A 10KVA-rated step-up isolation transformer is used to connect a 480v-delta
resistive heater to 208v-Y supply.
Measured currents are:
- primary (208v) current ~35A.
- secondary (480v) current ~15A
If my math is correct this transformer is undersized by 20 percent, or so:
- primary: 35A x sqrt(3) x 208v = 12.6KVA
- secondary: 15A x sqrt(3) x 480v = 12.3KVA
All wires and the breaker are sized to handle the measured currents. While
the windings seem quite cool, the laminations are uncomfortably warm to the
touch.
The load is turned on for periods of up to 15 minutes at a time and
disconnected for between 10 minutes to overnight, depending on the demand for
the equipment.
There doesn't seem to be evidence of over-temperature in the windings and
wires are sized properly. Is there a risk here?
Thanks.
> Not sure where you came up with the sqrt(3) in your equations.
> --
> JeffB
This was from a conversation with an engineer at the transformer
manufacturer's support department. (To be fair, I didn't mention the load as
purely resistive). He gave the formula as;
KVA / V / sqrt(3) = A
> The power factor for resistive heaters will be near 1. The transformer
> rating is just fine.
So the formula for a resistive load is:
KVA / V = A
meaning that this 10KVA transformer is good up to a maximum of:
- primary: 48.0A
- secondary: 20.8A
Is this right?
Thanks.
> You were right the first time. The sqrt(3) accounts for the fact that
> this is a 3-phase system.
So this 10KVA transformer is rated to handle a maximum current of:
- primary (208v): 27.8A
- secondary (480v): 12.0A
Is this right?
Because this is driving a purely resistive load, this doesn't help bring the
parameters within the transformer's rating, does it?
Thanks.
>> Because this is driving a purely resistive load, this doesn't help bring
>> the parameters within the transformer's rating, does it?
> No, a purely resistive load is best case.
But since the transformer is (presumably) designed with a worst-case PF (or
at least some PF value greater than 1), can't it be re-rated (what's the
opposite of "derated"?) if used with a strictly resistive load?
In other words, a transformer rated at 10KVA for a load with PF = x, should
be able to handle a greater KVA if driving a PF = 1 load. Shouldn't it?
On Jan 18, 9:52=A0pm, "Don Kelly" <[email protected]> wrote:
> -----------"whit3rd" <[email protected]> wrote in message
> The turnon transient isn't similar to normal AC operation
>[turnon can cause ] 150% of the normal flux in
> the core.
> A (rough) transient analysis of a 10kVA 1000/200 V transformer based on wh=
at
> appears to be reasonable data indicates in a worse case case ignoring
> saturation [the load doesn't help]
> The culprit is saturation.
Yep. It was my habit to do power surge testing on newbuilt
electronics with a switch and large nearly-unladen transformer,
because every tenth onswitch of the transformer generated
hum/arcing on the switch and significant powerline transients.
If my box survived that, I figured it'd work fine in the real-world
environment. HF from the switch arcing, 1-cycle dropouts
from the transformer going into saturation, maybe a
bit of overvoltage (until the arc formed)... that power
outlet was guaranteed hostile enough for a good test.
The fuse went open with no load, so I assumed the
problem to be saturation of the core, and just looked at the
scenario for that one effect, and found a likely treatment.
I understand that maximum-voltage switching is another
treatment (there are solid-state relays for this).
On Jan 17, 11:13 am, James T. <[email protected]> wrote:
> A 10KVA-rated step-up isolation transformer is used to connect a 480v-delta
> resistive heater to 208v-Y supply.
>
> Measured currents are:
> - primary (208v) current ~35A.
> - secondary (480v) current ~15A
>
> If my math is correct this transformer is undersized by 20 percent, or so:
> - primary: 35A x sqrt(3) x 208v = 12.6KVA
> - secondary: 15A x sqrt(3) x 480v = 12.3KVA
>
> All wires and the breaker are sized to handle the measured currents. While
> the windings seem quite cool, the laminations are uncomfortably warm to the
> touch.
>
> The load is turned on for periods of up to 15 minutes at a time and
> disconnected for between 10 minutes to overnight, depending on the demand for
> the equipment.
>
> There doesn't seem to be evidence of over-temperature in the windings and
> wires are sized properly. Is there a risk here?
