists a magnetic ux density B given by Bz Where A B ux density T tesla g1 ux in a component Wb weber A cross section area m2 example nd the ux density 360x10 6 Wb dh 001m002m 360ng 8282003 Electromechanical Dynamics 3 Relationship between BH in Free Space In free space the magnetic ux density B is directly proportional to the magnetic eld intensity H The constant of proportionality for free space is called the permeability constant 0 T B 0 H I30 in the SI system 3 15 yo 47tgtlt107 Hm 10 henrymeter 39 05 O0 200 400 600 800 10001200Am H 8282003 Electromechanical Dynamics 4 BH Characteristic of Magnetic Material The ux density is in uenced by the magnetic property of the material in which the ux passes instead of specifying a permeability for every material a relative permeability is de ned ur u uo relative permeability is unitless BZIUOIUFH 20T B 1395 silicon iron 1 for many materials the 10 relative permeability is 05 not constant but varies 0 ee space nonlinear Wrt B 0 1 2 3 4 5 6 mm H 8282003 Electromechanical Dynamics 5 cast iron etermining Relative Permeaility ne can find the relative permeability in a material by taking the ratio of the ux density in the material to the ux density that would have been produced in freespace T B B 25 o 111 H H 20 I Sllectron 1110 1395 Permallov r y 08 V Example B 06 39Vstee39 Determine the relative 39 r permeability of relay 1 l coban Amv steel 1 silicon at a 39 l ux density of 06 T j d 14 T 39 8282003 Electromechanical Dynamics 6 Faraday s Law Electromagnetic induction If the ux linking a loop or turn varies as a function of time a voltage is induced between its terminals The value of the induced voltage is proportional to the rate of change of ux In SI units E NAD Where At E induced voltage V N number of turns in the coil ACID change of ux inside the coil Wb At time interval of the ux changes s 8282003 Electromechanical Dynamics Faraday s Law Example a coil of 2000 turns surrounds a ux of 5 me produced by a permanent magnet the magnet is suddenly Withdrawn causing the ux inside the coil to drop uniformly to 2 me in 0100 s What is the induced voltage quot quotquot7 r N S l gt quotum y At1l105 8282003 Electromechanical Dynamics Voltage Induced in a Conductor It is often easier to calculate the induced voltage on a segment of conductor instead of the voltage on a coil E B v Where 120v E induced voltage V B uX density T l active length of I conductor in the 100 ms magnetic field m l v relative speed of e the conductor ms 8282003 Electromechanical Dynamics 9 Lorentz Force A currentcarrying conductor sees a force when placed in a magnetic eld fundamental principle for the operation of motors the magnitude of the force depends upon orientation of the conductor with respect to the direction of the eld force is greatest when the conductor is perpendicular to the eld The Lorentz or electromagnetic force F B I Sing where F force acting on the conductor N 6 angle between the ow directions of current and ux 82 82003 Electromechanical Dynamics 10 Lorentz Force on a Conductor Example a conductor of 15 m long is carrying a current of 50 A and is placed in a magnetic eld With a density of 12 T calculate the force on the conductor if it is perpendicular to the lines of ux calculate the force on the conductor if it is parallel to the lines of ux 8282003 Electromechanical Dynamics 1 l Direction of Force A CUI I EIlt C arrying a CUI I GIlt lS is222222222255222222222223222352222222523 n u a n n u n 4 v p u y c n y u v u n v 39 u u u u u u n a n g a o o n n c u n u u n u n u n o u u u c g u no 0 f 1 S d e d b y a g n C 1 e d j513I 33333333 3333 39 a I I u l u a I a u I I u I u a 39 n 39 u 39 o 39 u n 39 c o 39 I I u 39 D u I 0quot 2 T H 1 o f o elds never cross each other the flux lines of two magnet fields are vectorally added the generate d me chanic a1 f 0 re 6 tends to push the lines of flux back to an even distribution b 82 82003 Electromechanical Dynamics l2 Direction of Force Right hand rule Fingers point in the direction of current ow positive to negative Bend ngers into the direction of the magnetic eld north to south Thumb points in the direction of force 8282003 Electromechanical Dynamics 1 3 Residual Flux Metals that have a strong magnetic attraction can be modeled as being composed of many molecular size magnets orientation of the magnets are normally random by applying an external magnetic eld eg using a coil with a current ow the molecular size magnets will align themselves With the external eld 8282003 Electromechanical Dynamics Residual Flux When the external magnetic eld decreases the magnetic domains tend to retain their original orientation this is called residual induction OI o u o o o o a u a on o o o o o 00 C C O I o o o a o o col O o o I o o o no O o o a o n o a o o o o o n a o o no o o o o u a o a I o a o o a o a n o no o o o o o a o c o a o o n n a o c o o o o 0o o o o o n o n o o o c OJ 1 resid ual induction c I m V 0 gtmagnetlc field mtensuty H coercnve force 8282003 Electromechanical Dynamics 15 Hysteresis Loop To eliminate the residual ux a reverse coil current is required to generate a eld H in the opposite direction the magnetic domains gradually change their previous orientation flUX bGCOmGS 2610 W 0 gtmagnetic field intensity H f HC 1s the coerc1ve force me Bm Hm a energy 1s requ1red to overcome the 3 b molecular friction of the domains to 1 any changes in direction AC magnets have ac ux changing We 0Hc quot1 continuously and Will map out a i Br closed curve call a hysteres1s loop e 8282003 Electromechanical Dynamics d 16 Bm Hm Hysteresis Losses With an external ac ux the BH characteristics of a magnetic material traces out a curve from Bm Hm to Br O to 0 HC to Bm Hm to Br 0 to O HC magnetic material absorbs energy during each cycle and the energy is dissipated as heat the heat released per cycle Jm3 is equal to the area TAm of the hysteresis loop 8282003 Electromechanical Dynamics i 7 Eddy Currents 8282003 Electromechanical Dynamics An ac ux blinking a rectangularshaped conductor induces an ac voltage E across its terminals If the conductor terminals are shorted a substantial current ows The same ux linking smaller coils induce lesser voltages and lower currents A solid metal plate is basically equivalent to a densely packed set of rectangular shaped coils The induced currents owing inside the plate are eddy currents and ow to oppose E the change in ux Eddy Current Losses Eddy currents become a problem when iron must carry an ac ux eddy currents ow throughout the entire length of the iron core resistance in the iron causes energy to be dissipated as heat Losses can be reduced by splitting the core into sections lamination subdividing causes the losses to decrease progressively I varnish coatings insulate the laminates from current ows silicon in the iron increases the resistance 8282003 Electromechanical Dynamics Chapter 8 ThreePhase Circuits 2300 Electromechanical Dynamics Threephase circuits A 3phase generator has three identical windings on the stator placed at 1200 to each other at constant speed the voltages induced in the windings have the same effective value the voltage peaks occur at different times 2300 Electromechanical Dynamics 2 ThreePhase Generation The generation of threephase ac power can be Viewed as three separate power sources each with a 1200 degree angular shift in time of the sinusoidal waveform with respect to the other sources vat Vm sin27z 60 r vba Vm sin27r 60 r 7z 1260 Vm sin27z 60 r gm 120 120 2300 Electromechanical Dynamics E c3 E b2 ThreePhase Waveforms I I l If I I IVA llll l l I l I I llI ll I 1 AAAAAAAAAAA 39VVVVVVVVVVVV Wye connected circuits VaO C V290 Vcl Vab Va Vb Vline ab la lline a Vbc Vb Vc Vline bc lb lline b Vca Vc Va Vline ca lc lline c 2300 Electromechanical Dynamics Delta connected circuits a v00 W C b c V190 Vab Vline ab la lab lca lline a Vbc Vline bc lb lbc lab lline b i1 ca Vline ca c ca lbc lline c V 2300 Electromechanical Dynamics ThreePhase Circuits Example the 3phase generator shown below is connected to three 20 ohm load resistors The generator supplies a voltage of 120 V rms calculate the power dissipated in each resistor and in the 3phase load the peak power Pm and the ratio of P to the total load B a 6 3 S AffJJZ b C I b wp c 3 2300 Electromechanical Dynamics ThreePhase Circuits 0 Example a 3phase 60 Hz wye connected generator has a line voltage of 23900 V calculate the linetoneutral voltage voltage induced in the individual windings the time interval between the positive peak voltage of phases A and B the peak value of the line voltage 2300 Electromechanical Dynamics ThreePhase Circuits Example three identical impedances are connected in delta across a 550 V line The current drawn is 10 A calculate the current in each impedance 10 A the Impedance values L 550 v b 10A amp 10 A c gt on 2300 Electromechanical Dynamics Phase Sequence 39 3phase systems have a property called phase sequence it determines the direction of rotation of 3phase motors it means the order in Which the three line voltages become successively positive two sequences ABC amp ACB positive sequence ABC negative sequence ACB 0 Phase sequence is determined