When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.
--

Dan H.

Assuming equal loading, on 208 wye triple the amps times 120 will give you
the VA, or 360 times the amperage of one leg. 208 * SQROOT(3) = 360
"dan" wrote in message
...
When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.
--

Dan H.

Yes, but !! Be careful in caculating VA from current measured with a clamp-on type of ammeter,
though. In as much as E and I are out of phase, any leg current measured with clamp-ons will not be
accurate. Full load current will be equal to 3 x that of current in each leg. The formula is:

3-phase KVA = (Volts x Amps x Sq. root of 3) / (1000)

Bob Swinney
"dan" wrote in message ...
When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.
--

Dan H.

"Robert Swinney" wrote in message
...
Yes, but !! Be careful in caculating VA from current measured with a
clamp-on type of ammeter,
though. In as much as E and I are out of phase, any leg current measured
with clamp-ons will not be
accurate. Full load current will be equal to 3 x that of current in each
leg. The formula is:

3-phase KVA = (Volts x Amps x Sq. root of 3) / (1000)

In the formula above, Amps would be the circuit current. Full load current
(as in FLA) is the same as the current on one leg, if we're talking about
the same thing. The only way you can triple the amps in this case is if you
multiply by 120, which makes sense with 208 wye since it could be supplying
three 120 V circuits with a neutral. That math, however, doesn't work with
other three phase voltages.

Bob Swinney
"dan" wrote in message
...
When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.
--

Dan H.

The formula as given is for the full load rated current of a 3 phase motor, either delta or wye;
makes no difference. If one were to actually mesasure the KVA in each leg it would be 1/3 the KVA
as calculated for full load. The math works with all 3-phase.

Bob Swinney


"ATP*" wrote in message ...

"Robert Swinney" wrote in message
...
Yes, but !! Be careful in caculating VA from current measured with a
clamp-on type of ammeter,
though. In as much as E and I are out of phase, any leg current measured
with clamp-ons will not be
accurate. Full load current will be equal to 3 x that of current in each
leg. The formula is:

3-phase KVA = (Volts x Amps x Sq. root of 3) / (1000)

In the formula above, Amps would be the circuit current. Full load current
(as in FLA) is the same as the current on one leg, if we're talking about
the same thing. The only way you can triple the amps in this case is if you
multiply by 120, which makes sense with 208 wye since it could be supplying
three 120 V circuits with a neutral. That math, however, doesn't work with
other three phase voltages.

Bob Swinney
"dan" wrote in message
...
When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.
--

Dan H.

"Robert Swinney" wrote in message
...
The formula as given is for the full load rated current of a 3 phase
motor, either delta or wye;
makes no difference. If one were to actually mesasure the KVA in each leg
it would be 1/3 the KVA
as calculated for full load. The math works with all 3-phase.

Bob Swinney

KVA is not current, and the formula is not for current. Full load current is
not equal to 3x that of current in each leg.


"ATP*" wrote in message
...

"Robert Swinney" wrote in message
...
Yes, but !! Be careful in caculating VA from current measured with a
clamp-on type of ammeter,
though. In as much as E and I are out of phase, any leg current measured
with clamp-ons will not be
accurate. Full load current will be equal to 3 x that of current in each
leg. The formula is:

3-phase KVA = (Volts x Amps x Sq. root of 3) / (1000)

In the formula above, Amps would be the circuit current. Full load current
(as in FLA) is the same as the current on one leg, if we're talking about
the same thing. The only way you can triple the amps in this case is if
you
multiply by 120, which makes sense with 208 wye since it could be
supplying
three 120 V circuits with a neutral. That math, however, doesn't work with
other three phase voltages.

Bob Swinney
"dan" wrote in message
...
When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.
--

Dan H.

Hmmnnn - - what part of KVA did you find beneath your level of comprehension.? Actually the concept
is fairly simple for most people.

Bob Swinney
wrote in message ...

"Robert Swinney" wrote in message
...
The formula as given is for the full load rated current of a 3 phase
motor, either delta or wye;
makes no difference. If one were to actually mesasure the KVA in each leg
it would be 1/3 the KVA
as calculated for full load. The math works with all 3-phase.

Bob Swinney

KVA is not current, and the formula is not for current. Full load current is
not equal to 3x that of current in each leg.


"ATP*" wrote in message
...

"Robert Swinney" wrote in message
...
Yes, but !! Be careful in caculating VA from current measured with a
clamp-on type of ammeter,
though. In as much as E and I are out of phase, any leg current measured
with clamp-ons will not be
accurate. Full load current will be equal to 3 x that of current in each
leg. The formula is:

3-phase KVA = (Volts x Amps x Sq. root of 3) / (1000)

In the formula above, Amps would be the circuit current. Full load current
(as in FLA) is the same as the current on one leg, if we're talking about
the same thing. The only way you can triple the amps in this case is if
you
multiply by 120, which makes sense with 208 wye since it could be
supplying
three 120 V circuits with a neutral. That math, however, doesn't work with
other three phase voltages.

Bob Swinney
"dan" wrote in message
...
When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.
--

Dan H.

"Robert Swinney" wrote in message
...
Hmmnnn - - what part of KVA did you find beneath your level of
comprehension.? Actually the concept
is fairly simple for most people.

