MIC and Power factor: A question about units

Calahonda52

Formula corrected based on 2011 answer below

ESB say MIC is measured in kVA
A kilovolt-amp is similar to a kilowatt. The capacity level for customers is described as the Maximum Import Capacity (MIC).
https://www.esb.ie/esbnetworks/en/business-customers/mic/mic.jsp

I gather that for 3 Phase W = V*(A or I)* SQRT(3) *PF [ I had committed this last piece hence the confusion in rest of post]

The following is just one way of describing Power factor PF
Inductive loads require two kinds of current:
• Working power (kW) to perform the actual work of creating heat, light, motion, machine output, and so on .
• Reactive power (kVAR) to sustain the magnetic field
Working power consumes watts and can be read on a wattmeter . It is measured in kilowatts (kW) .
Reactive power doesn’t perform useful “work,” but circulates between the generator and the load .
It places a heavier drain on the power source, as well as on the power source’s distribution system .
Reactive power is measured in kilovolt-amperes-reactive (kVAR) .
Working power and reactive power together make up apparent power . Apparent power is measured in kilovolt-amperes (kVA) .

My question is: is the SQRT(3) reflected in the kVAR and kVA when taking about PF.
i.e. when kVAR is being measured is the SQRT(3) included?

Or looking at it another way is Apparent Power here kVA the same as in MIC?

Am assuming Volts for 3 PH: = 415V here in Ireland so if I have a MIC of 100 kVA can I calculate a Max Amps based on
3 Phase W = V*(A or I)* SQRT(3)

Thanks as always

2011

Calahonda52 wrote: »

ESB say MIC is measured in kVA
A kilovolt-amp is similar to a kilowatt.

I would not describe it that way. To make things simple let's consider a single phase circuit first.

Power is measured in Watts. It is a function of the voltage (unit volts), the current (unit amps) and the power factor (no unit < or = to 1).

VA (volt amps) is a function of voltage and current.

The capacity level for customers is described as the Maximum Import Capacity (MIC).
https://www.esb.ie/esbnetworks/en/business-customers/mic/mic.jsp

I gather that for 3 Phase W = V*(A or I)* SQRT(3)

I remember it this way:

3 Phase Power = Line voltage x Line current x Root 3 x Power Factor

The answer that you will get for the above will be in Watts. Divide by 1000 for kW.


3 Phase Volt Amps (which is the same as the apperent power) = Line volatge x Line current x Root 3

The answer that you will get for the above will be in VA. Divide by 1000 for kVA.

My question is: is the SQRT(3) reflected in the kVAR and kVA when taking about PF.
i.e. when kVAR is being measured is the SQRT(3) included?

I don't really understand this question. But if you put the correct values (line voltage, current and power factor) into the above formula you will get the correct answer. Once the power factor is not equal to one (it generally is not) the kVA will be > than the kW.

Or looking at it another way is Apparent Power here kVA the same as in MIC?

Yes, the apperent power is measured in kVA. The MIC should be based on kVA.

Bruthal

The square root of 3 is used because either the line current will divide into the loads in delta, or the line voltage divides in star. So instead of 3 x volts x amps for 3 phases, its root 3 x line volts x line amps.

Phase volts x phase amps x 3 x pf would give the kw too, because in delta the phase and line volts are the same. In star the line and phase currents are the same.

Calahonda52

Thanks guys: even more to learn!

Anyone got a layman's guide to how AC works.

I know how DC works and the flow of charged particles but clearly they don't come all the way to my light bulb in the AC supply

2011

Calahonda52 wrote: »

Thanks guys: even more to learn!

Anyone got a layman's guide to how AC works.

I know how DC works and the flow of charged particles but clearly they don't come all the way to my light bulb in the AC supply

Ask whatever AC related questions you like on this forum.

This is also pretty good :

http://www.allaboutcircuits.com/textbook/alternating-current/

brightspark

The simplest lay mans analogy for power factor is the beer one.

kVA is your full pint glass.
kW is the beer.
The foam is the reactive power.