by a simple circuit one lamp always burns brighter the sequence order is bright lamp dim lamp capacitor 2300 Electromechanical Dynamics 3phase source Eca A b horizontal axis Ebc I Eab Power Measurements Wattmeters measure active power singlephase wattmeters have two pairs of terminals voltage amp current each terminal pair has polarity markings polarity markings determine the direction of power ow 3phase 3wire circuits power is measured by two singlephase wattmeters total power is the sum of the two meters 3phase 4wire circuits power is measured by three singlephase wattmeters total power is the sum of the three meters 2300 Electromechanical Dynamics A l l Power Measurements 0 Example a fullload test on a 10 hp 3phs motor yields the following at VLine 600 V P1 5950 w P2 2380 w1Line 10A calculate the power factor of the motor 0 Example the same motor runs at noload ILine drops to 36 A and P1 1295 W P2 845 W calculate the noload losses and the power factor 2300 Electromechanical Dynamics Power Measurements Example three identical resistors connected as a wye dissipate a total power of 3000 W when energized by a 550 V line calculate the current in each line the resistor values E P P 3000 W A I 550 V 3phase O B R J 60 Hz line 39 C o R 2300 Electromechanical Dynamics l3 Power Measurements Example A 3phase 550 V 60 Hz line is connected to three identical capacitors in a delta con guration which draws 22 A calculate the capacitance in each capacitor 22A C 2300 Electromechanical Dynamics l4 Chapter 16 Synchronous Generators 3200 Electromechanical Dynamics Generator under Load 0 The behavior of a synchronous generator depends upon the connected load two basic load categories I I X isolated loads in n ebus isolated loads with a lagging pf current lags the terminal voltage E the voltage drop across the synchronous reactance E leads the current by 90 the induced voltage E0 generated by the ux 1 is equal to the phasor sum of E and EX 3200 Electromechanical Dynamics 2 Generator under Load isolated loads with a leading pf 1 Ex current leads the terminal voltage E the voltage drop across the synchronous reactance EX leads the current by 90 the induced voltage E0 generated by the ux 1 is equal to the phasor sum of E and EX note that EO always leads E by the angle 5 for lagging loads E0 is greater than E for leading loads E is greater than EO 3200 Electromechanical Dynamics Generator under Load 39 Example a 36 MVA 208 kV 3 phase generator 39 synchronous reactance is 9 ohms go 39 nominal current is 1000 A l 39 no load saturation curve is given adjust the excitation so that the terminal voltage is fixed at 21 kV 39 calculate the excitation current kV 18 16 14 nominal voltage linetoneutral 12 10 8 6 4 2 0o 100 200 300 gt 39 draw the phasor diagrams for the following load conditions no load resistive load of 36 MW capacitive load of 12 MVAr Electromechanical Dynamics 3 200 400tA Regulation Curves Voltage regulation is the behavior of the generator s terminal voltage as the load varies 0 Regulation is a function of the load current the regulation curve is a plot of the terminal voltage VT with respect to load current I ranging from no load to full load 0 for a fixed field excitation current 0 for a given load power factor family of curves are developed for various field excitation currents and for different load power factors percent regulation is defined as VRMX100 Trated 3200 Electromechanical Dynamics Regulation Curves Example consider the regulation curves for a 36 MVA 21 kV generator calculate the percent regulation corresponding to the unity power factor curve A v powerYacto s r39 aggmg m 16 t I F x rated load 3 14 10 12 kV 1000 A g 39IIIII lII 5 L 39 nun 7 r f m 09 leading a m 10 37 3 3 8 gt E 6 E E 4 m 2 0 0 250 500 750 1000 1250 A Load current I gt 3 200 Electromechanical Dynamics 6 Synchronization of a Generator Often two or more generators are connected in parallel to supply a common load in large utility systems connecting a generator to other generators is called paralleling many paralleled generators behaves like an infinite bus 0 voltage and frequency are constant and can not be easily altered before connecting a generator to an electrical grid it must be synchronized 0 the generator frequency is equal to the system frequency 0 the generator voltage is equal to the system voltage 0 the generator voltage is in phase with the system voltage 0 the phase sequence of the generator is the same as that of the system 3200 Electromechanical Dynamics 7 Synchronization of a Generator To synchronize a generator 3200 adjust the speed regulator of the prime mover so that frequencies are close adjust the excitation so that generator voltage and system voltage are equal observe the phase angle by means of a synchroscope which indicates the phase angle between two voltages 0 the pointer rotates proportional to the frequency difference 0 a zero mark indicates a zero degree phase angle 0 the speed regulator is adjusted so that the pointer barely creeps across the dial on the zero mark the line circuit breaker is closed Electromechanical Dynamics Connecting to an Infinite Bus 0 An infinite bus system is so powerful that it imposes its own voltage magnitude and frequency once an apparatus is connected to an infinite bus it becomes part of it for a synchronized generator the operator can only vary two machine parameters 0 the field excitation current IX 0 the prime mover s mechanical torque T 3200 Electromechanical Dynamics 9 Connecting to an Infinite Bus Varying the exciting current impacts the induced voltage E0 causes a current to ow that is 90 degrees out of phase due to the synchronous reactance I 2 E0 E sz does not affect the ow of active real power does cause reactive power to ow 3200 Electromechanical Dynamics 10 Connecting to an Infinite Bus Varying the mechanical torque by opening up the control valve of the prime mover an increase torque is developed the rotor will accelerate E0 will increase in value and begin to slip ahead of phasor E leading by a phase angle 5 Although both voltages have similar values the phase angle produces a difference of potential across the synchronous reactance 0 a current will ow but this time almost in phase with E 0 real active power will ow 3200 Electromechanical Dynamics ll Active Power Delivered The active power delivered by a synchronous generator is given by PE sin 6 S PE 2 3 phase power delivered by the generator E 2 induced generator voltage VT 2 generator terminal voltage XS 2 synchronous reactance per phase 6 phase angle between E and VT 3200 Electromechanical Dynamics 12 Active Power Delivered 0 30 60 90 120 5 degrees 3200 Electromechanical Dynamics 150 180 Active Power Delivered 0 Example a 36 MVA 21 kV 1800 rpm 3 phase 60 Hz generator is connected to the power grid 0 synchronous reactance of 9 2 per phase 0 line to neutral exciting voltage is 12 kV 0 line to line system voltage is 173 kV calculate 0 the active power delivered when the power angle 513 30 0 the peak power that the generator can deliver before losing synchronism 3200 Electromechanical Dynamics l4 Transient Reactance short X circuit A synchronous generator connected to a system is subject to switching events a short circuits load energization etc T load normal In many cases the equivalent circuit does not re ect the behavior of the machine the equivalent circuit is only valid for steady state operation normal load gtk shortcircuit gt Alternator reactance for sudden large current changes another reactance is needed X39d reactance X 39 whose value is T varies as a function of time i gt time the reactance for a short circuit 0 prior to the fault the reactance equals the synchronous value 0 at the instant of fault the reactance falls to a much lower value X 1 3200 Electromechanical Dynamics 15 Transient Reactance The reactance X is called the transient reactance can be as low as 15 of the synchronous reactance consequently the initial short circuit current is much higher than that corresponding to the synchronous reactance 3200 Electromechanical Dynamics 16 Transient Model 0 Example a 250 MVA 25 kV 3 phase generator delivers its rated output at unity power factor 0 a synchronous reactance of 16 pu 0 a transient reactance of 023 pu a short circuit suddenly occurs on the connecting transmission line close to the generator calculate 0 the induced voltage E0 prior to the short circuit 0 the initial value of the short circuit current 0 the final value of the short circuit current if the circuit breaker should fail to open 3200 Electromechanical Dynamics l7 Power Transfer 39 We are often interested in the active power that can be transmitted between source A and source B I using Kirchhoff s voltage law ix EA 2 EB jX AB A B the active power absorbed at source B is PB 2 EB AB cos6 B applying the geometry law of the sines for a triangle E X AB EA EA EA sin sing1 sin906 cos6 substitution results in PZEBEA B X sin 5 3200 Electromechanical Dynamics 18 Power Transfer 0 Example a transmission line connects two generators 0 generator A operates at E 20 kVA 5 0 generator B operates at E 15 kVA 42 0 the transmission line has a reactance of 14 ohms calculate the