THE PHYSICS OF... POWER

[winnie]

hmmm, Φ=0, so if you size the gen in VA where load is 100Vac_rms @ 100amps with no reactance, what size gen in VA do i build? hmmm, VA suddenly looks like kW.

you can twist the terms around all you want. kW and PF is just a twisting to make it easier for power folks to understand how much of their VA is not paid for by customer and how much overhead they need to account for. 100kW in yields 80kW paid for by cust, then the $kW cust better be larger than the $kW gen, otherwise might as well go back to basket weaving.

There is a key point here: kVA represents the generation capacity and the transmission capacity being used, but not the _power_ being generated.

The generator can only supply so much current. The transmission lines can only carry so much current.

With a unity power factor load, your 100kVA generator can supply 100kW. You probably need to dump 105kW of mechanical power into the generator to get this output.

With a 0.8 PF load, your 100kVA generator can only supply 80kW. You probably need to dump 85kW of mechanical power into the generator to get this output.

'Apparent power' is _not_ power, even though the units are the same. Apparent power is different from active power because it includes the transmission capacity being used for energy being shuttled back and forth.

Apparent power is a real concept, and is useful because it tells you how much of your capacity you are using, how much your wires will heat up, etc.

-Jon

 

[Ingenieur]

Once more, that's still nonsense.

but fact
what is nonsense to the undereducated is fact to the educated

 

[FionaZuppa]

BINGO!

all are joule/sec or va a power term
by convention
P Watt
Q var
S va

J/s by itself does not necessarily mean power.
you can express VAr in watt units, that does not mean work is done, J/s does not always mean work !!

everyone see the diff? if you want to describe it in proper physics then stop using "power" "Q" "S" "apparent" "real". switch to eV :thumbsup:

 

[Ingenieur]

Yes true. No ifs, not buts.

no if and or buts
va = joule/sec = watt
so P, Q and S are power
active, reactive and total/complex/apparent related by S = P + jQ
and they must all be in the same units to add up

 

[Ingenieur]

J/s by itself does not necessarily mean power.
you can express VAr in watt units, that does not mean work is done, J/s does not always mean work !!

everyone see the diff? if you want to describe it in proper physics then stop using "power" "Q" "S" "apparent" "real". switch to eV :thumbsup:

no one is saying Q does work
although over a 1/4 cycle it does
But P is 100% work
S is pf% work
Q is 0% over 1 cycle

 

[FionaZuppa]

There is a key point here: kVA represents the generation capacity and the transmission capacity being used, but not the _power_ being generated.

The generator can only supply so much current. The transmission lines can only carry so much current.

With a unity power factor load, your 100kVA generator can supply 100kW. You probably need to dump 105kW of mechanical power into the generator to get this output.

With a 0.8 PF load, your 100kVA generator can only supply 80kW. You probably need to dump 85kW of mechanical power into the generator to get this output.

'Apparent power' is _not_ power, even though the units are the same. Apparent power is different from active power because it includes the transmission capacity being used for energy being shuttled back and forth.

Apparent power is a real concept, and is useful because it tells you how much of your capacity you are using, how much your wires will heat up, etc.

-Jon

in real world LRC
do the same math but keep the load real watts the same. 100kW pure resistive, then 100kW that has a .8PF, what does gen kVA look like between the two. you have to upsize the gen to accommodate the 100kW with cos(Φ), right. what exactly are you upsizing, likely ampacity to accommodate the extra amps associated with jX? pushing extra amps does not come free, sorry. the 1st case of pure resistance requires 105kW input, then you upsize the gen kVA to accommodate the PF, so you think the gen input remains 105kW ??

if a customer load is 100% reactive with just a very small R(like 10^-400) (a huge inductor) and is wound to have a total Z of 10ohms @ 60Hz, 10kV is applied, the wires in ckt loop carry 1kA. dissipated watts at the load is 10^-34 watts, you are saying the gen needs a huge kVA rating but the input power only needs to be 10^-34 watts ??

no one is saying Q does work
although over a 1/4 cycle it does
But P is 100% work
S is pf% work
Q is 0% over 1 cycle

??
it does not, but then it does ?? you'll need to explain that. how does the jX component do work?

if i have portions of a cycle that do work, and other periods that do not, isnt the sum of that over time !=0 ??