Bob Swinney

It's pretty funny how you keep pressing a point when it's obvious you don't
have a clue. Your statement that full load current is equal to 3x that of
current in each leg is dead wrong. Stick to your original statement instead
of saying something else. How is full load current equal to 3x the current
in one leg?
wrote in message
...

"Robert Swinney" wrote in message
...
The formula as given is for the full load rated current of a 3 phase
motor, either delta or wye;
makes no difference. If one were to actually mesasure the KVA in each
leg
it would be 1/3 the KVA
as calculated for full load. The math works with all 3-phase.

Bob Swinney

KVA is not current, and the formula is not for current. Full load current
is
not equal to 3x that of current in each leg.


"ATP*" wrote in message
...

"Robert Swinney" wrote in message
...
Yes, but !! Be careful in caculating VA from current measured with a
clamp-on type of ammeter,
though. In as much as E and I are out of phase, any leg current
measured
with clamp-ons will not be
accurate. Full load current will be equal to 3 x that of current in
each
leg. The formula is:

3-phase KVA = (Volts x Amps x Sq. root of 3) / (1000)

In the formula above, Amps would be the circuit current. Full load
current
(as in FLA) is the same as the current on one leg, if we're talking about
the same thing. The only way you can triple the amps in this case is if
you
multiply by 120, which makes sense with 208 wye since it could be
supplying
three 120 V circuits with a neutral. That math, however, doesn't work
with
other three phase voltages.

Bob Swinney
"dan" wrote in message
...
When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.
--

Dan H.

What's that Lassie? You say that ATP* fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Sun, 28 Dec 2008 21:47:26 -0500:

"dan" wrote in message
...
When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Assuming equal loading, on 208 wye triple the amps times 120 will give you
the VA, or 360 times the amperage of one leg. 208 * SQROOT(3) = 360

I just remembered that I will need to know the power factor to figure
out what the watts used is.

Any way to do that without any fancy equipment? It probably changes
as the load changes.

I'm trying to figure out what it costs to run one of the shops air
compressors. They are 25hp screw type. The motor runs constant, and
a valve shunts the output when pressure is reached. I measured about
10A unloaded and 15A loaded on one leg with a clamp on ammeter.
I thought there would be more difference. Must be the power factor.

--

Dan H.

(dan) wrote:

When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.


Apples and oranges. Watts = power = work.

VA is volts x amps but ELI the ICE man comes into effect.

IOW the voltage and current waveforms are not in phase causing power to be lower than just
volts x amps. The amps that have to be carried down the utility are real to the utility
so in an industrial setting VA is what one is billed at for demand.

In a resistive circuit V * A * 1.732 or so would be the right number for watts or VA.

Wes

"dan" wrote in message
...
What's that Lassie? You say that ATP* fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Sun, 28 Dec 2008 21:47:26 -0500:

"dan" wrote in message
...
When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Assuming equal loading, on 208 wye triple the amps times 120 will give you
the VA, or 360 times the amperage of one leg. 208 * SQROOT(3) = 360

I just remembered that I will need to know the power factor to figure
out what the watts used is.

Any way to do that without any fancy equipment? It probably changes
as the load changes.

I'm trying to figure out what it costs to run one of the shops air
compressors. They are 25hp screw type. The motor runs constant, and
a valve shunts the output when pressure is reached. I measured about
10A unloaded and 15A loaded on one leg with a clamp on ammeter.
I thought there would be more difference. Must be the power factor.

--

Dan H.

A quick search on rotary air compressors turned up a 25 HP with an 18.5 KW
motor, which would be over 51 full load amps per phase before efficiencies
are even considered if my calcs are right. 15 amps would be pretty lightly
loaded. Take a look at this chart from WEG:

http://www.galco.com/techdoc/weg/02518ep3e284t_dat.pdf

They give efficiencies and power factors at different loads. I don't get
involved with large motors like this very often, maybe Bruce can comment.

"Wes" wrote in message
...
(dan) wrote:

When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.


Apples and oranges. Watts = power = work.

VA is volts x amps but ELI the ICE man comes into effect.

IOW the voltage and current waveforms are not in phase causing power to be
lower than just
volts x amps. The amps that have to be carried down the utility are real
to the utility
so in an industrial setting VA is what one is billed at for demand.

In a resistive circuit V * A * 1.732 or so would be the right number for
watts or VA.

Wes

Wes, does any meter with current transformers measure VA as opposed to a
smaller service with an inline wattmeter, which should measure actual watts?

One may look at 3 phases as originating in 3 separate generators. In fact, some early 3-phase was
distributed over 6 wires to illustrate the 3 generator concept. Of course the voltages and currents
in each phase (leg) were identical. The legs (phases) are 120 degrees apart. The formula for each
leg taken seperately is KVA = (Volts x Amps) / 1000. The three legs combined at 120 degrees have
KVA = (volts x amps x 1.73) / 1000.

Bob Swinney
"ATP*" wrote in message ...

"Wes" wrote in message
...
(dan) wrote:

When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.


Apples and oranges. Watts = power = work.

VA is volts x amps but ELI the ICE man comes into effect.

IOW the voltage and current waveforms are not in phase causing power to be
lower than just
volts x amps. The amps that have to be carried down the utility are real
to the utility
so in an industrial setting VA is what one is billed at for demand.