active power that ows over the line 0 which machine is receiving the power 3200 Electromechanical Dynamics Machine Efficiency The physical size of the synchronous machine has a profound effect upon efficiency power output relative cost and temperature rise losses in the machine 0 FR losses in the stator windings IdCZRflosses in the rotor field winding 0 iron core losses and mechanical losses keeping all machine parameters and materials the same 0 an increase in all linear dimensions causes voltage increases by the square output power increases by the 4th power losses increase by the 3rd power 3200 Electromechanical Dynamics 20 Chapter 14 Application of Induction Motors 102 52003 Electromechanical Dynamics Introduction 0 When selecting a 3phase induction motor for an application several motor types can fill the need manufactures often specify the motor class best suited to drive the load 0 3phase induction motors under 500 hp are standardized the frames have standard dimensions a 25 hp 1725 rpm 60 Hz motor from one manufacture can be replaced by that of any other manufacturer Without having to change the mounting holes shaft height or the type of coupling limiting values for electrical mechanical and thermal characteristics are standardized a motor must satisfy minimum requirements for starting torque lockedrotor current overload capacity and temperature rise 102 52003 Electromechanical Dynamics 2 Motor Classi cations 0 Environment explosionproof motor surroundings protected against dropping liquids and solids totally enclosed long anges splashproof motors 0 best for wet locations totally enclosed nonventilated 0 no shaftdriven fan 0 best for very wet and dusty locations totally enclosed fancooled motors 0 fan blows cooling air over the ribbed motor frame 102 52003 Electromechanical Dynamics 3 Motor Classi cations Electrical and mechanical properties highslip motor standard lockedrotor torque NEMA D 39 NEMA 39 B slip at rated load lies most motor applications between 5 and 15 locked rotor torque 13 to 07 pu highinertia loads ie for 20 hp to 200 hp mOtOI S ywheels and centrifuges 39 locked rotor current 300 I 64 pu maximum jab39aszt ilt high startingtorque motor 200 iiii 39 NEMA C m cage 1 desidl E cage 2W 9 d bl locked rotor torque 2 pu I ange 10 199 locked rotor current 64 pu maximum O nominal fullload torque copper or I 39 pump and pistontype O aluminumbari compressor appllcatlons 0 20 40 60 so gt speed 1 0252003 Electromechanical Dynamics 4 Motor Speed 10252003 The choice of motor speed is rather limited depends on the power frequency and the number of poles small variations due to slip quantum jumps between speed ranges Use of large slip values for desired speed is inefficient rotor losses are a function of slip P sPr loss rotor gearbox is often required to modify operating speed Low speed applications are often best served by using a high speed motor and a gearbox for a given output power a high speed motor compared with a lowspeed motor 0 costs less 0 is smaller sized has higher ef ciency and power factor Very highspeed applications gt 3600 rpm always require the use of a gearbox Electromechanical Dynamics Motor Characteristics under Load Most of the time a motor runs by taking the ratio for two different Close to Synchronous Speed operating conditions we get the following expression torque may vary from 0 to full T R E 2 load torque T n Sx Sn x x n this section of the torquespeed breakdown Tn Rn Ex torque curve is essentially a straight line 390 O O 0 O o o 0 39 o the slope of the line depends Cked39r t r curve corresponding to ma1nly on the rotor res1stance torque EN RN at rated power frequency the slip Corresponding toquot condition N s torque T line voltage E and m TN rotor resistance R are related by S T R TX s k 2 E condition where k is a motor construction dependent constant 1 0252003 Electromechanical Dynamics Motor Characteristics under Load Example a 3phase 208 V 6pole induction motor is connected to a 215 V supply and drives a constant torque load 0 the motor runs at 1140 rpm calculate the speed When the supply voltage is raised to 240 V Example a 3phase 460 V 8pole induction motor drives a compressor 0 just after starting the motor runs at 873 rpm with a cold rotor temperature of 230C 0 after several hours of operation the speed drops to 864 rpm calculate 0 the hot rotor resistance in terms of the cold resistance 0 the approximate hot temperature of the copper rotor bars 102 52003 Electromechanical Dynamics Starting an Induction Motor Highinertia load stains a motor by prolonging the starting period the starting currents in both the rotor and stator are high during starting overheating from PR losses becomes a problems prolonged starting of very large motors Will overload the utility transmission network Induction motors are often started on reduced voltage limits the current drawn by the motor reduces the heating rate of the rotor lengthens the starting period Heat dissipated in the rotor during starting from zero speed to rated speed is equal to the final kinetic energy stored in all the revolving components 102 52003 Electromechanical Dynamics Plugging an Induction Motor In some applications the motor and its load must come to a quick stop 10252003 such braking action can be accomplished by interchanging two stator leads the lead switching causes the revolving eld to turn in the opposite direction kinetic energy is absorbed from the mechanical load causing the speed to fall the absorbed energy is dissipated as heat in the rotor circuit the rotor also continues to receive power Pr from the stator plugging produces 12R losses that exceed the locked rotor losses 0 high rotor temperatures will result that may cause the rotor bars to overheat or even melt the heat dissipated in the rotor during plugging from initial speed to zero is three times the original kinetic energy of all revolving parts 3sz Electromechanical Dynamics 9 Plugging an Induction Motor Example a 100 kW 60 Hz 1175 rpm motor is coupled to a ywheel through a gearbox the kinetic energy of the revolving components is 300 k at rated speed the motor is plugged to a stop and allowed to run up to 1175 rpm in the reverse direction 0 calculate the energy dissipated in the rotor 102 52003 Electromechanical Dynamics 10 DC Braking An induction motor with highinertial load can also come to a quick stop by circulating dc current in the stator winding any two stator terminals can be connected to a dc source the dc current produces stationary NS poles in the stator 0 the number of stationary poles are the same at the number of rotating poles normally produced with ac currents as the rotor bars sweeps past the dc eld an ac rotor voltage is induced 0 the PR losses produced in the rotor circuit come at the expense of the kinetic energy stored in the revolving components the motor comes to a rest by dissipating as heat all the kinetic energy The benefit of dc braking is the far less heat that is produced dissipated rotor losses is equal to the kinetic energy of the revolving parts energy dissipation is independent of the dc current magnitude the braking torque is proportional to the square of the dc braking current 10252003 Electromechanical Dynamics 1 l DC Braking Example a 50 hp 1760 rpm 440 V 3phase induction motor drives a load with a total moment of inertia of 25 kg m2 0 dc resistance between two stator terminals is 032 ohms 0 rated motor current is 62 A the motor is stopped by connecting a 24 V battery across two of the motor s terminals calculate the dc current in the stator the energy dissipated in the rotor the average braking torque if the motor stopping time is 4 minutes 102 52003 Electromechanical Dynamics l2 Application of Induction Motors Homework Problems 1410 1414 1416 and 1422 102 52003 Electromechanical Dynamics Chapter 17 Synchronous Motors 32800 Electromechanical Dynamics Starting a Synchronous Motor A synchronous motor can not start by itself the motor is equipped with a squirrel case Winding so as to start as an induction motor during starting the dc eld Winding is short circuited when the motor has accelerated close to synchronous speed the dc excitation is then applied to produce the eld ux Pullin torque if the poles on the rotor at the moment the exciting current is applied happen to be facing poles of opposite polarity on the stator a strong magnetic attraction is set up between them the mutual attraction locks the rotor and stator poles together the rotor is literally yanked into step with the revolving field 32800 Electromechanical Dynamics Motor under Load axis of N pole 39 axis of S pole l of stator At noload conditions the rotor poles of rotor are directly opposite the stator poles and their axes coincide 0 As mechanical load is applied the rotor poles fall slightly behind the stator poles but continues to turn at synchronous speed axis oprolegt iFaxisofSpole of rotor m of stator greater torque is developed With increase separation angle there is a limit When the mechanical load exceeds the pullout torque the motor Will stall and come to a halt the pullout torque is a function of the dc excitation current and the ac stator current 