 

[wwhitney]

if a customer load is 100% reactive with just a very small R(like 10^-400) (a huge inductor) and is wound to have a total Z of 10ohms @ 60Hz, 10kV is applied, the wires in ckt loop carry 1kA. dissipated watts at the load is 10^-34 watts, you are saying the gen needs a huge kVA rating but the input power only needs to be 10^-34 watts ??

Indeed, in the steady state that is all true, and is required by conservation of energy. (Edit: I didn't check your computations). Getting to the steady state is a different matter.

Cheers, Wayne

 

[FionaZuppa]

Indeed, in the steady state that is all true, and is required by conservation of energy. (Edit: I didn't check your computations). Getting to the steady state is a different matter.

Cheers, Wayne

are you saying input to gen is only 10^-34 watts?
that's only true if poco side is lossless in all terms of LRC :thumbsup:, that example was already given back about 25 pages, it only holds true when jX is jX everywhere, which is not the case in real world. you must do work to push amps into X (penalty for that), X will release energy back via amps (penalty for that).

 

[wwhitney]

are you saying input to gen is only 10^-34 watts?
that's only true if poco side is lossless in all terms of LRC :thumbsup:, that example was already give back about 25 pages

Sure, you need to include any other resistive losses not specified. Since your R was so ridiculously low, I assumed everything was superconducting.

Nonetheless, take a look at post 353. The extra input power to the generator is only that required to overcome the additional transmission losses. It will be positive but much less than the VAr required.

C heers, Wayne

 

[Besoeker]

no if and or buts
va = joule/sec = watt

VA is VA. Not joules per second.

 

[winnie]

are you saying input to gen is only 10^-34 watts?
that's only true if poco side is lossless in all terms of LRC :thumbsup:, that example was already given back about 25 pages, it only holds true when jX is jX everywhere, which is not the case in real world. you must do work to push amps into X (penalty for that), X will release energy back via amps (penalty for that).

In a system with no losses (not the real world it takes 100kW in to produce 100kW out.

In a system with no losses, it takes 0kW in to produce 100kVA out in the steady state. There may be a transient load associated with 'charging up' the reactive system.

In the real world, with real losses, producing 100kVA of reactive power will require some amount of kW input, but only to overcome the losses.

In the question you posed, 100kW 1.0PF versus 100kW 0.8PF, then the generator would need to be be size to 125 kVA in order to supply the necessary load current, and the mechanical power input will likely be higher than in the 1.0PF case, but not 25kW more; rather by a small increase to cover the additional losses.

-Jon

 

[FionaZuppa]

Sure, you need to include any other resistive losses not specified. Since your R was so ridiculously low, I assumed everything was superconducting.

Nonetheless, take a look at post 353. The extra input power to the generator is only that required to overcome the additional transmission losses. It will be positive but much less than the VAr required.

C heers, Wayne

you are losing me.
a inductive motor is not worth crap is when you plug it in you get 100% work in terms of heat. a good motor has no R and 100% efficient in mag coupling. thus if i wanted a really good motor it has to be a really efficient mechanically dynamic xfrmer, it must couple the input mag field (reactive power) to the output (in essence, convert VAr to real work). if i use all of the mag field energy to do work then there is no VAr that the poco sees, and thus my inductor has a PF=1, but its a inductor.

motor A is 100kW PF .9 (R =10ohm Z=100ohm)
motor B is 100kW PF .95 (R =10ohm Z=100ohm)

why is that? likely because B is able to convert the reactive energy more so than A can, because the design of B is better.

it also explains why PF's change when motors are loaded. when you steal mag power into shaft output the PF goes up. the profile of that depends on motor design, slip #'s, yada yada yada.

all this doesnt change the fact that PF_{instantaneous} is a ratio that describes the losses (penalty) needed to deliver work at the load.