>
> Thanks.
From practical experience let me tell you what we found doing the same
thing to power a convection oven built for 480 volts from a 208 volt
distribution.
We asked the electrical contractor next door to do the installation.
He sized the transformer just like you did, then asked an expert who
told him he didn't consider the transformer loss, so he found a
transformer that covered it and then some. It was surplus to another
contractor and had been paid for by someone who didn't use it. So we
got it cheap. We have a 150 amp circuit breaker between the meter box
and the transformer. Works great. The transformer does get warm, but
not uncomfortably so.
The reason I am writing is to let you know that we initially turned
off the transformer on weekends using the 150 amp circuit breaker.
This was great for a while. Then one Monday morning the transformer
would not power up. We called the electrician and he eventually
discovered a 200 amp fuse in the meter box was open. He replaced it,
$50.. and it worked again. A few weeks later, the same thing happened
on a Monday morning. Same fuse problem. He said to stop turning off
the transformer. The high inrush current was blowing the fast acting
fuse, but not effecting the heat operated circuit breaker. The oven
was not powered up, so there was very little load on the transformer.
At $50 per fuse, that could quickly add up to more than the
electricity used over a weekend.
So, after you have done all your calculations, go see what fuses are
in the meter box for that circuit. They will be your limiting factor.
Paul
Enjoy
+++++++++++++++++++++++++++
SEX LIFE OF THE ELECTRON
(STORY OF MILLIE HENRY AND MICRO FARAD)
ONE NIGHT WHEN HIS CHARGE WAS PRETTY HIGH, MICRO FARAD DECIDED
TO
GET A CUTE LITTLE COIL TO LET HIM DISCHARGE.
HE PICKED UP MILLIE AND TOOK HER FOR A RIDE ON HIS POWER
AMPLIFIED MEGACYCLE. THEY RODE ACROSS THE WHEATSTONE BRIDGE,
AROUND THE SINE WAVE AND STOPPED IN A MAGNETIC FIELD BY A SMALL
FLOWING CURRENT.
MICRO FARAD, ATTRACTED BY MILLIE'S WAVES, SOON HAD HER AT
MINIMUM
RESISTANCE AND HER FIELD FULLY CHARGED. HE ALSO HAD HER
FREQUENCY LOWERED AND PULLED OUT HIS HIGH VOLTAGE PROBE. HE
INSERTED IT IN PARALLEL AND BEGAN TO SHORT CIRCUIT HER SHUNT.
FULLY CHARGED MILLIE SAID MHO, MHO, GIVE ME MHO. WITH TUBE AT
MAXIMUM CONDUCTION AND HER COIL VIBRATING FROM EXCESSIVE
CURRENT,
SHE SOON REACHED PEAK. THE EXCESSIVE CURRENT HAD HER SHUNT
PRETTY HOT AND MICRO FARAD'S CAPACITOR WAS RAPIDLY DISCHARGING
EVERY ELECTRON.
THEY FLUXED ALL NIGHT LONG, TRYING VARIOUS CONNECTIONS AND
CIRCUITS UNTIL HER MAGNETIC FIELD HAD LOST ALL OF ITS FIELD
STRENGTH.
AFTERWARDS MILLIE TRIED SELF INDUCTION AND CHARGED HER FIELD;
HOWEVER, MILLIE REVERSED HER POLARITY AND WHEN MICRO FARAD
STARTED FLUXING AGAIN, THEY BLEW EACH OTHERS FUSES.
POWER HAS BEEN CONSUMED.
WORK HAS BEEN DONE.
SO WATT.
Lew Hodgett wrote:
> Enjoy
> +++++++++++++++++++++++++++
> SEX LIFE OF THE ELECTRON
> (STORY OF MILLIE HENRY AND MICRO FARAD)
>
>
> ONE NIGHT WHEN HIS CHARGE WAS PRETTY HIGH, MICRO FARAD DECIDED
> TO
> GET A CUTE LITTLE COIL TO LET HIM DISCHARGE.
>
> HE PICKED UP MILLIE AND TOOK HER FOR A RIDE ON HIS POWER
> AMPLIFIED MEGACYCLE. THEY RODE ACROSS THE WHEATSTONE BRIDGE,
> AROUND THE SINE WAVE AND STOPPED IN A MAGNETIC FIELD BY A SMALL
> FLOWING CURRENT.