32800 Electromechanical Dynamics 3 Motor under Load source 6 pozZ JAOUU EX E EO 39 1 XS 1 EX E x EE0 J XS EE400 If EO EO45 ifE EO E 39 0 11160 uumcchanical Dynamics Motor under Load 0 Example a 500 hp 720 rpm synchronous motor connected to a 3980 V 3phase line generates an excitation voltage E0 of 1790 V linetoneutral when the dc exciting current is 25 A the synchronous reactance is 22 ohms the torque angle between E0 and E is 300 nd the value of EX the ac line current the power factor of the motor the developed horsepower the developed torque 32800 Electromechanical Dynamics Power and Torque When a synchronous machine operates as a motor under load the converted power is given by the same equation used for the synchronous generator EE PD 0s1n5 S As far as torque is concerned it is directly proportional the the mechanical power because of the xed rotor speed 955 PD ns TD 32800 Electromechanical Dynamics Maximum Torque The power equation shows that the mechanical power increases with the torque angle its maximum value is reached when 6 is 90 the poles of the rotor are then midway between the north and south poles of the stator P 2 E E0 max 32800 Electromechanical Dynamics Power and Torque 0 Example 150 kW 460 V 1200 rpm 60 Hz motor has a synchronous reactance of 08 2 per phase the excitation voltage is fixed at 300 V per phase determine the following the power versus the torque angle curve the torque versus the torque angle curve the pull out torque of the motor 32800 Electromechanical Dynamics Excitation and Reactive Power 0 Consider a wyeconnected synchronous motor connected to a power system with xed line voltage VL the line current I produces a mmf in the stator the dc eld current produces a dc mmf in the rotor the total ux 1 is created by the combined actions of the two mmf s 0 The total ux 1 induces the voltage Ea in the stator neglecting the very small voltage drop IRa Ea VL because VL is xed the ux 1 is also xed as in a transformer the constant ux 1 may be produced either by the stator or the rotor or by both 32800 Electromechanical Dynamics Excitation and Reactive Power 0 If the rotor exciting current IX is zero all the ux has to be produced by the stator the stator circuit absorbs considerable reactive power If the rotor exciting current is increased the rotor mmf helps produce part of the ux less reactive power is drawn from the ac power system Eventually by raising the rotor exciting current gradually the rotor produces all of the required ux the stator circuit draws no reactive power unity power factor 0 If the exciting current exceeds this critical level the stator delivers reactive power to the ac power system 32 800 Electromechanical Dynamics lO Effects of Excitation P VTAO EOA6jIz XS Eosin6XS gtP 2sin6 9i VTE0ccs6IXSsin VT 5 S OEosin6 IXS cos Eosin zXSizconStam P VTIccs P VT Q VTIsin I ccs 7 ccnstant T 32800 Electromechanical Dynamics 1 1 Effects of Excitation Power Locus I m 39 Leadlng Power Factor Lagging Power Factor m V VT i r 5 jXS 1m 5 E0 I can E0 1X5 1m Constant Power Locus Unity Power Factor 1m 7 Constant J XS 1m 32800 Electromechanical Dynamics Power Factor Rating 0 Most synchronous machines are designed to operate at unity power factor may be operated at fullload with a 08 leading power factor this is equivalent to a 06 leading reactive power factor the motor can deliver a reactive power equal to 75 of the rated mechanical power 32800 Electromechanical Dynamics 1 3 VCurves 0 Consider a synchronous motor operating at rated mechanical load examine the behavior as the excitation is varied mechanical power remains constant at unity power factor the motor current is at a minimum at unity power factor the total power equals the active power as excitation increases or decreases the motor current increases the total power increases by varying the excitation a plot of total power S with respect to the excitation voltage E0 is generated for a x load the family of active power curves are shaped as the letter V 32 800 Electromechanical Dynamics l4 VCurves 32800 Electromechanical Dynamics Effects of Excitation 0 Example 3000 kW 200 rpm 6600 V synchronous motor operates at fullload at a 80 leading power factor synchronous reactance is 11 Q calculate the following the apparent power of the motor the ac line current the value and phase angle of the induced voltage E draw the phasor diagram determine the torque angle 5 32 800 Electromechanical Dynamics l6 Stopping the Synchronous Motor Synchronous motors with their loads have large inertia may take several hours to stop after power has been disconnected from the power line to stop faster electrical or mechanical braking can be applied maintain full dc excitation on rotor and short the 3phase armature windings stator windings or maintain full dc excitation on rotor and connect the armature stator windings to a bank of external resistors or apply mechanical braking with electrical braking the motor slows because the stored energy is dissipated into the resistive elements of the circuit mechanical braking is usually applied only after the motor has reached half speed or less 32 800 Electromechanical Dynamics l7 Stopping the Synchronous Motor 0 Example a 1500 kW 4600 V 600 rpm motor is stopped by using the shortcircuit method E0 2400 XS 16 Q and RA 02 2 per phase moment of inertia 275 kg m2 calculate the power dissipated in the armature at 600 rpm the power dissipated in the armature at 150 rpm the kinetic energy at 600 rpm the kinetic energy at 150 rpm the time required for the speed to fall from 600 rpm to 150 rpm 32800 Electromechanical Dynamics 1 8 Machine Comparison Induction machines have excellent properties when speeds are above 600 rpm simple construction and maintenance at lower speeds induction machines become heavy and costly with relatively low power factors and efficiencies Synchronous motors are particularly attractive for low speed drives power factor can always be adjusted to 10 with high efficiencies and reduced weight and costs can improve the power factor of a plant while carrying its rated load can be designed to deliver a higher starting torque 32800 Electromechanical Dynamics 1 9 Machine Comparison a squirrelcage induction motor and a synchronous motor both rated at 4000 hp 1800 rmin 69 kV 60 Hz 98 25 synchronous motor I 97 4 synchronous motor induction 200 I 96 i motor I a I I C I a 2923 95 39 g 50 i l 8 I 8 II 94 I T f I 100 II induction motor x 93 quot quot 50 quotM 92 910 2 0 100 12 0 i 5 5 75 5 0 20 4o 60 80 100 gt mechanical power comparison of the ef ciency 32800 gt speed comparison of the starting torque Electromechanical Dynamics 20 Synchronous Condenser A synchronous condenser synchronous capacitor is a synchronous motor running at no load only purpose is to absorb or deliver reactive power in order to stabilize the system voltage the machine acts as an enormous 3phase capacitor or inductor the reactive power is varied by changing the dc eld excitation 32 800 Electromechanical Dynamics 21 Synchronous Condenser 0 Example a synchronous condenser is rated at 160 MVar 16 kV and 1200 rpm and is connected to 16 kV line the machine has a synchronous reactance of 08 2 per phase calculate the value of E0 so that the machine absorbs 160 Mvar delivers 120 Mvar 32800 Electromechanical Dynamics 22 Synchronous Motors Homework 1714 1715 1719 and 1720 32800 Electromechanical Dynamics 23 Chapter 13 ThreePhase Induction Motors 10 1 52003 Electromechanical Dynamics Introduction Threephase induction motors are the most common and frequently encountered machines in industry simple design rugged lowprice easy maintenance wide range of power ratings fractional horsepower to 10 MW run essentially as constant speed from zero to full load speed is power source frequency dependent not easy to have variable speed control requires a variablefrequency powerelectronic drive for optimal speed control two basic design types squirrelcage woundrotor 10 l 52003 Electromechanical Dynamics 2 Principal Machine Components An induction motor has two main parts a stationary stator 0 consisting of a steel frame that supports a hollow cylindrical core 0 core constructed from stacked laminations having a number of evenly spaced slots providing the space for the stator winding a revolving rotor composed of punched laminations stacked to create a series of rotor slots providing space for the rotor winding 0 one of two types of rotor windings conventional 3phase windings made of insulated wire woundrotor gtgt similar to the winding on the stator aluminum bus bars shorted together at the ends by two aluminum rings forming a squirrelcage shaped circuit squirrelcage 10 l 52003 Electromechanical Dynamics 3 Operating Principles As with all machines the induction motor operation is based on Faraday s laW and the Lorentz force on a conductor consider a series of conductors of length I Whose extremities are short circuited by two continuous bars A and B a permanent magnet is place just above this conducting ladder structure the magnet move rapidly to the right at a speed v such that the magnetic ux cuts across the conductors 10152003 Electromechanical Dynamics Operating Principles consequently 