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3 phase input power calculation
In Texas we have recently dereulated the Texas Utilities and they have decided to start a tarrif for power factors under .95.
By Alan Haston on 3 November, 2003 - 9:07 am

In Texas we have recently dereulated the Texas Utilities and they have decided to start a tarrif for power factors under .95.

So in an effort to show the company I work for what the future costs are going to be I was calculating the input power of our motors with several different power factors. What I saw shocked my little eyeballs. As the power factor got closer to .95 the more it costs to run it. Now how in the world did this happen? Is the formula for input power wrong? Seems not to be from what I have found So far. Did I use the wrong number somewhere? Thats possible, i used .70, .85 and .95 for my power factor instead 70, 85 and 95. But either way yields the same basic result, more costs as one gets a higher Pf. Or am I just seriously missing something?

I would appreciate someone helping clear this up for me.

I am using the standard P= V*I*PF*1.732/1000 for my power equation

What you are missing is that as the PF increases, current (I) is reduced because power (P) is constant.

hi Alan,
y u r using 1000 in this formula bcoz this formula will give u power in KW. while u want in power in only watts...now the formula will be P=1.732*V*I*pf
tryagin u will find better result than before...
c u

By Meir Saggie on 4 November, 2003 - 8:09 am

Formula is OK - but computation is (probably) wrong:
The constant value is the mechanical output on the motor shaft = the machine
load (P) - that presumably does not change in your mental exercise.
Working backwards from that (ignoring internal motor losses for a while),
you'll find that the lower the power factor, the higher the current (I)
drawn by the motor (inverse proportion to PF) - when P is constant.
If your utility charges only for kWh - no noticeable difference on the bill
If your utility also charges for kVAr - reading raises with PF getting lower
and you pay extra for the extra CURRENT drawn even though the POWER consumed
is the same.
kVAr is proportional to sin(arcos (PF))
Meir

By Johan Bengtsson on 4 November, 2003 - 8:16 am

It's correct!

For a given current and voltage the power needed is higher the higher pf.
If you have a low pf the current running tru the wires are not doing any
work, thereby no power and that's the problem.

For a given power usage the current are lower the higher pf you have and
that is what the power company wan't you to do something about (using their
wires for a lot of current doing no work and thereby not generating any
income).

/Johan Bengtsson

I think what you are overlooking is that from your utility company, you do not pay for 'Power', you pay for 'Volt-Amps' (VA). What you get out of your motors is Power, and is related to the VA by the power factor. P = VA * pf. So pf is much like efficiency. As you bring the pf up to .95, you will use less VA to generate the same amount of work, and so you will save money.

By Meir Saggie on 4 November, 2003 - 2:51 pm

To be precise, if you are a large consumer you pay for active energy (kWh) and for blind energy (kVArh) separately (and for Peak Demand also, in some cases - but this is a different story). So please do not confuse the readers here if you do not know the facts. There are no kVAh meters - only kWh and kVArh!

Meir

By Michael Griffin on 4 November, 2003 - 10:19 am

A sample calculation would have helped to tell us what you were doing. However, were you holding power constant? Power factor is a number which is used to account for current which is flowing, but doing no work. If you assumed that current is constant, then you are assuming that power is increasing; i.e. that your load has increased. Since this is not the case, current must fall in proportion (neglecting any changes in efficiency) to power factor increasing.

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************************
Michael Griffin
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After some reflection I think the power company is going to use something similar to the following to adjust the commercial rates. P= (V*I*(pf1-pf0/pf1)*1.732)/1000

pf0 = .95 power factor
pf1 = measured power factor

this would give a number greater than 1 when lower than .95 and would increase the further from .95 one gets. I will know more tomorrow I have a seminar to attend that should give some answers.

Any thoughts?

By Johan Bengtsson on 5 November, 2003 - 12:29 am

That particular formula look really wierd, so it's probably not the one. It produces negative values for some power factors... I think the company likes to be paid when you use their energy...

/Johan Bengtsson

By Bouchard, James \[CPCCA\] on 5 November, 2003 - 12:37 am

Our utility corrects the kW demand to 95% by using the ratio of the maximum kW demand /kVA demand to 0.95 if the ratio is less than 0.95 Note they do not reduce the kW is we are above 0.95 (which we usually are). In our case
the utility bills for kW demand and kWh consumption so the effect of a power factor below 0.95 is to increase the billing kW. We have automatic power factor correction equipment that maintains our PF above 0.95 so we usually have no problems.

James Bouchard

By Meir Saggie on 5 November, 2003 - 7:23 pm

I think you confused among demand (peak - kW), energy (active - kWh) and blind (reactive - kVArh). Peak demand is measured in power units (kW) as the average over a short (usually 15 minutes) period of time - the payment represents your contribution towards the capital investment of the utility in the generating equipment it needs to have in order to satisfy your demand. Active Energy is measured in real energy units (kWh) - stands for the energy your consumed (and the utility in turn paid for in terms of oil, gas, coal, uranium, etc. it consumed)
Reactive Energy is measured in reactive (blind) energy units (kVArh) - stand for the extra energy consumed in the transmission wiring (I*I*R) - that otherwise no consumer pays for because it is not metered - above the absolute minimum if your power factor were 1.00 or close to (say 0.95).

Meir

The cost of reactive current to the supplier includes the increased I*I*R losses, additional unit committment costs and increased infrastructure costs to supply the same maximum power demand.

The cost to the user to improve their power factor is capital to provide the PF correction capacitors and associated control equipment.

The cost to everyone of a very high power factor such as .95 is reduced stability and reliability.

I wonder what priority power authorities/companies place on reliability of supply.

Vince Dooley

Why does the high power factor reduce stability and reliability?