 

[mbrooke]

Can someone explain that to me, why PF changes as motors are loaded up/down?

 

[FionaZuppa]

Can someone explain that to me, why PF changes as motors are loaded up/down?

kinda simple, you are causing the mag field to couple, you put resistance against the shaft and that resistance is countered by the forces of mag field coupling, thus you are extracting (converting) what everyone is calling VAr power into work.

in essence, loading changes the mag angles which produces the torque. when the angles are the same you cannot develop a force.

 

[Ingenieur]

you are losing me.
a inductive motor is not worth crap is when you plug it in you get 100% work in terms of heat. a good motor has no R and 100% efficient in mag coupling. thus if i wanted a really good motor it has to be a really efficient mechanically dynamic xfrmer, it must couple the input mag field (reactive power) to the output (in essence, convert VAr to real work). if i use all of the mag field energy to do work then there is no VAr that the poco sees, and thus my inductor has a PF=1, but its a inductor.

motor A is 100kW PF .9 (R =10ohm Z=100ohm)
motor B is 100kW PF .95 (R =10ohm Z=100ohm)

why is that? likely because B is able to convert the reactive energy more so than A can, because the design of B is better.

it also explains why PF's change when motors are loaded. when you steal mag power into shaft output the PF goes up. the profile of that depends on motor design, slip #'s, yada yada yada.

all this doesnt change the fact that PF_{instantaneous} is a ratio that describes the losses (penalty) needed to deliver work at the load.

more Q is not 'converted'
less Q is required
less P is wasted

 

[GoldDigger]

kinda simple, you are causing the mag field to couple, you put resistance against the shaft and that resistance is countered by the forces of mag field coupling, thus you are extracting (converting) what everyone is calling VAr power into work.

NO! Because even though the reactive power is remaining more or less constant the resistive loading is increasing as the motor shaft is loaded.

Note that the mechanism involved in the change in PF (and in both resistive and reactive "power" component from locked rotor to rotating with anywhere from full to zero load is quite different.
In a startup situation the counter EMF is virtually zero and so the resistive and reactive components of the current may both be higher.

 

[Besoeker]

Can someone explain that to me, why PF changes as motors are loaded up/down?

Off load the motor supplies no power but needs exicitation. This is at low power factor with just no load losses being supplies. Typically, an induction motor off load takes around 30% of full load current for no power out. Hence the very low power factor.

As the motor load is increased the input and output power are greater but the excitation remains about the same so the power factor is higher.

 

[FionaZuppa]

NO! Because even though the reactive power is remaining more or less constant the resistive loading is increasing as the motor shaft is loaded.

Note that the mechanism involved in the change in PF (and in both resistive and reactive "power" component from locked rotor to rotating with anywhere from full to zero load is quite different.
In a startup situation the counter EMF is virtually zero and so the resistive and reactive components of the current may both be higher.

ok, show me.
i have a no-R motor, purely inductive motor. if i had electrical R in my motor that would just waste energy, this one is a real motor with copper that is super conducting. input amps is governed by 2πfL (only the jX portion of Z).

the input is only reactive, explain how power is transformed from electrical input amps to mechanical work output @ 100% efficiency. also show what VAr looks like for poco. me wait while you drum up the answer.

 

[GoldDigger]

ok, show me.
i have a no-R motor, purely inductive motor. if i had electrical R in my motor that would just waste energy. input amps is governed by 2πfL (only the jX portion of Z).

the input is only reactive, explain how power is transformed from electrical input amps to mechanical work output @ 100% efficiency. also show what VAr looks like for poco. me wait while you drum up the answer.

I am not going to go through the full modelling of what you describe.
I will just reiterate that the current through the motor will be affected by the back EMF produced by the combination of the rotation and the induced magnetism in the rotor.
And that when you load the motor down, even though if it could be statically represented as a pure inductor there will be a resistive component to the input current when it is loaded. That resistive input current is in addition to the magnetizing current, not the result of the magnetizing current somehow producing work.