0 a voltage E B I v is induced in each conductor as the ux cuts across 0 the induced voltage immediately drive a current 1 down the conductor underneath the magnet pole through the endbars and back through the other conductors 0 the currentcarrying conductors that lie in the magnetic eld of the permanent magnet experience a mechanical force F I X B 0 the mechanical force always acts in a direction to drag the conductor along With the magnetic eld movement With freedom to move the conducting ladder accelerates to the right 0 as the ladder structure picks up speed the conductors Will be out less rapidly by the moving magnet 0 the magnitudes of the inducted voltage E and the driven current I Will diminish 0 as a result the mechanical force Will also decrease 10 l 52003 Electromechanical Dynamics Operating Principles In an induction motor the ladder is closed upon itself to form a squirrelcase the moving permanent magnet is replace by a rotating magnetic eld the rotating eld is produced by the threephase ac current that ows in the stator windings 10152003 Electromechanical Dynamics The Rotating Field Consider a simple stator with 6 salient poles each with a coil coils on diametrically opposite sides are connected in series see previous page 0 the two coils are connected to produce mmf s that act in the same direction 0 creates three identical sets of windings labeled AN BN CN 0 mechanically spaced at 120 degrees to each other the three windings are wyeconnected with a common neutral 0 linetoneutral impedances are equal constituting a balanced load 0 the line currents are displaced in time by 120 degrees 0 assume a positive sequence rotation phases sequence ABC the magnetomotive force is inphase with the line currents 0 the mmf s are displaced in time by 120 degrees 0 the mmf s are displaced around the stator by 120 mechanical degrees 10 l 52003 Electromechanical Dynamics 7 The Rotating Field the total mmf Within the stator hollow space is the sum of the three phase mmf s the resulting mmf is a magnetic eld that varies in time and space the magnitude of the total mmf is constant the direction of the mmf revolves around the center aXis of the stator 10 13 1b C 5 I I o 60 120 180 240 300 360 degriees gtangle 6 5 lt one cycle gt 10 lt2 3 0 5 C 7 a 543 1 01 52003 Electromechanical Dynamics 8 Direction of Rotation 10152003 The positive crests of the currents in a positive ABC phase sequence follow each other in the order A a B a C a A this phase sequence produces a magnetic field that rotates clockwise for windings arranged ABC in a clockwise layout around the stator changing either the phase sequence or the winding layout will cause the magnetic field to rotate in the opposite direction phase group 1 phase A Ia 10 A Salient pole are never used instead the stator surface is smooth with slots cut into the stator to hold the windings in practice the use of a single coil per pole is subdivided into 2 3 or more coils lodged in adjacent slots constituting a phase group phase group 2 phase A Electromechanical Dynamics 9 Synchronous Speed 10152003 The number of poles controls the the synchronous speed each 3phase pole grouping covers a given mechanical angle as the number of poles increase the angular movement of the revolving ux decreases for one cycle of ac current the number of cycles for the revolving ux to make one complete mechanical rotation is proportional to the pole count The speed of rotation for the flux field depends upon both the frequency of the source and the number of poles on the stator Where p ns synchronous speed f power source frequency p number of poles the synchronous speed increases with frequency and decreases with the number of poles Electromechanical Dynamics lO Synchronous Speed Example a threephase induction motor has 20 poles the motor is connected to a 50 Hz power source calculate the synchronous speed 10 l 52003 Electromechanical Dynamics Starting Characteristics Consider the stator of an induction motor connected to a 3phase source let the rotor be locked in a stationary position 10152003 the revolving magnetic field produced by the stator cuts across the rotor bars and induces a voltage in all the conducting bars the induced voltage is ac because of the rapid succession of time varying ux from N to S to N etc transformer action the frequency of the induced voltage depends on the number of N and S poles that sweep across a conductor per second 0 at rest zero speed it is always equal to the source frequency the endrings form a threephase short circuit and the induced voltage drives a large current through all the bars typically 100 s of amps the large currents react with the stator magnetic eld creating strong forces the sum of all mechanical forces produce a torque on the rotor Electromechanical Dynamics 12 Rotor Acceleration As soon as the rotor is released the torque causes rapid acceleration in the direction of the rotating ux eld the rotor speed increases and the relative velocity of the magnetic eld with respect to the rotor progressively diminishes 0 the frequency and the magnitude of the induced voltage decreases because the rotor bars are being cut more slowly as a result the very large rotor current decreases rapidly with increased rotor speed 0 the mechanical forces and torque weaken the speed continues to increase asymptotically to the speed of the rotating ux synchronous speed 0 at synchronous speed the rotor would no longer cut any ux 0 zero induced voltages zero current zero force zero torque 10 l 52003 Electromechanical Dynamics l3 Applying Loads Consider a motor initially running at noload the produced torque is assumed equal to zero A mechanical load is connected to the rotor shaft a countertorque causes a decrease in rotor kinetic energy and a slow down of the rotor speed the relative speed between the rotor and the rotating ux at synchronous speed increases 0 the ux cuts the conducting bars at a higher and higher rate 0 frequency and magnitude of the induced voltage increases 0 the larger induced voltage drive more current in the rotor circuit 0 the current and ux react to produce greater drive torque The rotor speed and the mechanical load will reach a state of equilibrium the motor torque Will equal the load countertorque the speed Will be stable 10 l 52003 Electromechanical Dynamics l4 Slip Slip is the ratio of the speed difference between the synchronous speed and the actual rotor speed to the synchronous speed ns n Where 113 s slip ns synchronous speed n rotor speed for a rotor at standstill the slip is unity for a rotor spinning at the synchronous speed the slip is zero in most cases the motor slip ranges between 0 and 1 10 l 52003 Electromechanical Dynamics 15 Slip 0 Example a onehalf horsepower 6pole 3phase induction motor is connected to a 60 Hz power supply the fullload speed is 1140 rpm calculate the slip 10 l 52003 Electromechanical Dynamics Rotor Circuit The voltage and frequency induced in the rotor both depend upon the slip fZ S f E2 z S E00 Where f2 frequency of the voltage in the rotor f frequency of the power source s slip E2 voltage induced in the rotor at Slip 5 E00 opencircuit voltage induced in the rotor when at rest 10 l 52003 Electromechanical Dynamics l7 Rotor Circuit 0 Example a onehalf horsepower 6pole 3phase induction motor is connected to a 60 Hz power supply calculate the frequency of the rotor current under the following conditions at standstill at a shaft speed of 500 rpm rotating in the same direction as the revolving field at a shaft speed of 500 rpm rotating in the opposite direction of the revolving field at a shaft speed of 2000 rpm rotating in the same direction as the revolving field 10 l 52003 Electromechanical Dynamics 18 Threephase Induction Motors Homework Problems 1315 1316 1326 10 l 52003 Electromechanical Dynamics Chapters 14 amp15 Induction Generators 103 02003 Electromechanical Dynamics Introduction Induction motor operating as a generator consider a frictionless vehicle powered by a squirrelcage induction motor that is directly coupled to the Wheels as the vehicle climbs a hill the motor runs at slightly less than synchronous speed delivering a torque suf cient to overcome the force of gravity electric energy converts to kinetic energy then potential energy at the top of the hill or on level ground the force of gravity does not come into play and the motor runs unloaded and very close to synchronous speed as the vehicle descends a hill the motor runs slightly faster than synchronous speed and develops a counter torque that opposes the increase in speed potential energy converts to kinetic energy then electric energy 103 02003 Electromechanical Dynamics 2 Generator Operation 0 In generator operation 39 System the rotor spins above synchronous speed it develops a countertorque gasoline that opposes the overspeed 39 same effect as a brake the rotor returned the power as electrical energy instead of dissipating it as heat referred to as asynchronous generation squirrelcage kinetic energy is converted into electrical energy indUCtion mom the motor delivers active power to the electrical system the electrical system must provide reactive power to