Metering Formula
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 installed at 13.2 KV wye grounded system having a current transformer of 25:5 ratio. Determine the following; a). The internal multiplier b). The billing multiplier Solution: a). Internal Multiplier = Kh X Rr X Rs 10,000 For G.E.; Rs =50 for Type V-60, DS-50 and DS-60 Family Rs = 100 for Other family For Sangamo; Rs = 100 for J3 meters Rs = 50 for J4 and Polyphase meters For Westinghouse; Rs = 100 for all types of meter Internal Multiplier = 2.4 X 166-2/3 X 50 10,000 Internal Multiplier = 2 b). Billing Multiplier = Kh X Rr X Rs X CTR X VTR 10,000 Billing Multiplier = 2.4 X 166-2/3 X 50 X 25 X 70 10,000 5 1 Billing Multiplier = 700

2

Example #2: A single phase, 2-wire, bottom connected Harbin meter, type DD101x-6E, 15(100)A, having a 210 rev per KWh and 28 register ratio, determine the multiplier. Solution : Kh = 1000 watts/Kilowatt 210 rev/KWh Kh = 4.761904761905 Wh/rev. Or Kh = 4.762 Wh/rev. Based on Manufacturer certification issued to NEA gear ratio is 75 Multiplier = 4.762 Wh/rev. X 28 X 75 10,000 Multiplier = 1 2. Meter Watthour Constant (K h) Meter Watthour Constant (K h) - the number of watthour represented by one revolution of the disk. Kh = Voltage X Current X Number of Elements RPM X 60 Where; Voltage - rated voltage of the meter (nameplate voltage) Current - rated current of the meter (nameplate Test Amperes or TA) RPM - rated revolution per minutes of the meter based on manufacturers catalog. 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)

3

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

4

%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) Right stator, light load test @ one rev. of the meter disk resulted to 23.98 rev. of the standard. c) Series, light load test @ two rev. of the meter resulted to 24.04 rev. of the standard. d) Left stator, full load test @ ten rev. of the meter disk resulted to 24.07 rev. of the standard.

5

e) Right stator, full load test @ ten rev. of the meter disk resulted to 24.08 rev. of the standard. f) Series, full load test @ ten rev. of the meter resulted to 11.98 rev. of the standard. g). Indicate the percent error of the above test. h). Average accuracy of the above meter. Solution: Compute the Kh of the standard used For Light Load : Kh = 0.6 X 240 = 1.2 120 For Full Load : Kh = 0.6 X 240 X 50 = 12 120 5 a) Compute the Light Load rev. based on the meter and standard CR = Kh(MUT) X Rev.(MUT) X V(std)  = 28.8 X 1 = 24 Kh(std) X No. of Element X V(MUT) 1.2 X 1 %A = Computed Rev. X 100% = 24 X 100% Actual Rev. 24.05 %A = 99.79% b) Compute the Light Load rev. based on the meter and standard CR = Kh(MUT) X Rev.(MUT) = 28.8 X 1 = 24 Kh(std) X No. of Element 1.2 X 1 %A = Computed Rev. X 100% = 24 X 100% Actual Rev. 23.98 %A = 100.08% c) Compute the Light Load rev. based on the meter and standard CR = Kh(MUT) X Rev. (MUT) = 28.8 X 2 = 24 Kh(std) X No. of Element 1.2 X 2 %A = Computed Rev. X 100% = 24 X 100% Actual Rev. 24.04 %A = 99.83% d) Compute the Full Load rev. based on the meter and standard CR = Kh(MUT) X Rev. (MUT) = 28.8 X 10 = 24 Kh(std) X No. of Element 12 X 1 %A = Computed Rev. X 100% = 24 X 100% Actual Rev. 24.07 %A = 99.71%

6

e). Compute the Full Load rev. based on the meter and standard CR = Kh(MUT) X Rev. (MUT) = 28.8 X 10 = 24 Kh(std) X No. of Element 12 X 1 %A = Computed Rev. X 100% = 24 X 100% Actual Rev. 24.08 %A = 99.67% f). Compute the Full Load rev. based on the meter and standard CR = Kh(MUT) X Rev. (MUT) = 28.8 X 10 = 12 Kh(std) X No. of Element 12 X 2 %A = Computed Rev. X 100% = 12 X 100% Actual Rev. 11.98 %A = 100.17% g). Summary Computed Rev Standard Rev % Accuracy st LL, 1 Stator 24 24.05 99.79 LL, 2nd Stator 24 23.98 100.08 24 24.04 99.89 LL, Series st 24 24.07 99.71 FL, 1 Stator nd 24 24.08 99.68 FL, 2 Stator 12 11.98 100.17 FL, Series h). Average accuracy of the above meter. Ave. Accy. = 70% Full Load + 30% Light Load Ave. Accy. = 70%(100.17) + 30%(99.89) Ave. Accy. = 100.086% or Ave. Accy. = 75% Full Load + 25% Light Load Ave. Accy. = 75%(100.17) + 25%(99.89) Ave. Accy. = 100.10% 5. Demand Pointer Test Demand Pointer(KW)= Kh X Rev(std) X Meter Element 1000 X TI X Meter Multiplier Where; Kh = watthour constant of the meter. Rev(std) = desired rev. of the standard. Meter Element = number of element of the meter in series. TI = time interval of the meter in hour = TI 60 Test procedure

% Error -0.21 +0.08 -0.11 -0.29 -0.32 +0.17

1. Wired the meter (series all the stators), phantom load and standard and induced rated voltage. 2. Reset the demand pointer to zero. 3. Simultaneously, induced full load current on the meter and standard. 4. When the demand pointer register or moved, simultaneously turn-off the current on meter and the standard. 5. Get the initial reading of the demand pointer. 6. Reset the standard to zero. 7. Induced full load current, simultaneously, on the meter and the standard.

7

8. When the desired revolution of the standard has reached, simultaneously turnoff the current on the meter and the standard. 9. Get the final reading of the demand pointer. 10. Subtract the final reading to the initial reading and get the true demand pointer register. Example: A G.E. meter, three phase, class 10, form 5A, 240 V, type VM-63-A, TA 2.5, Kh 1.2, Time Interval 15 min., Rr 166-2/3. Determine the internal multiplier of the meter using the demand pointer test. Solution: b). Computation Demand Pointer(KW) = Kh X Rev(std) X Meter Element 1000 X TI X Meter Multiplier Assumed meter multiplier =1 Desired Rev. of the standard = 50 Demand Pointer(KW) = 1.2 X 50 X 2 1000 X (15/60) X 1 Demand Pointer(KW) = 0.48 c). Reading Initial reading = 0.05 KW Final Reading = 0.29 KW True reading = 0.29 - 0.05 = 0.24 KW d). Conclusion Therefore, the true meter internal multiplier is 2 and not 1. Assumed meter multiplier =2 Demand Pointer(KW) = 1.2 X 50 X 2 1000 X (15/60) X 2 Demand Pointer(KW) = 0.24 KW; which is the same KW as per test conducted. Using the meter multiplier formula M = Kh X Rr X Rs = 1.2 X 166-2/3 X 50 = 1 10,000 10,000 Which is not true since Rs is not really 50. Rs is 100.