>
> MICRO FARAD, ATTRACTED BY MILLIE'S WAVES, SOON HAD HER AT
> MINIMUM
> RESISTANCE AND HER FIELD FULLY CHARGED. HE ALSO HAD HER
> FREQUENCY LOWERED AND PULLED OUT HIS HIGH VOLTAGE PROBE. HE
> INSERTED IT IN PARALLEL AND BEGAN TO SHORT CIRCUIT HER SHUNT.
>
> FULLY CHARGED MILLIE SAID MHO, MHO, GIVE ME MHO. WITH TUBE AT
> MAXIMUM CONDUCTION AND HER COIL VIBRATING FROM EXCESSIVE
> CURRENT,
> SHE SOON REACHED PEAK. THE EXCESSIVE CURRENT HAD HER SHUNT
> PRETTY HOT AND MICRO FARAD'S CAPACITOR WAS RAPIDLY DISCHARGING
> EVERY ELECTRON.
>
> THEY FLUXED ALL NIGHT LONG, TRYING VARIOUS CONNECTIONS AND
> CIRCUITS UNTIL HER MAGNETIC FIELD HAD LOST ALL OF ITS FIELD
> STRENGTH.
>
> AFTERWARDS MILLIE TRIED SELF INDUCTION AND CHARGED HER FIELD;
> HOWEVER, MILLIE REVERSED HER POLARITY AND WHEN MICRO FARAD
> STARTED FLUXING AGAIN, THEY BLEW EACH OTHERS FUSES.
>
> POWER HAS BEEN CONSUMED.
>
> WORK HAS BEEN DONE.
>
> SO WATT.
>
>
pulse, throb.
Not sure where you came up with the sqrt(3) in your equations. Take a
look at http://www.powerstream.com/VA-Watts.htm (first page of many when
Googled). The power factor for resistive heaters will be near 1. The
transformer rating is just fine.
--
JeffB
remove no.spam. to email
James T. wrote:
> A 10KVA-rated step-up isolation transformer is used to connect a 480v-delta
> resistive heater to 208v-Y supply.
>
> Measured currents are:
> - primary (208v) current ~35A.
> - secondary (480v) current ~15A
>
> If my math is correct this transformer is undersized by 20 percent, or so:
> - primary: 35A x sqrt(3) x 208v = 12.6KVA
> - secondary: 15A x sqrt(3) x 480v = 12.3KVA
>
> All wires and the breaker are sized to handle the measured currents. While
> the windings seem quite cool, the laminations are uncomfortably warm to the
> touch.
>
> The load is turned on for periods of up to 15 minutes at a time and
> disconnected for between 10 minutes to overnight, depending on the demand for
> the equipment.
>
> There doesn't seem to be evidence of over-temperature in the windings and
> wires are sized properly. Is there a risk here?
>
> Thanks.
>
"Lew Hodgett" <[email protected]> wrote in message
news:[email protected]...
>
> "Don Kelly" wrote:
>
>> Haven't seen this one for about 50 years- Oh to be young and horny!
>
> You to huh.
>
> Lew
>
There was a similar story about the Military Radar Type AN4Q2
I can not remember the details now but it fitted the radar part very
well.
It had a resonant cavity and 2 tuning knobs and a whole lot of other
Radar jargon
----------------------------
"whit3rd" <[email protected]> wrote in message
news:[email protected]...
On Jan 18, 9:52 pm, "Don Kelly" <[email protected]> wrote:
> -----------"whit3rd" <[email protected]> wrote in message
> The turnon transient isn't similar to normal AC operation
>[turnon can cause ] 150% of the normal flux in
> the core.
> A (rough) transient analysis of a 10kVA 1000/200 V transformer based on
> what
> appears to be reasonable data indicates in a worse case case ignoring
> saturation [the load doesn't help]
> The culprit is saturation.
Yep. It was my habit to do power surge testing on newbuilt
electronics with a switch and large nearly-unladen transformer,
because every tenth onswitch of the transformer generated
hum/arcing on the switch and significant powerline transients.