It's a little difficult to explain without diagrams but I will try. Consider just a single ganerator connected to a large system.

The power equation for the generator is: P = (E0.VT/XS)sine(delta)
E0 is excitation
VT is terminal voltage
XS is synchronous reactance
delta is the angular difference between the rotor magnetic field and stator (system) magnetic field.

Note that power is zero when delta = 0 and maximum for any given E,V,X when delta = 90. Note also that maximum power is greater for increased E or V. That is, the higher the
reactice loads, (the lower PF) then the greater the maximum power.

From here on we really need a diagram but let's suppose there is a fault on the system which pulls the terminal voltage down. The angle delta cannot change instantaneously so the output power is immediately reduced. The mechanical input power from the turbine cannot change instantaneously so the machine starts to accelerate and delta increases. While this is happening the governor will start cutting back to try to counteract the acceleration. The greater the excitation E, the slower the acceleration and the more energy required to increase the speed of the machine. Similarly, the lower E is, then the faster the machine will accelerate. If the machine angle increases to a point beyond 90 degrees (maximum output) and reaches an angle on the downside of the curve where the mechanical input from the turbine is greater than the output power of the generator then the machine will accelerate uncontrollably, get out of synch and the protection systems will trip the machine. A lot depends on the moment of inertia of the machine and the excitation level.

Higher power factor, lower E = less stability.

Find a good text on power systems or synchronous machines. Study the stability sections or specifically, power swing curves.

Vince

By Bouchard, James \[CPCCA\] on 7 November, 2003 - 12:17 am

I have not confused demand and consumption. I have just reported our utility's billing practices . We pay about \$10 per kW maximum demand (on a 15 minute basis) corrected to at least 0.95 power factor per month plus an additional \$0.0242 per kWh consumed. The utility divides the maximum kW demand for the month by the maximum kVA demand for the month. If the result is 0.95 or higher they use the maximum demand kW for the billing. If it is less than 0.95 they multiply the kVA maximum demand by 0.95 and use this result for the kW demand billing. The kWh consumption is not adjusted.

So if the maximum kW demand for a month is 5000 and the maximum kVA demand is 6000 dividing we get 0.833 which is less than 0,95. So the utility would then multiply 6000 by 0.95 which gives 5700. They would then bill me for 5700 kW demand in that month instead of the 5000. In this case the penalty would be 700 x 10 or \$7,000.

In our case only the kW, kVA and kWh are shown on the bill. kVAR are not shown on the bill. The meter probably measures it (being an electronic one) but I do not recall it being reported. In the past they did have meters that reported the kVAR.

Actually everybody pays for reactive power in one way or another. The utility either includes it explicitly in the billing (usually for larger users) or implicitly in the way they calculate their rates. They do not give anything away.

James Bouchard

By KBPLSIDHPUR on 4 November, 2003 - 11:16 am

Dear Alan,
A higher power factor on the motor only means that it is doing more
useful work. More Work done == Higher power consumption. Higher Power
Consumption == Higher tariff. Slightly off the track, Induction motors due
to the inductance in their design have lagging power factors. Capacitors are
normally used to make up for the lag. This costs money and can also drive up
the Total Cost of Ownership.
I confess that I have not seen many motors but I am yet to see an
Induction motor having a name plate rating above 0.9 PF.
In India the utilities tend to look at the power factor of the total
distribution system and offer rebates if the power factor is nearer to
unity. Thus there is a tradeoff in the cost of installation of Capacitors
against Rebates from the utility and the fact that comparitively smaller
cable can be used within the premises.

Tomy Zacharia

hey dear! it's genuine question.
Actually power factor = actual voltage/ apparent voltage.Though we consider input voltage as 440V or 415V actually it is less due to load on transformer.As the load increases more current is drawn which forces impedance to exceed reactance voltage.
Now for any motor, power at the shaft is that which is drawn by the coupled machine.so the power drawn from the motor is motor effn multiply by power at the shaft which is constant for the machine.This power is put in to the formula as P = 1.73*V(act)*I=1.73*V(apparent)*power factor*I*motor efficiency
Now V aprnt=440V & motor effncy is constant for the motor Thus we get
Power drawn by the motor = constant*cos@*I here power drawn by the motor is also constant and depends on steady load on motor thus we get
I is inversly proportional to cos@ hence as power factor drops down current increases which is directly proportional to KWH.Hence you get higher billing amount.
hope u r convinced.

If 3 phase is considered as a more efficient power source, when then are we not running it in our homes, and/or on our appliances, and devices?

Is this so that the electric company can charge more?

By Curt Wuollet on 24 December, 2003 - 8:39 pm

No, it's simply because it would cost more. Not just for new 3 phase appliances, but thousands of miles of new wire would need to be installed, services converted, transformers replaced, etc. It is cost effective where large amounts of power are consumed in a concentrated area and motor loads are a large percentage of power consumed. Some loads like lighting and resistance heating are inherently two terminal and there is no benefit to three phase power for them. It's primarily a motor thing and is wonderful for bulk power conversion that involves rectification, like our induction furnaces. But there is little in the average residence that could use it to any advantage.

Regards

cww

Because:
1. Most appliances are manufactured as single phase (cheaper to manufacture - for the manufacturer).
2. Most homes are wired internally as single phase (sockets).
3. Power hungry devices - notably high capacity air conditioners - are already available as 3-phase machines.

A mental exercise for you - which appliances would you prefer to be 3-phase (including the burden of rewiring your house in order to accommodate them where you want to use them)?

Meir

hi. The power factor is pretty much telling you how much of the power the motors acually use is in "watts" or "usefull energy". so the higher the PF, the more effecient the motor is as far as "how much you're paying for the motor to run"