In a resistive circuit V * A * 1.732 or so would be the right number for
watts or VA.

Wes

Wes, does any meter with current transformers measure VA as opposed to a
smaller service with an inline wattmeter, which should measure actual watts?

Home meters have often not measured true watts. The expensive
ones do but they get close and don't play to much with phase angles.

Martin

ATP* wrote:
"Wes" wrote in message
...
(dan) wrote:

When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.

Apples and oranges. Watts = power = work.

VA is volts x amps but ELI the ICE man comes into effect.

IOW the voltage and current waveforms are not in phase causing power to be
lower than just
volts x amps. The amps that have to be carried down the utility are real
to the utility
so in an industrial setting VA is what one is billed at for demand.

In a resistive circuit V * A * 1.732 or so would be the right number for
watts or VA.

Wes

Wes, does any meter with current transformers measure VA as opposed to a
smaller service with an inline wattmeter, which should measure actual watts?

"Robert Swinney" writes:


Wes, does any meter with current transformers measure VA as opposed to a
smaller service with an inline wattmeter, which should measure actual watts?

A watt-hour-meter measures true watts.

Some utilities charge power factor penalties.
--
A host is a host from coast to
& no one will talk to a host that's close........[v].(301) 56-LINUX
Unless the host (that isn't close).........................pob 1433
is busy, hung or dead....................................20915-1433

"Martin H. Eastburn" writes:

Home meters have often not measured true watts. The expensive
ones do but they get close and don't play to much with phase angles.


I'd welcome a cite to same; with the tariff as well.

The Thomson moving coil watt hour meter, the one with the spinning
disk, measures honest-to-gosh watthours. It's the gold standard of the
industry.

Only recently have solid-state meters supplemented them, usually where
time-of-day billing and/or remote metering is wanted. They can also read
VARS I suppose; it's just a bit more firmware code. I expect they will
someday replace all Thomson meters.

Outside of those, I have never seen a residential installation that
even measured VARS, much less billed for them. Some do log a peak load
reading. [They had a needle that was pushed up and had friction against
falling back to zero...]

In some industrial installations, you can be charged a "power factor
penalty" but I've never seen any recording metering of same; merely
the utility tests it every so often. I suppose such is possible on the
largest [steel mill, auto plant] consumers that buy at the 132KV and
above level.


--
A host is a host from coast to
& no one will talk to a host that's close........[v].(301) 56-LINUX
Unless the host (that isn't close).........................pob 1433
is busy, hung or dead....................................20915-1433

Some confusion may come from the well known fact that current and voltage are out of phase in an
inductive circuit. As Wes pointed out ELI is the ICE man, a great memory aid which has voltage
leading current in L; current leading voltage in C. Power (heat) developed in an inductor is the
vectorial product of current and voltage. Power, however, is a sort of absolute quantity. Power
does not add vectorially. If 3 equal quantities comprise the whole, then each is equal to 1/3 of
the whole. If voltage is constant in 3 equal "packages" of power, then total power must be 3 x the
current in each package.

Bob Swinney


"ATP*" wrote in message ...

"Robert Swinney" wrote in message
...
Hmmnnn - - what part of KVA did you find beneath your level of
comprehension.? Actually the concept
is fairly simple for most people.

Bob Swinney

It's pretty funny how you keep pressing a point when it's obvious you don't
have a clue. Your statement that full load current is equal to 3x that of
current in each leg is dead wrong. Stick to your original statement instead
of saying something else. How is full load current equal to 3x the current
in one leg?
wrote in message
...

"Robert Swinney" wrote in message
...
The formula as given is for the full load rated current of a 3 phase
motor, either delta or wye;
makes no difference. If one were to actually mesasure the KVA in each
leg
it would be 1/3 the KVA
as calculated for full load. The math works with all 3-phase.

Bob Swinney

KVA is not current, and the formula is not for current. Full load current
is
not equal to 3x that of current in each leg.


"ATP*" wrote in message
...

"Robert Swinney" wrote in message
...
Yes, but !! Be careful in caculating VA from current measured with a
clamp-on type of ammeter,
though. In as much as E and I are out of phase, any leg current
measured
with clamp-ons will not be
accurate. Full load current will be equal to 3 x that of current in
each
leg. The formula is:

3-phase KVA = (Volts x Amps x Sq. root of 3) / (1000)

In the formula above, Amps would be the circuit current. Full load
current
(as in FLA) is the same as the current on one leg, if we're talking about
the same thing. The only way you can triple the amps in this case is if
you
multiply by 120, which makes sense with 208 wye since it could be
supplying
three 120 V circuits with a neutral. That math, however, doesn't work
with
other three phase voltages.

Bob Swinney
"dan" wrote in message
...
When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.
--

Dan H.

"Robert Swinney" wrote in message
...
Some confusion may come from the well known fact that current and voltage
are out of phase in an
inductive circuit. As Wes pointed out ELI is the ICE man, a great memory
aid which has voltage
leading current in L; current leading voltage in C. Power (heat)
developed in an inductor is the
vectorial product of current and voltage. Power, however, is a sort of
absolute quantity. Power
does not add vectorially. If 3 equal quantities comprise the whole, then
each is equal to 1/3 of
the whole. If voltage is constant in 3 equal "packages" of power, then
total power must be 3 x the
current in each package.