If my box survived that, I figured it'd work fine in the real-world
environment. HF from the switch arcing, 1-cycle dropouts
from the transformer going into saturation, maybe a
bit of overvoltage (until the arc formed)... that power
outlet was guaranteed hostile enough for a good test.
The fuse went open with no load, so I assumed the
problem to be saturation of the core, and just looked at the
scenario for that one effect, and found a likely treatment.
I understand that maximum-voltage switching is another
treatment (there are solid-state relays for this).
----------------------------
While it appears to me that the load will do little to reduce saturation
effects, the concept of maximum voltage switching could be very effective
as such switching will be at flux zeros corresponding to the initial
conditions in the core- eliminating the transient (assuming no residual
flux).
--
Don Kelly [email protected]
remove the X to answer
On Jan 17, 9:09=A0pm, "Don Kelly" <[email protected]> wrote:
> =A0Assuming the applied voltage is the same or nearly so, the inrush
> magnetising current will be the same as before. Adding load won't change
> this.
> For AC, unlike DC, it is the applied voltage that determines the peak flux=
> (E=3D4.44*f*N*phimax or phimax=3DE/(4.44*f*N) for sinusoidal excitation)
The turnon transient isn't similar to normal AC operation; a half-
cycle
of positive potential in normal AC conditions always occurs
right after the core achieves maximum negative flux. If the
core flux starts at zero (or even slightly positive due
to remnant magnetism) and THEN is powered at the beginning of
a half-cycle of positive potential, the end of that half-cycle of
excitation will occur with 150% of the normal flux in
the core.
If there is a substantial load attached, the flux will be less than
150%,
and (of course) if there is a short circuit attached to the output,
there will be zero flux (but despite the non-saturation of the core,
that won't help the blown-fuse situation much).
On Thu, 17 Jan 2008 22:27:35 GMT, James T. <[email protected]> wrote:
>>> Because this is driving a purely resistive load, this doesn't help bring
>>> the parameters within the transformer's rating, does it?
>
>> No, a purely resistive load is best case.
>
>But since the transformer is (presumably) designed with a worst-case PF (or
>at least some PF value greater than 1), can't it be re-rated (what's the
>opposite of "derated"?) if used with a strictly resistive load?
>
>In other words, a transformer rated at 10KVA for a load with PF = x, should
>be able to handle a greater KVA if driving a PF = 1 load. Shouldn't it?
No. kVA means what it says. i.e. kW/cos phi. The kVA rating of the transformer
is independent of the load PF. Load _power_ must be reduced for low PF loads.
But this transformer is probably reasonably rated if the duty factor is 50%
worst case and the on time 15 minutes worst case.
Mark Rand
RTFM
"Lew Hodgett" <[email protected]> wrote in message
news:[email protected]...
> Enjoy
> +++++++++++++++++++++++++++
> SEX LIFE OF THE ELECTRON
> (STORY OF MILLIE HENRY AND MICRO FARAD)
>
>
> ONE NIGHT WHEN HIS CHARGE WAS PRETTY HIGH, MICRO FARAD DECIDED
> TO
> GET A CUTE LITTLE COIL TO LET HIM DISCHARGE.
>
Haven't seen this one for about 50 years- Oh to be young and horny!
--
Don Kelly [email protected]
remove the X to answer
----------------------------
On Jan 17, 7:36 pm, whit3rd <[email protected]> wrote:
> On Jan 17, 12:43 pm, "[email protected]" <[email protected]> wrote:
>
> > [we used a transforme]
> > to power a convection oven built for 480 volts from a 208 volt
> > distribution.
> >... we initially turned
> > off the transformer on weekends using the 150 amp circuit breaker.
> > This was great for a while. Then one Monday morning the transformer
> > would not power up. We called the electrician and he eventually
> > discovered a 200 amp fuse in the meter box was open. He replaced it,
> > $50.. and it worked again. A few weeks later, the same thing happened
> > on a Monday morning. Same fuse problem. He said to stop turning off
> > the transformer. The high inrush current was blowing the fast acting
> > fuse, but not effecting the heat operated circuit breaker. The oven
> > was not powered up, so there was very little load on the transformer.
>
> That's a misconception. If you'd turned the oven ON before
> powering the transformer, the fuse would'nt blow. The 'load'
> that is important is the magnetic field in the transformer, which
> depends on input current MINUS scaled output current.
