Panel heaters, water heaters........
THE PHYSICS OF... POWER
- Thread starter Phil Corso
- Start date
- Status
FionaZuppa
Senior Member
- Location
- AZ
- Occupation
- Part Time Electrician (semi retired, old) - EE retired.
you have taking this outside of "The Physics..... of Power".
all conductors are inductors. i leave it up to the reader to go investigate why.
Last edited:
Really??
Is a water heater not an application of the physics of power??
Care to explain why it isn't?
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
Please allow me to push it over the top.
Of course, it doesn't hold a candle to The Music Game.
FionaZuppa
Senior Member
- Location
- AZ
- Occupation
- Part Time Electrician (semi retired, old) - EE retired.
you didnt read the posts. so such device with just R, reply was "panel heaters, water heaters" <--- NOT TRUE
put an LCR meter (using the R setting, most LCR's are really Z on the R setting). what do you get? now measure the heating element with a true DC (R) meter. if the LCR has some accuracy in low mH then you'll see Z vs R, etc.
lets stick with physics. lets ask the most basic of questions.
a load has a Z=10ohms @ 60Hz. the src-gen is 100Vrms 60Hz, how much work is required to push 10amps of charge through the voltage gradient of 100Vrms. for ease of math the gen itself has Z=0, and transmission losses =0.
if you wish to create varying PF's for the Z=10 feel free (but R>0<X), but if you understand the question you'll know why i say "feel free". you can even opt to stay in units of watts or switch to eV.
FionaZuppa
Senior Member
- Location
- AZ
- Occupation
- Part Time Electrician (semi retired, old) - EE retired.
how about this riddle
example
60Hz
L=1mH
2pi fL = 0.377
Z=100
10,000=R2+0.142
R=99.99
V=1,000Vac-rms
amps=10
Q=100*0.377 = 37.7watt
1/60=0.0166sec = period
0.625J per period
divide by 4 = 0.156J per ¼ period
peak voltage = 100/0.707 = 141.44v
peak amps = 141.44/10 = 14.14A
magnetic field energy is in-phase with current waveform
max mag energy when amps is max
mag field energy = ½LI2 = ½ 0.001H 200 = 0.1J
so from Q (0-90 degrees, ¼ period) we get 0.156J
but from mag field energy equation we get 0.1J
where does the 0.056J go ?
example
60Hz
L=1mH
2pi fL = 0.377
Z=100
10,000=R2+0.142
R=99.99
V=1,000Vac-rms
amps=10
Q=100*0.377 = 37.7watt
1/60=0.0166sec = period
0.625J per period
divide by 4 = 0.156J per ¼ period
peak voltage = 100/0.707 = 141.44v
peak amps = 141.44/10 = 14.14A
magnetic field energy is in-phase with current waveform
max mag energy when amps is max
mag field energy = ½LI2 = ½ 0.001H 200 = 0.1J
so from Q (0-90 degrees, ¼ period) we get 0.156J
but from mag field energy equation we get 0.1J
where does the 0.056J go ?
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
Nice example
Should be 37.7 VAr
You're missing a factor of 2/pi, as per my previous posts.
Conveniently, 2/pi * 0.156 = 0.1 (within rounding error). So with that, the discrepancy is gone.
Cheers, Wayne
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
If I remember correctly, copper does not superconduct but aluminium does.
Have you tried it yourself?
FionaZuppa
Senior Member
- Location
- AZ
- Occupation
- Part Time Electrician (semi retired, old) - EE retired.
didnt i already factor in period/4 ?
Q is based on integration over some period, energy built into the mag field is not, its instantaneous measurement when amps is max.
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
The energy associated with a magnetic field is given as a function of the current at that time and the inductance of the coil. It does not depend on the frequency or any other characteristic of how it got to that current.
It can also be calculated as the integral of the "power" flowing into the coil over time.
The two answers should be the same.
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
See post 356 for the reason that the factor of 2/pi is required when converting from VAr to joules in a quarter cycle.
Cheers, Wayne
Carultch
Senior Member
- Location
- Massachusetts
Is it supposed to be 2/pi? Or pi/2?
Because pi/2 is a quarter of a cycle, when the units are radians.
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
It's 2/pi. As per post 356,
Integral_(0, pi/2) sin(t)sin(t) = pi/4
Integral_(0, pi/2) sin(t)cos(t) = 1/2
The first integral is how much energy is transferred in a quarter cycle by 1 amp and 1 volt at (1/2pi) Hz in a quarter cycle with 0 phase shift. The second integral is for 90 degrees phase shift. The ratio of the second integral to the first integral is 2/pi. That ratio is going to be independent of frequency, I just picked the frequency (1/2pi) to make the functions in the integrand simple.
So the point is that if 1 watt transfers X joules in a quarter cycle (X = 1/240 for 60 Hz), then 1 VAr transfers 2/pi * X joules in a quarter cycle. FionaZuppa's calculation was off by this factor of 2/pi, because he/she was treating VArs as Watts when deriving the amount of energy transferred in a quarter cycle.
Cheers, Wayne
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
That should be 1/sqrt(2) amps and 1/sqrt(2) volts, of course, since we use RMS values, not peak values. But the ratio I was calculating is unaffected by this discrepancy.
Cheers, Wayne
FionaZuppa
Senior Member
- Location
- AZ
- Occupation
- Part Time Electrician (semi retired, old) - EE retired.
why would i integrate to find max energy from mag field, it happens at only two points for AC sin, 90 and 270, not the summation of instantaneous values over a period.
and another odd way to use the term power. didnt you say that Q was not power? the only flow in/out of an inductor (the X portion) is energy, and when frequency is known it may be expressed in terms of watts, but the interesting points will be 90 and 270 as this is where mag field energy is max, the max is will ever be.
i still did not see the math for how much work it takes to pump some amps of known voltage across a load of Z at set freq of 60Hz.
nobody ever really mentioned why the jX component presents ohms to the src.
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
You can find the max energy by integrating the power from the time when the energy is 0 to the time when the energy is maximum.
Q represents fluctuating power. On time scales of a 1/4 cycle or less, it can be unidirectional. On large times scales that are multiples of a 1/2 cycle, there will be no net power.
Suppose we have a constant AC voltage source of voltage V connected to a load with complex impedance Z * e^(i theta). [theta is in radians]. Then the current will have RMS magnitude V/Z and phase shift theta. The energy over a time period T that is a multiple of a 1/2 cycle will be:
integral_T I(t) V(t) dt
= integral_T 1/Z V(t shifted theta/2pi cycles) V(t) dt
= 1/Z integral_T (cos theta V(t) + sin theta V(t shifted 1/4 cycle)) V(t) dt
= 1/Z integral_T cos theta V(t) V(t) dt + 0
= (cos theta) V^2 T / Z
In other words, the power delivered is (V^2 / Z) * cos theta. That is also the rate of work required.
Cheers, Wayne
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
I used theta in the above, because it is common in mathematics, but I seem to recall that phi is more commonly used in physics. It's the same thing whatever letter you use, the phase angle of the impedance.
Cheers, Wayne
- Status