Bob Swinney

I'm quite aware of the math involved. I'm not confused. Your statement that
full load current is equal to 3x that of current in each leg is still wrong,
no matter how you try to rationalize it, because we don't express current in
a three phase circuit that way. If a three phase motor has a full load
current of 30 amps, that means each leg carries 30 amps, not 10. It's a
convention.

Your other assertion:

"Yes, but !! Be careful in caculating VA from current measured with a
clamp-on type of ammeter,
though. In as much as E and I are out of phase, any leg current
measured
with clamp-ons will not be
accurate"

is also wrong, because VA is not the same as watts. VA is apparent power and
it IS measured using ammeters and voltmeters.


"ATP*" wrote in message
...

"Robert Swinney" wrote in message
...
Hmmnnn - - what part of KVA did you find beneath your level of
comprehension.? Actually the concept
is fairly simple for most people.

Bob Swinney

It's pretty funny how you keep pressing a point when it's obvious you
don't
have a clue. Your statement that full load current is equal to 3x that of
current in each leg is dead wrong. Stick to your original statement
instead
of saying something else. How is full load current equal to 3x the current
in one leg?
wrote in message
...

"Robert Swinney" wrote in message
...
The formula as given is for the full load rated current of a 3 phase
motor, either delta or wye;
makes no difference. If one were to actually mesasure the KVA in each
leg
it would be 1/3 the KVA
as calculated for full load. The math works with all 3-phase.

Bob Swinney

KVA is not current, and the formula is not for current. Full load current
is
not equal to 3x that of current in each leg.


"ATP*" wrote in message
...

"Robert Swinney" wrote in message
...
Yes, but !! Be careful in caculating VA from current measured with a
clamp-on type of ammeter,
though. In as much as E and I are out of phase, any leg current
measured
with clamp-ons will not be
accurate. Full load current will be equal to 3 x that of current in
each
leg. The formula is:

3-phase KVA = (Volts x Amps x Sq. root of 3) / (1000)

In the formula above, Amps would be the circuit current. Full load
current
(as in FLA) is the same as the current on one leg, if we're talking
about
the same thing. The only way you can triple the amps in this case is if
you
multiply by 120, which makes sense with 208 wye since it could be
supplying
three 120 V circuits with a neutral. That math, however, doesn't work
with
other three phase voltages.

Bob Swinney
"dan" wrote in message
...
When figuring watts or VA on a three phase circuit, do you triple the
amps measured on one leg?

Thanks.
--

Dan H.

Apologies to Dan who started this long and annoying thread with an honest question.

The simple answer is No; it is not practical to measure current in the individual phases. This
pulsating current cannot easily be measured. I was wrong to try to show that the current was 1/3 of
rated load current. It can be seen, though, that Power in each phase must be equal to 1/3 total
Power if the phases are balanced. Simple logic supports this.

Instantaneous current is another matter. In a 3-Phase motor the phases are separated by 60
electrical degrees. At any point in time instantaneous current will vary in accordance with which
phase is being considered and with the point of time in the cycle. At say, time 0 of the sine wave,
Phase 1 current is at zero crossing (0 degrees), thus its instantaneous current would be I total x
sin 0. At this same time Phase 2 is at 60 degrees and its instantaneous current would be I total I
x sin 60. Also at this same time Phase 3 is at 120 degrees and its instantaneous current would be I
total x sin 120. Current of the 3 phases adds and the resultant could be stated: I instantaneous
at time zero = I total x (sin 0 + sin 60 + sin 120). I instantaneous at any time = I total x [sin
theta + sin (theta + 60) + sin (theta + 120)] These effects can be viewed by displaying 3 phases
with a common time base on an oscilloscope.

Bob Swinney




"David Lesher" wrote in message ...
"Martin H. Eastburn" writes:

Home meters have often not measured true watts. The expensive
ones do but they get close and don't play to much with phase angles.


I'd welcome a cite to same; with the tariff as well.

The Thomson moving coil watt hour meter, the one with the spinning
disk, measures honest-to-gosh watthours. It's the gold standard of the
industry.

Only recently have solid-state meters supplemented them, usually where
time-of-day billing and/or remote metering is wanted. They can also read
VARS I suppose; it's just a bit more firmware code. I expect they will
someday replace all Thomson meters.

Outside of those, I have never seen a residential installation that
even measured VARS, much less billed for them. Some do log a peak load
reading. [They had a needle that was pushed up and had friction against
falling back to zero...]

In some industrial installations, you can be charged a "power factor
penalty" but I've never seen any recording metering of same; merely
the utility tests it every so often. I suppose such is possible on the
largest [steel mill, auto plant] consumers that buy at the 132KV and
above level.


--
A host is a host from coast to
& no one will talk to a host that's close........[v].(301) 56-LINUX
Unless the host (that isn't close).........................pob 1433
is busy, hung or dead....................................20915-1433

"Robert Swinney" wrote in message
...
Apologies to Dan who started this long and annoying thread with an honest
question.

The simple answer is No; it is not practical to measure current in the
individual phases. This
pulsating current cannot easily be measured.

That will be a big surprise to electricians. It can and should be measured,
using an analog meter or a digital meter that calculates the root mean
square. AC voltage is also cyclic, does that mean we shouldn't measure it?