> With no output current, and possibly some remnant field,
> a newly-connected transformer takes its field to maximum,
> and that causes saturation and the fuse blows.
>
> If the core of a transformer staturates, the electrical circuit
> becomes
> just a copper resistor. Better to blow a $50 fuse than melt that
> copper resistor.
I am sure you are correct, in theory. However, the oven is computer
controlled and can't control until it gets power. Darn computers!
Paul
----------------------------
"[email protected]" <[email protected]> wrote in message
news:0d318c1c-c355-457e-96ac-9eb1345f9a98@j20g2000hsi.googlegroups.com...
> On Jan 17, 11:13 am, James T. <[email protected]> wrote:
>> A 10KVA-rated step-up isolation transformer is used to connect a
>> 480v-delta
>> resistive heater to 208v-Y supply.
>>
>> Measured currents are:
>> - primary (208v) current ~35A.
>> - secondary (480v) current ~15A
>>
>> If my math is correct this transformer is undersized by 20 percent, or
>> so:
>> - primary: 35A x sqrt(3) x 208v = 12.6KVA
>> - secondary: 15A x sqrt(3) x 480v = 12.3KVA
>>
>> All wires and the breaker are sized to handle the measured currents.
>> While
>> the windings seem quite cool, the laminations are uncomfortably warm to
>> the
>> touch.
>>
>> The load is turned on for periods of up to 15 minutes at a time and
>> disconnected for between 10 minutes to overnight, depending on the demand
>> for
>> the equipment.
>>
>> There doesn't seem to be evidence of over-temperature in the windings and
>> wires are sized properly. Is there a risk here?
>>
>> Thanks.
>
> From practical experience let me tell you what we found doing the same
> thing to power a convection oven built for 480 volts from a 208 volt
> distribution.
>
> We asked the electrical contractor next door to do the installation.
> He sized the transformer just like you did, then asked an expert who
> told him he didn't consider the transformer loss, so he found a
> transformer that covered it and then some. It was surplus to another
> contractor and had been paid for by someone who didn't use it. So we
> got it cheap. We have a 150 amp circuit breaker between the meter box
> and the transformer. Works great. The transformer does get warm, but
> not uncomfortably so.
>
> The reason I am writing is to let you know that we initially turned
> off the transformer on weekends using the 150 amp circuit breaker.
> This was great for a while. Then one Monday morning the transformer
> would not power up. We called the electrician and he eventually
> discovered a 200 amp fuse in the meter box was open. He replaced it,
> $50.. and it worked again. A few weeks later, the same thing happened
> on a Monday morning. Same fuse problem. He said to stop turning off
> the transformer. The high inrush current was blowing the fast acting
> fuse, but not effecting the heat operated circuit breaker. The oven
> was not powered up, so there was very little load on the transformer.
> At $50 per fuse, that could quickly add up to more than the
> electricity used over a weekend.
>
> So, after you have done all your calculations, go see what fuses are
> in the meter box for that circuit. They will be your limiting factor.
>
> Paul
------------------------------
Actually the input data given by James includes the transformer losses. The
difference between the KVA at 208V and that at 480V is 0.3KVA which
includes all real and reactive losses in the transformer- this is no big
deal (Real losses producing heat are probably below 0.1KW ). Also ratings
are based on output not input KVA so oversizing for losses isn't necessary.
If he was are switching the load on the secondary side, the transformer
should not have a problem. The core heating will be mainly due to core
losses which will be there even at no load. These should be normal.
Considering a duty cycle of 15 minutes on and 10 off, the 10KVA rating
appears adequate.
The advice that your contractor got meant that you had a larger transformer
with (most likely) higher core losses and inrush current than a smaller unit
sized transformer as well as a bigger initial price tag except in your case.
--
Don Kelly [email protected]
remove the X to answer
On Jan 17, 12:43=A0pm, "[email protected]" <[email protected]> wrote:
> [we used a transforme]
> to power a convection oven built for 480 volts from a 208 volt
> distribution.
>... we initially turned
> off the transformer on weekends using the 150 amp circuit breaker.
