Energy Meter

The multiplication factor is the PT Ratio X The CT ratio. From the nameplate the

PT ratio is 11000/110 = 100 CT ratio is 500/5 = 100 100 x 100 = 10000. If the MF has been change to 100, you have a change in the PT or CT ratios. You may not have a PT installed and thus the MF = 100

Thanks for the reply. But how do we derive the multiplication factor of an energy meter which doesn't use CT and/or PT. Is this some kind of gear ratio inside the energy meter.

There is a PT installed and the system has remained untouched for the past 25 years. The PT is 11000/110V and CT 500/5A.

The multiplication factor depends on 3 variables: PT ratio, CT ratio and Kh. Kh is the number of KWh per rotation of the meter disk. Even though solid-state meters do not actually have an induction disk, the Kh value is still used. In your case, Kh = 0.25.

Therefore, the multiplication factor is:

PTR x CTR x Kh = 100 x 100 x 0.25 = 2500

In electronic meters, there is an additional factor. The display can be programmed to better suit the load. For example, the calculation above indicates that the meter register will only change when 2.5 MWh of energy has been consumed. That's not very precise when most places charge by the KWh.

It is also possible that the service has been modified in the last 25 years, and the PTR and/or CTR is different now.

To determine the actual multiplier for the display, connect a constant resistance load to the meter. Measure the voltage and current at the meter inputs. Wait until the display ticks over, then time how long it takes to tick over again. The actual usage will be:

KWh = (Volts x Amperes (x √3 for 3 phase) x PTR x CTR) x time in seconds / 3600

Divide the actual usage by the display. That value will be the actual meter multiplier.

Now, 3 M.F values 10000, 2500 and 100. Which one is correct?

I worked with electrical commercial and used lot of energy meter. With my experience,the meter dial M.F value 100 is correct.

The 100 MF of the company should be correct, since as per the calculation for the MF it is CTRatio x PTRatio, which works out to be 10000 in the present case, the manufacturer by using some gear trains for the cyclometer for the energy meter must have used the MF 100.

Guest,

All of the multipliers in my previous post must be used for the meter to provide accurate energy data. In most cases, the meter dial assumes direct connection, meaning no instrument transformers (PT's or CT's) are involved. In that case, the PTR and the CTR both equal 1.0, and can be ignored.

In cases where the voltage or current is too high for direct connection (for example, a 3300V, 2,000A industrial service, potential and current transformers must be used between the line and the meter. The transformation ratios of these transformers must be included in the calculation.

FYI, I have worked in the electric utility field for over 30 years, as an electrician, instrument technician and engineer, and have designed, installed and commissioned literally thousands of metering systems on everything from 240V, 200A residential services to 138,000V, 3,000A utility grid interconnections.

hi there

some of electical energy meters has mechanical intenal MF ( 100/5 )

the result of elec. MF must multiplide to inverce mechanical one.

in this case if there is mech MF :

1-elec MF = (11000/110)X(500/5)=10000.

2-mech internal MF = 100/5=20 inverse = 5/100.

main Mf = 10000/200=10000x5/100=500.

I thaught that

Thanks