I was wrong to try to show that the current was 1/3 of
rated load current. It can be seen, though, that Power in each phase must
be equal to 1/3 total
Power if the phases are balanced. Simple logic supports this.

Yes, which was never in dispute.


Instantaneous current is another matter. In a 3-Phase motor the phases
are separated by 60
electrical degrees. At any point in time instantaneous current will vary
in accordance with which
phase is being considered and with the point of time in the cycle. At
say, time 0 of the sine wave,
Phase 1 current is at zero crossing (0 degrees), thus its instantaneous
current would be I total x
sin 0. At this same time Phase 2 is at 60 degrees and its instantaneous
current would be I total I
x sin 60. Also at this same time Phase 3 is at 120 degrees and its
instantaneous current would be I
total x sin 120. Current of the 3 phases adds and the resultant could be
stated: I instantaneous
at time zero = I total x (sin 0 + sin 60 + sin 120). I instantaneous at
any time = I total x [sin
theta + sin (theta + 60) + sin (theta + 120)] These effects can be viewed
by displaying 3 phases
with a common time base on an oscilloscope.

Bob Swinney


None of which is relevant to someone attempting to estimate the power use of
a compressor.







"David Lesher" wrote in message
...
"Martin H. Eastburn" writes:

Home meters have often not measured true watts. The expensive
ones do but they get close and don't play to much with phase angles.


I'd welcome a cite to same; with the tariff as well.

The Thomson moving coil watt hour meter, the one with the spinning
disk, measures honest-to-gosh watthours. It's the gold standard of the
industry.

Only recently have solid-state meters supplemented them, usually where
time-of-day billing and/or remote metering is wanted. They can also read
VARS I suppose; it's just a bit more firmware code. I expect they will
someday replace all Thomson meters.

Outside of those, I have never seen a residential installation that
even measured VARS, much less billed for them. Some do log a peak load
reading. [They had a needle that was pushed up and had friction against
falling back to zero...]

In some industrial installations, you can be charged a "power factor
penalty" but I've never seen any recording metering of same; merely
the utility tests it every so often. I suppose such is possible on the
largest [steel mill, auto plant] consumers that buy at the 132KV and
above level.


--
A host is a host from coast to
& no one will talk to a host that's close........[v].(301) 56-LINUX
Unless the host (that isn't close).........................pob 1433
is busy, hung or dead....................................20915-1433

ATP sez:

"That will be a big surprise to electricians. It can and should be measured,
using an analog meter or a digital meter that calculates the root mean
square. AC voltage is also cyclic, does that mean we shouldn't measure it?"

Sorry, ATP if I went way over your head with that one. I referred to the measurement of
instantaneous current, of course. Getchersef a good EE handbook and look that one up, OK ?

Bob Swinney

"Robert Swinney" wrote in message
...
ATP sez:

"That will be a big surprise to electricians. It can and should be
measured,
using an analog meter or a digital meter that calculates the root mean
square. AC voltage is also cyclic, does that mean we shouldn't measure
it?"

Sorry, ATP if I went way over your head with that one. I referred to the
measurement of
instantaneous current, of course. Getchersef a good EE handbook and look
that one up, OK ?

Bob Swinney


Like a kid who can't answer an essay question, you continue to top post out
of context, irrelevant information that you obviously don't understand. The
OP, Don wants to figure out how much it costs to run a 25 HP compressor. He
has taken current readings. These current readings and the characteristics
of his three phase service are enough to approximate the cost of running the
compressor. Instantaneous current is completely irrelevant. Here is what you
wrote:

"The simple answer is No; it is not practical to measure current in the
individual phases. This
pulsating current cannot easily be measured. I was wrong to try to show
that the current was 1/3 of
rated load current. It can be seen, though, that Power in each phase must
be equal to 1/3 total
Power if the phases are balanced. Simple logic supports this."

It's over, Bob. As in past technical arguments that you have attempted here,
you just keep digging the hole deeper.

"Robert Swinney" wrote in message
...
ATP sez:

"That will be a big surprise to electricians. It can and should be
measured,
using an analog meter or a digital meter that calculates the root mean
square. AC voltage is also cyclic, does that mean we shouldn't measure
it?"

Sorry, ATP if I went way over your head with that one. I referred to the
measurement of
instantaneous current, of course. Getchersef a good EE handbook and look
that one up, OK ?

Bob Swinney




From one:
http://picasaweb.google.com/coolflow...PhaseCircuits#

Less rigorous: http://picasaweb.google.com/coolflow...PhaseCircuits#

"Rick" wrote in message
...

"Robert Swinney" wrote in message
...
ATP sez:

"That will be a big surprise to electricians. It can and should be
measured,
using an analog meter or a digital meter that calculates the root mean
square. AC voltage is also cyclic, does that mean we shouldn't measure
it?"

Sorry, ATP if I went way over your head with that one. I referred to the
measurement of
instantaneous current, of course. Getchersef a good EE handbook and look
that one up, OK ?

Bob Swinney




From one:
http://picasaweb.google.com/coolflow...PhaseCircuits#

Less rigorous:
http://picasaweb.google.com/coolflow...PhaseCircuits#

That's nice, I have plenty of textbooks, it's just that measuring
instantaneous current has nothing to do with the thread. Bob is floundering
and keeps trying to introduce new material to cover up his previous
misstatements.