create the stator s rotating magnetic field 0 Active power delivered to the line is directly proportional to the slip higher engine speed produces greater electrical output rated output power is reached at very small slips ls l lt 3 103 02003 Electromechanical Dynamics 3 Generator Operation Reactive power sources capacitors across the motor terminals may supply the vars the motor supplies 3phase electrical loads Without an external 3phase source load the frequency generated is slightly less than corresponding to the speed of rotation the terminal voltage increases With capacitance but limit by iron saturation insuf cient capacitance causes the voltage not to build up capacitors must supply at least the vars normally absorbed When the machine operates as a motor 103 02003 Electromechanical Dynamics 4 Generator Operation 0 Example a 40 hp 1760 rpm 440 V 3phase squirrelcage induction motor is used as an asynchronous generator the rated current is 41 A and the fullload power factor is 84 calculate the capacitance required per phase if the capacitors are connected in delta the driving engine s shaft speed to generate a frequency of 60 Hz 103 02003 Electromechanical Dynamics Generator Equivalent Circuit o Iquot 39139 When an 1nductlon motor 1s A v 1 15 16 2 m dr1ven above synchronous I e e f A 1 speed the shp becomes negat1ve w n n 254 v 39110 900 MS S S ngtnsgtslt0 I 480 nS 39 H the value of RZs also becomes I A negative 3 4 3 R B 122 2 s lt O gt B lt O s the negative resistance indicates that power is owing from the rotor to the stator 1 03 0 2003 Electromechanical Dynamics 6 Complete Torquespeed Characteristics An induction machine can function as a motor a brake and an asynchronous motor all three operating modes can be seen from the torquespeed curve Pjs Pf Perv U 3 1m i MOTO R gzgz BRAKE gt a A Speed stator rotor nS P u ns 2 n 111 If st EGENERATOR S 7 2 Pe 1Dr Pm P8PrPm stator rotor stator rotor 103 02003 Electromechanical Dynamics Induction Motor Equivalent Circuit Homework Problems 1415 1426 103 02003 Electromechanical Dynamics Chapter 5 The DC Motor 91 52003 Electromechanical Dynamics Introduction 0 DC motors transform electrical energy into mechanical energy dc motors are found in many special industrial environments 0 Motors drive many types of loads from fans and pumps to presses and conveyors many loads have a de nite torquespeed characteristic other loads have a highly variable torquespeed characteristics motors must be adapted to the type of loads to be driven Motor types shunt series and compound connections 91 52003 Electromechanical Dynamics 2 Motor Operation DC motors are built the same way as generators Armature of a motor connected to a dc power supply Current ows through the armature winding Armature is within a magnetic eld A force is exerted on the windings F B Z I L The force causes a torque on the shaft 15 E The shaft rotates 5 9152003 Electromechanical Dynamics 3 Counter EMF Rotating armature cuts through the magnetic eld Voltage is induced in the armature windings E B I v The voltage opposes the ow of current ZnCI E 0 60 mil Power is taken from the electrical system i 4 P 2 E01 arm arm 9152003 Electromechanical Dynamics Acceleration of the Motor The net voltage acting on the armature circuit is E S E0 0 The resulting armature current I is limited only by the armature resistance 1 2 ES E0 a Ra 0 At rest the induced voltage is zero Eomt 0V the large current produces a large torque I 0 As speed increases the counter emf increases and the voltage difference diminishes resulting in a reduced current 91 52003 Electromechanical Dynamics Acceleration of the Motor 0 Example Separately excited dc motor has a resistance of 1 ohm and generates 50V at a speed of 500 rpm If the armature is connected to a 150V supply nd the starting current the counteremf when the motor runs at 1000 rpm the armature current at 1000 rpm the counteremf when the motor runs at 1460 rpm the armature current at 1460 rpm 91 52003 Electromechanical Dynamics Machine Power and Torque Power and torque characteristics can The mechanical power be determined over various shaft Pm EOIa speeds calculate the counter emf The developed torque Zn HT 0 Pm E0151 calculate the armature power 955 szPazEs 1a E0 2 calculate the voltage drop 1R losses 60 n T Z n 1 Es EO 151 Ra Ia 955 60 separate air gap power and losses Z I I PaZEsIa EOIaIR T 91 52003 Electromechanical Dynamics Machine Power and Torque 0 Example a 225 kW motor operates at 1200 rpm at 250 V calculate the rated armature current and developed torque 91 52003 Electromechanical Dynamics Speed of Rotation 0 We know that Z 111 0 60 The voltage drop across the armature resistance is always small compared to the supply voltage even as the load varies from noload to fullload E0 is approximately equal to E S n60E0 60ES ZCI ZCI 91 52003 Electromechanical Dynamics Armature Speed Control Speed is controlled by varying Motor speed changes the armature voltage E S proportionally to the armature voltage 60ES Th n z e armature voltage 1s Z I 1 controlled by an external quot variable power supply rl 113 the eld Winding is separately 11x excited by a constant voltage R source armature rheostat Es shunt field 9152003 Electromechanical Dynamics lO Field Speed Control Speed is controlled by varying p U the eld ux 39 39 Tvs n 60E S 2 n z T vs 1 Z I o assunnng a constant armature I supply voltage E S f rated 390 field 0 If rheostat 0 1 2 p39 U39 gt speed n I l gt armature current I x Motor speed changes in inverse shunt m proportion to the ux field The ux is controlled by a series rheostat Rf in the eld circuit 9152003 Electromechanical Dynamics 1 l Shunt Motor Under Load 9152003 Consider a motor at noload The application of a large mechanical load Speed of a shunt motor stays relatively constant from noload to fullload Example the small armature current does not produce enough torque to carry the load nd the armature current counter emf and mechanical power for a shunt motor running at 1500 rpm at 51A With a 120 V source 120 ohm eld Winding and 01 ohm armature resistance T 51 A 120V R 01SZ 12 lacking torque the shaft speed decreases the counter emf diminishes causing the armature current to increase higher armature current develops a larger torque 1209 When the load and motor torques are equal speed is constant Electromechanical Dynamics Series Motor 0 A series motor is identical in construction to a shunt motor except for the eld windings the eld is connected in series With the armature and must carry the full armature current the performance is completely different in a series motor the ux per pole depenr gt I upon the armature current and the load 0 the ux is proportional to the current 0 At fullload the ux per pole is the same the ux per pole is greater as that of the shunt motor the Starting torque is When the series motor starts the considerably greater than armature current is higher than normal for a shunt motor 9152003 Electromechanical Dynamics l3 Series Motor Speed Control Under light loading the armature current and ux per 3 pole are small the weak field causes the speed to rise over speed too small of loads may cause T 2 excess runaway speeds that can T destroy the rotor centrifugal force Speed can be controlled by the current in the field winding increase speed by placing a low resistance in parallel reducing the o 1 2 3 pu eld winding current 4 speed n a armature current I decrease speed by adding a series resistance increasing the IR drop 9152003 Electromechanical Dynamics 14 Series Motor Speed Control Example a 15 hp 240 V 1780 rpm DC series motor has a fullload rated current of 54 A its operating characteristics are given by the perunit curves T nd the current and speed when T the load torque is 24 Nm the ef ciency under these conditions O 1 2 3 pu gt speed 11 a armature current I 9152003 Electromechanical Dynamics 15 Compound Motor Compound motors have both a series and shunt eld windings the shunt eld is always stronger than the series eld in a cumulative compound motor the mmf of the two elds add in a differential compound motor the series eld is connected so the mmf opposes the mmf of the shunt eld 9152003 shunt l field TIX I 39 ES lll39r is 31 series 32 0 A1 F1 1 shunt field 1X F2 A2 Electromechanical Dynamics 1 6 Cumulative Compound Motor Under noload conditions 1396 V series the ser1es eld has a low current 4 ltlt QP W and the mmf is negligible 12 A sl um raLtedL load the shunt eld 1s fully exc1ted by quotg 10 shunt g 39 3 IX and the motor behaves hke a g a merngquot series a shunt machlne E 08 comp imriwd Ncompound 0 w As load 1ncreases 0396 the armature current passing 04 through the series eld generates 02 a larger mmf 0 the shunt eld rema1ns constant 0 02 04 05 03 10 12 14 15 Torque perunit and the total mmf is greater than at noload the motor speed falls With increasing load The motor speed changes from maximum at noload to a lO3 0 minimum at full load 9152003 Electromechanical Dynamics l7 Chapter 15 Induction Motor Equivalent Circuit 10292003 Electromechanical Dynamics Introduction 0 To gain better understanding of the properties and behaviors of the induction motor it is necessary to