Let me give a ridiculously oversimplified example that may still help:

Suppose you have a capacitor and a light bulb wired in parallel, but with a switch in series with the light bulb.
With the switch open only reactive current if flowing. When I close the switch there will be a high resistive component to the input current, but the VAR through the capacitor will not have changed at all and no part of the capacitive VAR has magically been converted into real watts.

I think that the biggest hurdle in understanding the motor problem is that no matter how perfect the stator coil and magnetic circuit may be it will not appear to be a pure inductor once the motor is rotating. (In fact when the motor is stationary there will be losses in the rotor unless it is a permanent magnet design, and hence not an induction motor in the first place.)
If you have a perfect inductor coupled by mutual inductance to second coil that is loaded down by a resistor, then the input current to the perfect inductor will no longer be purely reactive.

Thanks Carultch! Although I think my explanation may be easier to follow than the more persuasive actual numbers.
PS: If you put your tables into a CODE section in the post you have a better chance of getting the columns to line up.

PPS: With the 100% slip current (theoretically LRA) being only 130% of FLA, I think your model does not work well all the way to zero speed.

 

[Besoeker]

ok, show me.

Motor Data
Supply Voltage (V) 660
Stator nominal volts (V) 660
Rotor Voltage (V) 1000
No of Poles 4
Supply Frequency (Hz) 50
N sync (rpm) 1500
Power Rating (kW) (kW) 500
X1 (W) 0.0531
R1 (W) 0.0071
Xm (W) 2.00
Rm (W) 121
X2' (W) 0.075 Resistor stages Total
R2' (W) 0.007233333 0.704 0.194 0.08 0.036 0.027 1.041
Windage & Frict (kW) 8 0 1 0 0 0 Select
Effective rtr ind (mH) 1.10 0 0.194 0 0 0 0.194
1.235
Calculated Performance 0.215
External Rotor Res (W) 1.074 1.256 1.256 0.337 0.337 1.041 1.041 1.041 0.000
Slip (pu) 1.000 1.000 0.277 0.277 0.205 0.156 0.108 0.059 0.009
Speed (rpm) 0 0 1084 1084 1193 1266 1339 1412 1486
Stator Current (A) 796 696 268 695 536 228 207 193 524
Stator Current (%) 152% 133% 51% 133% 102% 43% 40% 37% 100%
Rotor Current (A) 492 426 122 425 319 83 57 31 311
Power Factor 0.888 0.888 0.683 0.888 0.874 0.552 0.424 0.258 0.872
Input Power (kW) 807.7 705.9 209.5 704.7 535.0 143.8 100.4 56.8 521.9
Gross Output (kW) 0.0 0.0 147.8 499.5 417.9 117.5 85.7 49.5 507.9
Wind & Frict (kW) 0.00 0.00 5.78 5.78 6.36 6.75 7.14 7.53 7.93
Net Output (kW) 0.0 0.0 142.1 493.8 411.5 110.8 78.6 42.0 500.0
Efficiency (%) 96.4% 96.8% 94.5% 96.0% 96.1% 91.9% 88.4% 79.3% 95.81%
Torque (kNM) 4.98 4.36 1.25 4.35 3.30 0.84 0.56 0.28 3.21
PU Torque (pu) 1.551 1.356 0.389 1.354 1.026 0.260 0.174 0.088 1.000
Load torque (pu) 0.00 0.00 0.388 0.39 0.52 0.62 0.73 0.86 1.00

Motor kVA (kVA) 910 795 307 794 612 261 237 220 599
Motor Current (A) 796 696 268 695 536 228 207 193 524
Motor Current (%) 152% 133% 51% 133% 102% 44% 40% 37% 100%
Motor input kVAr (kVAr) 418 366 307 794 612 261 237 220 599
PFC (kVAr) 265 265 265 265 265 265 265 265 265
Supply kVAr (kVAr) 153 101 42 529 347 -4 -28 -45 334
Supply kVA (kVA) 822 713 214 881 638 144 104 72 619
Supply Current (A) 719 624 187 771 558 126 91 63 542
Supply Current (%) 137% 119% 36% 147% 107% 24% 17% 12% 103%

You asked.