What's that Lassie? You say that Robert Swinney fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Tue, 30 Dec 2008 17:57:26 -0600:

Apologies to Dan who started this long and annoying thread with an honest question.
snip

That's OK. I'm reading every post and learning a few things along the
way.
--

Dan H.

On Thu, 01 Jan 2009 01:27:29 GMT, (dan) wrote:

What's that Lassie? You say that Robert Swinney fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Tue, 30 Dec 2008 17:57:26 -0600:

Apologies to Dan who started this long and annoying thread with an honest question.
snip

That's OK. I'm reading every post and learning a few things along the
way.

Robert's trying to overanalyze it - You can take a clamp-amp and
measure each phase (in case it is imbalanced somehow) and average it
out, and know what it costs roughly to run.

The wattmeter at the service is going to be reasonably accurate, or
the State Weights and Measures Commissioner will have words with the
power company - If the power company is using that meter to sell you
electricity they have a duty to be reasonably accurate - I'd say in
the +/- 1% to 2% range

Your average Clamp Amp is reading RMS, and is plenty close enough to
agree with what the power meter is seeing - unless you are running a
variable speed drive that distorts the heck out of the waveform.

Where it gets tricky is in measuring the start surges and quantifying
it's effects on the power bill, versus the lower but still real costs
of keeping a motor spinning unloaded. Oh, and the cost of a new motor
every year if you are short-cycling them.

There are more sophisticated solutions to reducing the start surges
in a shop, like electronic soft starters or a Delta-Wye starter on all
the large equipment, but the solution starts getting more expensive
than the problem....

And with a soft start solution you need the ability to positively
control the load till the motor is up to speed - often used on
elevator pumps where they can wait to open the hydraulic valves till
after the jack valve gets a 'full speed' signal from the motor
starter.

And there are other things to consider - quantifying that start
surge's effect, combined with all the other motors and welders and
such in the shop, and their effect on the instantaneous demand reading
on the meter and the Demand Charge adder on the bill.

Some shops with lots of huge motors can have their power bill
doubled because of the demand charges, because the utility metering
tells the engineers they have to yank out the 100KVA transformer and
place a 200KVA - your regular running loads might be under 100KVA,
but all the start surges will stress and take out the transformer if
they don't use an oversized one.

And when that pole pig in the parking lot (usually oil filled) blows
up, it blows up reel guud. (Get out the marshmallows!)

-- Bruce --

What's that Lassie? You say that Bruce L. Bergman fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Wed, 31 Dec 2008 18:29:35 -0800:

Robert's trying to overanalyze it - You can take a clamp-amp and
measure each phase (in case it is imbalanced somehow) and average it
out, and know what it costs roughly to run.

But what about the power factor? Or does it not matter that much?
I only read one leg, and I was surprised to see that when the motor
was loaded it was drawing 15A. And 10A when un-loaded.

Five amps seem low for the work that it's doing.
--

Dan H.

"dan" wrote in message
...
What's that Lassie? You say that Bruce L. Bergman fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Wed, 31 Dec 2008 18:29:35 -0800:

Robert's trying to overanalyze it - You can take a clamp-amp and
measure each phase (in case it is imbalanced somehow) and average it
out, and know what it costs roughly to run.

But what about the power factor? Or does it not matter that much?
I only read one leg, and I was surprised to see that when the motor
was loaded it was drawing 15A. And 10A when un-loaded.

Five amps seem low for the work that it's doing.
--

Dan H.

It's not doing 25 HP worth of work, that's for sure. The power factor in
this case will just mean it costs you less to run. I would use a PF from a
chart at a low loading, for example .75 (for a 75% PF) * the VA equals
watts. So 15 amps * 360 * .75 = watts, then you have to calculate hours used
and also your demand charge (KW) if the compressor comes on during a peak
demand window.

What's that Lassie? You say that ATP* fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Sat, 3 Jan 2009 21:22:49 -0500:

It's not doing 25 HP worth of work, that's for sure. The power factor in
this case will just mean it costs you less to run. I would use a PF from a
chart at a low loading, for example .75 (for a 75% PF) * the VA equals
watts. So 15 amps * 360 * .75 = watts, then you have to calculate hours used
and also your demand charge (KW) if the compressor comes on during a peak
demand window.

I'd have to agree, not doing 25 HP. The owner says that the
compressor(3) cost $10/ hour to run. But I know that he just pulled
that figure out of his ass. So I'm trying to figure a rough estimate
on what it costs to run one of them.

I was very surprised to see the unloaded current so high.
I'm going to check the other two legs on monday. I'll check the other
units too, just to see if they are close. I had checked on one that I
had just turned on, so maybe it wasn't fully warmed up yet.

As far as the demand charge, wouldn't even be 1% of our total load.
We have 21 big CNC machines. And in the summer, 10 big AC units on
the roof.
We have our own transformer, big as a small car, on a pad outside of
the power room.
--

Dan H.