> This was great for a while. Then one Monday morning the transformer
> would not power up. We called the electrician and he eventually
> discovered a 200 amp fuse in the meter box was open. He replaced it,
> $50.. and it worked again. A few weeks later, the same thing happened
> on a Monday morning. Same fuse problem. He said to stop turning off
> the transformer. The high inrush current was blowing the fast acting
> fuse, but not effecting the heat operated circuit breaker. The oven
> was not powered up, so there was very little load on the transformer.
That's a misconception. If you'd turned the oven ON before
powering the transformer, the fuse would'nt blow. The 'load'
that is important is the magnetic field in the transformer, which
depends on input current MINUS scaled output current.
With no output current, and possibly some remnant field,
a newly-connected transformer takes its field to maximum,
and that causes saturation and the fuse blows.
If the core of a transformer staturates, the electrical circuit
becomes
just a copper resistor. Better to blow a $50 fuse than melt that
copper resistor.
The sqrt(3) is the same as the tan 120 deg. for each leg of the 3 phase
transformer.
You were told right the first time by the engineer.
Most of those transformers run surprisingly hot. Take a look at the label
for the temp rise rating.
Tom
"James T." <[email protected]> wrote in message
news:[email protected]...
>> Not sure where you came up with the sqrt(3) in your equations.
>> --
>> JeffB
>
> This was from a conversation with an engineer at the transformer
> manufacturer's support department. (To be fair, I didn't mention the load
> as
> purely resistive). He gave the formula as;
>
> KVA / V / sqrt(3) = A
>
>> The power factor for resistive heaters will be near 1. The transformer
>> rating is just fine.
>
> So the formula for a resistive load is:
>
> KVA / V = A
>
> meaning that this 10KVA transformer is good up to a maximum of:
>
> - primary: 48.0A
> - secondary: 20.8A
>
> Is this right?
>
> Thanks.
>
On Thu, 17 Jan 2008 21:00:06 GMT, James T. <[email protected]>
wrote:
>> You were right the first time. The sqrt(3) accounts for the fact that
>> this is a 3-phase system.
>
>So this 10KVA transformer is rated to handle a maximum current of:
>
> - primary (208v): 27.8A
> - secondary (480v): 12.0A
>
>Is this right?
Looks right to me.
>
>Because this is driving a purely resistive load, this doesn't help bring the
>parameters within the transformer's rating, does it?
No, a purely resistive load is best case.
--
Ned Simmons
In alt.engineering.electrical JeffB <[email protected]> wrote:
| Not sure where you came up with the sqrt(3) in your equations. Take a
Actually he should have _divided_ by sqrt(3) ... after or before _multiplying_
by the number of conductors, which is 3. But since 3/sqrt(3) is sqrt(3), the
multiplication by sqrt(3) is actually the correct simpler formula.
If there is no neutral current, you can figure the power simply by multiplying
the currents by the L-N voltages. In a delta system, there is no neutral (or
at least not one supplied to or used by the load), so no neutral current, but
it works out the same as if you had one. Rather than typing in the value for
sqrt(3), I just do the arithmetic on things like this as: 35*3*120 = 12600 or
15*3*277 = 12465. Either way that is above the transformer rating.
--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------
In alt.engineering.electrical James T. <[email protected]> wrote:
|>> Because this is driving a purely resistive load, this doesn't help bring
|>> the parameters within the transformer's rating, does it?
|
|> No, a purely resistive load is best case.
|
| But since the transformer is (presumably) designed with a worst-case PF (or
| at least some PF value greater than 1), can't it be re-rated (what's the
| opposite of "derated"?) if used with a strictly resistive load?
|
| In other words, a transformer rated at 10KVA for a load with PF = x, should
| be able to handle a greater KVA if driving a PF = 1 load. Shouldn't it?
What is relevant for a transformer rating is both amps and voltage in an
independent way. There is a maximum amperage each winding can handle.
There is a maximum voltage each winding (and the core) can handle. That
is why they are rated in terms of VA, kVA, MVA, etc. They are not rated
in W, kW, or MW for this reason. If your PF is 1, then you can get the
most power through the transformer ... e.g. you can run 10 kW through a
10 kVA transformer. But if your PF is only 0.5, then the most power you
can run through a 10 kVA transformer is 5 kW. Low PF forces you to derate
the transformer. Unity PF lets you run it at maximum rated power.