examine the equivalent circuit develop an equivalent circuit from the basic principles 10292003 Electromechanical Dynamics Woundrotor Induction Motor 39 A 3phase WOUIldI OtOI inUCtiOIl model begins with the transformer s motor is very similar in construction equivalent circuit to a 3phase transformer circuit parameters the motor has 3 identical primary 0 Rm iron windage and friction losses windings 3 identical secondary Xm magnetizing reactance windings and perfect symmetry r1 stator winding resistance single phase analysis is possible x1 stator leakage reactance at standstill the motor acts exactly V 2 1 0t01 Winding I GSiStanCC like a conventional transformer 962 rotor leakage reactance l39 1 3 2 r2 i HA 0 01quot W 2 J Rm T R I 0 11 0 2 4 10292003 Electromechanical Dynamics 3 lt9 Woundrotor Induction Motor the magHGtiZing branCh can Qt be for a motor running at a slip S the Eglecmd rotor speed is nslS 39 19 may be 40 0f 1p due to th this will modify the values of ant gap E1 11 E2 and 12 for motors greater than 2 hp the magnetization branch can be shifted to the input terminals the frequency in the secondary Winding Will become Sf 10292003 Electromechanical Dynamics Woundrotor Induction Motor the induced rotOI voltage the effective value eff1 is equal to the effective E 2 S E1 value of 12 even though their frequencies are the rotor leakage reactance different 1x 2 S x2 the effect1ve value of E 1 1s equal to the effective value of E2 divided by the slip rotor res1stance 1s not frequency dependent R2 G R the model of the rotor circuit is E 2 S 1 R2 15x2 the phase angle between E2 and I2 is the same as the angle between E1 and I1 0 rl jxl 1 3 jsxz r2 T i L I stator l rotor J1 resistance I 2 resustance f lieu eney E if 2 4 WW frequency f frequency sf 10292003 Electromechanical Dynamics 5 Woundrotor Induction Motor 011 the Primary Side Without affecting the value of the stator I 1 Z S E1 current 1 the impact of the slip is shifted l 2 R2 1sz from the rotor voltage and reactance to the therefore rotor res1stance 12 E1 2 E1 greatly simpli es the analysis 2 S 1x2 10292003 Electromechanical Dynamics 6 Power Relationships The equivalent circuit enables the deduction of basic electromechanical power relationships By inspection active power absorbed P VTZRm 1121 1 122 RZs reactive power absorbed Q VTZXm 112x1122x2 r gt Pr RZS 10292003 apparent power absorbed S P2 Q2 power factor cos 6 P S line current 1p SVT active power supplied to the rotor Pr 122 R2 S 12R losses dissipated in the rotor 2 Pjr 11R2 S Pr developed mechanical power PmP13r131S I developed torque 955Pm 9SSB T o n n ef clency nszP S Electromechanical Dynamics Breakdown Torque amp Speed The developed torque is given The phasor diagram of this by the power delivered to the special case C rotor and the synchronous speed a 539 T 295513 nS the power delivered is a function of the resistance R2s MaX1mum torque occurs at 012 L maximum power transfer A f Ms B T maximum power transfer is fmm geOmetry3 found when RZs is equal to the AB and BC have the same length absolute value Of Z1 the angle ABC 2 180206 R2 S 21 angle CAB angle ACB 062 the voltage drops are equal h 61106 R a a V 11R2 sIIZ1 11 2cos 211Z1cos 2 T S 2 2 2 102 92003 Electromechanical Dynamics Breakdown Torque amp Speed the slip at breakdown torque is Sb 2R2 Zl the current at the breakdown torque is 1113 VTZZICOS the breakdown torque is 955VT 2 T 2 b M a arctanxr1 note that the magnitudes of both the breakdown torque and breakdown current are independent of the rotor current resistance R2 10292003 Electromechanical Dynamics Breakdown Torque Calculation 0 Example determine the maximum torque and corresponding speed of a 440 V 1800 rpm 60 Hz 5hp squirrel cage induction motor r1 15 ohms r2 12 ohmsjx 6 OthXm 110 ohms Rm 900 ohms 10292003 Electromechanical Dynamics 10 Torquespeed Curve 0 The complete torquespeed curve of a motor can be determined by selecting various values of slip and solving the circuit equations 0 Example determine the torque and speed of a 440 V 1800 rpm 60 Hz 5hp squirrel cage induction motor r1 15 ohms r2 12 ohmsjx 6 ohmsjXIn 110 ohms Rm 900 ohms 10292003 Electromechanical Dynamics Torquespeed Curve breakdown torque 223 Nm per phase N A I I l cb p 5hp440V3ph 20 1830 rmiln 601llz TORQUESPEED 16 CHARACTERISTIC Arotor torque quotan L 949LN mJ J TORQUE PER PHASE Nm m 3 rated torque 685 Nm 1753 amp per phase I I I I I 1 I l I I l c LI1J I l l rmin I Il 0 0 200 400 600 800 10001200 gtSPEED 10292003 Electromechanical Dynamics 1400 1600 1800 rmin 12 Test to Determine the Equivalent Circuit Noload and lockedrotor tests provide approximate values of r1 r2 Xm Rm and x Noload test small slip leads to high value of R2s with 11 being negligible compared to 10 noload circuit essentially consist of Xm and Rm measure the stator resistance RLL and assume a wye connection 7i RLL 2 run the motor at noload and at rated line voltage measuring the noload current and the total threephase active power SNL JgVNL NL QNL VSJZVL PJ L PfPVPNL 3ILr1 RmZEJZVLPf l39Pv szEJELQNL 10292003 Electromechanical Dynamics 1 3 Test to Determine the Equivalent Circuit Lockedrotor test under lockedrotor conditions stator current is almost 6 pu slip is unity and R2s equals r21 r2 magnetization branch can be neglected lockedrotor circuit consist of x r1 and r2 apply a reduced 3phase voltage such that the stator current is about equal to its rated value measure the line voltage line 10292003 current and 3phase power SLR JgVLR LR QLR V SLZR PL2R Xm QLR3IL2R r2 PLR3IL2R r1 Electromechanical Dynamics Test to Determine the Equivalent Circuit Example a noload test conducted on a 30 hp 835 rpm 440 V 3phase 60 Hz squirrelcage induction motor yields the following results noload linetoline voltage 440 V noload current 14 A noload power 1470 W resistance measured between two terminals 05 Q the results of lockedrotor test conducted at reduced voltage lockedrotor linetoline voltage 163 V lockedrotor power 7200 W lockedrotor current 60 A determine the equivalent circuit of the motor 10292003 Electromechanical Dynamics 1 5 Induction Motor Equivalent Circuit Homework Problems 152 154 155 157 10292003 Electromechanical Dynamics Chapter 13 ThreePhase Induction Motors 102 52003 Electromechanical Dynamics Squirrelcage Motors Typical performance over the power range 1 kW to 20 MW on a per unit basis fullload conditions IS 1 pu torque 1 pu 1 kW slip 0030 n 07 09 pf 080 085 20 MW slip 0004 n 096 098 pf 087 090 noload conditions torque 0 pu ef ciency 0 1 kW IS 05 pu slip z 0 pf 02 20 MW IS 03 pu slip z 0 pf 005 lockedrotor conditions slip 1 ef ciency 0 lkW IS56pu torquel53pu pf04 20MW IS46pu torque05lpu pf0l 102 52003 Electromechanical Dynamics 2 Squirrelcage Motors 10252003 Motor operating at noload the noload current is similar to the exciting current in a transformer composed of a magnetizing current that creates the revolving ux Pm small active power component that supplies the iron windage and frictional losses considerable reactive power is needed to create the revolving ux 0 a larger airgap requires greater amounts of reactive power Motor operating under load active power delivered from the source increases proportionally to the mechanical load 0 ef ciency at fullload is particularly high the stator and rotor currents produce mmf s that are similar to those found in the primary and secondary windings of a transformer leakage uxes are created in the stator and rotor circuits Electromechanical Dynamics 3 Squirrelcage Motors 0 Motor with a locked rotor power factor is low because the the lockedrotor current is 5 to leakage uxes are larger 6 times the fullload current tth Stator and rotor 12R losses are 25 to 36 Wlndmgs are net tlghtly times higher than normal coupled due to the airgap 0 the rotor must never remained locked for more than a few seconds a strong mechanical torque is developed 0 even though no power is delivered 102 52003 Electromechanical Dynamics 4 Estimating Motor Currents The fullload current Example approximate value based on estimate the fullload current empirical models locked rotor current and noload 6001 current of a 500 hp 1 E 2300 V 3phase induction motor Where estimate the apparent power 0 I fullload current drawn under lockedrotor 0 Ph output power in HP conditions E mth line voltage state the nominal rating of the 39 t 39 kW The startrng current m0 or 1H 5 to 6 times the fullload current The noload current 03 to 05 times the fullload current 102 52003 Electromechanical Dynamics AdWerM FbWS Rotor power the power owing across the air gap is the product of the rotating ux speed and the magnetic torque nS Tmagn r 9 55 Mechanical power the developed power is equal to the power transmitted to the rotor less the rotor copper losses and the windage and friction losses nTm 955 szPr P loss 10252003 Rotor copper losses for constant power and speed the magnetic torque and mechanical load torque must be equal 72 n nnsl SgtPm Prl S PIOSS Motor torque the torque developed at any speed is 955 Pm 955 Pr Tm magn n n S Electromechanical Dynamics 6 Active Power Flows 0 Example a 3phase induction motor draws 80 kW synchronous speed is 1200 rpm stator copper and iron