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Electric Machinery

by: Isaac Hauck

Electric Machinery EEL 4205

Isaac Hauck
University of Central Florida
GPA 3.7


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Class Notes
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This 18 page Class Notes was uploaded by Isaac Hauck on Thursday October 22, 2015. The Class Notes belongs to EEL 4205 at University of Central Florida taught by Staff in Fall. Since its upload, it has received 16 views. For similar materials see /class/227665/eel-4205-university-of-central-florida in Electrical Engineering at University of Central Florida.

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Date Created: 10/22/15
g Transformer Voltage Regulation 1 Ideal Nonideal g Transformer Voltage Regulation 2 F111 u oad EVS R ISjX IS 71 eqs eqs V V V n V Voltage regulation VR 2 M X 100 P S X 100 S S 6 Transformer Voltage Regulation 3 UCF V 7P VS ReqsIS JXeqsIS V n V Discussions VR X 100 S 1 Load is lagging inductive V1 lt 29 VA prwis I R 15 VPngtVS 3VRgtO H V PVs ReqSIS jX IS 1 eqs 2 Transformer Voltage Regulation 4 2 Load is resistive unit power factor VPngtVS 3VRgtO 3 Load is leading cal acitive V VP n maybe ltVs 3VRmaybegt 0 6 Transformer VR Calculation 1 1L VRMXIOO S Given 1 Vs xed 2 rated apparent power VA of the load 3 power factor of the load Find Is and thenVP Is VSIS 5 6V 613 Zgload rated VS ReqsIS JXeqsIS n a Transformer VR Calculation 2 2L VRMXIOO S further approximate calculation EZVS R IS jX IS n eqs eqs Ragga cos 6 LXWSIA sin 0 9 1 J V PzVSR I Z WIS cos6XeqsIS s1n6 6 Example 3 U CF For a 15 kVA 2300230 V 60 Hz transformer the equivalent circuit parameters are Reg SQXeq 6QRC 100 kQXM 10kQ Find voltage regulation at lll load when 1 load is 08 pflagging 2 load is 10 pf 3 load is 08 pf leading Process is in trans3m 6 W Per Unit System 1 I Define Base WWWW w I Normalization Actual quantity QuantIty 1n per un1t Base value of quant1ty 6 W Per Unit System 2 Analysis Procedure 1 select VA base and voltage base VA b 2 current base 2 A voltage base 39 lt b 3 res1stance base W current base 4 normalize all the quantities or get quantities per unit 5 perform a standard circuit analysis with all quantities in per unit 6 actual values of results 2 base value gtlt solved quantity in per unit UCF Ex mrl 4 1 A simple power system is shown in Lhe Figure This system contains a 480 V generator connected to an ideal 110 step up transformer a transmission line an ideal 201 step down transformer and a load The impedance of the transmission line is 20 j60 XL and the impedance of the load is 0 30 IL The base values for this system are chosen to be 480 V and 10 kVA at the generator 1 Find the base voltage current impedance and apparent power at every point in the power system 1 Convert this system to its per unit equivalent circuit C Find the power supplied to the load in this system J Find the power lost in the transmission line GI C 48010quot v W Region 1 Region 2 Region 3 lune 20 o 13960 2 1m Ziaad 10 4 30 l 201 zlme Vt UCF Ex ml 4 2 Ztot pu Ipu I Iiimi DU Ilaad pu 39 l 39 A Ivgv m 39 II v IG pu I Z line pu VG pu IG pu Iljne pu Ilaad pu Ipu Process is in trans4m 6 Video on Linear Machine U CF Linear Motor httpwwwyo utubecomwatchva4KUwFP42yo httpwwwyoutubecomwatchvggIgVNZSszampfeaturerelated German Maglev httpwwwyoutubecomwatchvhjdxC11rdsampfeaturerelated Linear Motion vs Ball Screw httpwwwyo utubecomwatchvaEdvuc2Qxyw httpwwwyoutubecomwatchvOuP1xT2BV8ampfeaturerelated Bose Active Suspension httpwwwyoutubecomwatchveSi6JQK1Iw 6 Timeharmonic Signal U CF V0 JEV 005wt 6V Vand I are rms values it J Icosat6i l t Va 2 ReJ Vej9vefw W i Z Phasor Domain I7 Vejev or I7 rms it RabEE BW I I 161a or rms 6 Instantaneous Power lm Definition pl vtit v t Timeharmonic Case Z vt ReNEVejevejm W REE5156 pt EVejeveja JETe jeve jm bElejevej le mi 6 U CF Average Power p0 UIejwv 91 VIe j9V 9i Vlejgv6iezjm VIej6V6ie2j Average Power p 1 190 2 1V1eflt9v 9igt VIe WV W T 2 2 VI cos6v 6 Note eizjwtdt O 6 Power Factor U CF P 0036V Unit Watt volt amp Power Factor pf 0056V I Always positive for passive load I If 0 lt 6V 61 lt 7 2 currentlagging If 7r2 lt 6V 61 lt O currentleading I From V 21eWV 6quot Z e19 aza Q V 2 Lagging inductive load Leading capacitive load 6 Reactive and Apparent Power U CF Average Power P 2 VI COS6V Reactive or Quadrature Power Q V sinwv Unit Var volt amp Apparent Power S Unit VA volt amp Power angle 6 6V Q A P 6 Complex Power U CF Complex Power E I7 139 VejevIej6quot VIej6V 6quot VICOS6V 6i jVIsin6V 61 P jQ Q A P


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