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Welding Machine commonly rated in amperes. And most of welding machine have 50V secondary winding or use a volt meter to measure it exactly. To calculate the kVA=(ampere rating x secondary voltage) / 1000.
Jerry said:
hi i am looking fr data regarding TSS and BMS any documents or discriptions are welcome
TSS turbine supervisory system
BMS burner management system
There is no any tangible proof that
state either of them fits all safety and protection requirements
under all circumstances. The selection totally depend on the type of
application. For eg. In most MV overhead applications where the MV/LV
X-formers are installed in the poles it feasible to install fuses
where can ease the replacement and can protect better in high
transient currents. If the applications types are tolerant to high
transient voltages and currents a circuit breaker shall be a
reasonable choice.Regards
jj
We implemented sensegrow energy management system last year. We have been monitoring 390 energy meters over the web. The meters send data to sense grow energy management platform. You can then log in on the http://www.ioeye.com to view energy or managment reports. The best part of the system is that you can access it from any where. It supports almost any kind of energy meters. Has support for carbon emission calculations. The only thing i found missing was manging other fuels on their platform. But they are working on that and would be released soon.
my mail is
Vidhyut's question is as I undertand that Capacitor when used in power factor improving applications,it supplies reactive power and so
compensates for gneration of reactive power by Inductors.
In Q=VISin(phi)–>when phi=-90 ie from if you can remember vector diagram of I & Ic –>it is (-Q )–>so again negative power.so capacitor is supplying reactive power,not consuming.
I dont think reactive power is an imaginary net power zero thing,it cause secondary heat losses for eg while considering a current
pasing through a cable whose reactance is high.
Hi, I suppose you're measuring power loss with a balanced 3-phase load on the secondary side of the transformer.
How much is the transformer size and the load power? and the difference between W1 W2 and W3? Can you give to us any numbers?
So we can evaluate, as correctly said by Smithy, if the difference is “on tollerance” or not.
As reactive power consumed is given as Q=VIsin(angle betn them)
In pure capacitor in which no resistance,angle between v and i is 90(clockwise angle +).
sin 90=1.That is reacive power is positive.and in power tringle we take it as suplying reactive power.
ABRAAHM said:
it is interesting to know that Capacitor consumes and inductor supplies.
I have understood negetive power iR-jIXc is due to fact that it is consuming power,but
phase angle of voltage is lagging behind current by 90degree.ie(sqrt(r2+x2) and phase
angle is -90 degree.
So it is just about phase angle,not touching amplitude.
laalini said
i think reactive power is negligible because of its complex nature .and it's all about a tendency but physically it may not have any applications.
The earlier response gives you a good idea on how to correctly size the cable to work reliably and comply with standards. However there was another element to your question ant that was energy efficiency and payback which was not really addressed.
This is a question of whether to oversize cables or not. Any cable has an loss due to its resistance. The correct selection of cable will result is a given loss. If you then select a cable of larger cross section, the resistance will be lower and thus the loss lower. To answer your question, you need to calculate the resitance of the minimum cable which will do the job. From this and the load current you can calculate the loss in the cable. From the operating cycle of the load over time you can calculate the kWHr consumption and from the $/kWhr cost you pay for electricity you can calculate how much it will cost over a period of time.
Do exactly the same for the cable of the next size up or even two sizes up.
You can then compare the two and you will know how much the bigger cables save you in energy costs and you can compare this with the capital cost of the larger cable to see if it is a value for money proposition. You have all the info that you need to run various payback scenarios on which to make a decision.
When you upsize cables make sure that the equipment you are feeding can accommodate the larger cables as equipment these days is often very tightly sized. Check what the fault current will now be because the larger cable will give less resistance and thus higher fault current. The cable wiill most probably handle it but will all other equipment in the chain.
Another aspect as to whether to choose aluminium or copper cable. You will need larger cross section aluminium cables than copper to achieve the same losses but often aluminium, even in the larger size is cheaper. Remember though here it is not just the cable cost to consider. The terminations are also more expensive as you often need bimetalic lugs, you need to make sure that the equipment you have is able to accommodayte the larger cross section on its teminations and the aluminium cables will occupy more area on your ladders or in conduits. Some atmospheres are also injurious to aluminium but not copper.
Hope that this satifactorily addresses your question.
There are always maunufacturing tolerances which will account for small differences between phases. If the differences are large then it is time to ask questions as there may be a problem with one winding.
To try to understand what is happening, I suggest that you stop thinking of reactive power as being consumed or supplied. Nett power is neither consumed or supplied by a capacitor or inductor. It is supplied in one half cycle of the AC wave and consumed in the other meaning that no net power is consumed. This is the reactive nature of capacitors or reactors as distinct to the passive resistor which consumes power on both the positive and negative half waves.
Now go back to your basic DC theory. When a step voltage is applied to a reactor it opposes the build up of current through it so the current builds up slowly to steady state governed by the RL time constant. When placed into an AC circuit it causes the current to lag the voltage. In contrast the capacitor resists change in voltage across it. You cannot instantaneously change the voltage across a capacitor without infinite current flow. When a voltage is applied to a series RC circuit the voltage instantly appears over the resistor and current =V/R flows. This charges the capacitor and the voltage across it builds up slowly. As the voltage on the capacitor rises, that across the resistor falls and as does the current. The rise in voltage (fall in current) is controlled bt the RC time constant. Thus when put in an AC circuit the current leads the voltage.
You can simply look at this in that the current slowly rises through an inductor when voltage is applied and current slowly falls through a capacitor. They perform in exactly the opposite way to oneanother. Thus application of either can be used to counteract the effect of the other but neither consume of supply real power. Saying that inductors consume reactive power and capacitors deliver it is only another way of saying the they do the opposite to each other.
Pls, How can I become a member?
Moreso, I wish to have access to responses to questions from other people. I tried to read some today, but I couldnt.
Thank you.
inrush current is refers to maximum, instantaneous input
current drawn by electrical device when first turned on.
When a transformer is first energized a transient current much
larger than the rated transformer current can flow for several
cycles.this ts caused because the transformer will always have some
residual flux and when transformer is energized the incoming flux
will add to the already existing flux which will cause the
transformer to move in to saturation.
Due to this the starting current is double the rated
value. this us not a fault, our protecting system is
designed such a way that it can with stand with this inrush
current without any tripping.it is interesting to know that Capacitor consumes and inductor supplies.
I have understood negetive power iR-jIXc is due to fact that it is consuming power,but
phase angle of voltage is lagging behind current by 90degree.ie(sqrt(r2+x2) and phase
angle is -90 degree.
So it is just about phase angle,not touching amplitude.
