Power Losses

Install heavier wiring, but only if the extra cost is less than the I²R losses that you can avoid.

If you know the wire gauge size , and therefore ohmic value, you can simply calculate I2R loss, both Live and return wires.

it means there is only line losses for the load. & wat about the lighting loss means the light having low power factor & three phase induction motor internal loss. have i to consider that as well.

You can't change those , can you?

yes can be changed. three phase induction motor taking more no load current is loss of energy. the motor can be changed.

Changed to what?

my mean induction motor internal losses by sevral times rewinding increases its no load current. increases its energy loss . its better to change the motor.

I don't understand this scenario. If the motor has been rewound to its original specifications, why would the no-load current increase? And if there is no load, why not turn the motor off? Why will things be different with a new motor?

It seems that you have done an excellent job of improving overall power factor (indeed, 0.99 might be too much), but I wonder if now you are "overthinking" things.

Re: I don't understand this scenario. If the motor has been rewound to its original specifications, why would the no-load current increase?

Typically during a rewind, little or nothing is done to repair the core laminations. (I'm not sure what you could do. ;-) So, over time, those laminations develop problems--looseness, maybe failure of insulation, etc., leading to increased no-load (and load) losses.

.99 PF seems to be over doing things.

New high efficiency motors may reduce your losses, but at what cost? One company I worked at had a policy of replacing any failed motors with HF motors. More expense was incurred when the starter protection had to be upgraded, due to the higher starting inrush current.

Your going to be spending a lot of money to save a little!

as i said the industry is in 1 km area . the distance between main lt panel & mcc panel is near about 300 mtr to 400 mtr. is it justified to install the capacitor at macc panel end to reduce the i2r losses

Moving the correct proportion of PF correction to the remote MCC panel is worth while. If it involves a new PF panel or a high capital outlay then no.

I'm sorry but it's up to you to work out the figures, we can only guess at them.

In energy efficient motors why should the starting/inrush current increase compared to the ordinary motor?. Will the no-load/FLA/locked rotor current also increase?. How about rotor inertia,will that too increase?. Does Energy efficient mean only improved efficiency/power factor?.

If you are concerned about the TINY amount of I²R losses in the cables feeding the motors, your PF correction should have been done AT THE MOTORS. If you have already corrected the ENTIRE FACILITY with bulk PF correction (a bad decision in the first place in my opinion), then adding more capacitors is not going to help and in fact will likely create destructive resonance situations as the capacitors interact with one another. So either remove the bulk PF correction you have and change to "at-load" capacitors, or forget this folly and accept that there is no way to eliminate ALL losses in an electrical distribution system.

Your apparent conclusion that .99PF is "not good enough" makes me fear that someone has sold you a bill of goods about correcting power factor for saving energy, something we rail on about ad nauseum in this forum, but few people listen. It's just a scam.

Oops! Sorry! Lost your post. Wasn't plagiarizing. S.M.

Powerfactor correction is too high.It seems that all the capacitive KVARS

have been added at receiving end and the effect is felt by bills. Suitable pf correction capacitors should be near the load i.e. Motors etc. This will certainly effect the operations.This will be possible as the load is scattered in 1 km. area. Line or cable losses will also improve.

A K Billore

.99 is TOO CLOSE to resonance situation. What jenious engineer instructed that application? S.M.

At this point, you may want to look at how all these loads are used to save energy.

Compressed air is extremely waste full (and dangerous) when used to blow dust that could be removed with a simple sweeper and a vacuum cleaner.

Leaky air lines waste a lot more than the I2R losses in the compressor wiring. So are plugged filters.

Lack of lubrication of rotating equipment can increase the power losses of the machine dramatically.

Air conditioner taking air from a hot location on the southern wall (or from the hot roof)use more power than from the shaded northern location.

Lack of building insulation in the air conditioned rooms is like trowing your money in the streets. Warm climate countries haven't discovered double pane windows and mineral wool insulation from what I gathered in the trips I had through my career.

Hi Marcot,

I am from warm climate country. is mineral wool better than air (in between t walls) for insulation ?

Thank you

Do you have a kVA or Wattmeter at the end of the line?. If you measure kW at either end of the line, you can calculate the power loss. Or else give length of line,Amp and Resistance of line, to calculate. Measurement is more accurate than calculation if you use a true RMS meter.

I may be wrong, but are you saying that your meter shows that you are operating at 99% of maximum power input, and that you are being billed at that rate? The first thing I would do is have the meter checked. Its almost impossible for a meter to continually function at such a high range. I suggest that you power down some curcuits and see if you get a reduction on the meter.

...

You might want to draw a simple line diagram so we can understand your situation better.

Thank you!

The distance to the factory from the point of metering is enough for a considerable line loss. Since increasing the size of ur wiring will incure a capital investment, the only solution left is to install another PF inprovement system at the factory end to take care Of the line loss. Calculating a loss will depend on the manufacturer's data about the cable. Usually the resistance per unit length multiplied by the expected load current will give you voltage drop along the cable. ie. Let R = Resistance of cable per unit length. I = expected load current L = total length of cable Therefore. RL = total resistance of cable from metering point to the factory Let. Vdt. = total voltage drop along the cable Vdt. = R x I VL(voltage at load end) = Vs(source voltage) - Vdt. With Vs and the load current, you can continue with ur calculation. Take care.

the p.f. correction devices should always be installed nearest to loads as the correction will result in reduced active current in distribution lines & consequently reduced line losses (watts), also thyristor controlled switchgears are more suitable for operation of motive loads, try to reduce contact resistance at various connection points/terminations/joints/jumpers in the distibution lines by proper tightenig/cleaning corrosion etc. the line losses could be measured with one 0.2 class metering at mains and at receiving/load ends.

Dear Ashutosh,

You have not referred about the voltage drop from the Sub-Station/Transformer to your metering point.

If the voltage drop is high, naturally the line losses will be more. High Power Factor will reduce the Line Current for the same power output, a vector diagram will explain this.

Thanks,

RAJESWARI.