--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------
In alt.engineering.electrical [email protected] <[email protected]> wrote:
| I am sure you are correct, in theory. However, the oven is computer
| controlled and can't control until it gets power. Darn computers!
Batteries! Capacitors!
--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------
----------------------------
"whit3rd" <[email protected]> wrote in message
news:[email protected]...
On Jan 17, 9:09 pm, "Don Kelly" <[email protected]> wrote:
> Assuming the applied voltage is the same or nearly so, the inrush
> magnetising current will be the same as before. Adding load won't change
> this.
> For AC, unlike DC, it is the applied voltage that determines the peak flux
> (E=4.44*f*N*phimax or phimax=E/(4.44*f*N) for sinusoidal excitation)
--------------------------------------------------------------------------------------------
REPLY
-------------------------------------------------------------------------------------------
The turnon transient isn't similar to normal AC operation; a half-
cycle
of positive potential in normal AC conditions always occurs
right after the core achieves maximum negative flux. If the
core flux starts at zero (or even slightly positive due
to remnant magnetism) and THEN is powered at the beginning of
a half-cycle of positive potential, the end of that half-cycle of
excitation will occur with 150% of the normal flux in
the core.
If there is a substantial load attached, the flux will be less than
150%,
and (of course) if there is a short circuit attached to the output,
there will be zero flux (but despite the non-saturation of the core,
that won't help the blown-fuse situation much).
True, the turn on isn't the same as steady state and I hope that I didn't
imply that. You are looking at the case when the transformer is energised at
the time when voltage passes through 0 which is a worst case situation. In
the case of a linear core transformer, the "steady state" flux will be at a
negative maximum so there is a bridging transient.
Krause and Wasynczulk, "Electromechanical motion devices" deals with a
transformer situation.
Linear model: no load-peak current max amplitude 2*root(2)
Short circuited: peak current 7.4*root(2)
Loaded somewhere in between.
Including saturation: no load 8*root(2) peak but only on the first half
cycle.
The primary flux linkages are essentially the same for all cases.
A (rough) transient analysis of a 10kVA 1000/200 V transformer based on what
appears to be reasonable data indicates in a worse case case ignoring
saturation and leakage reactance does show that the magnetising component of
current is offset by somethng under 200% from the steady state peak values
for no load. There is With rated load, and the same data, the analysis
shows a larger time constant and a smaller transient bridging current but
the maximum inrush current is still higher than the no load inrush.
The culprit is saturation. I am not going to try to do a numerical analysis
of this as it is messy,but if saturation occurs, the d(phi)/di or inductance
of the core becomes small, Leakage reactance isn't affected as much because
part of the leakage path is outside the core. The reference above shows an
initial sharp reduction in the secondary voltage due to saturation along
with the limiting effect of primary impedance. Initial load current will
then also be limited so that it may not have the effect that you suggest.
It is a bit messier than you have indicated but, if you have contradictory
references, I would like to see them -my observations are pretty much off
the cuff and if out to lunch, I would like to know why (and eat crow in that
case).
--
Don Kelly [email protected]
remove the X to answer
I see several things in your contention that I have problems with.
Assuming the applied voltage is the same or nearly so, the inrush
magnetising current will be the same as before. Adding load won't change
this.
For AC, unlike DC, it is the applied voltage that determines the peak flux
(E=4.44*f*N*phimax or phimax=E/(4.44*f*N) for sinusoidal excitation) and
this flux must be there independent of the core. The magnetising current is
that needed to provide the flux for the particular core. It will be high
with a lousy core or under saturation but the flux will be determined by the
voltage. The total input current will be increased by turning the load on
before energisation because you will have load current in parallel with the
magnetising current. However, given that, whenever switching occurs, there
is a transient which can produce a transient DC offset and this can lead to
saturation with a resulting current that can be excessive. The transformer
will appear as a non-linear reactor, rather than a resistor. Adding a load
would not change this as during this transient condition, the load voltage
may be quite small as would the load current because at high inrush
currents.
Rather than say that the magnetising current is the difference between the
load current equivalent and the primary current -which is true- consider
that the primary current consists of the load current equivalent + the
magnetising current which depends on the flux which depends in turn upon the
voltage in the AC case (Faraday's law, which applies, relates (rate of
change of) flux and voltage, not flux and current.
