lab_8_Voltage_Regulation_and_Efficiency

docx

School

Georgian College *

*We aren’t endorsed by this school

Course

2005

Subject

Electrical Engineering

Date

Feb 20, 2024

Type

docx

Pages

Uploaded by CoachScienceStork25

Lab 8 ELEC 2023 – PTD1 Lab #8 - Transformer Testing Thando Ncube (200537131), Zakrallah Shaker (200525699) Georgian College ELEC - 2023 Power Transmission and Dist Instructor: John Miller Date: November 11, 2023 In conformance with Section 8 of the Georgian College Academic Policies and Procedures the author(s) confirm that this work submitted for assessment is solely their work and is expressed in their own words. Any uses made within it of the works of any other author, in any form (ideas, equations, figures, texts, tables, programs, etc.), are properly acknowledged at the point of use. A list of the references used is include pg. 1

Lab 8 ELEC 2023 – PTD1 Lab 8: Voltage Regulation & Efficiency Lab Procedure : In this lab, we will perform a series of tests on a single-phase transformer to determine whether it meets a specified acceptance criterion. Equipmen t : - LabVolt AC Power Supply - Hammond Single Phase 300VA Transformer - LabVolt Data Acquisition and Control Interface - LabVolt Resistive Load Modules - Computer with LVDAC-EMS and LVDAM-EMS software pg. 2

Lab 8 ELEC 2023 – PTD1 TRANSFORMER DATA MFR: MODEL #: H301620 SERIAL #: 44 Data Table: Test V PRI I PRI W PRI VAR PRI VA PRI V SEC I SEC W SEC VAR SEC VA SEC FL 119.7 1.528 178.8 -37.24 183.8 234.4 0.685 160.5 0.426 160.7 OC 120 0.376 15.29 -39.80 45.15 239.9 SC 4.575 2.517 11.51 -0.977 11.60 1.258 Calculations: 1. From your measurements calculate the winding resistance (R COPPER LOSS ) & reactance (X COPPER LOSS ), as well as the value for R CORE (core loss resistance) and X CORE (magnetizing core loss reactance) 2. Draw the transformer equivalent circuit showing all component values. 3. (Assume the impedance is evenly divided between primary and secondary windings.) Voc=120 Vsc =4.575v Ioc= 0.376 Isc = 2.517A Poc=15.29 Psc = 11.51W Core Loss pf = Cos0 = W/VA = 15.29/45.15 x 0.376 = 0.9006 0 = Cos-1(0.901) = 25.7 Yoc = Ioc/Voc = 0.376<25.7/120 0.00313<25.7 0.00282i + 0.00135j R = 354.6 Xm =740.74 Copper Loss pf = -11.51/(4.575 x 2.517) = -0.99 cos-1(0.99) = 8.019 Zsc =4.5757/2.517<8.109=1.82<-8.109 Convert to rectangular to get R and X L Req1 1.8Ω Xeq1 .256Ω pg. 3

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version