losses are 5 kW windage and friction losses are 2 kW actual speed is 1152 rpm calculate the active rotor power the rotor copper losses the developed mechanical power the delivered mechanical power the motor s efficiency 102 52003 Electromechanical Dynamics Active Power Flows 0 Example a 3phase 60 Hz 8pole squirrelcage induction motor stator copper and iron losses are 5 kW and 1 kW respectively the motor absorbs 40 kW calculate the torque developed by the motor what is the relationship between the developed torque and the shaft speed 102 52003 Electromechanical Dynamics Active Power Flows 0 Example a 3phase 100 hp 600 V induction motor synchronous speed of 1800 rpm stator iron loss is 2 kW windage and friction losses are 12 kW stator copper resistance between two terminals is 034 ohms twowattmeter reading equal to 70 kW and line current of 78 A at a rotor speed of 1763 rpm calculate power supplied to the motor rotor copper losses mechanical power delivered to the load efficiency developed torque 102 52003 Electromechanical Dynamics 9 Torquespeed Characteristic The developed torque depends upon the speed the relationship is more complex than a single equation characteristics are provided by graphs With torquespeed curves 39 shown on a perunit base With the fullload torque T as the base torque im ortant oints p 0 p breakdown torque startmg 25 T 39 pullup 2 T breakdown lockedrotor torque g 15 wk 239 T nominal torque T I l 05 T f l I l 0 O 20 4O 60 80 100 rotational speed T I T B D 3 3 10252003 Electromechanical Dynamics 10 Torquespeed Characteristic 10252003 Starting torque torque at zero speed typically 15 times the fullload torque Pullup torque the minimum torque developed by the motor while accelerating from zero speed greater than fullload torque less than starting torque Breakdown torque the maximum torque that the motor can develop typically 25 times the fullload torque Normal operation at fullload the motor runs at 11 rpm rotor speed decreases slightly from synchronous speed with increasing load torque motor will stall when the load torque exceeds the breakdown torque for small motors lt15 kW the breakdown torque is about 80 of the synchronous speed for large motors gtlOOOkW the breakdown torque is about 98 of synchronous speed Electromechanical Dynamics 1 1 Rotor Resistance The torquespeed characteristics are greatly affected by the rotor resistance the rotor resistance of a squirrelcage rotor is essentially constant the rotor resistance of a wound rotor can be modi ed by adding an external threephase resistance load Is s i IR 3phase stator 1Q source mo brushes 39 h rotor K collector ring 6 I l 3 starting rheostat and 0 speed controller 10252003 7 Electromechanical Dynamics 12 Rotor Resistance 0 A high rotor resistance produces a high starting torque and a relatively small starting current desirable for motor starting problems appears as a rapid falloff in speed with increasing load a high slip at rated torque and large copper losses 0 Under running conditions it is preferable to have a low rotor resistance speed decreases much less with increasing load slip is small at rated torque ef ciency is high 102 52003 Electromechanical Dynamics Threephase Induction Motors Homework Problems 1317 1320 1323 and 1324 102 52003 Electromechanical Dynamics Chapter 1 Units of Measurement system of units System International SI English conversion of units scale conversions system of units conversions the perunit system of measurements ll H ll lli i i r 739 yi39 SI common units some used in this course are shown in this table three general areas we will be using electrical mechanical thermal working in SI units are often easier than in English units Quantity SI unit Symbol Mechanical Angle radian rad Energy joule J Force newton N Mass kilogram kg Power watt W Angular velocity radians per rads second Thermal Heat joule J Power watt W Specific Heat joule per kg K Jkg K Temperature kelvin K Thermal conductivity watt per meter K Wm K Electrical and Magnetic Energy joule J Frequency hertz Hz Inductance henry H Power watt W Magnetic field ampere per meter Am Magnetic flux weber Wb Magnetic flux density tesla T Magnetomotive force ampere A r 7 7ii ii i H s i ii w Table of conversion coefficients pages 866 868 in textbook Scale conversions SI prefixes T tera 1012 G giga 109 M mega 106 k kilo 103 m milli 10393 p micro 10396 n nano 10quot3 p pico 103912 39 v quot 9 a r 7 C V uL39 n n inf iii H urn 7 System of units conversions check for consistency stay within a given quantity eg impedance or power avoid crossing quantities always show conversion steps in your work example convert 850 Btu into kilowatthours 850Btu 1390551 h W39S025kwh Btu 36003 J 850 Btu 025 kWh hill 1itaslitmusnit A measurement system for giving a better idea of the size of something similar to the numbers used in baseball statistics comparable to how we use the gas gage on vehicles A per unit measurement always involves a base value this conversion unit is called the per unit base the converted value is expressed in terms of per unit example convert 115 kW to a per unit value on a 1 MVA base 115kW 0115per unit lMVA Chapter 5 DC Motors 91 82003 Electromechanical Dynamics Reversing the Rotation Direction The direction of rotation can be reversed by reversing the current ow in either the armature connection the shunt amp series eld windings shunt eld base series field comm 9182003 shunt field reversed reversed series series eld field Electromechanical Dynamics Motor Starting Full voltage applied to a starting motor can burn out the armature damage the commutator and brushes due to heavy sparking overload the supply feeder snapping off the shaft due to mechanical shock damage the mechanical load 0 Means must be provided to limit the starting current to reasonable values between 15 amp 2 pu of fullload current connect a rheostat in series with the armature as speed increases the counter emf increases the resistance can be reduced as the counter emf increases use power electronics to drive the armature current 91 82003 Electromechanical Dynamics Motor Starting Manual faceplate starter for a shunt motor contacts connect to currentlimiting resistors 9182003 aquot contact contact arm in off position m 1 contact arm 1 1 145 F manually move arm to 0 position n to start supply voltage causes U M armature is limited by four resistors full led current ow as speed increases EO builds When acceleration ceases arm is move to the next contact Where the motor begins to accelerate at last contact electromagnet holds arm in place Electromechanical Dynamics 4 Stopping the Motor Stopping a dc motor is a nontrivial operation large motors coupled to a heavy inertia load may take an hour or more to halt braking action is often required apply a braking torque to ensure rapid stop mechanical friction electrical braking reverse power ow dynamic braking transfer the armature circuit to a load resistor Plugging reversing the flow of armature current 91 82003 Electromechanical Dynamics 5 Dynamic Braking 39 The armature of a shunt motor is connected to a DPDT switch that connects the armature to either the line or external resistor R in normal operation the armature is connected to the source opening the switch the armature current a drops to zero and the rotor will spin until friction and windage losses brake the rotation the machine operates as a generator with noload closing the switch onto the resistor the induced voltage causes a reverse current to ow in R creating a counter torque the value of R is selected for twice the rated motor current braking at twice the drive torque 9182003 Electromechanical Dynamics Dynamic Braking The braking torque is 100 proportional to the braking T coasting res1stor s current Id 75 as the motor slows down dynamic braking EO decreases as well as a 50 P O consequently the braking m torque becomes smaller 2 r the torque goes to zero as 5 pluggihg the rotor halts i x o the speed drops qulckly at 0 To 27 0 3To seconds rst and then more slowly Time dynamic braking is an exponential decay 91 82003 Electromechanical Dynamics Plugging 9182003 The motor can be stopped more rapidly by plugging Plugging is the sudden reversing of the armature current l S S 11 accomplished by reversing the terminals to the armature circuit l f l under normal motoring conditions E 5 E0 Ra sudden reversing the terminals causes the net voltage acting on the armature circuit to become Es E0 resulting in a large reverse current 50X I a a limiting resistor in series is used to control the current to tWice fullload current Electromechanical Dynamics Plugging The braking torque is proportional to the armature current I a 9182003 initially the torque is twice the fullload torque and is limited by the current limiting resistor a reverse torque is developed even When the armature comes to a stop the reverse torque at zero speed is half of the initial braking torque 100 Coasting 75 dynamic braking 50 t m 25 n pluggmg l o x 0 T o 2TO 3To seconds gt Time as soon as the motor stops in two time constants the armature circuit must be opened Electromechanical Dynamics

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.