Calculating Elect. Consumption for 3 Phase Rectifier/Transformer

The figure is the power going into the rectifier minus the power going out of the rectifier. Unless I'm very much mistaken, that information is missing from the figures you have provided.

This is the AC power going into the rectifier. On the output of rectifier I am getting dc volts but that is another discussion.

You still can't do it.

Or you can think of this as an electric motor instead of a rectifier. I just need to know how to calculate the KW being consumed.

Thanks.

You still can't calculate the watts being dissipated by the rectifier because you don't have enough data to measure what's coming out. You only know what's going in.

Please advise how many watts the rectifier is using

He just did. Please learn abouit electricity before posting.

Thank you Spradhan. You understood my question correctly. I was only looking for KW consumption of my rectifier. Your answer seems correct to what I was thinking. For others "I am not interested in the efficiency of the rectifier so its got nothing to do with the power coming out in DC. Just plain and simple I was looking for the consumption of the rectifier." Thanks again Spradhan.

Then you should have asked about the power consumption of the electroplating system...

As others have stated above, you can't with the information you have provided. What's missing is the power consumed on the load (plating) side of the rectifier/transformer combination. It seems that you want to make a distinction between AC power and DC power, don't, because there isn't any; power is the instantaneous amount of work being done.

Simply take the DC Amps X DC Volts= Watts into the plating solution and subtract it from the Watts going into the transformer/rectifier combination. This will be the Watts lost in the conversion from AC to DC.

If you want to know the loss in only the transformer and rectifier separately you will have to measure the voltage drop across each and apply the appropriate formula, but this is definitely not something that you should do at these voltage and current levels.

One more thing, is the 0.8 PF read off a meter or is it from the equipment rating? If it is from the nameplate then there may be an error in the AC calculation due to the difference between ratings and actual usage conditions.

Assuming you have 20 x 3 amps going in you should have around 60 amps DC coming out so figuring the average rectifier diode has a .7 volt drop and that two are conducting at anyone time the rectifiers estimated power consumption would be around (60 x 1.4 =) 84 watts give or take.

If I'm correct, you are looking for the current loss through rectification of your 3Ø AC current to DC current....This will vary according to rectifier type and operating conditions....If you just need an approximate number you could quantify the loss at between 10% and 20%....If you need an accurate figure than you must use the method described above, by subtracting output of current from input...

Thank you Spradhan. You understood my question correctly. I was only looking for KW consumption of my rectifier. Your answer seems correct to what I was thinking. For others "I am not interested in the efficiency of the rectifier so its got nothing to do with the power coming out in DC. Just plain and simple I was looking for the consumption of the rectifier." Thanks again Spradhan.

Now the story gets a little more interesting. Now that the load is calculated via

\/3 X V-Line-Line X I X PF

=1.732 X 400 X 20 X 0.8

Total watts =11084.8

The energy meter supplied by the power company is giving a reading of 5KW per hour instead of 11KW. How is that possible?

The amps going into the rectifier were calculated using an accurate clamp meter and the voltage was also checked via an accurate multimeter.

The rectifier is a full wave diode type rectifier if that helps. Appreciate your comments on this one.

Shhh--we won't tell if you don't.

The values of Current and voltage given in your original post are instantaneous values at the moment of measurement by clamp meter.Hence the calculated power is also instantaneous power.

To find the average power consumption,you have to take the energy meter readings at regular intervals (say 24 hours/30 days etc.) and determine the difference between two consecutive readings.Dividing this value by the No. of hours in the said interval will give the average power consumption.

Hope this clarifies the matter.

My guess is that you have a constant voltage rectifier. This type of rectifier will vary the current produced depending on the applied load in the plating tank while maintaining the same output voltage. So at one moment when the load is in the tank it may be using 11 Kw but when the load is removed from the plating solution the current used is reduced dramatically. The 5Kw usage from your supplier is an average usage.

Elroy

Actually Elroy the irony is that the plating load is constant. We electrowin copper from copper sulphate solution and the plating area is the same. I guess the right question is that if you have an average clamp meter reading of 20 amps per phase and the voltage is 400 VAC and according to the formula you get 11kw consumption. But can real energy consumed be less by as much as 50% than the amps that are being shown on the clamp meter?

But can real energy consumed be less by as much as 50% than the amps that are being shown on the clamp meter?

I don't think so.

There are three possibilities that I can think of and I'll place them in order of likelihood.

1 You have a faulty amp-meter.

2 The amp-meter is of poor quality and is affected by the poor power factor and harmonics produced by your rectifier. ( I have done hundreds of amp-meter checks on rectifiers and have never had this issue although our power factor is kept at .98 to 1 Try a different meter.)

Remember the rectifier SCRs may only be using part of the sign wave. Your amp-meter may not know the difference but your service providers meter might.

3 Your electrical providers meter is malfunctioning and you are getting a sweet deal. ( Not likely)

Elroy. I will check the amp meter and also verify with another energy meter and revert to you.

Thanks bro.