1. Meter Multiplier Meter Multiplier - the multiplier applied to the register reading to obtain kilowatt-hours. M = Kh X Rr X Rs X CTR X VTR 10,000 Where ; Kh = watt-hour constant of the meter in watt-hour per revolution Rr = register ratio = the number of revolutions of the register worm wheel for a revolution of the first dial pointer (right hand). Rs = gear ratio = the number of revolution of the disk for one revolution of the first point CTR = current transformer ratio VTR = voltage transformer ratio Example #1: A G.E., three phase, three wire, 120 volts, type VM-63-A, Form 5A, Class 20, KWh meter indicates Kh of 2.4 at its nameplate and register ratio of 166-2/3 was ...view middle of the document...
60 - conversion from minutes to hour Example #1: A G.E., single phase, 2 wire, 240 volts, Type I-70-S, Form1S, Class 100 KWh meter indicates a Test Ampere (TA) of 30 at its nameplate. Determine the meter constant. Solution: Based on G.E. Catalog, this type of meter has a 16-2/3 rpm. Kh = 240 X 30 X 1 16-2/3 X 60 Kh = 7.2 3. Stop Watch Method a. Computed Watt (based on disk revolution) Watt = Kh X Disk Rev. X 3600 tsec Where ; Kh = watt-hour constant of the meter in watt-hour per revolution Disk Rev. = the number of revolution of the meter in tsec. tsec = time in second made by the disk revolution. b. Percent Accuracy (%A)
Percent Accuracy (%A) - the ratio of the actual registration of the meter to the true value of the quantity measured in a given time, expressed as a percentage. %A = Computed watt based on the disk rev. X 100% Measured watt load Example #1: A G.E., single phase, two wire, 240 volts, type I-70-S KWh meter indicates Kh of 7.2 at its nameplate. In 48 seconds, the meter disk makes 16 revolution having a load of 8.5 KW measured by wattmeter. Determine the accuracy of the meter. Solution: Watt = Kh X Disk Rev. X 3600 tsec Watt = 7.2 X 16 X 3600 48 Watt = 8640 watts or 8.64 KW %A = 8.64 X100% 8.5 %A = 101.65% Example #2 A Harbin, single phase, two wire, 240 volts, class 100 type DD101x-6E KWh meter indicates a 210 revolution per kilowatt-hour on its nameplate. With a known load of 940 Watts, the disk makes two revolutions for 36.5 seconds. Determine the accuracy of the meter. Solutions : Kh = 1000 watt/kilowatt Rev./KWh Kh = 1000 210 Kh = 4.76190 Wh/rev. Watt = Kh X Disk Rev. X 3600 tsec Watt = 4.76190 X 2 X 3600 36.5 Watt = 939.334
%A = Computed watt based on the disk rev. X 100% Measured watt load %A = 939.334 X 100% 940 %A = 99.93% 4. Meter Testing / Calibration a. Percent Accuracy (%A) %A = Computed Revolution X 100% Actual Revolution where; Actual Revolution - Actual revolution read on the meter standard b. Computed Revolution (CR) CR = Kh(MUT) X Rev. (MUT) X V(std) Kh(std) X No. of Element X V(MUT) where; Kh(MUT), Kh(std) - meter constant of Meter Under Test (MUT) and standard, respectively Rev.(MUT) - number of rev. of Meter Under Test (MUT) No. of Element - number of element or stator in used V(std), V(MUT) - Voltage induced to standard and Meter Under Test (MUT), respectively c. Average Accuracy = 70% Full Load + 30% Light Load or = 75% Full Load + 25% Light Load d. Percent Error = Computed Rev. - Actual Rev. X 100% Actual Rev. or Percent Error = Computed Rev. - 1 X 100% Actual Rev. or Percent Error = (%A - 1) X 100% where : %A - Percent Accuracy Example: Given : a) Meter - Three phase G.E. meter self-contained, class 200, Form 12S, Kh 28.8 Rr 166-2/3 240 volts, TA 30. b) Standard - G.E., Type IB-10, Kh 0.6 @120 volt @5 amp tap. ampere tap used 5 and 50. Determine the accuracy of the meter; a) Left stator, light load test @ one rev. of the meter disk resulted to 24.05 rev. of the standard. b)...