Recent content by Mondy

Hi Dan Thanks for your reply It is appreciated. Yes you are correct the phase angle error is very small in most measuring CTs and should be only a few minutes. I need some help if possible with the equations used in calculating the ratio and phase angle errors in measuring CTs as all the...

Is there anyone that can help me as I really could do with some assistance? Cheers Ray

Is there anyone that can help me as I really could do with some assistance? Cheers Ray

Can anyone help me with the accurate calculation of ratio and phase errors in measuring current transformers? I would like to get a spreadsheet or something similar together for the design of toroidal bushing transformers and so far have not come up with any equations or methods that agree with...

Can anyone help me with the accurate calculation of ratio and phase errors in measuring current transformers? I would like to get a spreadsheet or something similar together for the design of toroidal bushing transformers and so far have not come up with any equations or methods that agree with...

Just wanted to say thanks to both SparksAlot and ScottyUK for your invaluable help with this information and for the time you both spent helping me out. It is much appreciated and I owe you both a drink!! Cheers Ray

Hi SparksAlot Thanks my friend. From this am I right in saying that as the temperature goes up the core restivity increases and so the eddy currents reduce? Also what does the 0.00065 refer to? Cheers Ray

Hi Sparks Alot I think I have got most of it down, just a couple of questions. 1. How did you get to CORE = 23.7688+ from the measured was 24.7W? 2. The stray loss is 2.2163 WATTS @ 22deg.C and reduces due to the rise in resistance with temperature to 2.2163/k =1.8076W at 80deg.C is this...

Thats a good point, Is there no send facility via the forum as they have our addresses? Cheers buddy Ray

Hi SparkAlot Thanks for that my friend your effort is very much appreciated. I will have a good go with this. Your spreadsheet will be much appreciated if you would be so kind! Cheers Ray

Hi SparksAlot Many thanks for your time it is very much appreciated. Can you show me the equations you used to get your results. As I see it you came to the following using: Rated current @ 230v and 960VA = 960/230= 4.173 A Ratio up test voltage at rated current = (4.173A/2.188A)*...

Thanks SparksAlot If you could help me on this I would be eternally gratefull! OK here goes... all are measured values Hv (pri) resistance @ 22deg.C = 0.4978R Lv (sec) resistance as above = 0.0116R Voltage ratio Pri: 230v Sec: 25.5V = ratio 9.109 (24v rated on load at 40A or 960VA No-Load...

Hi Scotty Yes I agree with what your saying and this is almost definately the case the location is rural and +10% over volts is a regular occurrence for them apparantly, the iron saturation and hence the peaky input current is causing the issue, but there are other factors and that is why I...

Hi Dave & Scotty Thanks for your replies. No this is not a college assignment (although it feels a bit like one!) It is part of the analysis of a Norwegian transformer that is not performing at all well in the UK (at 240v + 10%) and is part of a large underfloor heating plan. These are the test...

Hi All Can anyone help me with the calculations involved in a transformer short circuit impedance test. I have the following data: Transformer rated voltage =230v Transformer rated power = 960VA Secondary short circuited during test ambient temperature during test = 22 deg.C Measured...