--
Don Kelly [email protected]
remove the X to answer
----------------------------
"whit3rd" <[email protected]> wrote in message
news:[email protected]...
On Jan 17, 12:43 pm, "[email protected]" <[email protected]> wrote:
> [we used a transforme]
> to power a convection oven built for 480 volts from a 208 volt
> distribution.
>... we initially turned
> off the transformer on weekends using the 150 amp circuit breaker.
> This was great for a while. Then one Monday morning the transformer
> would not power up. We called the electrician and he eventually
> discovered a 200 amp fuse in the meter box was open. He replaced it,
> $50.. and it worked again. A few weeks later, the same thing happened
> on a Monday morning. Same fuse problem. He said to stop turning off
> the transformer. The high inrush current was blowing the fast acting
> fuse, but not effecting the heat operated circuit breaker. The oven
> was not powered up, so there was very little load on the transformer.
That's a misconception. If you'd turned the oven ON before
powering the transformer, the fuse would'nt blow. The 'load'
that is important is the magnetic field in the transformer, which
depends on input current MINUS scaled output current.
With no output current, and possibly some remnant field,
a newly-connected transformer takes its field to maximum,
and that causes saturation and the fuse blows.
If the core of a transformer staturates, the electrical circuit
becomes
just a copper resistor. Better to blow a $50 fuse than melt that
copper resistor.
---------
I see some things in your contention that I have problems with.
Assuming the applied voltage is the same or nearly so, the inrush
magnetising current will be the same as before. Adding load won't change
this.
For AC, unlike DC, it is the applied voltage that determines the peak flux
(E=4.44*f*N*phimax or phimax=E/(4.44*f*N) for sinusoidal excitation) and
this flux must be there independent of the core. The magnetising current is
that needed to provide the flux for the particular core. It will be high
with a lousy core or under saturation but the flux will be determined by the
voltage. The total input current will be increased by turning the load on
before energisation because you will have load current in parallel with the
magnetising current. However, given that, whenever switching occurs, there
is a transient which can produce a transient DC offset and this can lead to
saturation with a resulting current that can be excessive. The transformer
will appear as a non-linear reactor, rather than a resistor. Adding a load
would not change this as during this transient condition, the load voltage
may be quite small as would the load current because at high inrush
currents.
Rather than say that the magnetising current is the difference between the
load current equivalent and the primary current -which is true- consider
that the primary current consists of the load current equivalent + the
magnetising current which depends on the flux which depends in turn upon the
voltage in the AC case (Faraday's law, which applies, relates (rate of
change of) flux and voltage, not flux and current.
--
Don Kelly [email protected]
remove the X to answer
On Thu, 17 Jan 2008 20:23:46 GMT, James T. <[email protected]>
wrote:
>> Not sure where you came up with the sqrt(3) in your equations.
>> --
>> JeffB
>
>This was from a conversation with an engineer at the transformer
>manufacturer's support department. (To be fair, I didn't mention the load as
>purely resistive). He gave the formula as;
>
>KVA / V / sqrt(3) = A
>
>> The power factor for resistive heaters will be near 1. The transformer
>> rating is just fine.
>
>So the formula for a resistive load is:
>
>KVA / V = A
>
>meaning that this 10KVA transformer is good up to a maximum of:
>
> - primary: 48.0A
> - secondary: 20.8A
>
>Is this right?
You were right the first time. The sqrt(3) accounts for the fact that
this is a 3-phase system.
--
Ned Simmons
"James T." <[email protected]> wrote in message
news:[email protected]...
>>> Because this is driving a purely resistive load, this doesn't help bring
>>> the parameters within the transformer's rating, does it?
>
>> No, a purely resistive load is best case.
>
> But since the transformer is (presumably) designed with a worst-case PF
> (or
> at least some PF value greater than 1), can't it be re-rated (what's the
> opposite of "derated"?) if used with a strictly resistive load?
>
> In other words, a transformer rated at 10KVA for a load with PF = x,
> should
> be able to handle a greater KVA if driving a PF = 1 load. Shouldn't it?
>
Sorry, the power factor can never exceed 1 because it is the cosine of the
phase angle between the current and the voltage.
Jim