Unlike in the pub you decide how much froth you want (there are methods of reducing it)


The reason why the MIC is charged in kVA is because you are paying for the current carrying capacities of the supply equipment and cables.
In the beer analogy the ESB needs a bigger glass to be able to serve your drink and you pay for the glasses that the ESB require to serve you drink whenever you ask. You will also be charged on every pint you buy that has a large frothy head, if it's just the regular head they don't bother charging you. (Usual pf greater than 0.9)




Supply voltage in Ireland is nominally 230V single phase, 400V three phase.

Risteard81

brightspark wrote: »

Supply voltage in Ireland is nominally 230V single phase, 400V three phase.

And is harmonised at that throughout all CENELEC countries.

Calahonda52

Thanks for all the input: the conceptual problem I have is whats happening at the electron or atomic level with AC.

From the book link above

"If a machine is constructed to rotate a magnetic field around a set of stationary wire coils with the turning of a shaft, AC voltage will be produced across the wire coils as that shaft is rotated, in accordance with Faraday’s Law of electromagnetic induction"
which is
a law stating that when the magnetic flux linking a circuit changes, an electromotive force [EMF] is induced in the circuit proportional to the rate of change of the flux linkage

Bruthal

In the machine rotating the magnetic field inside the sets of windings, first the north pole passes the windings in sequence, then the south pole. So as the north pole passes, current is generated in the windings, then the south pole passes, so the current generated is reversed. The magnitude is in the shape of a sine wave due to magnetic strength as the magnetic pole approaches and moves away from a winding.

Electrons flow through the windings to the load, and back. Then reverse. The electrons in a generator on the Irish grid will flow as far as the first transformer primary winding, and back to the generator. The secondary side of transformer then flows to next transformer primary and so on.

Alun

Bruthal wrote: »

Electrons flow through the windings to the load, and back. Then reverse. The electrons in a generator on the Irish grid will flow as far as the first transformer primary winding, and back to the generator. The secondary side of transformer then flows to next transformer primary and so on.

Kind of ... whilst electromagnetic energy travels along a conductor at very high speed, approaching the speed of light, the electrons themselves travel much, much more slowly, in the order of millimetres per hour.

In an AC situation, the electrons just shuffle back and forth over an extremely small distance indeed, in the order of micrometres.

An analogy my old physics teacher used to use was to imagine a very long (frictionless) hosepipe full of (again, frictionless!) peas. Pop an extra pea in one end of the pipe, and a pea will pop out of the other. The effect is very fast, almost instantaneous, but the peas in the hosepipe have only moved along by one 'pea length.'

Bruthal

Alun wrote: »

Kind of ... whilst electromagnetic energy travels along a conductor at very high speed, approaching the speed of light, the electrons themselves travel much, much more slowly, in the order of millimetres per hour.

In an AC situation, the electrons just shuffle back and forth over an extremely small distance indeed, in the order of micrometres.

An analogy my old physics teacher used to use was to imagine a very long (frictionless) hosepipe full of (again, frictionless!) peas. Pop an extra pea in one end of the pipe, and a pea will pop out of the other. The effect is very fast, almost instantaneous, but the peas in the hosepipe have only moved along by one 'pea length.'

Yes although the start looks like a Google quote there. Overall there is not much net movement. Or zero.

Bruthal

Alun wrote: »

Actually some knowledge dragged up from my admittedly ancient university degree in solid state electronics, but hey, what do I know?

No bother. I just meant I seen it written very similar before.

Calahonda52

Thanks for insights above.
looking at VFDs and VSD's etc and came across this for torque

Torque can be calculated in SI units as

T = Power (in watts)*9.554 / n

where

T = torque (N.m)

n = revolution per minute (rpm)


Any body know where the 9.554 comes from

wiki
https://en.wikipedia.org/wiki/Torque#Machine_torque
has

In the following formulas, P is power, τ is torque and ω is rotational speed.
P= τ (N.m) *2Pi (rad/rev)* ω (rev/sec)

Thanks

Got the answer over in engineering

9.554 is 60 / (2 x 3.14)

So the 60 comes from the conversion from revs per sec to per minute.