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Unformatted text preview: ELEC 352 Measurement Laboratory Experiment First semester 10/11 Simulation Three Phase Name ? ? ? Lo Wang Fu ? ? ? Chang Weng Lam Student No. DA828447 DA827696 DepartmentFaculty of Science and Technology MajorEEE Group B Lab Date18/09/2010 Handin Date18/10/2010 A. Procedure This simulation circuit would be basically constructed by two blocksets of subcircuits, like Fig. 1 below, Fig. 1 Logic structure of whole circuit dealing with different conditions And Blockset 1 would be changed according to varied conditions, and it is consisted of a three phase source, and the voltage and current from each one of three phase sources would be input into Blockset 2 (Fig. 2), it would sample the values of voltage and current supplied by Blocksets1, as well to compute for Active Power Meter, Reactive Power Meter, Apparent Power Meter, Distortion Power Meter and Power Factor, then sum up or average them to get the five Power parameters value for the whole system. Fig. 2 Architecture inside the Blockset 2 For the three subsystem circuits in Fig. 2, they are all the same, and their internal architecture was wired as Fig. 3 below, 1 Blockset 1 Blockset 1 Blockset 2 Blockset 2 Voltage Current A B C D Subsystem Fig. 3 Internal architecture of Subsystem circuit Conditions (1) Under the balance case: threephase sinusoidal voltage (220 v, 50 Hz) and sinusoidal current with a balanced inductive load ( R = 10 ohms, L = 10m H) (i) Measure power parameters, In order to check the current flowing through the wire between two neutral ports, there is a very small resistor added between them, to let the wire likes real. Simultaneously, by checking the current flowing through and voltage across this resistance, it would be learnt whether the system are working in balance or not. So, the peak amplitude values of AC Voltage Sources in the system would be V 13 . 311 2 220 = with frequency of 50 Hz, as well as R =10, L=10 mH and Blockset 1 would be built as Fig. 4, with R n = 0.01, Fig. 4 Architecture inside Blockset 1 for condition 1 After simulation, the results could be obtained in Blockset 2 immediately, as Fig. 5 shown, 2 A B C D A B C D Total Fig. 5 Results in Condition 1 in part of Blockset 2 By the handcalculating method, then impedance Z would be = + = + = + = 441 . 17 482 . 10 10 10 50 2 10 j m j X R Z L , and V V rms 220 2 / ) 10 912 . 5 2 220 ( 14  = for three phases, Then current I is computed as A Z V I rms rms  = = = 441 . 17 989 . 20 441 . 17 482 . 10 220 , since the loads to three phase sources are the same, so the five power parameters should be equal correspondingly, the power factor would be the difference angle between voltage and current, that is 9540 . ) 441 . 17 cos( cos = + = , and the total is the same; Active power P should be W I V P rms rms 222 . 4405 ) 441 . 17 cos( 989 . 20 220 cos = + = = ....
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 Spring '12
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