How do I calculate the multiplier for a high-voltage meter?
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I’m a beginner. As the title suggests, I want to know how to calculate the multiplier for high-voltage meters. For the high-voltage meter in my factory, the first reading was 411.02 kWh, and the second reading was 416.05 kWh. To find out how much electricity was consumed between these two readings, I use the formula (416.05 – 411.02) × multiplier. Now, I need to know how to calculate this multiplier. I know that the voltage is 6 kV, and the parameters of the high-voltage meter are 3×100 V, 5000 imp/kWh, with a current transformer set at 1000/5. Could anyone please tell me how to calculate this multiplier? I tried calculating it myself, but I think the result isn’t correct: 6000/100 = 60, 1000/5 = 200; so the multiplier would be 200 × 60 = 12,000
This post was last edited by myhx1668 on 2011-4-16 at 12:49. Is your current transformer incorrect? Are you sure it’s 1000/5? It should be 100/5, right? I’ve never seen a current transformer of this size for 6KV……
Your algorithm is correct: the multiplier of the voltage transformer times the multiplier of the current transformer. You said it was incorrect; I agree with what was said in the second comment – you must have entered the wrong value for the current transformer. For our units operating at 10 KV, the multiplier is 4000; for 6 KV, it’s…
This calculation is correct, so don’t worry.
I agree with what was said on the 2nd floor; currently, the rated current of switches is mostly 630A. It’s unlikely that the person who installed the system would use a current transformer with a rating of 1000/5
Reply to 3# amson: What does the parameter 3X100V on the high-voltage meter mean? Is 100V the value of the voltage transformer? If so, is it line voltage or phase voltage? 6kV should be line voltage; phase voltage is unlikely to occur at high voltages
This post was last edited and replied to by amson on 2011-4-18 at 18:35. Reply to 6# cqtb2012: This multiplier is the product of the transformation ratio of the current transformer (CT) and that of the voltage transformer (PT). For example, if the CT transformation ratio in a certain substation is 200/5 and the PT transformation ratio is 10000/100, then the multiplier for that substation is 40*100=4000. The multiplier multiplied by the reading on the electricity meter gives the amount of electricity used. Actually, your multiplication factor is indicated on the electricity bill – 3×100V means that the line voltage that each of the voltage circuits of the three-phase three-wire electricity meter connected through a voltage transformer can sustain over the long term is 100 volts; the 3 here represents the number of phases. Active energy meters are expressed in r(imp)/kWh ； Reactive power meters are expressed in r(imp)/kvarh. We determine the value that exceeds MD each month; you then need to estimate the electricity consumption for the following month. These values are usually around 0.x, and multiplying them by your multiplier gives you the actual amount of electricity used. It is based on this figure that a contract for electricity consumption is established with the power supply company. If the contracted amount is too low, the company will impose a fine; if it’s too high, you’ll have to pay extra money. For example, if my multiplier is 4000 and the highest value I experience is 0.4, then my contracted amount should be at least 1600 or more. The closer the value is to this figure, the more money you can save
Reply to 2# myhx1668: I went to check it today; it’s 1KA/5A； But the maximum current in the high-voltage section can’t possibly be only 100A, right?
This post was last edited and replied to by myhx1668 on 2011-4-19 at 11:49. Reply 8# cqtb2012: I’m speechless; your calculation method is correct. I wonder what the operating current is? How many kilowatt-hours were used in a few days?
Your calculation method is correct. I wonder what the capacity of your transformer is? I use a 2000KVA transformer. The rated current is 185; 1000/5 is also possible
Well, yes, the calculations should be correct actually. Thank you all. The multiplier is simply… But I have some doubts: is it okay to use a high-voltage primary line voltage of 6 kV per phase for the voltage ratio, and the current value from the current transformer of 1000/2?