"dan" wrote in message
...
What's that Lassie? You say that ATP* fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Sat, 3 Jan 2009 21:22:49 -0500:

It's not doing 25 HP worth of work, that's for sure. The power factor in
this case will just mean it costs you less to run. I would use a PF from a
chart at a low loading, for example .75 (for a 75% PF) * the VA equals
watts. So 15 amps * 360 * .75 = watts, then you have to calculate hours
used
and also your demand charge (KW) if the compressor comes on during a peak
demand window.

I'd have to agree, not doing 25 HP. The owner says that the
compressor(3) cost $10/ hour to run. But I know that he just pulled
that figure out of his ass. So I'm trying to figure a rough estimate
on what it costs to run one of them.

I was very surprised to see the unloaded current so high.

That's a pretty big motor, I would expect it to pull 10 amps just running.
Full load amps is close to 60.

I'm going to check the other two legs on monday. I'll check the other
units too, just to see if they are close. I had checked on one that I
had just turned on, so maybe it wasn't fully warmed up yet.

As far as the demand charge, wouldn't even be 1% of our total load.
We have 21 big CNC machines. And in the summer, 10 big AC units on
the roof.
We have our own transformer, big as a small car, on a pad outside of
the power room.
--

Dan H.

A drop in the bucket compared to all that, but could still be a significant
yearly cost.

What's that Lassie? You say that ATP* fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Mon, 5 Jan 2009 20:38:00 -0500:

snip


I was very surprised to see the unloaded current so high.

That's a pretty big motor, I would expect it to pull 10 amps just running.
Full load amps is close to 60.

I would too. But when the valve cycled, the load went to only 15A.


snip


As far as the demand charge, wouldn't even be 1% of our total load.
We have 21 big CNC machines. And in the summer, 10 big AC units on
the roof.
We have our own transformer, big as a small car, on a pad outside of
the power room.

A drop in the bucket compared to all that, but could still be a significant
yearly cost.

I agree. But I'm trying to get to an hourly cost. So when the owner
complains about the cost of having the air on for a few hours over the
weekend, I can counter with a realistic figure. And be able to 'show
my work' as well. He seems to think that since the motor runs
continuously, that 25hp of power is being used. So I want to have the
information about the loaded and unloaded current draw, and the duty
cycle when I'm using it. That way I can figure an hourly cost to run.

Thanks for all the help.

--

Dan H.

"dan" wrote in message
...
What's that Lassie? You say that ATP* fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Mon, 5 Jan 2009 20:38:00 -0500:

snip


I was very surprised to see the unloaded current so high.

That's a pretty big motor, I would expect it to pull 10 amps just running.
Full load amps is close to 60.

I would too. But when the valve cycled, the load went to only 15A.


snip


As far as the demand charge, wouldn't even be 1% of our total load.
We have 21 big CNC machines. And in the summer, 10 big AC units on
the roof.
We have our own transformer, big as a small car, on a pad outside of
the power room.

A drop in the bucket compared to all that, but could still be a
significant
yearly cost.

I agree. But I'm trying to get to an hourly cost. So when the owner
complains about the cost of having the air on for a few hours over the
weekend, I can counter with a realistic figure. And be able to 'show
my work' as well. He seems to think that since the motor runs
continuously, that 25hp of power is being used. So I want to have the
information about the loaded and unloaded current draw, and the duty
cycle when I'm using it. That way I can figure an hourly cost to run.

Thanks for all the help.

--

Dan H.
I haven't followed all of this thread so it may not be pertinent but you can
not tell much of anything about energy usage of loaded and unloaded motors
from the line currents and voltages alone. The power factor varies greatly
with load; I have seen motors with pretty small differences between idling
and full load currents. You need a clamp-on wattmeter or something similar
to get meaningful measurements.

Don Young

What's that Lassie? You say that Don Young fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Wed, 7 Jan 2009 20:55:36 -0600:

I haven't followed all of this thread so it may not be pertinent but you can
not tell much of anything about energy usage of loaded and unloaded motors
from the line currents and voltages alone. The power factor varies greatly
with load; I have seen motors with pretty small differences between idling
and full load currents. You need a clamp-on wattmeter or something similar
to get meaningful measurements.

Don Young

Damn. That's what I was afraid of.
--

Dan H.

(dan) wrote:

What's that Lassie? You say that Don Young fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Wed, 7 Jan 2009 20:55:36 -0600:

I haven't followed all of this thread so it may not be pertinent but you can
not tell much of anything about energy usage of loaded and unloaded motors
from the line currents and voltages alone. The power factor varies greatly
with load; I have seen motors with pretty small differences between idling
and full load currents. You need a clamp-on wattmeter or something similar
to get meaningful measurements.

Don Young

Damn. That's what I was afraid of.

As Don, I haven't read all this thread but as Don mentioned, power factor changes
depending on load.

Motors are inductive and the voltage and current waveforms are not in phase, they are most
out of phase when idling. That means P is not equal to I x E.

Outside of startup, amperage is fairly constant in an ac electric motor be it idling or
loaded.

Thus the comments on using a watt meter. That checks to see what P is and deals with I x
E being out of phase with each other.

Adding to the mix is your utility rate. Are you billed in KWH or KVA? If you have 3
phase service, there is a good chance you are billed for KVA. Home owners tend to be KWH.

Why KVA? Well an idling motor, doing no real work, draws current. Even though the
current and voltage wave forms are out of phase and thus not using a lot of power, the
heating effects of the ampere component is very real. The utilities see this in heating
of wires and their transformers, to them this is something that needs to be priced since
it does cost them money.

To add to this mess, an idling motor can have capacitors attached to normalize the load so
the power factor is closer to unity. Some factories have banks that are switched in and
out via monitoring systems if their power factor is under .80 or so. I assume inductors
could be used for highly capacitive loads. The reason for this is industrial users are
often charged for excessive power factor.

Wes

On 2009-01-09, dan wrote:
What's that Lassie? You say that Don Young fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Wed, 7 Jan 2009 20:55:36 -0600:

I haven't followed all of this thread so it may not be pertinent but you can
not tell much of anything about energy usage of loaded and unloaded motors
from the line currents and voltages alone. The power factor varies greatly
with load; I have seen motors with pretty small differences between idling
and full load currents. You need a clamp-on wattmeter or something similar
to get meaningful measurements.

Damn. That's what I was afraid of.

You can at least calculate the *worst-case* operating cost,
which assumes unity power factor. In any real situation, the cost will
be less (usually significantly less) than that calculated from clamp-on
ammeters. It *can't* cost more than your calculations show -- only
less.

Enjoy,
DoN.

--
Email: | Voice (all times): (703) 938-4564
(too) near Washington D.C. | http://www.d-and-d.com/dnichols/DoN.html
--- Black Holes are where God is dividing by zero ---

What's that Lassie? You say that DoN. Nichols fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by 10 Jan 2009 05:36:28 GMT:

You need a clamp-on wattmeter or something similar
to get meaningful measurements.

Damn. That's what I was afraid of.

You can at least calculate the *worst-case* operating cost,
which assumes unity power factor. In any real situation, the cost will
be less (usually significantly less) than that calculated from clamp-on
ammeters. It *can't* cost more than your calculations show -- only
less.

Thanks, that's good to know.
--

Dan H.

"dan" wrote in message
...
What's that Lassie? You say that DoN. Nichols fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by 10 Jan 2009 05:36:28 GMT:

You need a clamp-on wattmeter or something similar
to get meaningful measurements.

Damn. That's what I was afraid of.

You can at least calculate the *worst-case* operating cost,
which assumes unity power factor. In any real situation, the cost will
be less (usually significantly less) than that calculated from clamp-on
ammeters. It *can't* cost more than your calculations show -- only
less.

Thanks, that's good to know.
--

Dan H.

Do you have access to a good meter like a Fluke 87 and a clamp-on
attachment? If you can leave it running with the clamp-on attached you could
get max, min and average readings for the whole day.

What's that Lassie? You say that ATP* fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Sun, 11 Jan 2009 00:22:43 -0500:


"dan" wrote in message
...
What's that Lassie? You say that DoN. Nichols fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by 10 Jan 2009 05:36:28 GMT:

You need a clamp-on wattmeter or something similar
to get meaningful measurements.

Damn. That's what I was afraid of.

You can at least calculate the *worst-case* operating cost,
which assumes unity power factor. In any real situation, the cost will
be less (usually significantly less) than that calculated from clamp-on
ammeters. It *can't* cost more than your calculations show -- only
less.

Thanks, that's good to know.
--

Dan H.

Do you have access to a good meter like a Fluke 87 and a clamp-on
attachment? If you can leave it running with the clamp-on attached you could
get max, min and average readings for the whole day.

Nope. Just an old clamp-on that I picked up at a yard sale once.
It's a voltmeter too, but it needs funny leads that I don't have.
(they screw onto recessed studs)
--

Dan H.

I agree. But I'm trying to get to an hourly cost. So when the owner
complains about the cost of having the air on for a few hours over the
weekend, I can counter with a realistic figure. And be able to 'show
my work' as well. He seems to think that since the motor runs
continuously, that 25hp of power is being used. So I want to have the
information about the loaded and unloaded current draw, and the duty
cycle when I'm using it. That way I can figure an hourly cost to run.

Thanks for all the help.



I know of no 3-phase Kil-o-Watts, so do this. Find a surplus 3-phase
power meter, and mount a meter box on a board. Put it in the motor
supply leg, note the time/date/meter reading, and let it got for 5 hours
or a day.

That's the best way of finding the actual usage.

Your utility might even sell you an old meter, or try eBay. One gotcha
is finding a direct-reading one. Bigger ones will depend on external
voltage and current transformers.


--
A host is a host from coast to
& no one will talk to a host that's close........[v].(301) 56-LINUX
Unless the host (that isn't close).........................pob 1433
is busy, hung or dead....................................20915-1433

What's that Lassie? You say that David Lesher fell down the old
rec.crafts.metalworking mine and will die if we don't mount a rescue
by Tue, 27 Jan 2009 05:55:46 +0000 (UTC):

I know of no 3-phase Kil-o-Watts, so do this. Find a surplus 3-phase
power meter, and mount a meter box on a board. Put it in the motor
supply leg, note the time/date/meter reading, and let it got for 5 hours
or a day.

That's the best way of finding the actual usage.

Thanks, but that's a bit farther than I want to go.

I think I will just assume full nameplate amps when loaded and
multiply by the duty cycle.

BTW, I had been wrong on the voltage all along. It's being fed from
one of the 460V panels.
--

Dan H.